Marian FeckoDifferential Geometry and Lie Groups for Physicists

Some additional material to the bookVersion from April 15, 2021. In progress.

Here one finds some comments, improvements, more detailed hints and additional (solved) problems. All thisshould hopefully make the book more useful.

1.3 p.8: In the definition of a chart a homeomorphism φ : O → Rn is mentioned. In fact, a homeomorphismon an open subset of Rn might be more precise statement.

1.3.1 p.9: In (i) the general (n-dimensional) case might be treated as follows.

Let the sphere Sn be realized as

ρ2 + z2 = R2 ρ ≡ (x1, . . . , xn)

The South pole and the North pole, respectively, read

S = (0,−R) N = (0, R)

Let P ≡ (ρ, z) be a point on Sn (other then S or N). Then the line from S through P may be written as(ρ(t)z(t)

)=

(0−R

)+

[(ρz

)−(

0−R

)]t

i.e.

ρ(t) = ρt

z(t) = (R+ z)t−R

This line crosses, in a unique point (r, R), where r is to be determined, the (hyper)plane touching the Northpole, i.e. the plane z = R. A simple computation leads to

r =2R

R+ zρ

H The crossing happens for t = t1 such that z(t1) ≡ (R+ z)t1 −R = R, so that t1 = 2R/(R+ z). Then

r ≡ ρ(t1) = ρt1 =2R

R+ zρ

NFrom a similar computation we learn, that the line from N through P crosses, in a unique point (r′, R),

where r′ is to be determined, the (hyper)plane touching the South pole, i.e. the plane z = −R. Here, we get

r′ =2R

R− zρ

So, we see that the two vectors are parallelr′ = λr

and one only has to determine λ. In order to do that, notice that

rr′ =2R

R− z

2R

R+ zρ2 = (2R)2

ρ2

R2 − z2= (2R)2

from where we easily get λ = (2R)2/r2, or, finally

r′ =(2R)2

r

r

r1

2

1.3.2 p.10: In (iv) the fact S2 = CP 1 is to be proved. The following simple observation makes it easy:consider in general two n-dimensional manifolds M,N . Suppose each of them may be covered by exactly twocharts, (x1, x2) and (y1, y2) respectively, such that ”change of coordinates” expressions on the correspondingintersections, i.e. x2(x1) and y2(y1), happen to be given by the same formulas.

[Here x1 denotes the whole n-tuple (x11, x21, . . . , x

n1 ) and similarly for x2, y1, y2. Then x2(x1) means in detail

xa2(x11, x

21, . . . , x

n1 ) a = 1, . . . , n

and similarly for y2(y1).]

Then, M and N are diffeomorphic, the coordinate expression for the diffeomorphism being simply (check)

ya1 = xa1 yi2 = xi2

In particular, concerning CP 1 and S2, the coordinates introduced in 1.3.2(ii) (for p = 1) and 1.3.5 do havethe needed properties. (The common ”change of coordinates” formula is, in complex language, of the formz(w) = 1/w.)

The same trick clearly works for more than just two charts per manifold as well.

1.5 p.16: There is a flaw in the Theorem (assumptions are not stated correctly).

The Theorem deals with constraints in Cartesian space Rn as a tool for defining a manifold.

In particular it says that, given independent constrains ϕ1(x) = · · · = ϕm(x) = 0, the resulting manifold Mis (n−m)-dimensional, i.e.

each constraint reduces the dimension by one unit.

More generally, according to the theorem, if the rank of the Jacobian matrix

Jai (x) :=∂ϕa(x)

∂xi

is constant on the set M (i.e. at points satisfying the constrains), the dimension of the resulting manifold isn − k, where k := Rank J . (The case of maximum rank, i.e. k = m, corresponds to independent constraintsmentioned above.)

So we can state more precisely that, if the assumptions of the theorem are fulfilled,

each constraint reduces the dimension at most by one unit.

Well, consider the following simple example: Take n = 5 and

ϕ1(x1, x2, x3, x4, x5) = x1

ϕ2(x1, x2, x3, x4, x5) = x2

ϕ3(x1, x2, x3, x4, x5) = x3

ϕ4(x1, x2, x3, x4, x5) = x4

What we get is clearly M = the x5-axis in R5 as an one-dimensional manifold. (The constraints are indepen-dent.)

Now, on the same starting R5, consider just a single constraint

ϕ = (ϕ1)2 + (ϕ2)2 + (ϕ3)2 + (ϕ4)2

so thatϕ(x1, x2, x3, x4, x5) = (x1)2 + (x2)2 + (x3)2 + (x4)2

It is clear that this constraint alone produces the same one-dimensional resulting M = the x5-axis in R5. Sothis particular constraint reduces the dimension by as many as four units!

Then, perhaps, the assumptions of the theorem are not satisfied. Let us check:

3

The Jacobian matrix reduces to a single-raw matrix

J = 2(x1, x2, x3, x4, 0)

and, on M = the x5-axis in R5, it takes the value

J = (0, 0, 0, 0, 0) on M

so that k = rank J = 0 and dim M = n − k = 5 − 0 = 5. In spite of the fact that the rank of the Jacobianmatrix is constant on M (it is namely vanishing there, k = 0), the statement dim M = n − k does not hold,here. So,

the theorem (as it stands in the book) is not true in general.

It turns out that what one should really assume is that

the rank is constant on a neighbourhood of M

rather than just on M alone!

What this improved version of the theorem says for our example?

We can see that although the rank of the Jacobian matrix is k = 0 right on M , we have k = 1 at any pointoutsideM . So the rank k is not constant on a neighborhood ofM , the assumptions of the (correct version of the)theorem are not fulfilled and therefore the theorem says nothing about the number of dimensions reduced byour particular constraint. Reduction by four units is perfectly compatible with the theorem (although unusualand slightly counter-intuitive).

Actually, the situation, when a single function (used as a constraint) reduces dimension by more than oneunit, is well known in the context of the standard radial coordinate r (in polar coordinates (r, φ) in plane aswell as in cylindrical or spherical polar coordinates (r, φ, z) or (r, ϑ, φ) in space). Namely we all know that thecoordinate r is defective when r = 0. In particular, consider the family of constraints

ϕ(r, φ) = r − c

in the plane, where c is a nonnegative constant. For c positive, the constraint reduces dimension by one unit(resulting in circle with radius c). For c = 0, however, the dimension is reduced by as many as two units(resulting in a point). The same constraint

ϕ(r, ϑ, φ) = r − c

regarded as a function in the space reduces (for c = 0) the dimension even by three units (resulting in a pointagain). And this is exactly the reason why one should exclude the point where r = 0 from the part of (say)the plane, where r may be used as a coordinate. Coordinate functions should have the property of reducing thedimension by exactly one unit. This leads, then, to the fact that points (zero-dimensional subsets) are labeledby exactly n values of coordinate functions (we need n steps to reduce the dimension from n to zero).

Moral: One should be careful about the character of some innocently looking (smooth) constraints. Theymay behave rather unexpectedly.

1.5.9 p.19: One can check explicitly, that the formula leads (via the idea from the Exercise (1.5.6)) to thefollowing equivalences in the (u, v)-plane:

(u, v) ∼ (u, v + 2π)

(u, v) ∼ (u+ 2π, 2π − v)

This means that

- we can forget about the points of the (u, v)-plane outside the square ⟨0, 2π⟩ × ⟨0, 2π⟩- at the boundary of the square, there is equivalence of the type K2 in (1.5.11)

The only thing to check is that all points inside the square are inequivalent. Try to prove it yourself, i.e.prove that, inside the square, if (u, v) = (u′, v′), then resulting points in R4 are different (injectivity of the map(0, 2π)× (0, 2π) → R4).

4

2.2.1 - 2.2.3 p.24-25: Here we recapitulate general strategy of how to introduce, in terms of the concept ofmutually tangent curves, the tangent space TPM as a vector space canonically associated with a point P ∈M .

We think of curves γ(t) such that γ(0) = P (no other curves are of interest for us). In a neighborhood of P ,choose coordinates xi. Then, near P , we have first order approximation

xi(γ(t)) = xi(P ) + tai + . . . where ai =d

dt

∣∣∣∣0

xi(γ(t))

So, with respect to coordinates xi, the only data needed to first order approximation of a curve (within thefamily of curves γ(t) such that γ(0) = P ) consist in n-tuple of real numbers (a1, . . . , an).

Two curves with identical n-tuples ai (i.e. such that their first order approximations coincide) are calledtangent to one another.

H This fact does not depend on the choice of coordinates (i.e. the concept is well-defined). Indeed, if xi ↔ ai

and xi′ ↔ ai

′, then ai

′= J i

′

j aj , where J i

′

j = (∂xi′/∂xj)(P ) is the Jacobi matrix of the change of coordinates

x′(x), evaluated at P . Therefore, if for two curves γ1 and γ2 holds ai1 = ai2 (i.e. they are tangent to one another

with respect to x), then it also holds ai′

1 = ai′

2 (i.e. they are also tangent to one another with respect to x′; onejust multiplies the first equation by the Jacobi matrix). N

One easily checks that the concept of tangency of curves introduces equivalence relation into the family ofcurves of our interest. An equivalence class is denoted [γ] and the set of all equivalence classes is denoted TPM(the tangent space at P ∈M).

H All this is also well-defined: equivalent curves have equal n-tuples (a1, . . . , an) and equality of n-tuplesdoes not depend on the choice of coordinates. N

So, for each choice of coordinates, there is a bijection between the equivalence classes of curves and n-tuplesai (elements of Rn):

TPM ∋ [γ] ↔ a ∈ Rn w.r.t. x

↔ a′ ∈ Rn w.r.t. x′

Now recall that Rn has natural linear structure

(a1, . . . , an) + λ(b1, . . . , bn) := (a1 + λb1, . . . , an + λbn)

Therefore, we can steal, via a bijection, the linear structure from Rn and donate it to TPM (well, in officialparlance we induce the linear structure).

A small complication in this robbery presents the fact that there is no canonical bijection, but rather infinitelymany bijections, all of them of exactly the same value (each local coordinate system provides a bijection).Fortunately, it turns out that each choice of coordinates (and, consequently, of the corresponding bijection)leads to the same linear structure in TPM . So, there is the linear structure in TPM .

H Indeed, let [γ] and [σ] be two elements from TPM such that

[γ] ↔ a and [σ] ↔ b w.r.t. x

↔ a′ and ↔ b′ w.r.t. x′

Then, by definition,

[γ] + λ[σ] ↔ a+ λb w.r.t. x

↔ a′ + λb′ w.r.t. x′

Do both prescriptions (via x and via x′) lead to the same result? They do, if (and only if)

J(a+ λb) = a′ + λb′

(i.e. if first performing the linear combination via x and then translating the result into the language of thex′-tribe gives the same object as performing the linear combination directly via x′). But this is clearly so, since

J(a+ λb) = Ja+ λJb = a′ + λb′

5

This shows that we can induce linear structure into TPM using any bijection TPM ↔ Rn (i.e. using any localcoordinate system x). The result is always the same: there is canonical linear structure in TPM . N

Finally, let’s have a look at an appropriate basis in TPM . Well, since there is a standard (”natural”) basisEj in Rn (number 1 at the j-th place of the n-tuple and 0 at all remaining places), the bijection via coordinatesx gives us unique equivalence class [γj ] which corresponds to Ej . And then, for general [γ] ↔ a = ajEj , we get

[γ] = aj [γj ]

And what about the representative γj (i.e. the j-th curve itself) generating the class [γj ]? Well, we need toreturn back to the general formula

xi(γ(t)) = xi(P ) + tai + . . .

and set ai corresponding to the basis vector Ej , i.e. ai = δij . We get

xi(γj(t)) = xi(P ) + tδij + . . .

Any curve γj(t) with this first order approximation induces the class [γj ]. The simplest one is clearly the curve

xi(γj(t)) = xi(P ) + tδij

(i.e. such that there are ”no dots at all” at the end of the expansion). Such curves γj(t) (for all j = 1, . . . , n)are known as coordinate curves. (For j-th coordinate curve, γj(t), the only coordinate whose value changes as tincreases is xj , namely the coordinate xj increases linearly.) So, in this approach to vectors on a manifold, wecan take as a basis of the tangent space TPM the vectors [γj(t)], equivalence classes generated by coordinatecurves γj(t).

Recall that there is an alternative notation γ for the equivalence class [γ] (see p.24 in the book; actually γis used much more frequently than [γ]). Using both notations, we can write for a general vector v

v = γ ≡ [γ] = ai[γi] = aiγi

2.4.3 and 2.4.4 p.35: A slightly more detailed discussion of higher dual spaces might be useful. In particular,we show here, first, how natural bases are introduced in all those spaces. And second, how one easily gains,from elementary properties of those bases, an alternative insight into the canonical isomorphism L→ L∗∗ from(2.4.3).

Recall that the n-th dual space of L is simply the dual space of the (n−1)-st one. That’s why the constructionof the dual basis from (2.4.2) may be repeated for each pair of adjacent duals. So, as soon as we pick up a basisea in L, the whole (infinite) chain of bases in all higher duals is canonically defined. Now, a change of basisea 7→ e′a = Abaeb in L by means of a non-singular matrix A results, according to (2.4.2), in the change of basisea 7→ e′

a= (A−1)abe

b by means of A−1 in L∗. Then, however, we also get Ea 7→ E′a = ((A−1)−1)baEb ≡ AbaEb,

Ea 7→ E′a = (((A−1)−1)−1)abEb ≡ (A−1)abE

b etc.

So, the overall situation reads as follows:

space L L∗ L∗∗ L∗∗∗ L∗∗∗∗ L∗∗∗∗∗ etc.

basis ea ea Ea Ea Ea Ea etc.

scrambled by A A−1 A A−1 A A−1 etc.

whereδab = ⟨ea, eb⟩ = ⟨Eb, ea⟩ = ⟨Ea, Eb⟩ = ⟨Eb, Ea⟩ = ⟨Ea, Eb⟩ = etc.

Now it is clear that the correspondence

ea ↔ Ea ↔ Ea ↔ etc.

provides the canonical isomorphisms between members of the (infinite) chain of the spaces

L ↔ L∗∗ ↔ L∗∗∗∗ ↔ etc.

6

First, it is clearly a bijection (i.e. a linear isomorphism, when extended by linearity from the basis to the entirespace L). Moreover, and this is crucial observation, it does not depend on the choice of a basis in L, since thebases within all members of the chain are scrambled in the same way - in terms of the matrix A.

[Compare this with the similarly looking correspondence ea ↔ ea. It gives a perfect isomorphism of L andL∗ as long as we keep the basis ea fixed. However, as soon as we change the basis ea in L, the isomorphism isreplaced by a different one, given by the new correspondence Abaeb ↔ (A−1)abe

b.]

And, perhaps, one should not be overly surprised that the isomorphism L↔ L∗∗ introduced here is nothingbut the one proposed in the hint to (2.4.3), i.e. the isomorphism f given by the formula

f : L→ L∗∗ ⟨f(v), α⟩ := ⟨α, v⟩

Indeed, in terms of the bases ea and Ea in L and L∗∗ respectively we have

f(ea) = BbaEb

with a matrix B. Then

⟨f(ea), eb⟩1.= ⟨BcaEc, eb⟩ = Bca⟨Ec, eb⟩ = Bba2.= ⟨eb, ea⟩ = δba

so thatf(ea) = Ea

2.4.18 p.44: For a general linear map A : L1 → L2, the statement is only true for positive (or negative) definitemetric tensor h in L2. If h fails to be positive (or negative) definite, induced tensor A∗h still may be ”good”,i.e. non-degenerate, but one must be careful in choosing ”good” map A.)

Consider, to illustrate the matter, the two-dimensional ”Minkowski” (linear) space, i.e. the (indefinite) space(L2, h) with

h = e1 ⊗ e1 − e2 ⊗ e2

Let L1 be one-dimensional, spanned by e1. A general linear map A : L1 → L2 is given as

A : e1 7→ ue1 + ve2 u, v ∈ R

Then (check)A∗h = · · · = (u2 − v2) e1 ⊗ e1

So the resulting induced tensor A∗h is

- positive definite (= non-degenerate) for u2 − v2 > 0

- negative definite (= non-degenerate) for u2 − v2 < 0

- vanishing (most degenerate possible)) for u2 − v2 = 0 (i.e. for u = ±v)The problem lies in the third possibility. Although, say,

A : e1 7→ e1 + e2

is a perfect maximum-rank linear map L1 → L2 (the rank being 1 :-), the resulting induced tensor A∗h vanishes,i.e., it is definitely not a metric tensor in L1. (Actually, it is induced from an isotropic subspace of L2.)

2.6 p.52: At the end of the page, gradf , the gradient of f , is introduced as a vector field (rather than thecovector field df). It’s important property should be mentioned: at any point x, the vector gradf is orthogonalto the level surface of f passing through x (i.e. the hypersurface f = const.). [Meaning it is orthogonal to anyvector v tangent to the surface.] Indeed, for any vector v we have

g(gradf, v) = ⟨df, v⟩ = vf

due tog(gradf, v) = gij(gradf)

ivj = gijgik(df)kv

j = (df)kvk = ⟨df, v⟩

7

(the first equal sign) and 2.5.3(i) (the second equal sign). Then

g(gradf, v) = 0 ⇔ vf = 0 ⇔ v is tangent to the hypersurface f = const.

