discrete morse theory and generalized factor order · discrete morse theory and generalized factor...

Discrete Morse Theory and GeneralizedFactor Order

Bruce SaganDepartment of Mathematics, Michigan State U.

East Lansing, MI 48824-1027, sagan@math.msu.eduwww.math.msu.edu/˜sagan

Robert WillenbringDepartment of Mathematics, Michigan State U.

East Lansing, MI 48824-1027, willenb4@msu.edu

March 12, 2011

Introduction to Forman’s Discrete Morse Theory (DMT)

The Mobius function and the order complex ∆(x , y)

Babson and Hersh apply DMT to ∆(x , y)

Generalized Factor Order

References

Outline

References

Let σ be a simplex and τ be a face of σ with dim τ = dimσ − 1.

A collapse is a strong deformation retract of σ onto ∂σ − int τ .Ex.

στ → →

Let ∆ be a simplicial complex. Denote the reduced homologygroups, Betti numbers, and Euler characteristic of ∆ by

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

A Morse matching (MM) on ∆ is a matching of the simplices of∆ such that, for every matched pair (σ, τ), the correspondingcollapses can all be done to form a new (cell) complex ∆c . Theunmatched simplices in ∆ are called critical and these are theonly ones surviving in ∆c . We also have ∆ ' ∆c , soHd (∆) ∼= Hd (∆c) for all d .

TheoremLet ∆ have a MM with cd critical simplices of dimension d.

1. (Weak Morse inequalities) βd (∆) ≤ cd for all d.2. χ(∆) =

∑d (−1)dcd .

Let σ be a simplex and τ be a face of σ with dim τ = dimσ − 1.A collapse is a strong deformation retract of σ onto ∂σ − int τ .

στ → →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

∑d (−1)dcd .

Let σ be a simplex and τ be a face of σ with dim τ = dimσ − 1.A collapse is a strong deformation retract of σ onto ∂σ − int τ .Ex.

→ →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

∑d (−1)dcd .

στ →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

∑d (−1)dcd .

στ → →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

∑d (−1)dcd .

στ → →

Let ∆ be a simplicial complex.

Denote the reduced homologygroups, Betti numbers, and Euler characteristic of ∆ by

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

∑d (−1)dcd .

στ → →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

∑d (−1)dcd .

στ → →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

A Morse matching (MM) on ∆ is a matching of the simplices of∆ such that, for every matched pair (σ, τ), the correspondingcollapses can all be done to form a new (cell) complex ∆c .

Theunmatched simplices in ∆ are called critical and these are theonly ones surviving in ∆c . We also have ∆ ' ∆c , soHd (∆) ∼= Hd (∆c) for all d .

∑d (−1)dcd .

στ → →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

A Morse matching (MM) on ∆ is a matching of the simplices of∆ such that, for every matched pair (σ, τ), the correspondingcollapses can all be done to form a new (cell) complex ∆c . Theunmatched simplices in ∆ are called critical and these are theonly ones surviving in ∆c .

We also have ∆ ' ∆c , soHd (∆) ∼= Hd (∆c) for all d .

∑d (−1)dcd .

στ → →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

∑d (−1)dcd .

στ → →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

∑d (−1)dcd .

στ → →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

1. (Weak Morse inequalities) βd (∆) ≤ cd for all d.

2. χ(∆) =∑

d (−1)dcd .

στ → →

Hd (∆), βd (∆) = rk Hd (∆), χ(∆) =∑

d (−1)d βd (∆).

∑d (−1)dcd .

Outline

References

Let P be a poset.

The Mobius function of [x , y ] ⊆ P is

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

The order complex of [x , y ] is the abstract simplicial complex

∆(x , y) = {C : C is a chain in (x , y)}.

Theoremµ(x , y) = χ(∆(x , y)).

Ex. The µ(x ,w) in the following interval are purple.

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

Let P be a poset. The Mobius function of [x , y ] ⊆ P is

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

µ(x , x) = 1, µ(x , y) = −∑

x≤z<y

µ(x , z).

[x , y ] =

−1 −1

µ(x , y) = 0

∆(x , y) = a b

χ(∆(x , y)) = 0

Outline

References

Let C = C(x , y) be the set of containment-maximal chains in(x , y).

