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### Transcript of Rho Invariants and Knots - uni-regensburg.detoe09424/rho_invariants_and… · Rho Invariants and...

Rho Invariants and Knots∗

Universität Regensburg, SS 2020

June 10, 2020

1 Lecture 1: Preliminaries

1.1 Some Linear Algebra

Let H be a �nite-dimensional vector space over K = R or C and ϕ : H×H →K a Hermitian or skew-Hermitian form.

De�nition 1. 1. The radical of ϕ is de�ned as

rad(ϕ) := {a ∈ H : ϕ(a, b) = 0 ∀ b ∈ V }.

2. The orthogonal complement of a subpsace V ⊆ H is de�ned as

V ⊥ := {a ∈ H : ϕ(a, b) = 0 ∀ b ∈ V }.

De�nition 2. A subspace V ⊆ H is called...

1. isotropic if V ⊆ V ⊥;

2. Lagrangian if V = V ⊥.

Example 3. Let H = R2 and

ϕ ∼(

0 11 0

).

Then span (1, 0) is Lagrangian.

Remark. If ϕ is non-degenerate, then:

1. dimH = dimV + dimV ⊥;

∗Notes written by Damian Iltgen, based on a series of lectures given by Enrico To�oli.

1

2. V ⊆ H is Lagrangian i� V is isotropic and dimV = 12 dimH.

De�nition 4. Suppose that ϕ is Hermitian, so that ϕ(a, a) ∈ R. Thena ∈ H is called...

1. positive if ϕ(a, a) > 0;

2. negative if ϕ(a, a) < 0;

3. isotropic if ϕ(a, a) = 0.

A subspace V ⊆ H is called positive/negative/isotropic, if every a ∈ V ispositive/negative/isotropic.

De�nition 5. Let ϕ be Hermitian. Then the signature of ϕ is de�ned as

sign(ϕ) := maxV+⊆H

dimV+ − maxV−⊆H

dimV−,

where V+ is a positive and V− is a negative subspace.

De�nition 6. ϕ is called metabolic if it admits a Lagrangian.

Proposition 7. 1. ϕ metabolic ⇐⇒ sign(ϕ) = 0.

2. If A is a matrix representing ϕ, then

sign(ϕ) = # pos. eigenvalues - # neg. eigenvalues.

1.2 Intersection Forms

Let Mm be a compact orientable manifold of dimension m and consider theform

IjM : Hj(M,∂M ;R)×Hm−j(M,∂M ;R)→ R, (a, b) 7→ 〈a ∪ b, [M ]〉.

Here, [M ] denotes the fundamental class of M . Note that via Poincaréduality, IjM can be equivalently de�ned on Hm−j(M ;R) × Hj(M ;R). Wewill drop from now on the notation of coe�cients and always assume thathomology is taken with real coe�cients.

De�nition 8. The form IjM is called the intersection form of M .

Remark. IjM (b, a) = (−1)m−jIm−jM (a, b).

2

Suppose from now on that m = 2k and write IkM = IM . Then

IM : Hk(M,∂M ;R)×Hk(M,∂M ;R)→ R

is symmetric if k is even, and skew-symmetric if k is odd.

De�nition 9. Let M4l be of dimension 4l. The signature of M is de�nedas σ(M) := sign(IM ).

Remark. 1. If M ∼= N via an orientation preserving homeomorphism,then σ(M) = σ(N).

2. Let −M be M with orientation reversed. Then σ(−M) = −σ(M)since reversing orientation changes the sign of the fundamental class[M ].

Proposition 10. IfM admits an orientation reversing self-homeomorphism,then σ(M) = 0.

Proof. By assumption, the exists an orientation preserving homeomorphismf : M → −M . Then σ(M) = σ(−M) = −σ(M), so σ(M) = 0.

Example 11. 1. D4 admits a re�ection, so σ(D4) = 0.

2. Similarly σ(M) = 0 for M = S4, S2 ×Σ, D2 ×Σ, S1 ×N3 (here, Σ isa surface and N3 is a 3-manifold).

Example 12. Let H∗(CP 2) be the cohomology ring of CP 2. It is knownthat H∗(CP 2) = R[t]/(t3). If t ∈ H2(CP 2), then t ∪ t = t2 ∈ H4(CP 2), soIM (t, t) 6= 0. Therefore IM can be represented by a non-zero 1 × 1-matrixand it follows that σ(CP 2) 6= 0.

Consider the composition

Hk(M,∂M)i∗→ Hk(M)

PD→ Hk(M,∂M),

where i∗ is the map induced by the inclusion i : (M, ∅)→ (M,∂M) and PDis the Poincaré duality isomorphism.

Proposition 13. IM (a, b) = 〈b,PDi∗a〉.