There is a ”skew” analog of all this (see Chapter 14): the ”skew gradient” of f (= the ”hamiltonian field” ζfgenerated by f) is ”skew-orthogonal” to the hypersurface f = const., meaning

ω(ζf , v) = 0 ⇔ vf = 0 ⇔ v is tangent to the hypersurface f = const.)

(”Skew-orthogonality” is defined with respect to ω.)

2.6 p.52: The components of the gradient (as a vector field) are given by (gradf)i = gik∂kf . You maybe disappointed with this result, however, trying to compute it, say, in spherical polar coordinates and thenchecking it versus the result in your favorite book on, say, electrodynamics. The two results differ. Why?Because in your electrodynamics book orthonormal components are displayed whereas the result mentionedhere refers to coordinate components, respectively. See more details in Chapter 8 (e.g. 8.5.10).

3.1.7 p.59: In order to provef∗ C = C f∗

for a diffeomorphism f :M → N , one should realize that contractions are performed on different manifolds (Mand N , respectively) where, consequently, different frame fields are used.

Assume that ei and ei is a (mutually dual) frame and co-frame field on (a part of) M and the same is true

for ea and ea on (a part of) N . Then

(f∗ Ct)(V, . . . ;α, . . . ) = (Ct)(f∗V, . . . ; f∗α, . . . )

= t(f∗V, . . . , ea, . . . ; f∗α, . . . , ea, . . . )

whereas

(C f∗t)(V, . . . ;α, . . . ) = (f∗t)(V, . . . , ei, . . . ;α, . . . , ei, . . . )

= t(f∗V, . . . , f∗ei, . . . ; f∗α, . . . , f∗ei, . . . )

So, the second result may be obtained from the first one by the replacement ea 7→ f∗ei and ea 7→ f∗e

i. Noticehowever that, f being a diffeomorphism, f∗ei and f∗e

i is an equally good mutually dual frame/co-frame fieldon N as is ea and ea. And since in (2.4.8) we learned that contraction does not depend on the choice of theframe/co-frame field (provided that they satisfy the duality condition), we are done.

4.3.6 p.74: A useful fact about commutator of vector fields is that it is preserved by maps. Let us explain inmore detail what does it mean and prove it is really the case.

Consider two vector fields U, V on M . Let f :M → N be a map and suppose the fields U, V are projectablewith respect to f (see the text before 3.1.3). Denote the images U , V (they are vector fields on N)

f∗U = U f∗V = V

Notice that the fact that U is projectable (and f∗U = U) may also be expressed in terms of action of fields onfunctions as follows (recall that (Uχ)(m) = Umχ)

(Uψ) f = U(ψ f) (≡ U(f∗ψ))

or Uf(m)ψ = Um(ψ f) (≡ Um(f∗ψ))

where ψ is a function on N . The same is clearly true for V . Then, the straightforward calculation shows that

(f∗[U, V ]m)ψ = [U, V ]m(ψ f)= Um(V (ψ f))− Vm(U(ψ f))

= Um((V ψ) f)− Vm((Uψ) f)

= Uf(m)(V ψ)− Vf(m)(Uψ)

= [U , V ]f(m)ψ

8

i.e.f∗[U, V ]m = [U , V ]f(m)

or, finally, we get the preservation of commutator of (f -projectable, otherwise the r.h.s. makes no sense) vectorfields in the form

f∗[U, V ] = [f∗U, f∗V ]

Note: this fact is (implicitly) also present for the particular case of a diffeomorphism f : M → M (wheref∗ = f−1

∗ ) in the result f∗LV = Lf∗V f∗ of 8.3.7. Indeed, when applied on U we get

f∗(LV U) ≡ f∗[V,U ] = Lf∗V f∗U ≡ [f∗V, f∗U ]

The computation displayed above shows its validity for a general (smooth) map f :M → N .

6.2.8, 6.2.9 p.131: the hint for the proof might be even simpler. We are to prove (in 6.2.8, say) the Cartan’s

”magic formula”LV = iV d+ diV

It is written in the hint, that the formula to be proved is an equality of two derivations (of degree 0) of thealgebra Ω(M) and therefore it is enough to verify it on degrees 0 and 1, ”where it is easy (e.g. in components”).Now, for the degree 1 case, no components are needed. Actually 1-forms of the structure dψ (i.e. exact 1-forms)suffice. (Observe that all p-forms are sums of expressions of the structure fdψ∧· · ·∧dχ, i.e. products of 0-formsand exact 1-forms. This is, btw. used in the hint for 6.2.11!) So, one is only to prove

LV f = (iV d+ diV )f

LV df = (iV d+ diV )df

which is evident (just definitions are needed).

7.2.4 p.148: Some more details to the hint: Denote Pjµ and Qjµ the j-th coordinate of the points Pµ andQµ respectively (j = 1, . . . , n, µ = 0, 1, . . . , n). Then the fact, that x 7→ Ax + a maps Pµ into Qµ (for eachµ = 0, 1, . . . , n) leads to the system of equations

AijPjµ + ai = Qiµ µ = 0, 1, . . . , n

We are to show that, given the points Pµ and Qµ, a unique pair (A, a) emerges. If we choose P0 = (0, . . . , 0), P1 =(1, . . . , 0), . . . , Pn = (0, . . . , 1), so that Pj0 = 0, Pjk = δjk, the equations read

ai = Qi0 Aijδjk + ai = Qik

so thatai = Qi0 Aij = Qij −Qi0

(one can also obtain the same result without any computation after some contemplation). So, any simplex maybe obtained from the standard one by means of a unique affine transformation. Consequently, any two simplicesmay be mapped into one another by means of a unique affine transformation. [Map the first simplex to thestandard one (by the inverse of (A, a) given above) and then, in the second step, map the standard one to thesecond simplex.]

7.2.6 p.149: In part (ii) the third, more explicit (and useful), characterization of the standard simplex sp maybe mentioned:

sp : (x1, . . . , xp) such that 0 ≤ x1 ≤ 1

0 ≤ x2 ≤ 1− x1

0 ≤ x3 ≤ 1− x1 − x2

. . .

0 ≤ xp ≤ 1− x1 − · · · − xp−1

Therefore, integration over sp amounts to∫sp

(. . . ) =

∫ 1

0

dx1∫ 1−x1

0

dx2· · ·∫ 1−(x1+···+xp−1)

0

dxp (. . . )

9

Notice that ∫sp+1

(. . . ) =

∫sp

∫ 1−(x1+···+xp)

0

dxp+1 (. . . )

where sp at the r.h.s. is the base sp of the sp+1 (such face of the boundary, where xp+1 = 0). So, as an example,

∫s3

(. . . ) =

∫ 1

0

dx

∫ 1−x

0

dy

∫ 1−(x+y)

0

dz (. . . ) =

∫s2

∫ 1−(x+y)

0

dz (. . . )

where s2 at the r.h.s. is the base s2 of the s3 (such face of the boundary, where z = 0; the ”floor” of the3-simplex).

7.5.3 p.153: more details for arbitrary p:1.: The boundary of the standard simplex sp+1 ≡ (P0, P1, . . . , Pp+1) contains p+ 1 terms. Among them, thereare p− 1 ”internal” terms, where the points P1, . . . , Pp are omitted (in each term a single point). Realize thatwhen Pk (the k-th one) is omitted, the corresponding p-simplex does not ”stick into the k-th dimension” (forall the points xk = 0 holds). This means, however, that the pull-back of the differential dxk with respect tothe map Φk (into the standard simplex sp) vanishes (Φ

∗kdx

k = d0 = 0). This leads to vanishing of the completeform Φ∗

kη = Φ∗k(fdx

1 ∧ · · · ∧ dxp) to be integrated over this particular sp. Since this is true for all ”internal”points P1, . . . , Pp mentioned above, there are just two terms of the boundary which do contribute to the integral- those in which the first and the last points are omitted respectively. They read explicitly

∂sp+1 = ∂(P0, P1, . . . , Pp+1)

= (P1, . . . , Pp+1) + (−1)p+1(P0, P1, . . . , Pp) (plus irrelevant terms)

= (−1)p+1[(P0, P1, . . . , Pp)− (Pp+1, P1, . . . , Pp)]

2.: So, the ”boundary” side of the Stokes theorem is∫∂sp+1

η = (−1)p+1[

∫(P0,P1,...,Pp)

η −∫(Pp+1,P1,...,Pp)

η]

whereη := f(x1, . . . , xp, xp+1)dx1 ∧ · · · ∧ dxp

3.: The ”bulk” side of the Stokes theorem may be expressed, as mentioned in the book (see also comment to(7.2.6) in this text), in the form∫

sp+1

dη = (−1)p+1[

∫sp

dx1 . . . dxpf(x1, . . . , xp, 0)− f(x1, . . . , xp, 1− (x1 + · · ·+ xp))]

where ∫sp

dx1 . . . dxp (. . . ) ≡∫ 1

0

dx1∫ 1−x1

0

dx2· · ·∫ 1−(x1+···+xp−1)

0

dxp (. . . )

4.: Then, the statement∫sp+1

dη =∫∂sp+1

η is true if

∫(P0,P1,...,Pp)

f(x1, . . . , xp, xp+1)dx1 ∧ · · · ∧ dxp =∫sp

dx1 . . . dxpf(x1, . . . , xp, 0)∫(Pp+1,P1,...,Pp)

f(x1, . . . , xp, xp+1)dx1 ∧ · · · ∧ dxp =∫sp

dx1 . . . dxpf(x1, . . . , xp, 1− (x1 + · · ·+ xp−1))

5.: The p-simplex (P0, P1, . . . , Pp) forms the base (floor) of the standard (p + 1)-simplex sp+1 (so xp+1 = 0 onit). It is clearly the Φ-image of the standard p-simplex sp:

(P0, P1, . . . , Pp) = Φ(sp) Φ : (u1, . . . , up) 7→ (u1, . . . , up, 0)

10

Using the definition∫Φ(sp)

η =∫sp

Φ∗η we immediately get the first equality in 4.

6.: The p-simplex (Pp+1, P1, . . . , Pp) forms the slant face of the standard (p + 1)-simplex sp+1. (One obtains(Pp+1, P1, . . . , Pp) from (P0, P1, . . . , Pp) by lifting the vertex P0 to the height 1 along xp+1 and a general point(x1, . . . , xp) inside to the height 1 − (x1 + · · · + xp); draw the pictures for p = 1, 2.) It is the Φ-image of thestandard p-simplex sp:

(Pp+1, P1, . . . , Pp) = Φ(sp) Φ : (u1, . . . , up) 7→ (u1, . . . , up, 1− (u1 + · · ·+ up))

Using again the definition∫Φ(sp)

η =∫sp

Φ∗η for this Φ, we get the second equality in 4.

8.1.3 p.165: Here we learned the (generalized) ”integration by parts” formula∫D

dα ∧ β = −∫D

ηα ∧ dβ +

∫∂D

α ∧ β

which holds for any two forms α and β such that their degrees fulfil degα + deg β + 1 = n and D being ann-dimensional domain, and, as a simplest special case, the ”ordinary” 1-dimensional formula∫ b

a

f ′(x)g(x)dx = −∫ b

a

f(x)g′(x)dx+ [fg]ba

Let us mention (here) what the general formula offers in vector analysis context (see Section 8.5. for treatmentof vector analysis in terms of forms).

Here (in E3), we have degα+ deg β = 2, leaving just two cases of interest,

1. degα = 0 deg β = 2

2. degα = 1 deg β = 1

From (8.5.2) then

1. α = f β = A · dS2. α = A · dr β = B · dr

so that (from (8.5.4))

1. dα = (∇f).dr dβ = (divA) dV

2. dα = curlA · dS dβ = curlB · dS

Therefore, taking int account

(A.dr) ∧ (B.dr) = (A×B).dS

(A.dr) ∧ (B.dS) = (A.B)dV

from (8.5.8), the ”general by parts” formula gives us the following two integral identities:

1.

∫D

(∇f ·A) dV = −∫D

f(divA) dV +

∫∂D

(fA) · dS

2.

∫D

(B · curlA) dV =

∫D

(A · curlB) dV +

∫∂D

(A×B) · dS

8.1.4 p.166: see additional material to (8.5.6) here

8.3.6 p.174: It is useful to realize that jV depends actually on as many as two tensor fields, V and g. So, themore precise way is to write it as jV,g:

jV,gα := g(V, . ) ∧ α

Then, the fact that it is natural with respect to diffeomorphisms reads

f∗jV,gα = jf∗V,f∗gf∗α

11

This is indeed readily verified:

f∗(jV,gα) = f∗(g(V, . ) ∧ α) = f∗(g(V, . )) ∧ f∗α = (f∗g)(f∗V, . ) ∧ f∗α ≡ jf∗V,f∗gf∗α

Or, alternatively, one can use the flat symbol g to express

jV,gα := (gV ) ∧ α

and write

f∗(jV,gα) = f∗((gV ) ∧ α) = (f∗(gV )) ∧ (f∗α) = (f∗gf∗V ) ∧ (f∗α) ≡ jf∗V,f∗gf

∗α

where we used naturalness of g (see 8.3.8)

8.5.4 p.181: Two-dimensional vector analysis (in E2 rather than E3) is mentioned (in a footnote, p.179) as asimpler realization of the usual one. Perhaps some additional remarks might be useful concerning this topics.(See also additional material to (8.5.6) here.)

First, rewrite the diagram (8.5.4) (valid for three dimensions)

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)d−−−−→ Ω3(M)

id

y y♯ y♯∗ y∗

F(M) −−−−→grad

X(M) −−−−→curl

X(M) −−−−→div

F(M)

in the form

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)d−−−−→ Ω3(M)

a0

y ya1 ya2 ya3F(M) −−−−→

A1

X(M) −−−−→A2

X(M) −−−−→A3

F(M)

where a0, a1, a2, a3 denote isomorphisms encoding scalar and vector fields into forms of various degrees andA1, A2, A3 are the effective differential operators from vector analysis. Then, clearly

A1 = a1 d a−10

A2 = a2 d a−11

A3 = a3 d a−12

so thatA2 A1 = 0 A3 A2 = 0

These are, in reality, the statements

curl grad = 0 div rot = 0

Now, the corresponding general counterpart in two dimensions reads

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)

a0

y ya1 ya2F(M) −−−−→

A1

X(M) −−−−→A2

F(M)

A1 = a1 d a−10 A2 = a2 d a

−11 A2 A1 = 0

Notice, however, that on a two-dimensional M the star operator ∗, when applied to ”middle degree forms”,is an operator on Ω1 (so it does not connect forms of different degrees, Ω1(M) ↔ Ω2(M), as is the casein E3). Therefore, there are as many as two distinct natural identifications of vector fields with one-forms,Ω1(M) ↔ X(M): via ♯ : Ω1(M) → X(M) and via ♯∗ : Ω1(M) → X(M). Or, when looking from the other

12

side, there are two ways how vector fields may be encoded into one-forms, via : X(M) → Ω1(M) and∗ : X(M) → Ω1(M).

Therefore we get, instead of the single diagram in three dimensions, as many as two corresponding generaldiagrams in two-dimensional vector analysis:

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)

a0

y ya1 ya2F(M) −−−−→

A1

X(M) −−−−→A2

F(M)

and

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)

b0

y yb1 yb2F(M) −−−−→

B1

X(M) −−−−→B2

F(M)

A1 = a1 d a−10 A2 = a2 d a

−11 A2 A1 = 0

B1 = b1 d b−10 B2 = b2 d b

−11 B2 B1 = 0

When concrete isomorphisms ai and bi are inserted, we get

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)

id

y y♯ y∗

F(M) −−−−→grad

X(M) −−−−→D2

F(M)

and

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)

id

y y♯∗ y∗

F(M) −−−−→D1

X(M) −−−−→−div

F(M)

and

grad = ♯ d D2 = ∗ d D2 grad = 0

D1 = ♯ ∗ d div = ∗−1 d ∗ div D1 = 0

(We have, on vector fields in E2,

∗d(♯∗)−1 = ∗d ∗−1 = − ∗ d ∗ = − ∗ ηd ∗ = − ∗−1 d ∗ = ∗−1d ∗ η = δ = −div

due to ∗−1 = ∗η, see 5.8.2, and δV = −divV , see 8.3.4).

So, in comparison with the three-dimensional case, there are still, in two dimensions, the two good oldoperations grad and div , but, first, curl (as an operation vector field 7→ vector field) is missing and, second,two ”new” differential operations (denoted here as D1 and D2) emerged.1 The first, D1, is of type scalar 7→vector (like the gradient) whereas the second, D2, is of type vector 7→ scalar (like the divergence). In particular,in Cartesian coordinates in E2 we have

grad : f 7→ (∂1f, ∂2f)

D1 : f 7→ (−∂2f, ∂1f)D2 : (A1, A2) 7→ (∂1A2 − ∂2A1)

div : (A1, A2) 7→ (∂1A1 + ∂2A2)

Observe that the composition of two operators belonging to the same diagram indeed vanishes (since it is justmasked dd = 0)

D2 grad : f 7→ (∂1f, ∂2f) 7→ (∂1∂2f − ∂2∂1f) = 0

div D1 : f 7→ (−∂2f, ∂1f) 7→ (∂1(−∂2f) + ∂2(∂1f)) = 0

Composition of two operators belonging to distinct diagrams leads (for both possibilities) to the Laplace operator.Abstractly

div grad = D2 D1 = ∗d ∗ d =

1Notice on component expressions below, however, that D2A is just the 3-rd component of the three-dimensional curl f . AlsoD1f may be regarded as hamiltonian vector field generated by f , since the (metric) volume form ω can be viewed as a symplectic

form (see below; also Chapter 14).