By convention, list the elements of C ∈ C from smallestto largest including x , y . If � is a total order on C then a newface of C ∈ C is C′ ⊆ C with C′ 6⊆ B for all B ≺ C.Ex. If B : x ,a, c,d , y and C : x ,b, c,d , y then the new faces ofC are those containing b.We wish to construct a MM inductively by

matching all new faces of each C except perhaps one. (1)

B : x = x0, x1, . . . , xm = y and C : x = y0, y1, . . . , yn = y

agree to level k if x0 = y0, . . . , xk = yk . They diverge from levelk if they agree to level k but xk+1 6= yk+1.Ex. B : x ,a, c,d , y and C : x ,b, c,d , y diverge from level 0.Call � a poset lexicographic (PL) order if, whenever C,Ddiverge from some level k and C′,D′ agree with C,Drespectively to level k + 1, then

C � D ⇐⇒ C′ � D′.Proposition (Babson and Hersh)If � is an PL-order then it has a MM satisfying (1).

Let C = C(x , y) be the set of containment-maximal chains in(x , y). By convention, list the elements of C ∈ C from smallestto largest including x , y .

If � is a total order on C then a newface of C ∈ C is C′ ⊆ C with C′ 6⊆ B for all B ≺ C.Ex. If B : x ,a, c,d , y and C : x ,b, c,d , y then the new faces ofC are those containing b.We wish to construct a MM inductively by

Let C = C(x , y) be the set of containment-maximal chains in(x , y). By convention, list the elements of C ∈ C from smallestto largest including x , y . If � is a total order on C then a newface of C ∈ C is C′ ⊆ C with C′ 6⊆ B for all B ≺ C.

Ex. If B : x ,a, c,d , y and C : x ,b, c,d , y then the new faces ofC are those containing b.We wish to construct a MM inductively by

Let C = C(x , y) be the set of containment-maximal chains in(x , y). By convention, list the elements of C ∈ C from smallestto largest including x , y . If � is a total order on C then a newface of C ∈ C is C′ ⊆ C with C′ 6⊆ B for all B ≺ C.Ex. If B : x ,a, c,d , y and C : x ,b, c,d , y then the new faces ofC are those containing b.

We wish to construct a MM inductively by

Let C = C(x , y) be the set of containment-maximal chains in(x , y). By convention, list the elements of C ∈ C from smallestto largest including x , y . If � is a total order on C then a newface of C ∈ C is C′ ⊆ C with C′ 6⊆ B for all B ≺ C.Ex. If B : x ,a, c,d , y and C : x ,b, c,d , y then the new faces ofC are those containing b.We wish to construct a MM inductively by

agree to level k if x0 = y0, . . . , xk = yk .

They diverge from levelk if they agree to level k but xk+1 6= yk+1.Ex. B : x ,a, c,d , y and C : x ,b, c,d , y diverge from level 0.Call � a poset lexicographic (PL) order if, whenever C,Ddiverge from some level k and C′,D′ agree with C,Drespectively to level k + 1, then

agree to level k if x0 = y0, . . . , xk = yk . They diverge from levelk if they agree to level k but xk+1 6= yk+1.

Ex. B : x ,a, c,d , y and C : x ,b, c,d , y diverge from level 0.Call � a poset lexicographic (PL) order if, whenever C,Ddiverge from some level k and C′,D′ agree with C,Drespectively to level k + 1, then

agree to level k if x0 = y0, . . . , xk = yk . They diverge from levelk if they agree to level k but xk+1 6= yk+1.Ex. B : x ,a, c,d , y and C : x ,b, c,d , y diverge from level 0.

Call � a poset lexicographic (PL) order if, whenever C,Ddiverge from some level k and C′,D′ agree with C,Drespectively to level k + 1, then

C � D ⇐⇒ C′ � D′.

Proposition (Babson and Hersh)If � is an PL-order then it has a MM satisfying (1).

How do we identify new faces?

A closed interval in a chainC : x0, . . . , xn is a subchain of the form

I = C[xk , xl ] : xk , xk+1, . . . , xl .

Open intervals C(xk , xl) are defined similarly. A skipped interval(SI) is I ⊆ C with C − I ⊆ B for some B ≺ C. A minimal skippedinterval (MSI) is a SI which is minimal with respect tocontainment.Ex. If B : x ,a, c,d , y & C : x ,b, c,d , y then C has MSI {b}.

I(C)def= {I : I is an MSI of C}.