Proof. IM (a, b) = 〈a ∪ b, [M ]〉 = 〈b, i∗a ∩ [M ]〉 = 〈b,PDi∗a〉.

Consider the commutative diagram below.

3

Hk(M,∂M) Hk(M)

Hk(∂M) Hk(M) Hk(M,∂M).

PD

i∗

PD

j∗ i∗

One can see from this diagram that IM (α, β) = 〈PD−1i∗α, β〉 is yet anotherpossible de�nition for the intersection form in homology.

Corollary 14. We have that

rad(IM ) = ker(Hk(M ; ∂M)i∗→ Hk(M))

PD∼= im(Hk(∂M)j∗→ Hk(M)).

Corollary 15. M closed =⇒ IM is non-degenerate.

Example 16. If Σ is a closed surface, then H∗(Σ;R) is a symplectic space.

1.3 A Knot Theoretical Example

Let K ⊂ S3 be a knot in the 3-sphere S3 = ∂D4. Further, let F ⊂ D4

be an oriented, properly embedded, compact, connected, surface such that∂F = K ⊂ S3. De�ne

WF := D4\νF,where νF is a tubular neighborhood of F . Observe that

D4 = WF ∪ νF, WF ∩ νF ∼= F × S1.

Consider the following part of the Mayer-Vietoris sequence:

H3(D4)︸ ︷︷ ︸=0

→ H2(F × S1)→ H2(WF )⊕H2(νF )︸ ︷︷ ︸=0

→ H2(D4)︸ ︷︷ ︸=0

.

It follows that H2(F × S1) ∼= H(WF ) ∼= Z2g, where g denotes the genus ofF .

Proposition 17. σ(WF ) = 0. In fact, IWFis trivial.

Proof. Consider the following commutative diagram:

H2(∂WF ) H2(WF )

H2(F × S1)

∼=

One can conclude from this diagram that rad(IM ) ∼= H2(WF ), so IWFis

trivial.

Problem: How can you get some non-trivial signature of of WF ?

4

2 Lecture 2: Twisted (Co-)Homology

Let X be a topological space with a universal cover p : X → X, π = π1(X)and V some Abelian group. A representation α : π → Aut(V ) induces anatural (left) Zπ-module structure on V . We will write Vα for V with thestructure induced by α.

Remark. There is a 1:1-correspondence

Hom(π,Aut(−))↔ (left) Zπ-modules.

Recall that if V is a left R-module and W is a right R-module, then thetensor product W ⊗R V is a well-de�ned Abelian group.

De�nition 18. Let X e a topological space with universal cover p : X → Xand V a left Zπ-module. De�ne

H∗(X;V ) := H∗(C∗(X)⊗Zπ V )

H∗(X;V ) := H∗(HomZπ(C∗(X)t, V )),

where C∗(X)t denotes C∗(X) with the right Zπ-module structure changed tothe left Zπ-module structure. H∗(X;V ) resp. H∗(X;V ) is called the twisted(co-)homology of X with coe�cients in V .

Remark. We will drop from now on the upper t notation that indicatesthe switch from right- to left-module structure and assume that it will beunderstood from the context.

Example 19. Let τ : π → Aut(V ) be the trivial representation and let Vτbe V with the trivial Zπ-module structure given by τ (i.e. elements of π actas the identity and elements of Z act as usual by multiplication). Then

ϕ : C∗(X)⊗Zπ Vτ → C∗(X)⊗Z V, σ ⊗ v 7→ p∗(σ ⊗ v)

is an isomorphism (to see the inverse direction, take a lift σ′ ⊗ v′ of σ ⊗ v,multiple lifts are possible (given by π), but σ′ · g ⊗ v′ = σ′ ⊗ v′ for allg ∈ π since τ acts trivially). Hence H∗(X;Vτ ) ∼= H∗(X;V ). SimilarlyH∗(X;Vτ ) ∼= H∗(X;V ).

Example 20. Let α : π → Aut(Zπ) be the tautological representation (i.e.the induced Zπ-structure on a module is just ring multiplication). Then

C∗(X)⊗Zπ Zπ ∼= C∗(X) =⇒ H∗(X;Zπ) ∼= H∗(X;Z).

Careful: this is not true for cohomology!

5

Proposition 21. 1. H0(X;Vα) ∼= Vα/〈{v − g · v : g ∈ π}〉.

2. H0(X;Vα) ∼= {v ∈ Vα : g · v = v}.

Proof (Cohomology case). Consider the map

ϕ : HomZπ(C0(X), Vα)→ HomZπ(C1(X), V )

f 7→(γ 7→ f(γ(1))− f(γ(0))

).