13

and in particular, in Cartesian coordinates

div grad : f 7→ (∂1f, ∂2f) 7→ (∂1∂1f + ∂1∂1f) = fD2 D1 : f 7→ (−∂2f, ∂1f) 7→ (∂1(∂1f)− ∂2(−∂2f)) = f

There is still another (and perhaps the most natural) point of view: the manifold E2 is, at the same time,Riemannian and symplectic manifold (see Chapter 14).2 The symplectic form is just dx ∧ dy, the standard(metric) volume form in E2. Therefore, there are as many as two natural ways of identifying vectors with1-forms (i.e. two natural ways of raising and lowering of indices) - in terms of the metric tensor g and in termsof the symplectic form ω. Therefore,

- two vector fields may be associated with gradient df as a covector field, the gradient ∇f ≡ grad f as wellas the Hamiltonian field ζf ; the latter is nothing but D1f :-)

- two ways of lowering indices on a vector field W result in two distinct 1-forms associated with the vectorfield; then the exterior derivative d produces two distinct 2-forms, both of them proportional, however, tothe (metric) volume form; so, the ”proportionality coefficients” may be regarded as (two distinct) functionsassociated to the vector field; they turn out to be (check) D2W (for the metric version) and divW (for thesymplectic version) respectively.

As a simple (and perhaps useful) application, consider 2-dimensional incompressible fluid flow. Incom-pressibility gives (from continuity equation)

div v = 0

So, at least locallyv = D1ψ ≡ ζψ

for some function ψ (see the right diagram above). This function (”potential” of the velocity field) is known asstream function in 2D-hydrodynamics. Why? Since D1ψ is nothing but the Hamiltonian field generated by ψ,we get

vψ = ζψψ = ψ,ψ = 0

(Poisson bracket of two equal entries). So ψ is constant along stream-lines. Put it differently, the fluid flowis everywhere parallel to contours (level lines) of ψ. So, by plotting level sets (lines, here) of ψ, we can ”see”,modulo direction, the fluid flow.

Notice also that when Laplace operator is applied on ψ, we get vorticity function ω

∆ψ = ω

Indeed, we get∆ψ = D2 D1ψ = D2v = ∗ d v = ∗ (dv)

Now dv is just vorticity form, which, being a two-form, is necessarily proportional to the volume form, theproportionality factor being (by definition) the vorticity function ω. And since Hodge star gives just 1 whenapplied on the volume form, we get the desired result.

8.5.6 p.182: When the general Stokes theorem∫Ddα =

∮∂D

α is applied on statements obtained from threeindividual blocks (three appearances of d) of the combined diagram

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)d−−−−→ Ω3(M)

id

y y♯ y♯∗ y∗

F(M) −−−−→grad

X(M) −−−−→curl

X(M) −−−−→div

F(M)

we get the following triple of ”classical” integral theorems

Ω0(M)d−−−−→ Ω1(M)

id

y y♯F(M) −−−−→

gradX(M)

↔ (∇f).dr = df ↔∫C

(∇f).dr = f(B)− f(A)

2Manifolds like E2, where both Riemannian and symplectic structure is available (and, in addition, these structures are related

in a specific way) are known as Kahler manifolds.

14

Ω1(M)d−−−−→ Ω2(M)y♯ y♯∗

X(M) −−−−→curl

X(M)

↔ (curlA).dS = d(A.dr) ↔∫S

(curlA).dS =

∮∂S

A.dr

Ω2(M)d−−−−→ Ω3(M)y♯∗ y∗

X(M) −−−−→div

F(M)

↔ (divA)dV = d(A.dS) ↔∫D

(divA)dV =

∮∂D

A.dS

(the gradient one, the Stokes one and the Gauss one). Now, we can mimic the procedure for both two-dimensionaldiagrams (see comment to 8.5.4 here in this text)

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)

id

y y♯ y∗

F(M) −−−−→grad

X(M) −−−−→D2

F(M)

and

Ω0(M)d−−−−→ Ω1(M)

d−−−−→ Ω2(M)

id

y y♯∗ y∗

F(M) −−−−→D1

X(M) −−−−→−div

F(M)

What we get is the following quadruple of ”classical” (two-dimensional) integral theorems:

Ω0(M)d−−−−→ Ω1(M)

id

y y♯F(M) −−−−→

gradX(M)

↔ (∇f).dr = df ↔∫C

(∇f).dr = f(B)− f(A)

Ω0(M)d−−−−→ Ω1(M)

id

y y♯∗F(M) −−−−→

D1

X(M)

↔ ∗(D1f).dr = −df ↔∫C

∗(D1f).dr = f(A)− f(B)

Ω1(M)d−−−−→ Ω2(M)y♯ y∗

X(M) −−−−→D2

F(M)

↔ (D2A)dS = d(A.dr) ↔∫S

(D2A)dS =

∮∂S

A.dr

Ω1(M)d−−−−→ Ω2(M)y♯∗ y∗

X(M) −−−−→−div

F(M)

↔ (divA)dS = d(∗(A.dr)) ↔∫S

(divA)dS =

∮∂S

∗(A.dr)

The first and the second integral statements just say (in cartesian coordinates) the same (and well-known)thing, namely that (check) ∫

C

(∂xf)dx+ (∂yf)dy = f(B)− f(A)

The third and the fourth integral statements say the following (check):∫S

(∂xAy − ∂yAx)dxdy =

∮∂S

Axdx+Aydy

∫S

(∂xAx + ∂yAy)dxdy =

∮∂S

Axdy −Aydx

15

Although these two statements perhaps look different at first sight, a bit closer look shows (just rename(Ax, Ay) 7→ (−Ay, Ax) in the first one and You get the second one) that it is actually a single statementagain, namely the classical Green theorem∫

S

(∂xg − ∂yf)dxdy =

∮∂S

fdx+ gdy

(see 8.1.4). So, there are altogether no more than two independent statements emerging from∫Ddα =

∮∂D

α intwo-dimensional vector analysis.

8.5.6 p.182: see additional material to (8.1.3) here

8.5.8 p.183: The following immediate consequences from the results listed in (ii) might be useful:

(A.dr)(B) = A ·B(A.dS)(B,C) = A · (B ×C)

(hdV )(A,B,C) = h(A · (B ×C))

10.2.6 p.212: The group of the upper-triangular matrices

A(a, b, c) =

1 a c0 1 b0 0 1

mentioned here, with the composition law (check)

A(a, b, c)A(a, b, c) = A(a+ a, b+ b, c+ c+ ab)

is also known as the Heisenberg group. Why? Let us compute its Lie algebra. Using the method of computationintroduced in (11.7.5) we can immediately see that it is a vector space consisting of the matrices of the form

X(x, y, z) =

0 x z0 0 y0 0 0

= xE1 + yE2 + zE3

where

E1 =

0 1 00 0 00 0 0

E2 =

0 0 00 0 10 0 0

E3 =

0 0 10 0 00 0 0

The commutation relations read

[E1, E2] = E3 [E1, E3] = 0 [E2, E3] = 0

Now compare these relations with the celebrated Heisenberg (”canonical”) commutation relations (CCR)

[p, x] = −i~1 [p, i~1] = 0 [x, i~1] = 0

We can see that this is the same (abstract) Lie algebra.

In order to match the Heisenberg commutation relations for more canonical variables

[pi, xj ] = −i~δij 1 [pj , i~1] = 0 [xj , i~1] = 0

we promote x, y to become n-dimensional columns (z remains to be a number). Namely, a Lie algebra elementis, in general,

X(x, y, z) =

0 xT z0 0n y0 0 0

= xiEi + yiFi + zG

16

Then one checks that[Ei, Fj ] = δijG [Ei, G] = 0 [Fi, G] = 0

and so we are done.

In the coordinates introduced so far (for n = 1, to be specific), we have the following obvious possibility torelate the coordinates on the group and on the Lie algebra respectively:

A(a, b, c) = I+X(x, y, z)

i.e. 1 x z0 1 y0 0 1

!=

1 a c0 1 b0 0 1

(*)

i.e.

a(x, y, z) = x

b(x, y, z) = y

c(x, y, z) = z

So, the coordinates on the Lie algebra can directly serve as coordinates on the group.

There is also another standard parametrization of the group. Note that for a general Lie algebra element wehave

X =

0 x z0 0 y0 0 0

X2 =

0 0 xy0 0 00 0 0

X3 =

0 0 00 0 00 0 0

so that the exponential of X is simply

eX = I+X +1

2X2 =

1 x z + 12xy

0 1 y0 0 1

!=

1 a c0 1 b0 0 1

(**)

which provides an alternative parametrization of upper-triangular matrices: We can, again, use (x, y, x) ascoordinates on the group (so we can use coordinates on the Lie algebra as coordinates on the group). However,their relation to “original” (a, b, c)-type coordinates is more complicated, namely

a(x, y, z) = x

b(x, y, z) = y

c(x, y, z) = z +1

2xz

One can check that the composition law in the group within these particular (“exponential”) coordinates (x, y, z)reads

(x, y, z) (x, y, z) = (z + x, y + y, z + z +1

2(xy − yx))

11.2.2 p.223: We introduced the commutator [X,Y ] of two vectors from G ≡ TeG by the formula

[X,Y ] := [LX , LY ](e) (*)

Bilinearity and skew-symmetry is clear. Here, let us discuss in more detail how Jacobi identity emerges (i.e.why the commutator is well-defined by this formula :-)

So, let us have X,Y, Z ∈ G. Construct corresponding left-invariant fields LX , LY , LZ . Since Jacobi indentitydoes hold for commutator of vector fields (in general, see 4.3.6), we have

[[LX , LY ], LZ ] + [[LZ , LX ], LY ] + [[LY , LZ ], LX ] = 0

Evaluating both sides in e ∈ G we get

[[LX , LY ], LZ ](e) + [[LZ , LX ], LY ](e) + [[LY , LZ ], LX ](e) = 0 (**)

17

Now, [LX , LY ] is a left-invariant field due to 8.3.7 (or 4.3.6 - see the additional material to the latter). Then,due to 11.1.4i), it is LW for some W ∈ G. From (*) we see, that W = [X,Y ], so [LX , LY ] = L[X,Y ]. Thereforewe can rewrite (**) as

[L[X,Y ], LZ ](e) + [L[Z,X], LY ](e) + [L[Y,Z], LX ](e) = 0 (***)

Using once more (*) we are done:

[[X,Y ], Z] + [[Z,X], Y ] + [[Y, Z], X] = 0

11.4.4 p.229: It might be instructive to treat in parallel both additive and multiplicative groups of real numbers,i.e. the groups

(R,+) : m(x, y) = x+ y

and GL(1,R) : m(x, y) = xy

The corresponding basis left-invariant vector fields are ∂x and x∂x respectively, so we get

group left-invariant fields one-parameter subgroups exponential map

(R,+) k∂x x(t) = kt k 7→ x(1) = k

GL(1,R) kx∂x x(t) = ekt k 7→ x(1) = ek

Now think of the additive group and its Lie algebra. Note that the Lie algebra structure just adds the multi-plication by scalars and the (trivial) commutator to the structure of the group. Therefore, there is a naturalidentification (!) of the group manifold R with the Lie algebra manifold R, namely x↔ x. After the identifica-tion, the exponential map comes out to be just the ”identity” map k 7→ k on R.

For the multiplicative group, the Lie algebra manifold R is mapped (via the ”ordinary” exponential mapk 7→ ek, as mentioned in the assignment of the exercise) onto the subgroup GL+(1,R) of the group GL(1,R)(the connected component of the identity).

11.5.2 p.230: in (ii) we are to inspect that Lf ′(X) = f∗LX . It is computed straightforwardly. First note, thathomomorphism property of f may be written as f Lg = Lf(g) f . Then

Lf ′(X)(f(g)) = Lf(g)∗f′(X) = Lf(g)∗f∗X = (Lf(g) f)∗X = (f Lg)∗X = f∗LX(g)

which is just detailed version of the statement Lf ′(X) = f∗LX .

11.7.20, 21 p.241: Let us discuss the method of computation of the canonical 1-form on matrix groups even

more explicitly (the method described here is very convenient so one should not feel sorry for additional fewminutes spent on full understanding of the matter).

So, first, consider Lie groups G and H and a smooth injective homomorphism

f : G→ H

Let

f ′ : G → H

be the corresponding derived homomorphism of Lie algebras. If Ei is a basis in G and Ea is a basis in H, thenthe matrix fai of f ′ is given by

f ′(Ei) = fai Ea (1)

The fact that f is homomorphism

f(g1g2) = f(g1)f(g2)

may be rewritten as a condition

f Lg = Lf(g) f (2)

relating left translations Lg and Lh on G and H respectively.

18

Now recall (see 11.2.6) that the definition of the canonical 1-forms θG and θH (on G and H respectively)reads

θGg := Lg∗ : TgG→ TeGG ≡ GθHh := Lh∗ : ThH → TeHH ≡ H

(Here θGg denotes the 1-form θG at point g ∈ G and eG is the unit element on G. The Lie algebra G of G isidentified with the tangent space TeGG at eG.) Therefore, taking (. . . )∗ of both sides of (2) we get

f∗ θGg = θHf(g) f∗ (3)

The canonical 1-forms may be decomposed (see 11.2.6) as follows:

θG = eiEi (4)

θH = eaEa (5)

Then (3) givesf∗ (eigEi) = (eaf(g)Ea) f∗ (6)

Notice that both side are mappings TgG→ H. On argument v ∈ TgG we obtain

(f∗ (eigEi))(v) := f∗(⟨eig, v⟩Ei) = ⟨eig, v⟩f ′(Ei) = (eigf′(Ei))(v)

((eaf(g)Ea) f∗)(v) := ⟨eaf(g), f∗v⟩Ea = ⟨(f∗ea)g, v⟩Ea = ((f∗ea)gEa)(v)

oreif ′(Ei) = f∗(eaEa) (7)

This can be briefly written as

f∗θH = f ′(θG) where f ′(θG) ≡ f ′(eiEi) := eif ′(Ei) (8)

So, performing pull-back f∗ of the canonical 1-form θH we get ”almost” canonical 1-form θG, the difference liesin that the resulting 1-form takes values in f ′(G) ⊂ H rather than in G itself.

Now, let us specify H to be GL(n,R). Then f ≡ ρ becomes representation of G in Rn

ρ : G→ GL(n,R)

Natural (”Weyl”) basis in the Lie algebra gl(n,R) is labeled by as many as two indices, so (1) becomes

ρ′(Ei) = ρbaiEab (9)

and (5) takes the form

θGL(n,R) = eabEba ≡ (x−1dx)abE

ba

where we already used the explicit coordinate expression of the canonical 1-form on GL(n,R) known from11.7.19. In this particular case we get from (7)

(ρ∗(x−1dx)ba)Eab = eiρ′(Ei) = (ρbaie

i)Eab (10)

Taking into account that the representation ρ : G→ GL(n,R) has coordinate expression

z 7→ x(z) i.e. in detail zµ 7→ xba(z) (11)

we get from (10)

(x−1)bc(z)dxca(z) = ρbaie

i (12)

Finally, recalling thatρbai = (Ei)ba (13)

19

are nothing but (. . . )ba-elements of generators

Ei := ρ′(Ei) (14)

of the representation ρ (see 12.1.4), we can rewrite (12) as

x−1(z)dx(z) = eiEi (15)

This is the final formula enabling one to compute basis left-invariant 1-forms ei (and, consequently, the canonical1-form θG = eiEi on matrix Lie group G). One has to perform the following steps

- write down the formula (11) (i.e. parametrize the matrix group G)

- compute the expression x−1(z)dx(z)

- identify coefficients (1-forms) ei standing by the generators EiNormally, one uses the ”identity” representation ρ = id , ρ(A) = A, of the matrix group G. See explicit

examples in 11.7.22 and 11.7.23.

12.1.13 p.251: More general and elegant result may be obtained virtually with the same effort: If we start

with a bilinear form (not necessarily symmetric!) given by a general real matrix, the procedure (1/2π)∫ 2π

0dα

(averaging over SO(2)) leads to (check)(a bc d

)7→ a+ d

2

(1 00 1

)+b− c

2

(0 1−1 0

)This nicely shows that the only invariant bilinear forms are (any multiple of) the ordinary scalar product and(any multiple of) the ordinary volume form. You can also check that repeated averaging already does nothing(the map behaves as a projector) (

a b−b a

)7→

(a b−b a

)See also 12.2.8.

12.2.5 p.254: Here, it might be not clear what precisely Nj mean (although a reader may deduce it fromdescription of so(r, s) in 11.7.6) so I add a more detailed account of so(1, 3), which is important in its own right.

Consider matrices C from the Lie algebra so(1, 3) (the Lie algebra of the Lorentz group), i.e. (see 11.7.6)real 4× 4 matrices, satisfying

(ηC)T + ηC = 0 η ≡(1 0T

0 −I

)(0 is the 3-dimensional null column and I is the 3× 3 unit matrix). Check thati) in the 1+3 block form (similar to the form of the matrix η) we can parameterize the matrices C as follows

C(u,X) =

(0 uT

u X

)XT = −X

(u is a 3-dimensional real column and X is a 3× 3 real antisymmetric matrix, i.e. X ∈ so(3))ii) the natural decomposition of the resulting most general matrix(

0 uT

u X

)=

(0 0T

0 X

)+

(0 uT

u 0

)i.e.