Lemma (Babson and Hersh)C′ ⊆ C is new ⇐⇒ C′ ∩ I 6= ∅ for all I ∈ I(C).Proof “ =⇒ ” If C′ ∩ I = ∅ for some I then C′ ⊆ C − I. And I askipped interval implies C − I ⊆ B for some B ≺ C. But thenC′ ⊆ B for B ≺ C, contradicting the fact that C′ is new.

How do we identify new faces? A closed interval in a chainC : x0, . . . , xn is a subchain of the form

I = C[xk , xl ] : xk , xk+1, . . . , xl .

Open intervals C(xk , xl) are defined similarly.

A skipped interval(SI) is I ⊆ C with C − I ⊆ B for some B ≺ C. A minimal skippedinterval (MSI) is a SI which is minimal with respect tocontainment.Ex. If B : x ,a, c,d , y & C : x ,b, c,d , y then C has MSI {b}.

I = C[xk , xl ] : xk , xk+1, . . . , xl .

Open intervals C(xk , xl) are defined similarly. A skipped interval(SI) is I ⊆ C with C − I ⊆ B for some B ≺ C.

A minimal skippedinterval (MSI) is a SI which is minimal with respect tocontainment.Ex. If B : x ,a, c,d , y & C : x ,b, c,d , y then C has MSI {b}.

I = C[xk , xl ] : xk , xk+1, . . . , xl .

Open intervals C(xk , xl) are defined similarly. A skipped interval(SI) is I ⊆ C with C − I ⊆ B for some B ≺ C. A minimal skippedinterval (MSI) is a SI which is minimal with respect tocontainment.

Ex. If B : x ,a, c,d , y & C : x ,b, c,d , y then C has MSI {b}.

I = C[xk , xl ] : xk , xk+1, . . . , xl .

Lemma (Babson and Hersh)C′ ⊆ C is new ⇐⇒ C′ ∩ I 6= ∅ for all I ∈ I(C).

Proof “ =⇒ ” If C′ ∩ I = ∅ for some I then C′ ⊆ C − I. And I askipped interval implies C − I ⊆ B for some B ≺ C. But thenC′ ⊆ B for B ≺ C, contradicting the fact that C′ is new.

I = C[xk , xl ] : xk , xk+1, . . . , xl .

Lemma (Babson and Hersh)C′ ⊆ C is new ⇐⇒ C′ ∩ I 6= ∅ for all I ∈ I(C).Proof “ =⇒ ”

If C′ ∩ I = ∅ for some I then C′ ⊆ C − I. And I askipped interval implies C − I ⊆ B for some B ≺ C. But thenC′ ⊆ B for B ≺ C, contradicting the fact that C′ is new.

I = C[xk , xl ] : xk , xk+1, . . . , xl .

Lemma (Babson and Hersh)C′ ⊆ C is new ⇐⇒ C′ ∩ I 6= ∅ for all I ∈ I(C).Proof “ =⇒ ” If C′ ∩ I = ∅ for some I then C′ ⊆ C − I.

And I askipped interval implies C − I ⊆ B for some B ≺ C. But thenC′ ⊆ B for B ≺ C, contradicting the fact that C′ is new.

I = C[xk , xl ] : xk , xk+1, . . . , xl .

Lemma (Babson and Hersh)C′ ⊆ C is new ⇐⇒ C′ ∩ I 6= ∅ for all I ∈ I(C).Proof “ =⇒ ” If C′ ∩ I = ∅ for some I then C′ ⊆ C − I. And I askipped interval implies C − I ⊆ B for some B ≺ C.

But thenC′ ⊆ B for B ≺ C, contradicting the fact that C′ is new.

I = C[xk , xl ] : xk , xk+1, . . . , xl .

Call a C containing an unmatched face critical . How do weidentify critical chains?

We need to turn I(C) into a set ofdisjoint intervals J (C) as follows. Since I(C) has nocontainments, the intervals can be ordered I1, . . . , Il so that

min I1 < . . . < min Il and max I1 < . . . < max Il .

Let J1 = I1. Construct I′2 = I2 − J1, . . . , I′l = Il − J1 and throw outany which are not containment minimal. Let J2 = I′j where j isthe smallest index of the intervals remaining. Continue in thisway to form J (C).

Theorem (Babson and Hersh)Let [x , y ] be an interval and let ≺ be an PL order on C(x , y).

1. C ∈ C(x , y) is critical ⇐⇒ J (C) covers C.2. The critical face of a critical chain C is obtained by picking

the smallest element from each J ∈ J (C).3. We have

µ(x , y) =∑

(−1)#J (C)−1

where the sum is over all critical C ∈ C(x , y).