Since

HomZπ(C0(X), Vα) ∼= {f : X → V : f(x · g) = g−1 · f(x)},

we have that

H0(X;Vα) = kerϕ = {f : X → V : f constant, f(x · g) = g−1 · f(x)}∼= {v ∈ V : v = g−1 · v ∀ g ∈ π}.

Example 22. Let α : π → GL(C) ∼= C∗ ⊂ Aut(C) be a representation.Then

H0(X;Cα) = C/spanC{1− α(g) · 1} =

{C, α trivial,

0, otherwise.

(If α is not trivial, then we quotient by something that is at least of dimension1). Similarly

H0(X;Cα) = {z ∈ C : α(g) · z = z ∀ g ∈ π} =

{C, α trivial,

0, otherwise.

Example 23. X = S1, α : π = 〈t〉 → C∗, t 7→ ω ∈ C∗. If ω 6= 1, thenH0(S1;Cα) = H0(S1;Cα) = 0.

Remark. Cellular homology also works in the twisted case.

Example 24. Consider X = S1 with the usual CW-structure consisting ofone 0-cell and one 1-cell. Let p : R → S1 be the universal cover of S1 andα : π = 〈t〉 → C∗, t 7→ ω ∈ C∗ a representation. Note that Zπ ∼= Z[t±1].Consider the twisted cellular complex

0→ Ccell1 (R)⊗Zπ Cα

∂→ Ccell0 (R)⊗Zπ Cα → 0.

The two non-trivial terms �t into a commutative diagram as follows:

6

Ccell1 (R)⊗Zπ Cα Ccell

0 (R)⊗Zπ Cα

Z[t±1]⊗Zπ Cα Z[t±1]⊗Zπ Cα

C C.

∼=

∼=

∼=

∂′

∼=∂′′

Choose the vertical isomorphisms in the upper square such that ∂′ is givenby ∂′(tk⊗z) = (tk+1−tk)⊗z. The vertical isomorphisms in the lower squareare given by tk ⊗ z 7→ ωkz. Finally, ∂′′ is given by ∂′′(ωkz) = (ωk+1 − ωk)z.Looking at the image of 1⊗ z under the maps in the lower square, we get

1⊗ z (t− 1)⊗ z

z (ω − 1)z.

Thus we see that if ω 6= 1, then H∗(S1;Cα) = 0.

Exercise. If ω 6= 1, then also H∗(S1;Cα) = 0.

Remark. Basic properties of twisted (co-)homology:

1. Induced maps: f : X → Y induces a composition

β : π1(X)f∗→ π1(Y )

α→ Aut(V )

and therefore also an induced map on twisted (co-)homology:

f∗ : H∗(X;Vβ)→ H∗(Y ;Vα)

f∗ : H∗(Y ;Vα)→ H∗(X;Vβ).

2. Relative (co-)homology: given a subspace Y ⊆ X, we can use Y =p−1(Y ) ⊆ X in order to de�ne twisted relative (co-)homology.

3. Long-exact sequences in twisted (co-)homology.

4. Mayer-Vietoris-type sequence for twisted (co-)homology.

7

3 Lecture 3: Twisted Signature

3.1 Subaces and Relative Homology

Let X be a (path-connected) topological space with universal cover p : X →X and a representation α : π → Aut(V ). Further, let A ⊂ X be a subspaceand write A =

⊔iAi, where Ai are the path-connected components of A. Af-

ter taking care of base points, we obtain inclusions π1(Ai) ↪→ π and inducedrepresentations

αi : π1(Ai) ↪→ πα→ Aut(V )

for all i. In this way, we can de�ne the (twisted) subspace homologyH∗(Ai;Vαi)using the universal cover of Ai. However, by using the universal cover of Ai

we don't have canonical inclusions H∗(Ai;Vαi)i∗→ H∗(X;Vα). A possible

workaround is an alternative de�nition of subspace homology (which turnsout to be equivalent, see Proposition 26 below).

De�nition 25 ([FK06]). Let X, A ⊂ X and α be as above. De�ne

H∗(A ⊂ X;Vα) := H∗(C∗(AX)⊗ Vα),

where AX = p−1(A).

It turns out that H∗(A ⊂ X;Vα) decomposes into the direct sum of thepreviously described homology groups H∗(Ai;Vαi).

Proposition 26. H∗(A ⊂ X;Vα) ∼=⊕

iH∗(Ai;Vαi).

Proof. Assume that A is connected (otherwise do it for every path-connectedcomponent of A). We have

C∗(AX)⊗Zπ Vα ∼= C∗(A)⊗Zπ1(A) Vα1 .

Here, A is the universal cover of A and α1 is the map

α1 : π1(A) ↪→ πα→ Aut(V ).