C(u,X) = C(0, X) + C(u, 0)

may be regarded as a decomposition of the total 6-dimensional linear space so(1, 3) into the direct sum of two3-dimensional subspaces

so(1, 3) = so(1, 3)rot + so(1, 3)boo

and the elements of the corresponding subspaces generate rotations and boosts (= ,,genuine” Lorentz transfor-mations). So, if x = xµ = t, x1, x2, x3 is a point in the Minkowski space, then the rules

x 7→ (1 + ϵC(0, X))x and x 7→ (1 + ϵC(u, 0))x

20

describe infinitesimal rotations and infinitesimal boosts respectivelyiii) the commutation relations in the Lie algebra read

[C(u1, X1), C(u2, X2)] = C(X1u2 −X2u1, [X1, X2] + u1uT2 − u2u

T1 )

and, in particular

[C(0, X), C(0, Y )] = C(0, [X,Y ])

[C(u, 0), C(v, 0)] = C(0, uvT − vuT )

[C(0, X), C(u, 0)] = C(Xu, 0)

From this we see that so(1, 3)rot is a (Lie) subalgebra (isomorphic to so(3)), but so(1, 3)boo fails to be a subalgebra(being just a subspace)

[so(1, 3)rot, so(1, 3)rot] ⊂ so(1, 3)rot

[so(1, 3)boo, so(1, 3)boo] ⊂ so(1, 3)rot

[so(1, 3)rot, so(1, 3)boo] ⊂ so(1, 3)boo

iv) if we choose natural bases in the spaces of X’s and u’s (for X’s the matrices lj from exercise 11.7.13 and foru’s the columns ej with a single 1 and all others 0, formally (ej)k := δjk), we get as a basis of the Lie algebraso(1, 3) the matrices

sj = C(0, lj) =

(0 0T

0 lj

)Nj = C(ej , 0) =

(0 eTjej 0

)j = 1, 2, 3

Their commutation relations read

[si, sj ] = ϵijksk

[Ni, Nj ] = −ϵijksk[si, Nj ] = ϵijkNk

v) to find a representation of the Lie algebra so(1, 3), then, means to find any operators (matrices) si, Ni whosecommutation relations exactly copy those for si, Ni from iv)

Now, I hope, the problem 12.2.5 should be crystal clear (and easy).

12.3.16 , 13.1.10 pp.267,292: More on conjugation (and conjugation classes, in particular for the group SO(3)).

Let G act on a set M . Let an element g perform x 7→ y ≡ gx. What kind of transformation does then theelement hgh−1? Since

(hgh−1)(hx) = h(gx) = hy

we can deduce the following (simple albeit important) piece of wisdom:

if action of g performs x 7→ y

then action of hgh−1 performs hx 7→ hy

and read it in a more wordy way as follows: In general, the h-conjugated element hgh−1 does ”the same job”on h-transformed set M as the original element g does on the original set M .

Now, let us apply it to the group SO(3) acting in the standard way (via rotations) in the 3-dimensional Euclideanspace:

if action of A performs rotation x 7→ y

then action of BAB−1 performs rotation Bx 7→ By

21

Let A perform the rotation through angle α about n. So A = eαn.l. If (e1, e2,n) is an orthonormal right-handedbases, then

A performs e1 7→ e1 cosα+ e2 sinα

e2 7→ −e1 sinα+ e2 cosα

n 7→ n

Then, according to the general wisdom,

BAB−1 performs Be1 7→ Be1 cosα+Be2 sinα

Be2 7→ −Be1 sinα+Be2 cosα

Bn 7→ Bn

Since (Be1, Be2, Bn) is an orthonormal right-handed bases, too, this means that BAB−1 performs the rotationthrough the same angle α about the transformed vector Bn. In formulas:

Beαn.lB−1 = eα(Bn).l or B(n.l)B−1 = (Bn).l

We see, that the conjugacy class of an element A ≡ eαn.l consists of all rotations through the same angle αabout arbitrary vectors n.

12.4.6 p.272, an alternative (and instructive) hint: Modify the solution of 12.2.5 presented here (see above).Actually, one can treat both cases, so(1, 3) as well as so(4), in parallel starting from the following more generalcondition

(ηC)T + ηC = 0 η ≡(1 0T

0 −λI

)where

λ = 1 for so(1, 3)

λ = −1 for so(4)

Then, modifying the calculations to the λ-case we obtain

[si, sj ] = ϵijksk

[Ni, Nj ] = −λϵijksk[si, Nj ] = ϵijkNk

In order to split the Lie algebra to the direct sum of two Lie subalgebras (spanned by Ai and Bi respectively)we introduce )

Ai := si + aNi

Bi := si + bNi

Then[Ai, Bj ] = ϵijk ((1− λab) sk + (a+ b) Nk))

If this is to vanish (as it should for the direct sum), one has to choose

a+ b = 0 1− λab = 0

or, equivalently

Ai := si + aNi

Bi := si − aNi λ = −a2

Then the complete set of commutation relations becomes

[Ai, Aj ] = 2 ϵijk Ak

[Bi, Bj ] = 2 ϵijk Bk

[Ai, Bj ] = 0

22

so that (s,N) 7→ (A,B) provides a desired direct sum decomposition. This only occurs, however, when

λ = −a2

is satisfied.

In the case of so(4), i.e. when λ = −1, this is easily fulfilled by a = 1

Aj := sj +Nj

Bj := sj −Nj

and we get the decomposition needed in 12.4.6 .

In the case of so(1, 3), however, when λ = +1, we badly need a = i (or a = −i)

Aj := sj + iNj

Bj := sj − iNj i ≡√−1

and this is not acceptable as long as we understand so(1, 3) to denote the (usual) real Lie algebra (where, ofcourse, complex linear combinations are not allowed). So, so(1, 3) (regarded as a real Lie algebra) cannot be(contrary to so(4)) decomposed in this way.

[The needed complex combinations can be used if we think of so(1, 3) as a complex Lie algebra (the complexi-fication of the real so(1, 3)). (This is also standardly done in physics, when 2-component relativistic (”dotted”and ”undotted”) spinors are introduced. The matrices, representing si and Ni act in complex spaces and,consequently, their complex linear combinations are as good as real ones.)]

12.5.4 p.281: Maybe a bit more systematic formulation: ηab and ηab are invariant tensors for O(r, s), ϵa...b andϵa...b are invariant tensors for SL(n). Then, consequently, all of them are invariant tensors for SO(r, s).

12.6.2 p.283: The complex constructed here is often called the Chevalley-Eilenberg complex (and correspondingLie algebra cohomologies then become Chevalley-Eilenberg cohomologies)

13.1.10 p.292: see 12.3.16, p.267 (in this text)

13.2.7 p.296: In the hint to (i) the map f given as

π(g) 7→ π′(g) or, in another notation [g] 7→ [g]′

is, unfortunately, not well defined. Indeed, take two elements of gH, say g and gh. They both, by definition ofπ, project to π(g). So they both can play the role of ”g” in the formula (the role of a representative). However,π′(g) = π′(gh), since h /∈ H ′ ≡ kHk−1 in general. So we have to invent another map to prove the statement :-(

A way to find it might be based on using more general result mentioned in ii): each homogeneous space

(M, Lg) is isomorphic to some canonical space (G/H,Lg) (with canonical action Lg[g] ≡ g · g := [gg]).

H Indeed, choose m ∈ M an denote Hm its stabilizer (Lhm = m,h ∈ Hm). Arbitrary point in M is g ·m.(In addition to g, the same job does any gh, h ∈ Hm, see additional material to 13.2.7). Define

φm :M → G/Hm g ·m 7→ [g] ↔ gHm

(Because of (gh) ·m 7→ [gh] = [g], φm is well-defined.) Then it is clearly bijective and it also fulfills

Lg φm = φm Lg

since(Lg φm)(g ·m) = Lg[g] = [gg] = φm((gg) ·m)) = φm(g · (g ·m)) = (φm Lg)(g ·m)

So, φm provides isomorphism of an arbitrary homogeneous space (M, Lg) to the canonical homogeneous space(G/H,Lg) with H := Hm. N

Notice that the isomorphism φm depends on the choice of m ∈M . If

m′ ≡ k ·m

23

is another point in M , we get, in the same way, another isomorphism

φm′ :M → G/Hm′ g ·m′ 7→ [g]′ ↔ gHm′

Now it is clear, that we can obtain an isomorphism of two canonical homogeneous spaces simply by compositionof φ−1

m and φm′

ψ : G/Hm → G/Hm′ ψ := φm′ φ−1m

or

Midentity−−−−−→ M

φm

y yφm′

G/Hm −−−−→ψ

G/Hm′

Let us compute explicit formula. We have

φm : g ·m 7→ gHm

φm′ : g ·m′ 7→ gHm′ m′ ≡ k ·mφ−1m : gHm 7→ g ·m

(we use notation gHm instead of [g] for the point of G/Hm in order to make computation more transparent).Then

ψ ≡ φm′ φ−1m : gHm 7→ g ·m ≡ g · (k−1 ·m′) ≡ (gk−1) ·m′ 7→ (gk−1)Hm′

so thatψ(gHm) = (gk−1)Hm′

or, in square bracket notation,

ψ[g] = [gk−1]′ (rather than ψ[g] = [g]′ from the hint in the book!)

It is clearly a bijection from G/Hm to G/Hm′ . Let us check it is also equivariant:

g · (ψ(gHm)) = g · ((gk−1)Hm′)) = (ggk−1)Hm′ = ((gg)k−1)Hm′ = ψ((gg)Hm) = ψ(g · (gHm))

so thatL′g ψ = ψ Lg

H Let us also check directly that the ”k−1-correction” to the original (wrong) formula makes the mappingalready well-defined (see the beginning of this additional text): Indeed, if we use representative gh instead of g,we get

[gh] 7→ [(gh)k−1]′ = [ghk−1]′ = [(gk−1)(khk−1)]′ = [(gk−1)h′]′ = [gk−1]′

regardless of h ∈ H (since khk−1 = h′ ∈ H ′ ≡ kHk−1). NThe map ψ may be expressed even more simply. Notice that ψ is fully specified by ascribing its value to

a single point (the value on any other point is already fixed by equivariancy). Then, taking this point to beHm ↔ [e], we get

ψ(Hm) = k−1Hm′ or, equivalently ψ[e] = [k−1]′

We can also proceed in the following alternative way: Let H,H ′ be any two subgroups of G and let

ψ : G/H → G/H ′

be an isomorphism. Because of equivariancy of ψ, it is enough to tell the result of ψ[e]. Since the result is inG/H ′, it has to have the form

ψ[e] = [g]′

for some (yet unknown) g ∈ G. Now, the map ψ should not depend on representatives (should be well-defined).This gives

ψ[e] = ψ[h] = ψ(h · [e]) = h · (ψ[e]) = h · [g]′ = [hg]′!= [g]′

24

for any h ∈ H. This gives the following restriction on g: for each h ∈ H,

hg = gh′ h′ ∈ H ′

or, equivalently,g−1hg = h′ h′ ∈ H ′

This means, that conjugation of H by g−1 (which gives clearly a subgroup of G isomorphic to H itself) is to bea subgroup of H ′. Since we know, however, that #H = #H ′ (otherwise ψ : G/H → G/H ′ cannot be bijection),3

we get equalityg−1Hg = H ′

This may be reformulated as follows: ψ : G/H → G/H ′ is an isomorphism (if and) only if H ′ is a conjugatedsubgroup w.r.t. H, i.e. H ′ = kHk−1 for some k ∈ G, and then

ψ[e] = [k−1]′

13.2.7 p.296: Here we learned that any homogeneous space (M, Lg) is isomorphic to the canonical homogeneousspace (G/Gx, Lg). The following (simple) useful fact remained, however, somehow hidden (it is not mentionedexplicitly):

Let x ∈ Ox and let y ≡ gx ∈ Ox be another point of the orbit Ox. As we see, starting from x, we can reachy by the action of g. The question then arises: Are there also other group elements which do the same job(perform x 7→ y)? The answer says that the set of all those elements is just the coset gGx. (This gives a usefulinterpretation of the coset gGx.)

[Indeed, it is clear that elements from gGx do shift x to gx, since Gx ”does nothing” and g shifts x to gx. Now,letk be another group element with the same property, so that kx = gx. Then g−1k ∈ Gx and, consequently,k ∈ gGx.]

It might seem that we can enlarge the set gGx to GygGx (since Gy at the end ”does nothing” when acting ony in the same way as Gx ”does nothing” when acting on x). However, we learned in 13.1.10 that Gy = gGxg

−1

and thereforeGygGx = (gGxg

−1)gGx = gGxGx = gGx

13.2.9 p.297: we learned in this problem that the formula

z 7→ az + b

cz + d≡ LAz A ≡

(a bc d

)∈ GL(2,C)

(the Mobius or linear-fractional transformation) defines a left action of GL(2,C) in the (extended) complexplane and that A and λA result in the same action (so that ”only SL(2,C) matters”). Here we present auseful interpretation of the formula in more geometric terms. The new point of view shows clearly why thestrange looking formula works (why it provides an action). It also leads straightforwardly to a more dimensionalgeneralization.

Consider the complex vector space Cn (the needed modification for the real case will be evident). There isa natural action of GL(n,C) on Cn given by

LAu := Au

We know from 1.3.2(ii) that the rays in Cn (complex lines in Cn passing through the origin) form a new set(manifold) called the complex projective space CPn−1. Now it is intuitively clear that the action on vectorspasses to the action on rays (by means of the representatives). Namely, if [u] denotes the ray given by the vectoru, we have

LA[u] := [LAu] ≡ [Au]

Indeed, the formal check reads

LAB [u] = [(AB)u] = [A(Bu)] = LA[Bu] = (LALB)[u]

3Or, using symmetry argument, by the same computation with H and H′ interchanged we get that by conjugation of H′ we

get a subgroup of H :-)

25

Note thatLλA[u] = [(λA)u] = [λ(Au)] = [Au] = LA[u]

so thatLλA = LA

So, the lesson is that

i) GL(n,C) (as well as any subgroup) not only naturally acts on the vector space Cn (the action being linear,i.e. a representation), but also on a more complicated manifold, the projective space CPn−1 (here, of course,the action is not linear)

ii) in this action, A produces the same result as λA does

Let us compute explicitly the coordinate expression of the action in the simplest case, on CP 1. The originalaction of GL(2,C) on C2 reads (

z1

z2

)7→

(a bc d

)(z1

z2

)Introduce the standard two charts on CP 1 given by (complex) coordinates ξ1 or η1 (see the hint to 1.3.2)(

z1

z2

)∼

(1ξ1

)∼

(η1

1

)where η1 = 1/ξ1

We get (a bc d

)(1ξ1

)=

(a+ bξ1

c+ dξ1

)= (a+ bξ1)

(1ξ1

)∼

(1ξ1

)ξ1 =

c+ dξ1

a+ bξ1(a bc d

)(η1

1

)=

(aη1 + bcη1 + d

)= (aη1 + b)

(η1

1

)∼

(η1

1

)η1 =

aη1 + b

cη1 + d

So, effectively,

ξ1 7→ ξ1 =c+ dξ1

a+ bξ1

η1 7→ η1 =aη1 + b

cη1 + d

We see that the expression of the transformation in terms of the variable η1 is given by the Mobius (linear-fractional) transformation. Concerning ξ1, we obtained a modification of the Mobius (linear-fractional) trans-formation and one can easily check that it (i.e. the composition rule) really works.

It is also instructive to check, that both formulas describe the same abstract transformation

LA : CP 1 → CP 1 (or, equivalently, S2 → S2)

We have to check thatη1 = 1/ξ1 implies η1 = 1/ξ1

which is straightforward:

η1 =aη1 + b

cη1 + d=a+ b/η1

c+ d/η1=a+ bξ1

c+ dξ1=

1

ξ1

The reader is invited to show that the explicit formula for the transformation of CPn−1 in terms of the coordi-nates (η1, . . . , ηn−1) reads

ηa =Aabη

b +AanAnb η

b +Ann

[Note: we obtained an action of SL(2,C) on S2. One can check that the action of the unitary subgroup, SU(2),gives nothing but rotations of the (standard round) sphere. This can be ultimately traced back to the fact thatunitary group preserves the natural ”hermitian” scalar product in Cn and, consequently, the induced so calledFubini-Study scalar product on CPn (not mentioned in the book)]

13.3.1 p.300: in (ii) a method is described in which a covering group steals a representation from the coveredgroup (by composition with the covering homomorphism f , i.e. ρ = ρ f). It should be clear that any

26

homomorphism can be used for this purpose (if f is a homomorphism G → G and ρ is a representation of G,then ρ f is a representation of G; if ρ is irreducible, the same is true for ρ).

13.4.7 p.315 In the hint to 13.4.7iv) a reference to 4.6.10 might be useful

13.4.7− 10 p.315-16 Prove that the trace and the determinant are the only (independent) invariants of a

symmetric 2 × 2 matrix A with respect to A 7→ B−1AB, B special orthogonal.

Solution: The problem is from linear algebra. Nevertheless, a useful idea might be to treat the probleminfinitesimally, introduce the fundamental field of the action and find (in this way!) the most general invariantfunction.

For infinitesimal B = I+ ϵC, where CT = −C, we have

A 7→ B−1AB = (I− ϵC)A(I+ ϵC) = A+ ϵ(AC − CA) + . . .