Call a C containing an unmatched face critical . How do weidentify critical chains? We need to turn I(C) into a set ofdisjoint intervals J (C) as follows.

Since I(C) has nocontainments, the intervals can be ordered I1, . . . , Il so that

µ(x , y) =∑

(−1)#J (C)−1

Call a C containing an unmatched face critical . How do weidentify critical chains? We need to turn I(C) into a set ofdisjoint intervals J (C) as follows. Since I(C) has nocontainments, the intervals can be ordered I1, . . . , Il so that

µ(x , y) =∑

(−1)#J (C)−1

Let J1 = I1.

Construct I′2 = I2 − J1, . . . , I′l = Il − J1 and throw outany which are not containment minimal. Let J2 = I′j where j isthe smallest index of the intervals remaining. Continue in thisway to form J (C).

µ(x , y) =∑

(−1)#J (C)−1

Let J1 = I1. Construct I′2 = I2 − J1, . . . , I′l = Il − J1 and throw outany which are not containment minimal.

Let J2 = I′j where j isthe smallest index of the intervals remaining. Continue in thisway to form J (C).

µ(x , y) =∑

(−1)#J (C)−1

Let J1 = I1. Construct I′2 = I2 − J1, . . . , I′l = Il − J1 and throw outany which are not containment minimal. Let J2 = I′j where j isthe smallest index of the intervals remaining.

Continue in thisway to form J (C).

µ(x , y) =∑

(−1)#J (C)−1

µ(x , y) =∑

(−1)#J (C)−1

µ(x , y) =∑

(−1)#J (C)−1

1. C ∈ C(x , y) is critical ⇐⇒ J (C) covers C.

2. The critical face of a critical chain C is obtained by pickingthe smallest element from each J ∈ J (C).

3. We haveµ(x , y) =

(−1)#J (C)−1

the smallest element from each J ∈ J (C).

3. We haveµ(x , y) =

(−1)#J (C)−1

µ(x , y) =∑

(−1)#J (C)−1

Outline

References

Let A be a set (the alphabet) and let A∗ be the set of words wover A.

Call u ∈ A∗ a factor of w if w = xuy for some x , y ∈ A∗.Ex. u = abba is a factor of w = baabbaa.Factor order on A∗ is the partial order u ≤ w if u is a factor ofw . The inner and outer factors of w = a1a2 . . . an arei(w) = a2 . . . an−1.

o(w) = longest word which is a proper prefix and suffix of w .Ex. w = abbab has i(w) = bba and o(w) = ab.Call w = a1 . . . an flat if a1 = . . . = an. Let |w | be w ’s length.Theorem (Bjorner)In factor order on A∗

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

1 if |w | − |u| = 2, w not flat, u ∈ {o(w), i(w)};(−1)|w |−|u| if |w | − |u| < 2;0 otherwise.

Also, ∆(u,w) ' ball or sphere when µ(u,w) = 0 or ±1, resp.Ex. µ(a,abbab) = µ(a,ab) = −1.

Let A be a set (the alphabet) and let A∗ be the set of words wover A. Call u ∈ A∗ a factor of w if w = xuy for some x , y ∈ A∗.

Ex. u = abba is a factor of w = baabbaa.Factor order on A∗ is the partial order u ≤ w if u is a factor ofw . The inner and outer factors of w = a1a2 . . . an arei(w) = a2 . . . an−1.

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

Let A be a set (the alphabet) and let A∗ be the set of words wover A. Call u ∈ A∗ a factor of w if w = xuy for some x , y ∈ A∗.Ex. u = abba is a factor of w = baabbaa.

Factor order on A∗ is the partial order u ≤ w if u is a factor ofw . The inner and outer factors of w = a1a2 . . . an arei(w) = a2 . . . an−1.

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

Let A be a set (the alphabet) and let A∗ be the set of words wover A. Call u ∈ A∗ a factor of w if w = xuy for some x , y ∈ A∗.Ex. u = abba is a factor of w = baabbaa.Factor order on A∗ is the partial order u ≤ w if u is a factor ofw .

The inner and outer factors of w = a1a2 . . . an arei(w) = a2 . . . an−1.