The construction of twisted subspace homology as in De�nition 25 hasseveral advantages. Since H∗(A ⊂ X;Vα) is de�ned via the preimage of Aunder the covering map p, we indeed have now canonical inclusions

H∗(A ⊂ X;Vα)i∗→ H∗(X;Vα).

In particular, this allows us to obtain a long exact sequence of a pair (X,A)and a Mayer-Vietoris sequence in twisted homology.

8

Exercise. Show that the previous discussion also works for twisted coho-mology.

Example 27. Consider X = S1 ∨ S1, the wedge of two circles. Let U ⊂X be an open neighborhood around the wedge point and A,B ⊂ X openneighborhoods of the individual circles in X such that A ∩ B = U . Notethat π = 〈t, s〉 and consider the representation

α : : 〈s, t〉 → C∗, s 7→ ω, t 7→ ξ, ω, ξ 6= 1.

We would like to compute H∗(X;Cα) using the Mayer-Vietoris sequence.The sequence is as follows (we omit the coe�cients Cα to shorten notation):

0→ H1(A)︸ ︷︷ ︸=0

⊕H1(B)︸ ︷︷ ︸=0

→ H1(X)→ H0(U)︸ ︷︷ ︸=C

→ H0(A)︸ ︷︷ ︸=0

⊕H0(B)︸ ︷︷ ︸=0

→ H0(X).

Hence

H0(X;Cα) = 0, H1(X;Cα) ∼= C, Hi(X;Cα) = 0, i > 1.

Exercise. 1. Describe a generator in C1(X) ⊗Zπ Cα). What if ω 6= 1,ξ = 1?

2. Compute H∗(S1;Cα) using Mayer-Vietoris.

3.2 Products in twisted (co-)homology

Let X be a topological space and V,W Zπ-modules. We would like to havea product on Hj(X;V )×Hk(X;W ), similar to the cup product in singularhomology. On the level of chain complexes, we can de�ne

HomZπ(Cj(X), V )×HomZπ(Ck(X),W )→ HomZπ(Cj+k(X), V ⊗Z W )

(f, g) 7→(σ 7→ f(σ|0,...,j)⊗ g(σ|j,...,j+k)

),

where σ|i,...,i+j means σ restricted to the vertices i, . . . , i + j. This extendsto a well de�ned operation on twisted cohomology, yielding the product

∪ : Hj(X;V )×Hk(X;W )→ Hj+k(X;V ⊗Z W )

Similarly, we can de�ne a cap product for twisted (co-)homology. On thelevel of chain complexes, we can de�ne

HomZπ(Cj(X), V )×(Cj+k(X)⊗Zπ W )→ Ck(X)⊗Zπ (V ⊗Z W )

(f, σ ⊗ w) 7→ σ|j,...,k+j ⊗(f(σ|0,...,j)⊗ w

)9

which yields us

∩ : Hj(X;V )×Hj+k(X;W )→ Hk(X;V ⊗Z W ).

Remark (Kronecker pairing and evaluation). Let R be a commutativering. Recall from ordinary homology theory that there is the Kroneckerpairing

〈·, ·〉 : Hj(X;R)×Hj(X;R)∩→ H0(X;R)

ε→ R

(f, σ ⊗ r) 7→ f(σ)r.

Here ε is the augmentation map [∑

i σi ⊗ ri] 7→∑

i ri (which is an isomor-phism if X is path-connected). Associated to the pairing is the evaluation

ev : : Hj(X;R)→ Hom(Hj(X;R), R)

f 7→(σ ⊗ r 7→ 〈f, σ ⊗ r〉 = f(σ)r

)This pairing carries over to twisted homology with some slight adjustments.Imitating 〈·, ·〉 on the level of chain complexes in twisted homology, an ele-ment (f, σ⊗w) gets mapped to f(σ)⊗w. However, σ⊗w = σg−1 ⊗ gw forg ∈ π, and (f, σg−1 ⊗ gw) gets mapped to gf(σ)⊗ gw = g(f(σ)⊗w), so weneed the identi�cation

g(f(σ)⊗ w) ∼ f(σ)⊗ w ∀ g ∈ π.

Observe that this is exactly the identi�cation used in Example 22 to describethe 0-th twisted homology group. In other words,

V ⊗Z W/ ∼ ∼= H0(X;V ⊗Z W ).

Thus, the Kronecker pairing in twisted homology is

〈·, ·〉 : Hj(X;V )×Hj(X;W )∩→ H0(X;V ⊗Z W )

ε→ V ⊗Z W/ ∼(f, σ ⊗ w) 7→ f(σ)⊗ w

together with the evaluation

ev : : Hj(X;V )→ Hom(Hj(X;W ), V ⊗Z W/ ∼)

f 7→(σ ⊗ w 7→ 〈f, σ ⊗ w〉 = f(σ)⊗ w

)We will use the Kronecker pairing later to de�ne the twisted intersectionform.