Then,

A ≡(u vv w

)7→

(u vv w

)+ ϵ

((u vv w

)(0 k−k 0

)−(

0 k−k 0

)(u vv w

))=

(u vv w

)+ ϵk

(−2v u− wu− w 2v

)so that

(u, v, w) 7→ (u(ϵ), v(ϵ), w(ϵ)) = (u+ kϵ(−2v), v + kϵ(u− w), w + kϵ(2v))

The generator (fundamental field) ξ of the action reads

ξ = −2v∂u + (u− w)∂v + 2v∂w

= 2v(∂w − ∂u) + (u− w)∂v

We are to find the most general function Φ(u, v, w), which is invariant with respect to the flow of ξ, i.e. suchthat

ξΦ!= 0

This greatly simplifies in appropriate coordinates. We see from the structure of ξ that variables u − w, v playan important role. So try a change of coordinates (u, v, w) → (x, y, z) given by

λ(u− w) = x

v = y

λ(u+ w) = z

Then we getξ = (1/λ)x∂y − (4λ)y∂x

or, for λ = 1/2,ξ = 2(x∂y − y∂x)

This field clearly generates rotations in xyz-space about z-axis. So, introducing further standard cylindricalcoordinates r, φ, z according to

x = r cosφ

y = r sinφ

z = z

we come to the most simple coordinate form of the field ξ:

ξ = 2∂φ

Then, the invariance condition reads

ξΦ!= 0 ⇔ ∂Φ(r, φ, z)

∂φ

!= 0

27

and the obvious general solution isΦ(r, φ, z) = f(r, z)

But note, that

TrA ≡ T (u, v, w) = u+ w = 2z ≡ T (r, z)

detA ≡ D(u, v, w) = uw − v2 = (z + x)(z − x)− y2 = z2 − (x2 + y2) = z2 − r2 ≡ D(r, z)

so that T , D are just two independent combinations of r, z.

[Since dT ∧ dD = · · · = −4rdz ∧ dr, the combinations T , D are independent. One can easily express explicitly

r, z in terms of T , D.]

This means, that any invariant function of matrix elements of A is of the form f(r, z) or, equivalently, ofjust two particular combinations of matrix elements of A, namely TrA and detA. Q.E.D.

14.2.4 p.339 In this exercise, we study canonical transformations as such coordinate tranformations,

(qa, pa) 7→ (Qa(q, p), Pa(q, p))

which preserve canonical (Darboux) expression of the symplectic form, i.e. such that

ω = dpa ∧ dqa!= dPa ∧ dQa ≡ dPa(q, p) ∧ dQa(q, p)

holds. This is, however, equivalent to several alternative ways of expressing the same fact. First, the (canonical)form of the corresponding Poisson tensor is preserved

P =∂

∂pa∧ ∂

∂qa=

∂

∂Pa∧ ∂

∂Qa

(since P ω = −1), resulting then in equally simple (canonical) explicit formulas of Poisson bracket in old andnew coordinates

f, g =∂f

∂pa

∂g

∂qa− ∂f

∂qa∂g

∂pa=

∂f

∂Pa

∂g

∂Qa− ∂f

∂Qa∂g

∂Pa

This is, in turn, equivalent to the statement, that the component matrix Pij has, in both coordinates, (qa, pa)and (Qa, Pa) the canonical form

Pij(q, p) =(0n −InIn 0n

)= Pij(Q,P )

Now, because

- components are just the values of a tensor on basis arguments, i.e. Pij(z) = P(dzi, dzj)

- the Poisson bracket is defined as f, g = P(df, dg), so that Pij(z) = zi, zjwe see that we can rewrite the equation above in the form

Pij(q, p) =(qa, qb qa, pbpa, qb pa, pb

)=

(0 −δabδba 0

)=

(Qa, Qb Qa, PbPa, Qb Pa, Pb

)= Pij(Q,P )

And this, at last, can be rephrased as the statement, that canonical transformations are such coordinate tran-formations,

(qa, pa) 7→ (Qa(q, p), Pa(q, p))

which preserve canonical Poisson brackets between the coordinate functions, i.e. for which

qa, qb = Qa, Qb = 0

pa, qb = Pa, Qb = δba

pa, pb = Pa, Pb = 0

28

In quantum mechanics, after replacing of the Poisson bracket (of functions on phase space) by a constantmultiple of the commutator (of operators in the corresponding Hilbert space), canonical transformations ofoperators are such transformations,

(qa, pa) 7→ (Qa(q, p), Pa(q, p))

which preserve the canonical commutation relations (CCR)

[qa, qb] = [Qa, Qb] = 0

[pa, qb] = [Pa, Q

b] = −i~δba[pa, pb] = [Pa, Pb] = 0

Or, finally, when expressed in terms of annihilation and creation operators

aa :=1√2~

(qa + ipa) a+a :=1√2~

(qa − ipa)

Aa :=1√2~

(Qa + iPa

)A+a :=

1√2~

(Qa − iPa

)as such transformation of operators

(aa, a+a ) 7→ (Aa(a, a

+), A+a (a, a

+))

which preserve the CCR in the form

[aa, ab] = [Aa, Ab] = 0

[aa, a+b ] = [Aa, A

+b ] = δab

[a+a , a+b ] = [A+

a , A+b ] = 0

14.3.4 p.342 The concept of relative invariance of a form was introduced in 14.3.2 and 14.3.3, but I somehowforgot to mention the corresponding version of integral invariants :-(. Here the gap is filled.

Consider an invariant p-form α and a relative invariant p-form β. This means

LV α = 0

LV β = dρ

for some (p− 1)-form ρ. Let Φt ↔ V be the corresponding flow. Then, because of

Φ∗t = 1 + tLV + (t2/2)LV LV + . . .

we have

Φ∗tα = α

Φ∗tβ = β + dϕt ϕt = tρ+ (t2/2)LV ρ+ . . .

Upon integration over a p-chain c we get ∫Φt(c)

α =

∫c

α∫Φt(c)

β =

∫c

β +

∫∂c

ϕt

So, we see that α leads to an integral invariant, but the situation with β is more delicate. There is a term whichspoils the ”integral invariant property”. But still there is a possibility to construct an integral invariant. Weshould just restrict ourself to closed chains c. Indeed, clearly

∂c = 0 ⇒∫Φt(c)

β =

∫c

β

29

So, whenever we find a form β obeying LV β = d(. . . ) for some V , there is a relative integral invariant given bythe integral ∫

c

β c such that ∂c = 0

Now back to Hamiltonian mechanics (with time-independent Hamiltonian H(q, p)). We know (from 14.1.6iiiand 14.3.3) that

Lζfω = 0

Lζf (ω ∧ ω) = 0

Lζf (ω ∧ ω ∧ ω) = 0

etc.

So, we have the (absolute) integral invariants described in the book. Now, for θ such that ω = dθ, we get

Lζf θ = iζf dθ + diζf θ = d(iζf θ − f) ≡ d(. . . )

But then, we have

Lζf θ = d(. . . )

Lζf (θ ∧ ω) = d(. . . )

Lζf (θ ∧ ω ∧ ω) = d(. . . )

Lζf (θ ∧ ω ∧ ω ∧ ω) = d(. . . )

etc.

since, for exampleLζf (θ ∧ ω) = d(iζf θ − f) ∧ ω = d[(iζf θ − f) ∧ ω] ≡ d(. . . )

Therefore we have, in time-independent Hamiltonian mechanics, absolute integral invariants

I2 ≡∫D2

ω for any D2

I4 ≡∫D4

ω ∧ ω for any D4

I6 ≡∫D6

ω ∧ ω ∧ ω for any D6

etc.

and relative integral invariants

I1 ≡∫c1

θ for any closed c1 (∂c1 = 0)

I3 ≡∫c3

θ ∧ ω for any closed c3 (∂c3 = 0)

I5 ≡∫c5

θ ∧ ω ∧ ω for any closed c5 (∂c5 = 0)

etc.

Interested reader might find useful to consult my (mostly tutorial) article ”Modern geometry in not-so-highechelons of physics: Case studies”, Acta Physica Slovaca 63, No.5, 261 359 (2013) (99 pages) (available onlineat http://www.physics.sk/aps or http://arxiv.org/abs/1406.0078)

14.5.3 p.351 Here we show a concrete example of a moment map. Namely, the moment map of a standard

action of the rotation group SO(3) on the common two-dimensional ”round” sphere S2 (treated as a symplecticmanifold, see 14.2.3).

The first way of computation:

30

The generators of the action are displayed in 13.4.6 (or in any textbook of Quantum Mechanics :-):

ξl1 = − sinφ∂ϑ − ctgϑ cosφ∂φ

ξl2 = cosφ∂ϑ − ctgϑ sinφ∂φ

ξl3 = ∂φ;

(Here Ej = lj is the standard basis in so(3), see 11.7.13). The volume = symplectic form (on unit sphere) reads

ω = sinϑdϑ ∧ dφ

Therefore

iξl1ω = · · · = −d(− sinϑ cosφ)

iξl2ω = · · · = −d(− sinϑ sinφ)

iξl3ω = · · · = −d(− cosϑ)

The functions in the brackets are, however, nothing but (restriction to the sphere of) minus standard coordinatefunctions (x, y, z) ≡ (x1, x2, x3) in the ambient space E3. So, we get

iξEjω = −d(−xj(ϑ, φ))

Since Pj(ϑ, φ) should be after −d symbol, we get

Pj(ϑ, φ) = −xj(ϑ, φ)

orP ≡ PjE

j = −xj(ϑ, φ)Ej

The second way of computation:

Recall that orbits of the co-adjoint action for G = SO(3) are just spheres (see 14.6.7). On the unit sphere,the canonical symplectic form (see 14.6.3) is a 2-form

ω = f(ϑ, φ)dϑ ∧ dφ

obeyingωX∗(ξl1 , ξl2) = ⟨X∗, [l1, l2]⟩ = ⟨X∗, l3⟩ = x3 = cosϑ

Plugging explicit expressions for ξl1 and ξl2 gives f(ϑ, φ) = sinϑ. So, this is exactly the standard volume form.

But there is an explicit result for the moment map for any co-adjoint orbit in 14.6.5iv):

P : O → G∗ X∗ 7→ X∗

So, a point on the sphere with spherical coordinates (ϑ, φ), i.e. sitting in

X∗ ↔ (x1(ϑ, φ), x2(ϑ, φ), x3(ϑ, φ) ≡ (sinϑ cosφ, sinϑ sinφ, cosϑ)

of G∗ ≡ so(3)∗ ≡ R3 maps to the point

X∗ ↔ xjEj

in G∗ ≡ so(3)∗ ≡ R3. Put it another way, the three components of the moment map are just the three

Cartesian coordinates of the point. This is, modulo sign, in agreement with the first way of computation. (Thesign probably comes from the freedom in the sign of the symplectic form on S2, i.e. ω is equally good as −ω.But I am not sure, the reader is invited to fill this gap herself :-)

Two simple checks of the result:

1. Consider a Hamiltonian on S2 invariant with respect to the action. So, it is (completely!) rotational-invariant. So constant. There are three conserved quantities, x1, x2, x3. This means, however, there is nomotion on the sphere. But this is OK for constant Hamiltonian :-)

31

2. Consider a Hamiltonian on S2 invariant just with respect to rotation about the x3-axis. So, it only dependson ϑ, H = H(ϑ). The relevant part of the action is the action of SO(2) generated by ξl3 . Then P3 = x3 is

conserved (and nothing more). But this is OK for Hamiltonian dependent on just ϑ, since ϑ = · · · = 0, so thatϑ(t) = const., so that x3 is const.

14.5.7 p.354 We learned that, for each dimension (each basis element Ei) of the Lie algebra G, we get conservedquantity Pi(x). A natural question arises whether these conserved quantities are functionally independent. Putit another way, whether each dimension of the symmetry Lie algebra G adds a fully-fledged conserved quantityrather than quantities, which are just functions of those already obtained before.

Imagine f(x) and g(x) are two conserved quantities (f = 0 = g). Then h(f(x), g(x)) is clearly conservedquantity as well. Indeed,

h =∂h

∂ff +

∂h

∂gg = 0

Note that, in this case,

dh =∂h

∂fdf +

∂h

∂gdg

so the differentials of the three functions f, g, h are linearly dependent. On the contrary, linear independenceof df1, . . . , dfn implies functional independence of f1(x), . . . , fn(x), which means there is no F such that

F (f1(x), . . . , fn(x)) = 0

or equivalently, one cannot express any particular fk in terms of the rest members fj .

Therefore, in the case of our interest, we are interested in (point-wise) linear independence of the differentials

dP1(x), . . . , dPn(x) n = dimG

Now, hamiltonian character of symmetry generators

iξEjω = −dPj

(see the text between 14.5.1 and 14.5.2, p.350) and non-degeneracy of ω shows that linear independence ofthe differentials is the same thing as linear independence of the generators ξEj . So, when the generators are(point-wise) linearly independent? The answer: the action is to be free. Indeed, otherwise elements X of theLie sub-algebra, which corresponds to (non-vanishing) stabilizer of a point x result in vanishing generatorsξX . So, in an adapted basis Ej , some of ξEj

vanish and the (whole set of) generators ξEjcease to be linearly

independent.

14.7.1 p.361 Some more details concerning the proof of the statement are presented here.

First, let us see in more detail, why ”the complement of the complement turns out to be the initial subspaceonce again”. (This is a well-known fact in Euclidean case, but perhaps it is less clear for the case of symplecticorthogonality.)

So, let L1 ⊂ L be a k-dimensional subspace of an n-dimensional linear space L. Then, there is a distinguished(n− k)-dimensional subspace L1 ⊂ L∗ in the dual space, the annihilator of L1 (see 2.4.9, 10.1.13)

⟨L1, L1⟩ = 0 L1 ⊂ L , L1 ⊂ L∗

(If (ea, ei) is an adapted basis in L (i.e. ea ∈ L1), then i-part of the dual basis (ea, ei) of L∗ serves as a basis of

L1.)

Now, we have two (mutually inverse) linear isomorphisms

ω : L→ L∗ v 7→ ω(v, . )

ω−1 : L∗ → L α 7→ ω−1(α, . )

(They are isomorphisms because of non-degeneracy of ω.) Let L2 ⊂ L be the image of L1 w.r.t. ω−1

L2 := ω−1(L1) ⊂ L

32

Since ω−1 is isomorphism, the dimension of L2 is the same as the dimension of L1, i.e. (n− k) (complementarydimension w.r.t. L1)

dimL1 = k ⇒ dimL2 = n− k (n ≡ dimL)

Now, let v ∈ L1 and w ≡ ω−1(α, . ) ∈ L2. Then

ω(v, w) = ω(v, ω−1(α, . )) = ωijviωkjαk = −αkvk ≡ −⟨α, v⟩

This shows that L2 is nothing but the symplectic-orthogonal complement L⊥1 of L1. (Vanishing of the LHS

holds for exactly those w which are images of α ∈ L1.) So, the dimension of the complement L⊥1 is always

complementary to the dimension of L1

dimL = dimL1 + dimL⊥1 n = k + (n− k)

Consequently, the dimension of the complement of the complement is complementary to the dimension of thecomplement and this is nothing than the dimension of the original subspace L1:

dim(L⊥1 )

⊥ = dimL− dimL⊥1 = dimL1

(since n− (n− k) = k :-). Now, by definition (of L⊥1 )

ω(L1, L⊥1 ) = 0

The same formula, however, shows (!) that L1 ⊂ (L⊥1 )

⊥ (only the fact that it is a subspace is clear at once).But since we already know that L1 and (L⊥

1 )⊥ have the same dimension, the spaces are actually equal:

(L⊥1 )

⊥ = L1

So, in general, making orthogonal complement (of a subspace) twice is doing nothing, we return to the subspaceagain

U⊥⊥ = U

[Looking at the proof we see, that this fact is true for both symmetric and skew-symmetric ”scalar products” inL. The symmetric case needs not be positive definite, it works in pseudo-Euclidean cases as well. What reallymatters is non-degeneracy of the product. Notice, however, that only these two cases - symmetry and skew-symmetry - make the very notion of orthogonality (and, consequently, the orthogonal complement) meaningful!For a ”general” (even though non-degenerate) bilinear form, the value of the ”scalar product” B(v, w) dependson the order of the pair, so it can happen B(v, w) = 0 and B(w, v) = 0. So, orthogonality depends on theorder.]

B.t.w., the fact that the dimensions of L1 and L⊥1 add to the full dimension of L (i.e. dimL = dimL1 +

dimL⊥1 ) does not mean, that the subspaces themselves do add (i.e. that L = L1 ⊕ L⊥

1 )! Consider, as a simpleexample, two dimensional symplectic space with ω = e1 ∧ e2. Let L1 = Span e1. Then L⊥

1 = L1 - it is thesubspace itself (!). Such subspaces are called Lagrangian and they play important role in the theory.

Finally, let us mention, just for fun, a direct proof of the opposite implication (not based on the fact thatU⊥⊥ = U ; notice that this very implication is crucial in the proof of 14.7.6). What we want to prove is thefollowing statement: any vector which is symplectic-orthogonal to the subspace tangent to Mp is necessarilyof the form ξX for some X ∈ G. Well, being tangent to Mp is the same thing as annihilate all 1-forms dPi.So, the annihilator of the subspace of vectors tangent to Mp is spanned by the covectors dPi. The orthogonalcomplement to the subspace of vectors v tangent to Mp consists of all vectors w such that ω(w, v) = 0 (bydefinition). Equivalently, of such w that the 1-form ω(w, . ) belongs to the annihilator. Since the latter isspanned by dPi, there are numbers Xi such that

ω(w, . ) = XidPi

Applying ”raising of indices” operation on both sides of the equation we get

w = XiζPi = XiξEi = ξX

33

And that’s all :-)

15.3.4 p.385 The coordinate-free expression of the RLC-connection from (iii) is called the Koszul formula

15.3.4 p.385, 15.3.14 p.389, 15.6.9 p.411 and 22.5.4 p.665:

In 15.3.4, the classical (coordinate) expression

Γijk =1

2gil(glj,k + glk,j − gjk,l) i.e. 2Γijk = gij,k + gik,j − gjk,i (a1)

for Christoffel symbols of the metric and symmetric (= RLC) connection is derived.