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

Let A be a set (the alphabet) and let A∗ be the set of words wover A. Call u ∈ A∗ a factor of w if w = xuy for some x , y ∈ A∗.Ex. u = abba is a factor of w = baabbaa.Factor order on A∗ is the partial order u ≤ w if u is a factor ofw . The inner and outer factors of w = a1a2 . . . an arei(w) = a2 . . . an−1.

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

o(w) = longest word which is a proper prefix and suffix of w .

Ex. w = abbab has i(w) = bba and o(w) = ab.Call w = a1 . . . an flat if a1 = . . . = an. Let |w | be w ’s length.Theorem (Bjorner)In factor order on A∗

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

o(w) = longest word which is a proper prefix and suffix of w .Ex. w = abbab has i(w) = bba and o(w) = ab.

Call w = a1 . . . an flat if a1 = . . . = an. Let |w | be w ’s length.Theorem (Bjorner)In factor order on A∗

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

o(w) = longest word which is a proper prefix and suffix of w .Ex. w = abbab has i(w) = bba and o(w) = ab.Call w = a1 . . . an flat if a1 = . . . = an.

Let |w | be w ’s length.Theorem (Bjorner)In factor order on A∗

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

o(w) = longest word which is a proper prefix and suffix of w .Ex. w = abbab has i(w) = bba and o(w) = ab.Call w = a1 . . . an flat if a1 = . . . = an. Let |w | be w ’s length.

Theorem (Bjorner)In factor order on A∗

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w);

µ(u,w) =

µ(u,o(w)) if |w | − |u| > 2, u ≤ o(w) 6≤ i(w); ⇐1 if |w | − |u| = 2, w not flat, u ∈ {o(w), i(w)};(−1)|w |−|u| if |w | − |u| < 2;0 otherwise. ⇐

µ(u,w) =

Also, ∆(u,w) ' ball or sphere when µ(u,w) = 0 or ±1, resp.

Ex. µ(a,abbab) = µ(a,ab) = −1.

µ(u,w) =

Also, ∆(u,w) ' ball or sphere when µ(u,w) = 0 or ±1, resp.Ex. µ(a,abbab)

= µ(a,ab) = −1.

µ(u,w) =

Also, ∆(u,w) ' ball or sphere when µ(u,w) = 0 or ±1, resp.Ex. µ(a,abbab) = µ(a,ab)

= −1.

µ(u,w) =

Write chains in [u,w ] dually from largest to smallest element.

An embedding of u in w is η ∈ (A ] {0})∗ obtained by zeroingout the positions of w outside of a given factor equal to u.Ex. If u = abba and w = baabbaa then η = 00abba0If y covers x then there is a unique embedding of x in y , unlessy is flat in which case we choose the embedding starting with 0.So any maximal chain C : w = w0,w1, . . . ,wm = u determinesa chain of embeddings with labels l(C) = (l1, . . . , ln)

C : η0l1→ η1

l2→ η2l3→ . . .

lm→ ηm

where the li give the position of the new zero in ηi .Ex. C : baabbaa,aabbaa,aabba,abba becomesC : baabbaa 1→ 0aabbaa 7→ 0aabba0 2→ 00abba0,l(C) = (1,7,2).

Lemma (S and Willenbring)The total order on C(w ,u) given by B � C iff l(B) ≤lex l(C) is aPL-order

Write chains in [u,w ] dually from largest to smallest element.An embedding of u in w is η ∈ (A ] {0})∗ obtained by zeroingout the positions of w outside of a given factor equal to u.

Ex. If u = abba and w = baabbaa then η = 00abba0If y covers x then there is a unique embedding of x in y , unlessy is flat in which case we choose the embedding starting with 0.So any maximal chain C : w = w0,w1, . . . ,wm = u determinesa chain of embeddings with labels l(C) = (l1, . . . , ln)

C : η0l1→ η1

l2→ η2l3→ . . .

lm→ ηm

Write chains in [u,w ] dually from largest to smallest element.An embedding of u in w is η ∈ (A ] {0})∗ obtained by zeroingout the positions of w outside of a given factor equal to u.Ex. If u = abba and w = baabbaa then η = 00abba0

If y covers x then there is a unique embedding of x in y , unlessy is flat in which case we choose the embedding starting with 0.So any maximal chain C : w = w0,w1, . . . ,wm = u determinesa chain of embeddings with labels l(C) = (l1, . . . , ln)

C : η0l1→ η1

l2→ η2l3→ . . .

lm→ ηm

Write chains in [u,w ] dually from largest to smallest element.An embedding of u in w is η ∈ (A ] {0})∗ obtained by zeroingout the positions of w outside of a given factor equal to u.Ex. If u = abba and w = baabbaa then η = 00abba0If y covers x then there is a unique embedding of x in y , unlessy is flat in which case we choose the embedding starting with 0.