10

Remark. As in the untwisted case, there are relative versions of the cupproduct, of the cap product and of the Kronecker pairing. These behaveexactly as for ordinary (co-)homology. For details, see [FNOP19] of [Her19].

The famous Poincaré duality theorem also holds in the twisted setting.

Theorem 28 (Poincaré Duality). LetM be a compact oriented manifoldof dimensionm, and let V be a left Zπ-module. Then, there are isomorphisms

Hj(M,∂M ;V )∩[M ]→ Hm−j(M ;V )

Hj(M ;V )∩[M ]→ Hm−j(M,∂M ;V ).

3.3 Technical Subtleties

Given a manifoldM2k, we would like to de�ne an intersection form on twistedcohomology Hk(M ;V ) similar to the non-twisted case. As we are interestedin extracting a signature out of it, a natural choice for V is the n-dimensionalcomplex vector space Cnα twisted by some representation α : π → GL(n,C),as this makes homology and cohomology spaces complex vector spaes. Us-ing the construction of the previous section, the cup product between twocohomology classes with values in Cnα gives a class with values in Cnα ⊗Z Cnα.

In order to imitate what we did in the untwisted case, we would like toevaluate this against the (complex-valued) fundamental class of M , whichlives in H2k(M,∂M ;C). We need thus a map

ϕ : Cnα ⊗Z Cnα → C

which is a homomorphism of Zπ-modules (where C carries the Zπ-structuregiven by the trivial representation). In fact, such a map would induce a mapon cohomology

ϕ : H2k(X;Cnα ⊗Z Cnα)→ H2k(X;C).

Postcomposing the cup product with ϕ yields a new cup product that willbe suitable to de�ne a twisted intersection form.

A natural choice for ϕ is the complex standard inner product, de�nedby v ⊗ w 7→ vtw. However, it is not yet a homomorphism of Zπ-modules,because

ϕ(α(g)(v ⊗ w)) = ϕ(α(g)v ⊗ α(g)w) = vtα(g)tα(g)w

for g ∈ π, and vtα(g)tα(g)w 6= vtw in general. We can �x this by requiringthat α is a unitary representation, i.e. α : π → U(n) ⊂ GL(n,C). This yields

11

us the desired maps ϕ and ϕ. Postcomposing with ϕ, we obtain the followingsesquilinear form

∪α : Hk(M,∂M ;Cnα)×Hk(M,∂M ;Cnα)→ H2k(M,∂M ;C)

In a similar way, from the twisted Kronecker pairing, for a pair of spaces(X,A) we can de�ne a map

〈·, ·〉α : Hj(X,A;Cnα)×Hj(X,A;Cnα)→ C

in such a way that it is anti-linear in the �rst coordinate and linear in thesecond coordinate.1 Moreover, it can be easily proved that this pairing isnon-degenerate.

Note that there are two ways do de�ne 〈·, ·〉, either via direct evaluationagainst a homology class or via cap product and augmentation map. Sincewe only plan to evaluate against the fundamental class [M ], we do not needto adjust the cap product because we can simply take [M ] to be containedin homology with integer coe�cients (so that W = Z in the de�nition of thecap product above).

3.4 Twisted Intersection Pairing

De�nition 29. Let M2k be a compact manifold with a unitary represen-tation α : π → U(n). We de�ne the twisted intersection pairing to be thesesquilinear form

IαM : Hk(M,∂M ;Cnα)×Hk(M,∂M ;Cnα)→ C(a, b) 7→ 〈a ∪α b, [M ]〉.

As in the non-twisted case, IαM is Hermitian if k even, and skew-Hermitianif k odd. Using Poincaré duality, we obtain an isomorphism Hk(M ;Cnα) ∼=Hk(M,∂M ;Cnα), so we also have an intersection pairing on twisted homology.

Proposition 30. IαM (a, b) = 〈b,PDi∗(a)〉α.1This choice is taken in order for the following formula (leading to Proposition 30) to

be satis�ed:〈a ∪α b, γ〉 = 〈b, a ∩ γ〉α.

12

Corollary 31. We have

rad(IαM ) = ker(Hk(M,∂M ;Cnα)→ Hk(M ;Cnα))

PD∼= im(Hk(∂M ;Cnα)→ Hk(∂M ;Cnα)).

De�nition 32. De�ne

σα(M) :=

{sign(IαM ), k even,

sign(i · IαM ), k odd.

Here, i denotes the imaginary unit in C. σα(M) is called the twisted signa-ture of M .