In 15.3.14, it is generalized to just metric connection (torsion allowed, still coordinate frame).

In 15.6.9, metric connection in orthonormal frame is discussed (torsion allowed).

In 22.5.4, metric connection coefficients in orthonormal frame enter Dirac operator in ”curved space”.

It might be of interest to learn what the situation looks like for the most general case, i.e. when

i) torsion as well as non-metricity may not vanish (connection is neither metric nor symmetric)

ii) the frame field ea is general (neither orthonormal nor coordinate; we have [ea, eb] = ccabec).

Recall that we found the above mentioned expressions for Γijk (for the RLC case!) by solving system ofequations

Γijk + Γjik = gij,k since it is metric

Γijk − Γikj = 0 since it is symmetric

Now, the system is to be generalized to

Γabc + Γbac = Aabc Aabc = Abac (a2)

Γabc − Γacb = Babc Babc = −Bacb (a3)

where the following abbreviations are used:

Aabc ≡ gab,c − gab;c

Babc ≡ −(cabc + Tabc)

and

gab,c := ecgab

gab;c := (∇cg)ab non-metricity tensor

H Indeed:

gab;c = (∇cg)(ea, eb)

= ∇c(g(ea, eb))− g(∇cea, eb)− g(ea,∇ceb)

= ecgab − Γdacgdb − Γdbcgad

= gab,c − Γbac − Γabc

T cabec = T (ea, eb)

= ∇aeb −∇bea − [ea, eb]

= Γcbaec − Γcabec − ccabec

so thatTcab ≡ gcdT

dab = gcd(Γ

dba − Γdab − cdab) = Γcba − Γcab − ccab

N

34

Notice that the non-metricity tensor is a natural measure of deviation from metricity of a connection. Wewill not assume this tensor to vanish, now. Rather we will only assume this tensor to be given.

Now, what was the trick which enabled us to solve the equations (a2) and (a3)? We see from the result (a1)that what very much helps is to make the following combination

gij,k + gki,j − gjk,i

from (a2) and then use (a3) whenever needed. (Here, the expression is written in a form when each term beginswith a different index.)

In order to formalize (and then to repeat) the trick, one can introduce, for any three-index quantity Qabc,the contortion operation (also known4 as the Schouten braces)

Qabc := Qabc +Qcab −Qbca (a4)

Then, clearly, (a1) reads2Γijk = gij,k for the RLC-connection (a5)

But the real goal to introduce Schouten braces, as indicated above, is to use them to solve the general system(a2) and (a3). So, let us repeat the trick there:

Apply the Schouten braces to (a2) and, when needed, use (a3).

In this way we get unique solution:2Γabc = Aabc −Bacb

H Indeed:

Aabc ≡ Aabc +Acab −Abca

= (Γabc + Γbac) + (Γcab + Γacb)− (Γbca + Γcba)

= (Γabc + Γacb) + (Γbac − Γbca) + (Γcab − Γcba)

= (Γabc + (Γabc −Babc)) + (Bbac) + (Bcab)

= 2Γabc −Babc +Bbac +Bcab

so that2Γabc = Aabc +Babc −Bbac −Bcab

The last three terms clearly resemble Schouten braces of (perhaps) Babc. Unfortunately,

Babc −Bbac −Bcab = Babc : −(

(check). But, fortunately, there still is a possibility to write it as

Babc −Bbac −Bcab = −Bacb : −)

(check, use skew symmetry Babc = −Bacb). So, in order to express the result in terms of Schouten braces,one also has to perform some index reshuffling. This is a darker side of life, since the resulting formula is thenharder to remember. N

Inserting concrete expressions for Aabc and Babc we get the following expression for Christoffel symbols ofgeneral linear connection on Riemannian manifold (M, g):

2Γabc = gab,c − gab;c + Tacb + cacb (a6)

The third term is often expressed in terms of specifically introduced contortion tensor

Kabc :=1

2Tacb ≡ 1

2(Tacb + Tbac − Tcba) = −1

2(Tabc + Tbca − Tcab) (a7)

4See page 132, formula 3.7 in J.A.Schouten: Ricci Calculus, Springer-Verlag 1954; Schouten braces are not to be confused with

a completely different (and much more profound!) object, the Schouten bracket (known also as Schouten-Nijenhuis bracket).)

35

(for the last equality sign we used skew-symmetry Tabc = −Tacb). Then, the decomposition formula (a6) reads

Γabc =1

2(gab,c + gac,b − gbc,a)−

1

2(gab;c + gac;b − gbc;a) +Kabc −

1

2(cabc + cbca − ccab) (a8)

So, in general, coefficients of arbitrary linear connection may be uniquely written as a sum of four terms:

The first plus the fourth term correspond to good old RLC-connection.

The second term is given by (minus) Schouten braces of non-metricity tensor. (It vanishes for metricconnection).

The third term is given by the contortion tensor, i.e. by a combination of Schouten braces and indexreshuffling of the torsion tensor (it vanishes for symmetric connection).

Symbolically (in words)

general = RLC + non-metricity part + contortion part (a9a)

so thatmetric = RLC + contortion part (a9b)

andmetric and symmetric = RLC (a9c)

Particular cases of (a8):

1. In coordinate (= holonomic) frame, the last term (containing anholonomy coefficients cabc) vanishes andwhat remains is

Γijk =1

2(gij,k + gik,j − gjk,i)−

1

2(gij;k + gik;j − gjk;i) +Kijk (a10)

2. In orthonormal frame ecgab = 0 (since gab = ηab = const.). If the connection is, in addition, metric, i.e.gab;c = 0, we are left with

Γabc = −1

2(cabc + cbca − ccab) +Kabc (a11)

(This is the correct result of 15.6.20iii) - the contortion term is forgotten, there. In 15.6.9 it is ok, since onlyRLC case is discussed.)

Btw, from (a10) we see, that Christoffel symbols of any linear connection are given as a sum of a uniqueconnection (namely the RLC one) plus a tensor. Therefore, ”the difference of any two connections is a tensor”.

Let us remark, finally, that Schouten braces are ”invertible” - one can express original quantity Qijk in termsof appropriate combinations of its Schouten braces. Namely, the explicit formula reads

Qijk =1

2(Qijk +Qjki) (a12)

In particular, it means that one can reconstruct the ”original” torsion tensor from ”its” contortion tensor:

Tijk = −(Kijk +Kkij) (a13)

H Indeed, we can write a matrix relationQijkQkijQjki

=

1 1 −1−1 1 11 −1 1

QijkQkijQjki

(The first line is definition (a4), remaining lines arise by just renaming indices.) The inverse relation isQijk

QkijQjki

=1

2

1 0 11 1 00 1 1

QijkQkijQjki

The first line gives (a12).

36

Similarly, from (a7) we haveKijk

Kkij

Kjki

= −1

2

1 −1 11 1 −1−1 1 1

TijkTkijTjki

⇒

TijkTkijTjki

=

−1 −1 00 −1 −1−1 0 −1

Kijk

Kkij

Kjki

The first line gives (a13). N

15.3.9 p.386 Remarkable insight is gained through the approach discussed in the additional material to theproblem (15.6.10). It is shown there that the resulting angle of rotation of the vector may be expressed in termsof the solid angle subtended by the path. (Here the angle is clearly 1/8 of the total solid angle 4π.)

15.3.14 p.389: see the comment to 15.3.4, p.385 (here, in the Additional material).

15.4.3 p.391 There is an error in part (i), both in the assignment and in the hint. A more detailed generaltreatment of how (velocities and) accelerations are related for two curves, differing just in parametrization,might be useful to understand the situation.

Ok, let γ(t) = γ(σ(t)). So, we are given two curves, γ and γ, passing a common set of points on M , butdiffering in ”times” visiting particular point. (As an example, if σ(t) = t3, then γ(1) = γ(1), γ(2) = γ(8)),γ(3) = γ(27), etc.). In a point P ≡ γ(t) = γ(σ(t)), two relevant vectors may be associated with each curve,their velocity and acceleration. So, altogether, as many as four vectors are to be considered in (each) P :

v := ˙γ(t) a := (∇ ˙γ˙γ)(t) v := γ(σ(t)) a := (∇γ γ)(σ(t))

The task is to express hatted quantities in terms of original (unhatted) ones. From (2.3.5) we know how hattedvelocity is expressed in terms of the original one:

v = σ′v σ′ ≡ σ′(t)

so that v is just an appropriate multiple of v.

Hvf =

d

dtf(xi(σ(t))) =

∂f

∂xidxi

dσ

dσ

dt=

(σ′(t)

dxi

dσ∂i

)f = (σ′(t)γ(σ(t))) f = (σ′(t)v)f

NLet us proceed to the acceleration, now. We get

a = ∇v v = ∇v(σ′v) = (vσ′)v + σ′∇vv = (vσ′)v + σ′∇σ′vv = (vσ′)v + (σ′)2∇vv = σ′′v + (σ′)2a

or, in the notation used in the book,∇ ˙γ

˙γ = σ′′γ + (σ′)2∇γ γ

H We can regard σ(t) (and, consequently, also σ′(t) and σ′′(t)) as a function on the curve γ(t). Then,

applying ˙γ on σ′, giving by definition derivative of σ′ along the curve γ(t), results in nothing but the derivativeof σ′ with respect to t

(vσ′)(t) = ( ˙γσ′)(t) = (d/dt)σ′(t) = σ′′(t)

NSo, the answer to the question, how the hatted quantities are related to the original ones, reads:

v = σ′v

a = σ′′v + (σ′)2a

Then, the answer to (15.4.3) part (i) follows readily: if γ is an affinely parameterized geodesics (i.e. a = 0),then a = σ′′v. Therefore, requiring γ be an affinely parameterized geodesics as well, needs σ′′ = 0 (i.e.σ(t) = k1t+ k2).

Example 1: Dimensional check. Let dimension of γ, γ (or corresponding coordinates xi) be A, dimension of tbe B and dimension of σ be C. Then the dimensions of relevant quantities read:

v ↔ A/B a↔ A/B2 v ↔ A/C a↔ A/C2 σ′ ↔ C/B σ′′ ↔ C/B2

37

and we see that both equations are dimensionally correct.

Example 2: A simple random (and dull) check of the validity of the two formulas above: Consider a one-dimensional motion of a point on a straight line given by x(t) = t4 and a reparametrization of the formσ(t) = t3. Then

σ(t) = t3 σ′(t) = 3t2 σ′′(t) = 6t

x(t) = t4 v ≡ x(σ(t)) = 4t9 a = x(σ(t)) = 12t6

x(t) = t12 v ≡ ˙x(t) = 12t11 a = ¨x(t) = 122t10

The question is whether

12t11?= 3t24t9

122t10?= 6t4t9 + (3t2)212t6

is true. Yes, it is :-)

Example 3: Consider (as γ(t)) a uniform motion of a point along a circle. Then, as we know, the acceleration ais oriented to the center (and the velocity is tangent to the circle, the vectors are perpendicular). The magnitudeof both velocity and acceleration is constant. Now let, say, σ(t) = t2. Then γ(t) describes a motion which is(still along the same circle but) non-uniform - we get v = 2tv, so the magnitude of the new velocity grows. Thisis compatible with a = 2v + 4t2a - we see that the new acceleration exhibits tangent (constant) component 2v,so the magnitude of the velocity grows linearly.

15.4.4 , 18.4.6 pp.392,511: In 15.4.4 we learned how to use Lagrange equations for finding geodesics (of theRLC-connection) - simply take

L(x, v) = (1/2)gab(x)vavb

(the kinetic energy alone) and write down Lagrange equations. From 18.4.6 we know how to compute theHamiltonian ”corresponding” to a given Lagrangian. For our L we get

H(x, p) = (1/2)gab(x)papb

(the same kinetic energy alone). So, we can also write geodesic equations in the form of appropriate Hamiltonianequations - simply take H(x, p) = (1/2)gab(x)papb and write down Hamiltonian equations. We get explicitly(check)

xa = gab(x)pb

pa = −(1/2)gbc,a(x)pbpc

Once we solve these 2n first order equations, we get curves on T ∗M , in coordinates xa(t), pa(t). Their projectionsto M , i.e. xa(t), are then geodesics on (M, g).

One can also easily check explicitly, that these (first order) equations are indeed equivalent to the (second order)geodesic equations. Indeed, from the first equation we have pa = gab(x)x

b. Plugging this into the second oneleads to

gabxb + gab,cx

bxc = −1

2gbrg

bc,agcsx

rxs

Differentiating the identity gbcgcd = δbd with respect to xa we get

−gbrgbc,agcs = grs,a

and, therefore

gabxb + Γabcx

bxc = 0 Γabc :=1

2(gab,c + gac,b − gbc,a)

15.5.4i p.403: There is a tricky point (potential sign error) in the action of the(11

)-tensor A of a general

derivation D ≡ LV +A.

Recall that a general(11

)-tensor A was, up to now (see 2.4.5ii), considered in three roles:

38

- as a mapping v 7→ A(v; . ) ≡ A(v) (vector to vector)

- as a mapping α 7→ A( . ;α) ≡ A(α) (covector to covector)

- as a mapping (v, α) 7→ A(v;α) (vector and covector to a number)

Here it is considered in a fourth role:

- as a mapping t 7→ A · t ((pq

)-tensor to

(pq

)-tensor)

In particular, the tensor t may be a vector or a covector. In these two particular cases we get

A · v = A(v) (1)

A · α = −A(α) (2)

This means, on a basis vectors an (dual) covectors

A · ea = Abaeb (1a)

A · ea = −Aabeb (2a)

So, there is an important difference between A(α) and A · α, a potential danger to forget about the minus signin A · ( )-operation on covectors.

H By definition, the mappingt 7→ A · t

is to be a derivation of tensor algebra which, in addition, vanishes on(00

)-tensors (scalars) and commutes with

contractions. Then we get

A · ⟨α, v⟩ 1.= 0

2.= ⟨A · α, v⟩+ ⟨α,A · v⟩

The first line comes from vanishing on scalars, the second line holds because of

A · ⟨α, v⟩ ≡ A · (C(α⊗ v)) = C(A · (α⊗ v)) = C((A · α)⊗ v + α⊗ (A · v)) = ⟨A · α, v⟩+ ⟨α,A · v⟩

So, we get⟨A · α, v⟩ = −⟨α,A · v⟩

or, choosing v = ea and α = eb,(A · eb)a = −(A · ea)b

This means that if we chooseA · v := A(v)

we inevitably have to acceptA · α = −A(α)

since definitions of A(v) and A(α), see (1) and (2), result in

⟨A(α), v⟩ = ⟨α,A(v)⟩

(Btw. we could equally well choose plus sign on covectors and then accept minus sign on vectors. So aninvariant statement is that there is a sign difference on either vectors or covectors in action of A as a derivationof tensor algebra in comparison with its natural action on the latter stemming from mere definition of A as a(11

)-tensor.) N

15.5.4ii p.403: The formula ∇V = LV + (∇V ) to be proved in part (ii) is not completely general. Rather, iftorsion does not vanish, it should read

∇V = LV + (∇V ) + T (V, . )

Indeed, let A := ∇V − LV . Then, from the definition of the torsion tensor

T (V,W ) := ∇VW −∇WV − [V,W ]

39

we infer

A(W ) = (∇V − LV )(W ) = ∇VW − [V,W ] = ∇WV + T (V,W ) ≡ (∇V + T (V, . ))(W )

So, in components,Aµν = V µ;ν + TµρνV

ρ

15.5.6i p.404: From the result derived in i)

R(∂i, ∂j)∂k = Rlkij∂l

combined withτA,A = (1− ϵ2R(U, V ))

from 15.5.1 one can envisage a situation in which the Rlkij-component of the Riemann tensor is ”in action”.

Namely, perform parallel transport of ∂k around the ϵ × ϵ coordinate loop in the ij-plane (spanned on ∂i and∂j). Then the resulting vector differs from the original ∂k by a vector, whose l-th component is just −ϵ2Rlkij :

∂k 7→ ∂k − ϵ2Rlkij∂l

15.5.6ii p.404: Here, explicit formula for Ak...lr...s;i;j −Ak...lr...s;j;i is presented. As an example it says that for vectorfield W we get

W k;i;j −W k

;j;i = −RkmijWm

Well, there is a problem with this formula. Namely, one can interpret the expression present at the l.h.s. in (atleast) two ways:

W k;i;j

1.= (∇j(∇iW ))k my book (1)

W k;i;j

2.= (W k

;i) ;j virtually all other sources :-( (2)

And, unfortunately, it makes a difference provided that there is a non-vanishing torsion. Let’s compute:

In my interpretation, ∇iW =W k;i∂k is still a vector and the same is true for

∇j(∇iW ) = ∇j(Wk;i ⊗ ∂k) = (∂jW

k;i)∂k +W k

;i∇j∂k = ((W k;i) ,j + ΓkmjW

m;i)∂k (A)

So a single Γ-term occurs (coming from ∇j∂k = Γmkj∂m).