So any maximal chain C : w = w0,w1, . . . ,wm = u determinesa chain of embeddings with labels l(C) = (l1, . . . , ln)

C : η0l1→ η1

l2→ η2l3→ . . .

lm→ ηm

Write chains in [u,w ] dually from largest to smallest element.An embedding of u in w is η ∈ (A ] {0})∗ obtained by zeroingout the positions of w outside of a given factor equal to u.Ex. If u = abba and w = baabbaa then η = 00abba0If y covers x then there is a unique embedding of x in y , unlessy is flat in which case we choose the embedding starting with 0.So any maximal chain C : w = w0,w1, . . . ,wm = u determinesa chain of embeddings with labels l(C) = (l1, . . . , ln)

C : η0l1→ η1

l2→ η2l3→ . . .

lm→ ηm

where the li give the position of the new zero in ηi .

Ex. C : baabbaa,aabbaa,aabba,abba becomesC : baabbaa 1→ 0aabbaa 7→ 0aabba0 2→ 00abba0,l(C) = (1,7,2).

C : η0l1→ η1

l2→ η2l3→ . . .

lm→ ηm

where the li give the position of the new zero in ηi .Ex. C : baabbaa,aabbaa,aabba,abba becomesC : baabbaa 1→ 0aabbaa 7→ 0aabba0 2→ 00abba0,

l(C) = (1,7,2).

C : η0l1→ η1

l2→ η2l3→ . . .

lm→ ηm

C : η0l1→ η1

l2→ η2l3→ . . .

lm→ ηm

DMT gives a proof of Bjorner’s formula which explains thedefinitions of i(w) and o(w) and the inequality between them.

Ex. Let u = a, w = abbab and consider all chains in C(w ,u)passing through ab = o(w):

B : abbab 1→ 0bbab 2→ 00bab 3→ 000ab 5→ 000a0,

C : abbab 5→ abba0 4→ abb00 3→ ab000 2→ a0000.

Note that C(abbab,ab) is an SI of C and is, in fact, an MSI.

Proposition (S and Willenbring)Let u ≤ o(w) 6≤ i(w) and let C ∈ C(w ,u) be thelexicographically first chain passing through the prefixembedding of o(w) in w. Then I = C(w ,o(w)) is an MSI.Proof Let B be the chain which goes from w to the suffixembedding of o(w) and then continues to u as does C. ThenB ≺ C and C − I ⊆ B so I is an SI. Because o(w) 6≤ i(w) thereare only two embeddings of o(w) in w : prefix and suffix. Thus,since C is lexicographically first through the prefix embedding,any other chain prior to C must agree with the portion of B upto the suffix embedding of o(w). So I is an MSI.

DMT gives a proof of Bjorner’s formula which explains thedefinitions of i(w) and o(w) and the inequality between them.Ex. Let u = a, w = abbab and consider all chains in C(w ,u)passing through ab = o(w):

Proposition (S and Willenbring)Let u ≤ o(w) 6≤ i(w) and let C ∈ C(w ,u) be thelexicographically first chain passing through the prefixembedding of o(w) in w.

Then I = C(w ,o(w)) is an MSI.Proof Let B be the chain which goes from w to the suffixembedding of o(w) and then continues to u as does C. ThenB ≺ C and C − I ⊆ B so I is an SI. Because o(w) 6≤ i(w) thereare only two embeddings of o(w) in w : prefix and suffix. Thus,since C is lexicographically first through the prefix embedding,any other chain prior to C must agree with the portion of B upto the suffix embedding of o(w). So I is an MSI.

Proposition (S and Willenbring)Let u ≤ o(w) 6≤ i(w) and let C ∈ C(w ,u) be thelexicographically first chain passing through the prefixembedding of o(w) in w. Then I = C(w ,o(w)) is an MSI.

Proof Let B be the chain which goes from w to the suffixembedding of o(w) and then continues to u as does C. ThenB ≺ C and C − I ⊆ B so I is an SI. Because o(w) 6≤ i(w) thereare only two embeddings of o(w) in w : prefix and suffix. Thus,since C is lexicographically first through the prefix embedding,any other chain prior to C must agree with the portion of B upto the suffix embedding of o(w). So I is an MSI.