Remark. If τ : π → U(n) denotes the trivial representation, then

στ (M) = n · σ(M)

(also works if k is odd if we de�ne σ(M) = 0 in this case).

Example 33. Let M be a pair of pants (i.e. a 2-sphere with three disksremoved). Note that M is homotopy equivalent to a wedge of two circles.Consider the representation α : π → U(1) ⊆ C∗. Get H1(M ;Cα) ∼= C. Ifω, z, ωz 6= 1, then H∗(∂M ;Cα) = 0, so rad(IαM ) = 0, so sign = ±1 (can'tdecide yet).

4 Lecture 4: Some Calculations

4.1 Computing the Twisted Intersection Pairing and Signa-ture

Remark (Change of Convention). Recall that we initially consideredC∗(X) as a right Zπ-module. We will change this convention now and con-sider C∗(X) as a left Zπ-module, so that H∗(X;V ) = H∗(V ⊗Zπ C∗(X)). Inparticular, a representation α : π → Aut(V ) now induces a right Zπ-modulestructure on V by viewing elements of Cn as row vectors and making matricesact on them from the right.

We decided to do this change of convention because deck transformationnaturally act from the left rather from the right on the universal cover X.

Recall the de�nition of the twisted intersection pairing: given a compactoriented manifold M2k and a representation α : π → U(n), we de�ned

IαM : Hk(M,∂M ;Cnα)×Hk(M,∂M ;Cnα)→ C(a, b) 7→ 〈a ∪α b, [M ]〉.

13

Further, we de�ned the twisted signature of M as

σα(M) :=

{sign(IαM ), k even,

sign(i · IαM ), k odd.

Example 34. Let τ : π → U(1) be the trivial representation. If k is even,then στ (M) = σ(M), and if k is odd then στ (M) = 0 =: σ(M) (becausethere is a Lagrangian).

Example 35. Let M be a pair of pants and let

α : π ∼= F2 = 〈s, t〉 → U(1) ⊆ C, s 7→ ω, t 7→ ξ.

Here, F2 denotes the free group on two generators. If α is non-trivial, thenH1(M ;Cα) ∼= C. If ω, ξ, ωξ 6= 1, then rad(IαM ) = 0, so σα(M) = ±1. Ifeither ω = 1, ξ = 1, or ωξ = 1, then σα(M) = 0. Problem: What's thevalue of σα(M) as a function of ω, ξ?

In order to answer the question in the previous example, we have to passfrom algebra to geometry. For this we recall the geometric interpretation ofthe ordinary (untwisted) intersection form:

Theorem 36. Let M2k be a compact oriented manifold. Then the intersec-tion form counts the algebraic intersection of chains, i.e.

IM (PD−1([σ]),PD−1([ν]) = σ · ν.

In order to obtain a similar statement for twisted homology, we �rstconsider ordinary homology as integer homology twisted by the trivial rep-resentation, so that

C∗(M) ∼= Z⊗Zπ C∗(M).

Then, given a simplex σ ∈ C∗(M), we obtain a lift in the universal cover

σ ∈ C∗(M), where two lifts di�er by multiplication with an element g ∈ π.Thus we obtain

Corollary 37. Let M2k be a compact oriented manifold. Then

IM (PD−1([σ]),PD−1([ν]) = σ · ν =∑g∈π

σ · gν.

We would like to transform this to the case of twisted homology. So letα : π → U(1) be a representation, and consider

Hk(M ;Cα) = Hk(Cα ⊗Zπ C∗(X)).

14

Theorem 38. The twisted intersection form satis�es

IαM (PD−1[z ⊗ σ],PD−1[w ⊗ ν]) =(∑g∈π

σ · gνα(g))zw.

Recall our problem: we would like to determine the twisted signature σαof a pair of pants as a function in ω, ξ. By the above theorem, we will haveto consider intersections in the universal cover of the pair of pants. To geta better picture of this, we �rst consider the situation for a wedge of twocircles, which is homotopy equivalent to a pair of pants.

Example 39. Let M = S1 ∨ S1 and consider a non-trivial representation

α : π = F2 = 〈s, t〉 → U(1), s 7→ ω, t 7→ ξ.

ThenH1(M ;Cα) = H1(Cα ⊗Zπ C∗(M)) ∼= C.