In the ”rest of the world” interpretation, W k;i already corresponds to a

(11

)-tensor B = ∇W

B = Bijdxj ⊗ ∂i Bij :=W i

;j (B)

so that(W k

;i) ;j ≡ Bki ;j = Bki ,j + ΓkmjBmi − ΓmijB

km (C)

and as many as two Γ-terms do occur (also the new lower index i adds its Γ-term).

Now, from (A), (B) and (C) we see that

(∇j(∇iW ))k = Bki ,j + ΓkmjBmi

(W k;i) ;j = Bki ,j + ΓkmjB

mi − ΓmijB

km

i.e. the two interpretations ((1) versus (2)) are related as follows:

(W k;i) ;j = (∇j(∇iW ))k − ΓmijW

k;m

Since we know from 15.5.6i) that antisymmetrization leads in my interpretation to

(∇j(∇iW ))k − (∇i(∇jW ))k = −RkmijWm

40

we can write(W k

;i) ;j − (W k;j) ;i = −RkmijWm − (ΓmijW

k;m − ΓmjiW

k;m)

And because ofT kij = −(Γkij − Γkji)

(see 15.3.3), we get finally

(W k;i) ;j − (W k

;j) ;i = −RkmijWm + Tmij Wk;m

We see that, sadly enough, that there is a torsion term present in the expression, in general. (In ”virtually allother sources” interpretation of the expression W k

;i;j)

15.6.9 p.411: see the comment to 15.3.4, p.385 (here, in the Additional material).

15.6.10 p.411: We learned how, within the Cartan formalism, the Gaussian curvarure K arises through theonly independent curvature 2-form

β = dα = Ke1 ∧ e2 ≡ KdS

dS being the (canonical metric) area element on the surface under consideration. Here we add another usefulinterpretation of this 2-form in terms of holonomy.

Consider a loop γ which is the boundary of a domain S, γ = ∂S. Fix an orthonormal frame field ea on S.Then, with respect to the frame field, ωab = ϵabα. Take a unit vector v and perform the parallel transport ofthe vector around the loop. The parallel transport equations for v read

va + ⟨ωab , γ⟩vb = va + ϵab⟨α, γ⟩vb = 0

or in detail

v1 + ⟨α, γ⟩v2 = 0

v2 − ⟨α, γ⟩v1 = 0

Let φ(t) denote the angle between e1(t) and v(t). Since v is unit vector, we may parameterize the componentsof v as follows

v1 = cosφ v2 = sinφ

Then we get a single equation for φ(t)φ = ⟨α, γ⟩

So, in time dt the increment of dφ isdφ = ⟨α, γ⟩dt

an the total net angle summed in the course of the loop γ is

[φ] =

∫ 1

0

⟨α, γ⟩dt ≡∮γ

α =

∫S

dα =

∫S

KdS

The net angle is, however, nothing but the holonomy for the loop. (The group element of the rotation groupSO(2) is given in terms of the angle [φ] modulo 2π.) So, the holonomy is given simply as the total Gaussiancurvature

holonomy for ∂S =

∫S

KdS ≡ total Gaussian curvature (over S)

A remarkable observation indeed! Take, in particular, constant Gaussian curvature K0. Then

holonomy for ∂S = K0

∫S

dS = K0S

So the holonomy is now proportional to the total area of the domain S. This is the case if we take the sphereS2, where (see 15.6.11) K(x) = K0 = 1/ρ2. We get

holonomy for ∂S = K0S = (1/ρ2)S = (1/ρ2)ρ2S1 ≡ S1

41

where S1 denotes the area of the corresponding unit sphere, i.e. the solid angle subtended by the loop.

holonomy for loop γ = solid angle subtended by γ

The reader is invited to check this elegant result for the particular case of the Foucault angle [(15.3.10, 15.6.22)- here the solid angle subtended by the parallel ϑ0 = const. is 2π(1− cosϑ0)] as well as for the geodesic trianglein (15.3.9) [here we get 1/8 of the unit sphere’s area, i.e. 4π/8 = π/2].

15.6.16 p.415: Here, various versions of Bianchi identity and Ricci identity are discussed.

In part i) we are asked to prove their “form version”, i.e.

dΩ+ ω ∧ Ω− Ω ∧ ω = 0 Bianchi identity

dT + ω ∧ T = Ω ∧ e Ricci identity

H This is remarkably easy: Let’s apply d on Cartan structure equations (15.6.7):

de+ ω ∧ e = T ⇒ dω ∧ e− ω ∧ de = dT

dω + ω ∧ ω = Ω ⇒ dω ∧ ω − ω ∧ dω = dΩ

Using, on the right, Cartan structure equations themselves, we get

(Ω− ω ∧ ω) ∧ e− ω ∧ (T − ω ∧ e) = dT

(Ω− ω ∧ ω) ∧ ω − ω ∧ (Ω− ω ∧ ω) = dΩ

or, after some canceling,

Ω ∧ e− ω ∧ T = dT

Ω ∧ ω − ω ∧ Ω = dΩ

which is exactly what we need. NIn part iv) interesting point is coordinate (index) version of both identities for RLC-connection (used stan-

dardly in general relativity), i.e.

dΩ+ ω ∧ Ω− Ω ∧ ω = 0 ⇔ Rij[kl;m] = 0 Bianchi identity

Ω ∧ e = 0 ⇔ Ri[jkl] = 0 Ricci identity

H We want to derive coordinate (index) versions of the identities from the corresponding form versions. Let’sexpress the forms w.r.t. coordinate coframes (15.6.1), (15.6.3):

ei = dxi ωij = Γijmdxm Ωij =

1

2Rijkldx

k ∧ dxl

Then

dΩij =1

2Rijkl,mdx

k ∧ dxl ∧ dxm

ωia ∧ Ωaj =1

2ΓiamR

ajkldx

k ∧ dxl ∧ dxm

−Ωia ∧ ωaj = −1

2ΓajmR

iakldx

k ∧ dxl ∧ dxm

and therefore, summing all three pieces, we get

(dΩ+ ω ∧ Ω− Ω ∧ ω)ij =1

2(Rijkl,m + ΓiamR

ajkl − ΓajmR

iakl)dx

k ∧ dxl ∧ dxm

Now comparing the terms in the bracket with

Rijkl;m ≡ Rijkl,m + ΓiamRajkl − ΓajmR

iakl − ΓakmR

ijal − ΓalmR

ijka

42

we see that two terms are missing in the bracket to become the complete covariant derivative Rijkl;m. However,

combining of the two missing terms with dxk ∧ dxl ∧ dxm actually gives zero

(−ΓakmRijal − ΓalmR

ijka)dx

k ∧ dxl ∧ dxm = 0

because of the symmetry of Christoffel symbols w.r.t. lower indices! So we can freely add them to the bracketwith no consequence and get

(dΩ+ ω ∧ Ω− Ω ∧ ω)ij =1

2Rijkl;mdx

k ∧ dxl ∧ dxm

Then, vanishing of the l.h.s. (i.e. the form version of Bianchi idendity) may be also expressed as

Rijkl;mdxk ∧ dxl ∧ dxm = 0

or, finally, asRij[kl;m] = 0

which is exactly what we need.

Ricci identity is even simpler: we have

(Ω ∧ e)i = Ωim ∧ dxm =1

2Rijkldx

k ∧ dxl ∧ dxm

Then vanishing of the expression (i.e. Ricci idendity) may be also expressed as

Rijkldxk ∧ dxl ∧ dxm = 0

or, finally, asRi[jkl] = 0

which is, again, exactly what we need. NIn part iii) coordinate (index) version of both identities for a general linear connection (so non-vanishing

torsion may be present) is found to be

dΩ+ ω ∧ Ω− Ω ∧ ω = 0 ⇔ Rij[kl;m] +Rijr[kTrlm] = 0 Bianchi identity

Ω ∧ e− (dT + ω ∧ T ) = 0 ⇔ Ri[jkl] = T i[jk;l] + T ir[jTrkl] Ricci identity

H Let’s show the difference (w.r.t. part iv) discussed above) on Bianchi identity (Ricci identity is thenproved in a similar way). So, we proceed as it is discussed in part iv) above. There we come to the point wheretwo terms are missing in the bracket to become the complete covariant derivative Rijkl;m. Combining them

with dxk ∧ dxl ∧ dxm actually gave zero

(−ΓakmRijal − ΓalmR

ijka)dx

k ∧ dxl ∧ dxm = 0

because of the symmetry of Christoffel symbols w.r.t. lower indices. Now the difference starts, since thesymmetry is no longer true. Rather, we have

Γijk = Γi(jk) + Γi[jk] = Γi(jk) −1

2T ijk (see (15.3.3))

Therefore, what we get now is

(ΓakmRijal + ΓalmR

ijka)dx

k ∧ dxl ∧ dxm = (Γa[km]Rijal + Γa[lm]R

ijka)dx

k ∧ dxl ∧ dxm

= −1

2(T akmR

ijal + T almR

ijka)dx

k ∧ dxl ∧ dxm

= −1

2(Rija[lT

akm] −Rija[kT

alm])dx

k ∧ dxl ∧ dxm

=1

2(Rija[kT

alm] +Rija[kT

alm])dx

k ∧ dxl ∧ dxm

= Rija[kTalm])dx

k ∧ dxl ∧ dxm

43

Therefore we can write, now,

(dΩ+ ω ∧ Ω− Ω ∧ ω)ij =1

2(Rijkl;m + ΓakmR

ijal + ΓalmR

ijka)dx

k ∧ dxl ∧ dxm

=1

2(Rijkl;m +Rija[kT

alm])dx

k ∧ dxl ∧ dxm

and, consequently,Rij[kl;m] +Rija[kT

alm] = 0

For Ricci identity, the new part (namely dT + ω ∧ T ) reads

(dT + ω ∧ T )i = dT i + ωij ∧ T j T i =1

2T ikldx

k ∧ dxl

=1

2(T ikl,m + ΓijmT

jkl)dx

k ∧ dxl ∧ dxm

=1

2(T ikl;m + ΓakmT

ial + ΓalmT

ika)dx

k ∧ dxl ∧ dxm

=1

2(T ikl;m + Γa[km]T

ial + Γa[lm]T

ika)dx

k ∧ dxl ∧ dxm

=1

2(T ikl;m − T akmT

ial)dx

k ∧ dxl ∧ dxm

=1

2(T ikl;m − T ialT

akm)dxk ∧ dxl ∧ dxm

=1

2(T i[kl;m] + T ia[kT

alm])dx

k ∧ dxl ∧ dxm

Then, because of still valid fact

(Ω ∧ e)i = Ωim ∧ dxm =1

2Ri[klm]dx

k ∧ dxl ∧ dxm

we come toRi[klm] = T i[kl;m] + T ia[kT

alm]

which is, as usually, exactly what we need. N

15.6.20 p.417: see the comment to 15.3.4, p.385 (here, in the Additional material).

16.4.6 p.453: Here we learned that a term of the form

⟨α, α⟩

in an action leads to energy-momentum tensor

Tab = 2(iaα, ibα)− gab(α, α)

Now, let us concentrate on a rather specific property of the latter, namely the sign of energy density correspond-ing to a general ⟨α, α⟩ term, i.e. the sign of T00 component of the energy-momentum tensor. From the generalformula we get, in particular

T00 = 2(i0α, i0α)− (α, α)

In order to decide whether this is positive or negative definite (or perhaps indefinite), the results of (16.3.6)come in handy. The problem teaches us that for

α = dt ∧ s+ r

we get(α, α)E1,3 = (ηs, s)E3 + (ηr, r)E3

Recall that both expressions (scalar products) on the right-hand side of the expression for T00 are of E1,3-type.We need, using the formula above, to express them in terms of E3-type scalar products. (The reason is, that

44

E1,3-type scalar product is indefinite, whereas E3-type scalar product is positive definite: (s, s)E3 ≥ 0 for anys.) So, we need to express (i0α, i0α) in terms of E3-type scalar products. Now,

α = dt ∧ s+ r ⇒ i0α = s

and so(i0α, i0α)E1,3 = (ηs, s)E3

Therefore,

T00 = 2(i0α, i0α)E1,3 − (α, α)E1,3

= 2(ηs, s)E3 − ((ηs, s)E3 + (ηr, r)E3)

= (ηs, s)E3 − (ηr, r)E3

The last formula clearly displays all we need.5 Let the degree of α be even. Then ηr = r, ηs = −s and so

T00 = −(s, s)E3 − (r, r)E3 ≤ 0

On the contrary, for the degree of α being odd, ηr = −r, ηs = s and

T00 = (s, s)E3 + (r, r)E3 ≥ 0

So, altogether, we see the following simple picture:

degree of α is even T00 ≤ 0

degree of α is odd T00 ≥ 0

Since physics only likes the case T00 ≥ 0, the result above is to be understood as follows: the term ⟨α, α⟩ shouldenter the action with plus sign iff α has odd degree and with minus sign iff α has even degree:

degree of α is even −⟨α, α⟩ in action integral

degree of α is odd +⟨α, α⟩ in action integral

This concise rule is indeed satisfied by all field theory action integrals mentioned in the book. Check, inparticular, (each individual term in) the following cases:

free (massive) scalar field S =1

2⟨dϕ, dϕ⟩ − 1

2m2⟨ϕ, ϕ⟩ (16.3.7)

free (massive) vector field S = −1

2⟨dW, dW ⟩+ 1

2m2⟨W,W ⟩ (16.3.8)

free electromagnetic field S = −1

2⟨dA, dA⟩ (16.3.2)

16.5.3 p.460: Here, the Hilbert action is discussed, and in particular its form in two-dimensional case (whichis exceptional) is shown.

And what about the three-dimensional case? Well, in orthonormal frame we have

Ωab ∧ ∗(ea ∧ eb) = Ωab ∧ ωab cec = Ωab ∧ ωabcec = ϵcabΩab ∧ ec = Ωa ∧ ea

so thatRgωg = Ωa ∧ ea

where we used, in 3D, a possibility to express curvature forms in terms of three two-forms

Ωa := ϵabcΩbc

5Except for love, of course. Sorry, John/Paul.

45

(not to be confused with Ricci 1-forms Ra ≡ ibΩba = Rabe

b, see 16.5.1).

18.4.6 p.511: see 15.4.4, p.392 (in this text)

18.4.10 p.514 The resultLVA = dχ

might look unnatural (one should expect LVA = 0 as the ”invariance condition” for A). Note, however, thataccording to 18.4.9(iv) there is a (”gauge”) freedom

A 7→ A′ = A+ df

in choosing A. So, if some particular A obeys LVA = dχ, then the ”new” one, A′ obeys

LVA′ = d(χ− V f)

So when we choose f such thatV f = χ

(in coordinates, where V straightens out, we are to solve the equation ∂f/∂x1 = χ) we get

LVA′ = 0

The moral is that we cannot ask ”nice” symmetry condition LVA = 0 being valid for all A’s from the equivalenceclass A ∼ A+df , since the transformation A 7→ A+df spoils it. In general, just the weaker condition LVA = dχholds and only for a carefully chosen representative A this simplifies to LVA = 0. (Since there is no comparablefreedom in g and ϕ, we have always ”nice” invariance conditions LV g = 0 and LV ϕ = 0 respectively for thosefields.)

19.3.10 p.534. Just prior to this problem the following formulation of Frobenius integrability criterion is

presented without proof: In terms of constraint 1-forms θi, D is integrable if and only if there are (n − k)2

1-forms σij such that dθi = σij ∧ θj , i.e.

D is integrable ⇔ ∃σij : dθi = σij ∧ θj

Let us have a look how this particular formulation follows from the one mentioned after 19.3.7:

D is integrable ⇔ θi∣∣D = 0 ⇒ dθi

∣∣D = 0 i.e. U, V ∈ D ⇒ dθi(U, V ) = 0

Well, consider a co-frame (ei, ea) adapted to D, i.e. ei := θi and ea realize a completion to a co-frame. Alsoconsider the dual frame (ei, ea). So, D = Span ea.

Then, since dθi is a 2-form, we can decompose it w.r.t. (ei, ea) as follows:

dθi = Aikjek ∧ ej +Biaje

a ∧ ej + Ciabea ∧ eb Aikj , B

iaj , C

iab functions

Now, the previous formulation says (for U = ea, V = eb)

D is integrable ⇔ dθi(ea, eb) = 0

Butdθi(ea, eb) = 2Ciab

So integrabilty kills Ciab (and vice versa) and we have

D is integrable ⇔ dθi = Aikjek ∧ ej +Biaje

a ∧ ej ≡ (Aikjek +Biaje

a) ∧ ej

So, definingAikje

k +Biajea =: σij

(and recalling that ej := θj) we finally get

D is integrable ⇔ dθi = σij ∧ θj

46

19.3.12 p.535. In part (iii) we study, as an example to integrability conditions

f i[α,β] = f j[αfiβ],j (a)

for system of partial differential equations

yi,α = f iα(x, y) xα independent, yi dependent

the following simple system

∂f

∂x= f sin y (b1)

∂f

∂y= λfx cos y (b2)

(So that y1 ≡ f , (x1, x2) ≡ (x, y), f11 ≡ f sin y, f12 ≡ λfx cos y.)

Let’s have a look on the system in elementary way, without using (a).

First, notice that we have a solution f = 0 for any λ. (So, the statement in the book that the system has asolution only for λ = 1 is indeed wrong, see Errata.)