Proposition (S and Willenbring)Let u ≤ o(w) 6≤ i(w) and let C ∈ C(w ,u) be thelexicographically first chain passing through the prefixembedding of o(w) in w. Then I = C(w ,o(w)) is an MSI.Proof Let B be the chain which goes from w to the suffixembedding of o(w) and then continues to u as does C.

ThenB ≺ C and C − I ⊆ B so I is an SI. Because o(w) 6≤ i(w) thereare only two embeddings of o(w) in w : prefix and suffix. Thus,since C is lexicographically first through the prefix embedding,any other chain prior to C must agree with the portion of B upto the suffix embedding of o(w). So I is an MSI.

Proposition (S and Willenbring)Let u ≤ o(w) 6≤ i(w) and let C ∈ C(w ,u) be thelexicographically first chain passing through the prefixembedding of o(w) in w. Then I = C(w ,o(w)) is an MSI.Proof Let B be the chain which goes from w to the suffixembedding of o(w) and then continues to u as does C. ThenB ≺ C and C − I ⊆ B so I is an SI.

Because o(w) 6≤ i(w) thereare only two embeddings of o(w) in w : prefix and suffix. Thus,since C is lexicographically first through the prefix embedding,any other chain prior to C must agree with the portion of B upto the suffix embedding of o(w). So I is an MSI.

Proposition (S and Willenbring)Let u ≤ o(w) 6≤ i(w) and let C ∈ C(w ,u) be thelexicographically first chain passing through the prefixembedding of o(w) in w. Then I = C(w ,o(w)) is an MSI.Proof Let B be the chain which goes from w to the suffixembedding of o(w) and then continues to u as does C. ThenB ≺ C and C − I ⊆ B so I is an SI. Because o(w) 6≤ i(w) thereare only two embeddings of o(w) in w : prefix and suffix.

Thus,since C is lexicographically first through the prefix embedding,any other chain prior to C must agree with the portion of B upto the suffix embedding of o(w). So I is an MSI.

Proposition (S and Willenbring)Let u ≤ o(w) 6≤ i(w) and let C ∈ C(w ,u) be thelexicographically first chain passing through the prefixembedding of o(w) in w. Then I = C(w ,o(w)) is an MSI.Proof Let B be the chain which goes from w to the suffixembedding of o(w) and then continues to u as does C. ThenB ≺ C and C − I ⊆ B so I is an SI. Because o(w) 6≤ i(w) thereare only two embeddings of o(w) in w : prefix and suffix. Thus,since C is lexicographically first through the prefix embedding,any other chain prior to C must agree with the portion of B upto the suffix embedding of o(w).

So I is an MSI.

Let P be any poset.

Define generalized factor order on P∗ bysaying u = a1 . . . ak ≤ w = b1 . . . bn if w has a factorbi+1 . . . bi+k with

a1 ≤P bi+1, . . . ,ak ≤P bi+k .

Ex. If P = P (positive integers) in the normal ordering then352 ≤ 175614 in P∗ because comparing 352 with the factor 756gives 3 ≤ 7, 5 ≤ 5, and 2 ≤ 6.Note that if A is an antichain then generalized factor order on Ais Bjorner’s factor order. Using DMT we have been able todetermine µ for P∗. The analogues of i(w) and o(w) are notobvious and the proof is an order of magnitude harder than foran antichain. It would not have been possible without Babsonand Hersh’s adaptation of DMT.

Let P be any poset. Define generalized factor order on P∗ bysaying u = a1 . . . ak ≤ w = b1 . . . bn if w has a factorbi+1 . . . bi+k with

a1 ≤P bi+1, . . . ,ak ≤P bi+k .

Ex. If P = P (positive integers) in the normal ordering

then352 ≤ 175614 in P∗ because comparing 352 with the factor 756gives 3 ≤ 7, 5 ≤ 5, and 2 ≤ 6.Note that if A is an antichain then generalized factor order on Ais Bjorner’s factor order. Using DMT we have been able todetermine µ for P∗. The analogues of i(w) and o(w) are notobvious and the proof is an order of magnitude harder than foran antichain. It would not have been possible without Babsonand Hersh’s adaptation of DMT.

a1 ≤P bi+1, . . . ,ak ≤P bi+k .