The universal cover of M is the Cayley graph Γ on two generators s, t. Letσ denote the 1-simplex as shown in Figure 1 below, and consider 1 ⊗ σ ∈Cα ⊗Zπ C1(M). If p denotes the origin of Γ, then

∂(1⊗ σ) = 1⊗ (st− ts)p = (ωξ − ξω)⊗ p = 0,

so 1⊗ σ de�nes an element in H1(M ;Cα) and is in fact a generator.-

Figure 1: The universal cover of M = S1 ∨ S1 with the ele-ment σ ∈ C1(M)

Observe that the abelianization of F2 is Z2. It follows from Shapiro'slemma that

Cα ⊗ZF2 C∗(M) ∼= Cα ⊗Z[Z2] C∗(Mab),

15

where Mab denotes the universal abelian cover of M (i.e. the cover corre-

sponding to the commutator subgroup [π, π] ⊆ π). It is easy to see that Mab

is the lattice Z2 ⊂ R2 and that σ corresponds to one square in Mab, whichwe will denote by σ as well (see Figure 2 below). We will use this observationto determine the twisted signature of a pair of pants in the next example.

Figure 2: The universal abelian cover of M = S1 ∨ S1 withthe element σ ∈ C1(Mab) in red

Example 40. Let us turn back to the case where M is a pair of pants. Wecan visualize σα(M) in terms of ω, ξ by considering its values on the squarein R2 spanned by the arguments of ω and ξ (based at the origin). By ourprevious discussion, σα(M) = 0 if one of the arguments is either 0 or 2π, or ifthe arguments add up to 2π. In all other cases, we know that σα(M) = ±1,and it remains to determine when the signature is +1 and when it is −1.

By thickening up the universal abelian cover of S1∨S1 and transformingσ appropriately, we obtain a model of the universal abelian cover of the pairof pants with a generator of its �rst homology, which we will denote by σ aswell.

16

Figure 3: The universal abelian cover of a pair of pants withthe element σ ∈ C1(M) in red

The element 1 ⊗ σ is a generator of H1(Cα ⊗Z[Z2] C∗(Mab)), which is

isomorphic to H1(M ;Cα) ∼= C. It follows that

IαM (PD−1[1⊗ σ],PD−1[1⊗ σ]) =∑g∈Z2

σ · (n,m)σωnξm

= ωξ − ωξ − ω + ω − ξ + ξ.

Using that z − z = 2iIm(z) for z ∈ C, we get

σα(M) = sign(Im(ω) + Im(ξ)− Im(ωξ)).

Therefore, we obtain

σα(M) =

{+1, argω + arg ξ < 2π

−1, argω + arg ξ > 2π.

4.2 Knot Signature

Let K ⊂ S3 be a knot and F ⊆ S3 a Seifert surface. There exists a tubularneighborhood of F homeomorphic to [−1, 1]×F ⊂ S3. Given a curve b ⊂ F ,we denote by b+ the image of b under the map F 7→ {1}×F ⊂ [−1, 1]×F ⊂S3 (this is sometimes called the positive push-o� of b).

De�nition 41 (Seifert Pairing). The Seifert pairing is de�ned as

VF : H1(F ;Z)×H1(F ;Z)→ Z, ([a], [b]) 7→ lk(a, b+).

17

Note that the Seifert pairing is not symmetric, so in order to get a sig-nature out of V , we need to consider the symmetrization V + V t.

De�nition 42 (Knot Signature). The signature of K is deifned as

σ(K) := sign(V + V t).

Given ω ∈ U(1) ⊆ C∗, there is a more re�ned version of the knot signa-ture known as the Levine-Tristram signature.

De�nition 43 (Levine-Tristram Signature). The Levine-Tristram sig-nature of a knot K ⊂ S3 is de�ned as

σω(K) := sign((1− ω)V + (1− ω)V t) ∈ Z.

This de�nes a function σK : U(1)→ Z, σK(ω) := σω(K) which is called theLevine-Tristram signature function.

Remark. Note that

sign((1− ω)V + (1− ω)V t) = sign((1− ω)V + (1− ω)V t),

i.e. it doesn't matter where we put the complex conjugate of ω.

5 Lecture 5: Atiyah-Patodi-Singer ρ-Invariants (PartI)

Let K ⊆ S3 be a knot with Seifert surface F and Seifert pairing VF , and letσK : U(1)→ Z denote the Levine-Tristram signature function.

Example 44. Let K be the trefoil with Seifert surface and �rst homologybasis as depicted in Figure 4 below.

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Figure 4: The Levine-Tristram signature of the trefoil

The Seifert matrix is

V =

(−1 10 −1

).

For θ ∈ [0, 1), the Levine-Tristram signature is

σK(e2πiθ) =

−2, 1

6 < θ < 56

−1, θ = 16 ,

56

0, otherwise.

Now let F denote F with its interior pushed into int(D4). De�ne

WF := D4\νF .

Note that H1(WF ;Z) ∼= Z (generated by a meridian of K). A representationα : π1(WF )→ U(1) factors throughH1(WF ;Z) and can therefore be speci�edby a map Z = 〈t〉 → U(1), t 7→ ω.