Now, imagine f = 0. Then, for F ≡ ln f , we get slightly simpler equations

∂F

∂x= sin y (c1)

∂F

∂y= λx cos y (c2)

So, if F (x, y) exists, its dF readsdF = sin y dx+ λx cos y dy

Then, necessarily,0 = ddF = d(sin y dx+ λx cos y dy) = (λ− 1) cos y dx ∧ dy

Therefore we inevitably need λ = 1. If λ = 1 holds, we easily get from (c1) and (c2)

F (x, y) = x sin y +A(y) = x sin y +B(x)

so that A = B = k = const. and finally (only for λ = 1 !)

F (x, y) = x sin y + k i.e. f(x, y) = eF = K ex sin y

If we want to use criterion (a), it reads, here (check!)

(1− λ)f cos y = 0 (d)

for each (x, y, f). This is equivalent to λ = 1.

19.4.4 p.540 Correct version of the expression of Hi reads

Hi ≡ ∂hi := ∂i − ⟨ωab , ∂i⟩ybc∂ca ≡ ∂i − ⟨ωab , ∂i⟩ξEba

Notice the hat symbol on the first ωab - it is very important. Indeed, according to 19.1.4

ξEba= yca∂

bc

so that the above equality cannot be true with two unhatted ωab (the summation within the y∂ part is different).However, the relation between ωab and ωab is (see 19.2.1)

ωab := (y−1)ac (π∗ωcd)y

db + (y−1)acdy

cb

47

so that⟨ωab , ∂i⟩ = (y−1)ac ⟨ωcd, ∂i⟩ydb

Then,

Hi = ∂i − ⟨ωab , ∂i⟩ξEba

= ∂i − ⟨ωab , ∂i⟩yca∂bc= ∂i − (y−1)ar⟨ωrd, ∂i⟩ydb yca∂bc= ∂i − ⟨ωab , ∂i⟩ybc∂ca

or, consequently⟨ωab , ∂i⟩yca∂bc = ⟨ωab , ∂i⟩ybc∂ca

We see that the replacement ωab 7→ ωab indeed produces the change in summation in the y∂ expression.

19.6.1− 3 p.545-547: Here we present coordinate versions (in coordinates (x, y) introduced in 19.1.1) of the

“abstract” functions ΦB : LM → (V, ρ).

Let’s start with a vector field. We have, then, equivariant Φv : LM → (Rn, ρ10), i.e. in coordinates some

va(x, y) obeying va(x, yA) = (A−1)ab vb(x, y)

In particular, for y = I (the unit matrix), we get va(x,A) = (A−1)ab vb(x, I), i.e.

va(x, y) = (y−1)abVb(x) V b(x) := vb(x, I)

So we see that the family of functions va(x, y) obeying va(x, yA) = (A−1)ab vb(x, y) is parametrized by functions

V a(x) of variables x alone. (There is already no freedom in the dependance on variables y.)

What is the meaning of those functions V a(x)? Well, recall that the space (Rn, ρ10) is just the space ofcomponents of vectors, so (think it over) V a(x) are just components

- of the vector field V ≡ V a(x)ea ↔ Φv ↔ va(x, y)

- w.r.t. the frame ea ↔ y = I, i.e.- w.r.t. the frame ea ↔ the coordinates yab (see 19.1.1)

In a similar way we get, as an example, the following functions on LM :

αa(x, y) = ybaαb(x) satisfying αa(x, yA) = Abaαb(x, y)

corresponding to a covector field α = αa(x)ea,

gab(x, y) = ycaydb gcd(x) satisfying gab(x, yA) = AcaA

db gcd(x, y)

corresponding to metric tensor field g = gab(x)ea ⊗ eb,

tcab(x, y) = (y−1)ciyjaykb T

ijk(x) satisfying tcab(x, yA) = (A−1)ciA

jaA

kb tijk(x, y)

corresponding to torsion tensor field T = T cab(x)ea ⊗ eb ⊗ ec or

rdabc(x, y) = (y−1)di yjaykb ylcR

ijkl(x) satisfying rdabc(x, yA) = (A−1)diA

jaA

kbA

lcrijkl(x, y)

corresponding to Riemann curvature tensor field R = Rdabc(x)ea ⊗ eb ⊗ ec ⊗ ed.

19.6.6 p.549: Here we speak of an equivariant function Φ which is defined on the fibers over the curve γ(t)and which corresponds to field of quantities of type ρ (e.g. a tensor field B) on γ. We are to check that itsderivative along the horizontal lift γh of the curve γ corresponds (just in the sense in that Φ corresponds to B)to the covariant derivative ∇γB of the field B

Φ ↔ B ⇒ (γ)hΦ ↔ ∇γB

48

Let us check, first, that if Φ is of type ρ, the same holds for the derivative (γ)hΦ.

Recall (see 19.5.2 i)) that the horizontal lift γhe (i.e. the lift from x to e over x) is uniquely given by

π γhe = γ γhe (0) = e ⟨ω, ˙(γhe )⟩ = 0

Now it is useful to notice that it is GL(n,R)-invariant in the sense

RA γhe = γheA

(so that RA-image of the lift starting at e is a lift again, starting at eA).

H Indeed, let us denote Γ := RA γhe . It is clearly a curve on LM enjoying the following properties:

π Γ = γ Γ(0) = eA ⟨ω, Γ⟩ = 0

(We used π RA = π for the first property and Γ = RA∗˙(γhe ) as well as R

∗Aω = A−1ωA for the last one.) The

three properties define, however, exactly the curve γheA. NNow, define (again on the fibers over the curve γ(t)) the function

χ := ˙(γh)Φ i.e. χ(e) := ˙(γhe )Φ

(So, at each point e of the fiber over x, the corresponding lift of γ is applied on Φ.)

Then

χ(eA) = ˙(γheA)Φ = RA∗˙(γhe )Φ = ⟨dΦ, RA∗

˙(γhe )⟩ = ⟨d(R∗AΦ),

˙(γhe )⟩ = ρ(A−1)⟨dΦ, ˙(γhe )⟩ = ρ(A−1)χ(e)

So χ ≡ (γ)hΦ is an equivariant function (of type ρ) as well, i.e. it indeed corresponds to a quantity “of thesame type” as B does.

Which one? The derivative of Φ along γh corresponds to

- derivative of components (since the space (V, ρ) of values of Φ is the space of components of B)

- of the quantity B (since Φ ↔ B)

- w.r.t. autoparallel frame (since γh corresponds to autoparallel frames).

And this corresponds (see 19.6.4) to covariant derivative ∇γB of B.

20.2.5 p.562: Here we learn that two crucial properties of connection form read

R∗gω = Ad g−1ω ⟨ω, ξX⟩ = X

In 20.2.6 we add that any ω possessing the two properties is a connection form. And then, in 20.2.7, we alsocame to infinitesimal version of the former property, namely

LξXω = −[X,ω]

Here, in the Additional material, we just mention that the two properties may also be equivalently rewritteninto the following convenient form:

Lξiωj = −cjikωk ⟨ωi, ξj⟩ = δij (a)

where ξj ≡ ξEj are generators of the action Rg and cjik are the structure constants of the Lie algebra w.r.t. Ei(i.e. [Ei, Ek] = cjikEj).

H Indeed, set X = Ei; then

LξEiω = (Lξiωj)Ej

!= −[Ei, ω] = −[Ei, ω

kEk] = −[Ei, Ek]ωk = −cjikω

kEj

⟨ω, ξi⟩ = ⟨ωj , ξi⟩Ej!= Ei

49

NExample: consider a GA(1,R)-principal bundle π : P → M and let ω be a connection form. Since a possiblebasis E1, E2 in the Lie algebra ga(1,R) is

E1 =

(1 00 0

)E2 =

(0 10 0

)(see 11.7.10 - notice however an error mentioned in Errata), any connection form may be written as

ω = ωjEj = ω1E1 + ω2E2 =

(ω1 ω2

0 0

)The right translation on GA(1,R) is

R(a,b)(u, v) = (u, v) (a, b) = (ua, ab+ a)

(see 10.2.6). One easily finds the corresponding generators:

ξ1 = u∂u ξ2 = u∂v (b)

So, if we use a local trivialization of the bundle, we have P ∼ O ×GA(1,R) with coordinates (xµ, u, v), wherexµ are local coordinates on O ⊂ M , and the generators of the action of GA(1,R) on (the corresponding partof) P look exactly like those in (b).

Therefore if we want to find the most general connection form on this particular principal bundle, we canstart with the most general expressions

ω1 = Aµ(x, u, v)dxµ +B(x, u, v)du+ C(x, u, v)dv

ω2 = Aµ(x, u, v)dxµ + B(x, u, v)du+ C(x, u, v)dv

and then restrict to those of them, which satisfy (a), i.e., in our particular case,

Lξ1ω1 = 0 ⟨ω1, ξ1⟩ = 1

Lξ1ω2 = −ω2 ⟨ω1, ξ2⟩ = 0

Lξ2ω1 = 0 ⟨ω2, ξ1⟩ = 0

Lξ2ω2 = ω1 ⟨ω2, ξ2⟩ = 1

The right (”algebraic”) column alone quickly restricts ω1 and ω2 to

ω1 =du

u+Aµ(x, u, v)dx

µ

ω2 =dv

u+ Aµ(x, u, v)dx

µ

The left (”differential”) column then restricts Aµ(x, u, v) and Aµ(x, u, v) further to

Aµ(x, u, v) = fµ(x) Aµ(x, u, v) =1

u(vfµ(x) + gµ(x))

(check :-). So, the most general ω1 and ω2 are

ω1 =du

u+ fµ(x)dx

µ

ω2 =dv

u+

1

u(vfµ(x) + gµ(x))dx

µ

i.e. the most general connection form looks (locally) as follows:

ω =

(ω1 ω2

0 0

)=

1

u

(du+ ufµ(x)dx

µ dv + (vfµ(x) + gµ(x))dxµ

0 0

)(c)

50

It is parameterized by arbitrary functions fµ(x) and gµ(x). For example, for all of them vanishing we get

ω =

(ω1 ω2

0 0

)=

1

u

(du dv0 0

)Locally we can use ”canonical” section

σ : xµ 7→ (xµ, u(x), v(x)) = (xµ, 1, 0)

(abstractly m 7→ (m, g(m)) = (m, e), where e is the unit element of the group; see the proof of 20.1.3). In thisparticular gauge, the gauge potential reads

A ≡ σ∗ω =

(A1 A2

0 0

)=

(fµ(x)dx

µ gµ(x)dxµ

0 0

)≡

(A1µ(x)dx

µ A2µ(x)dx

µ

0 0

)So, the functions fµ(x) and gµ(x) parameterizing the general expression (c) may be identified with gaugepotentials in the canonical gauge. In particular, the example with vanishing fµ(x) and gµ(x) corresponds toA = 0 (in the particular gauge) and, therefore, to F ≡ dA+A∧A = 0. This means (due to the fact, that F = 0is already gauge-independent statement), that the example corresponds to flat connection (Ω = 0).

21.3.3 p.602: A useful information about holonomy may be gained here.

We learned here, that the quantity of type ρn (with respect to the group U(1)) gets multiplied, when parallel

transported along γ, by the factor e−in∫γA

z(0) 7→ z(1) = e−in∫γAz(0)

Let γ be a loop, now. Then γ(0) = γ(1) and the holonomy associated with the loop is the group elementg ∈ U(1) such that

γh(1) = Rgγh(0) ≡ γh(0)g

We assumed z to be a quantity of type ρn. Therefore it behaves in general as follows

z(pg) = ρn(g−1)z(p)

(R∗gΦ = ρ(g−1)Φ). In our particular case

p = γh(0) g = eiα (= unknown, wanted, the holonomy) ρn(eiα) = einα

and alsoz(0) = z(γh(0))

Therefore

z(1) ≡ z(γh(1))1.= z(γh(0)eiα) = e−inαz(γh(0)) = e−inαz(0)

2.= e−in

∮γAz(0)

whencethe holonomy for γ ≡ g ≡ eiα = ei

∮γA

If γ is the boundary of a 2-dimensional surface S, then∮γA =

∫SF and so, alternatively, we can write down

the holonomy in terms of the curvature 2-form

the holonomy for γ ≡ g ≡ eiα = ei∫SF

Particularly interesting is the case of a bundle over two-dimensional Riemannian base manifold (M, g) (a surface)such that the curvature form coincides (possibly up to a constant multiple) with the metric area (= volume)form dS

F = dS

51

Then, clearly ∫S

F = the area of S

so thatthe holonomy for ∂S = ei (area of S)

[This is to be compared with the additional text to 15.6.10.]

To give a concrete (and important)6 example, consider the Hopf bundle π : S3 → S2 studied in 20.1.7-10.We can introduce a U(1)-connection by defining the subspace spanned by left-invariant fields (e1, e2) on SU(2)(see 11.7.23) to be the horizontal subspace Hor pP (whereas the span of e3 is the vertical one Ver pP , see thefigure in 20.1.10; this means that the horizontal subspace is defined as the orthogonal complement, w.r.t. theKilling metric, to the vertical one). Alternatively (check), we can fix the connection via the connection form

ω := ie3

(it is to be u(1) ≡ iR-valued, that’s why the i). Then

Ω = dω = ide3 = i(− sinϑdϑ ∧ dφ)

and soiF ≡ σ∗Ω = i(− sinϑdϑ ∧ dφ)

(for any σ, check) so thatF = − sinϑdϑ ∧ dφ ≡ −dS

where dS is the standard round area form on S2. Therefore

the holonomy for ∂S = e−i (area of S) = e−i (solid angle subtended by γ)

22.5.4 p.665: In the 4-th expression of the Dirac operator, Γabc in the 2-nd expression was replaced by theexpression

Γabc = −1

2(cabc + cbca − ccab)

from 15.6.20iii. From Additional material to 15.6.20iii we know, however, that the full expression reads

Γabc = −1

2(cabc + cbca − ccab) +Kabc

where

Kabc := −1

2(Tabc + Tbca − Tcab)

is the contortion tensor. So, the contortion term is to be included also in the Dirac operator, in general.

Recall that spin connection corresponds to a metric, but not necessarily symmetric connection on (M, g),see the book, p.658 in Sec.22.4. Therefore, whenever we need covariantly differentiate spinor fields (and wedo need it within the construction called the Dirac operator), torsion of our connection is to be specified. Ifthe connection under consideration has non-vanishing torsion, the contortion term has to be present in thecorresponding Dirac operator:

D/ψ 2.= γa(eaψ +

1

4Γbcaγ

bγcψ)

4.= γa

(eaψ +

1

8((cabc + ccba − cbca) + 2Kbca) γ

bγcψ

)

Appendix A p.673: The concept of the direct sum V ⊕W of vector spaces V and W was introduced in the

beginning of Appendix A. An interesting particular case is the space V ⊕ V ∗ (i.e. when W is the dual of V ).

6This stuff is used in the context of the Berry phase for the spin 1/2. The phase is intimately connected with the holonomy of

this particular bundle.

52

Why? Because there are two important canonical structures available on this space: a metric tensor (with npluses and n minuses in canonical form) g and a symplectic form ω. Both are constructed in terms of canonicalpairing ⟨ . , . ⟩ between V and V ∗ alone. They are given as follows:

g((v, α), (w, β)) := ⟨α,w⟩+ ⟨β, v⟩ω((v, α), (w, β)) := ⟨α,w⟩ − ⟨β, v⟩

In particular, if we choose the basis Ea, Ea in V ⊕ V ∗, where

V ∋ ea ↔ (ea, 0) ≡ Ea ∈ V ⊕ V ∗

V ∗ ∋ ea ↔ (0, ea) ≡ Ea ∈ V ⊕ V ∗

where, of course,⟨ea, eb⟩ = δab

we get the component expressions

g ↔(g(Ea, Eb) g(Ea, E

b)g(Ea, Eb) g(Ea, Eb)

)=

(0 δbaδab 0

)ω ↔

(ω(Ea, Eb) ω(Ea, E

b)ω(Ea, Eb) ω(Ea, Eb)

)=

(0 −δbaδab 0

)So, the symplectic form is already in canonical form in the basis. Metric tensor turns out to be indefinite, withn pluses and n minuses.

H Determinant of the matrix of g is

det

(0 II 0

)= det

(I II 0

)= det

(I 0I −I

)= det I det(−I) = (−1)n = 0

so that there are no zeroes on diagonal in canonical form. What is on the diagonal is a (non-zero :-) solution ofthe characteristic equation

0 = det

(−λI II −λI

)= det

(−λI 0I (1/λ− λ)I

)= det(−λI) det(1/λ− λ)I = (λ− 1)n(λ+ 1)n

Therefore, the canonical form is diag (+1, . . . ,+1,−1, . . . ,−1). NIf V carries a representation ρ of a group G, then (see 12.1.8) V ∗ carries the corresponding dual representation

ρ and, consequently, V ⊕ V ∗ carries their direct sum ρ ⊕ ρ (see 12.4.10). Now, remarkably (albeit technicallytrivially), both g and ω are invariant with respect to ρ⊕ ρ.

H Indeed, for the metric tensor g, say, we have (for any group element k ∈ G)

g((ρ⊕ ρ)(k)(v, α), (ρ⊕ ρ)(k)(w, β)) := g(ρ(k)v, ρ(k)α), (ρ(k)w, ρ(k)β)) (see 12.4.10)

= ⟨ρ(k)α, ρ(k)w⟩+ ⟨ρ(k)β, ρ(k)v⟩= ⟨α,w⟩+ ⟨β, v⟩ (see 12.1.8)

= g((v, α), (w, β))

and similarly for the symplectic form ω. NSo, starting with just a representation (V, ρ) of G, we get, in an appropriately enlarged space, much stronger

structure, the G-invariant metric g as well as symplectic form ω.