Ex. If P = P (positive integers) in the normal ordering then352 ≤ 175614 in P∗ because comparing 352 with the factor 756gives 3 ≤ 7, 5 ≤ 5, and 2 ≤ 6.

Note that if A is an antichain then generalized factor order on Ais Bjorner’s factor order. Using DMT we have been able todetermine µ for P∗. The analogues of i(w) and o(w) are notobvious and the proof is an order of magnitude harder than foran antichain. It would not have been possible without Babsonand Hersh’s adaptation of DMT.

a1 ≤P bi+1, . . . ,ak ≤P bi+k .

Ex. If P = P (positive integers) in the normal ordering then352 ≤ 175614 in P∗ because comparing 352 with the factor 756gives 3 ≤ 7, 5 ≤ 5, and 2 ≤ 6.Note that if A is an antichain then generalized factor order on Ais Bjorner’s factor order.

Using DMT we have been able todetermine µ for P∗. The analogues of i(w) and o(w) are notobvious and the proof is an order of magnitude harder than foran antichain. It would not have been possible without Babsonand Hersh’s adaptation of DMT.

a1 ≤P bi+1, . . . ,ak ≤P bi+k .

Ex. If P = P (positive integers) in the normal ordering then352 ≤ 175614 in P∗ because comparing 352 with the factor 756gives 3 ≤ 7, 5 ≤ 5, and 2 ≤ 6.Note that if A is an antichain then generalized factor order on Ais Bjorner’s factor order. Using DMT we have been able todetermine µ for P∗.

The analogues of i(w) and o(w) are notobvious and the proof is an order of magnitude harder than foran antichain. It would not have been possible without Babsonand Hersh’s adaptation of DMT.

a1 ≤P bi+1, . . . ,ak ≤P bi+k .

Ex. If P = P (positive integers) in the normal ordering then352 ≤ 175614 in P∗ because comparing 352 with the factor 756gives 3 ≤ 7, 5 ≤ 5, and 2 ≤ 6.Note that if A is an antichain then generalized factor order on Ais Bjorner’s factor order. Using DMT we have been able todetermine µ for P∗. The analogues of i(w) and o(w) are notobvious and the proof is an order of magnitude harder than foran antichain.

It would not have been possible without Babsonand Hersh’s adaptation of DMT.

a1 ≤P bi+1, . . . ,ak ≤P bi+k .

Outline

References

1. Babson, E., and Hersh, P., Discrete Morse functions fromlexicographic orders, Trans. Amer. Math. Soc. 357, 2(2005), 509–534 (electronic).

2. Bjorner, A., The Mobius function of factor order, Theoret.Comput. Sci. 117, 1-2 (1993), 91–98, Conference onFormal Power Series and Algebraic Combinatorics(Bordeaux, 1991).

3. Forman, R., A discrete Morse theory for cell complexes, inGeometry, topology, & physics, Conf. Proc. Lecture NotesGeom. Topology, IV. Int. Press, Cambridge, MA, 1995,pp. 112–125.

4. Forman, R., A user’s guide to discrete Morse theory, Sem.Lothar. Combin. 48 (2002), Art. B48c, 35 pp. (electronic).

5. Sagan, B., and Willenbring, R., The Mobius function ofgeneralized factor order, in preparation.

6. Sagan, B., and Vatter, V., The Mobius function of acomposition poset, J. Algebraic Combin. 24, 2 (2006),117–136.

THANKS FORLISTENING!

discrete morse theory and generalized factor order · discrete morse theory and generalized factor...

Documents

discrete morse theory and persistent...

generalized discrete fourier transform with nonlinear...

a generalized discrete strong discontinuity...

discrete morse theory & persistent homotopy · morse theory...

rewriting systems and discrete morse...

discrete morse theory and localization - university of...

approximating persistent homology with discrete morse...

discrete morse functions on infinite complexes...discrete...

computing persistent homology via discrete morse...

parameterized complexity of discrete morse...

a generalized endogenous grid method for discrete ... · a...

discrete morse theory - École...

graph topology and discrete morse theory - github...

methods for analyzing continuous, discrete, and incomplete...

discrete morse theory and applications combinatorics seminar...

simplifying complicated simplicial complexes: discrete morse...

smooth and discrete morse theory - tqft...smooth morse...

discrete morse theory on simplicial...

mesh parameterization with generalized discrete conformal...

the good news and the really bad news about discrete...