Proposition 45. σα(WF ) = σK(ω).

Proof (Sketch). The universal abelian cover of WF can be built explicitlyby cutting D4 along F and gluing in�nitely many copies together along theboundary components. Then H2(WF ;Cα) ∼= H1(F ;Z) ⊗Z C with inverseisomorphism given by [a] 7→ [ca]. It follows that

IαWF(a, b) =

∑n∈Z

ca · tncbωn

= lk(a, b+) + lk(b, a+)− ωlk(a, b+)− ωlk(b, a+)

= [(1− ω)V + (1− ω)V t](a, b).

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5.1 Signature Theorems

Theorem 46 (Novikov Additivity). Let M2k = M1 ∪∂ M2 be the closedmanifold given by the union of the manifolds M1, M2 along their commonboundary. Then

σ(M) = σ(M1) + σ(M2).

The same is true for the twisted signature σα associated to some represen-tation α : π1(M)→ U(n).

Theorem 47 (Signature Theorems). Let M2k be a smooth, closed andoriented manifold and α : π1(M)→ U(n) a representation.

1. σ(M) = 〈L(M), [M ]〉 (Hirzebruch signature theorem).

2. σα(M) = n〈L(M), [M ]〉 = nσ(M) (Atiyah-Singer).

Here, L(M) denotes the n-th L-class of M (i.e. the n-th polynomial of theL-genus with indeterminates replaced by the Pontryagin classes pi(M)) and〈·, ·〉 is the evaluation pairing.

Corollary 48. M closed =⇒ σα(M) = nσ(M).

Let M2k be a closed oriented Riemannian manifold. Then an equivalentform of the above signature theorem is

σ(M) =

∫ML∇(M), σα(M) = n

∫ML∇(M),

where L∇(M) is the L-genus of M seen as a 2k-di�erential form. If wereplace the assumption closed with compact, then

σ(M) =

∫ML∇(M) + C(N), σα(M) = n

∫ML∇(M) + Cα(N),

where N = ∂M and C(N) and Cα(N) are correction terms.

Remark. 1. If the manifold is a product near ∂M (i.e. of the form I ×∂M), we see that C and Cα only depend on ∂M (and on the restrictionα|∂M ). Indeed, let M and M ′ have the same boundary and considerW = M ∪∂ (−M ′). Then

C(M)− C(M ′) =

[σ(M)−

∫ML∇(M)

]−[σ(M ′)−

∫M ′L∇(M ′)

]= σ(W )−

∫WL∇(W ) = 0,

and similarly for σα.

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2. In particular, the di�erence nC(N)−Cα(N) = nσ(M)− σα(M) is aninvariant of ∂M . However, it still requires the existence of a boundingmanifold M in order to be de�ned.

Theorem 49 (Atiyah-Patodi-Singer, '75). LetM be a compact orientedRiemannian manifold with boundary ∂M = N , with metric of product formnear N . Then

σ(M) =

∫ML∇(M)− η(N),

where η(N) is the value at 0 of a meromorphic extension of

η(s) :=∑

λ∈spectrumDN

signλ|λ|−s.

In the twisted case,

σα(M) = n

∫ML∇(M)− ηα(N),

where ηα(N) is de�ned as η(N) but with DN replaced by a twisted di�eren-tial operator Dα

N .

De�nition 50 (Rho Invariant). Let N2k+1 be a closed oriented manifoldwith a representation α : π1(N)→ U(n). The ρ-invariant of N is de�ned as

ρα(N) := ηα(N)− nη(N) ∈ R.

Corollary 51. 1. If there exists M such that ∂M = N and α extends,then ρα(N) = nσ(M)− σα(M).

2. ρα(N) is independent of the metric.

Proof. 1. Immediate from a comparison of the twisted and untwistedsignature formula (see also 2. of the previous remark).

2. M := [0, 1] ×N with two di�erent metrics on the ends and apply the�rst statement of the corollary.

Example 52. Let N = S1 and α : π = Z → U(1), 1 7→ ω = e2πiθ (θ ∈[0, 1])). Then

ρα(S1) =

{0, θ = 0

2θ − 1, θ 6= 0.

Exercise. Check 1. of Corollary 51 for M a pair of pants.

21

Example 53. ρ(∂WF ) = σ(WF ) − σα(WF ) = −σK(ω). where ω = α(µ) isthe image of the meridian under α.

Proposition 54. Let F be a closed oriented surface. Then

ρα(F × S1) = 0

for all α : π1(F × S1)→ U(1).

6 Lecture 6: Atiyah-Patodi-Singer ρ-Invariants (PartII)

6.1 The Natural Lagrangian

Let Σ2k be a closed oriented manifold with Σ = ∂Y (Y compact oriented)and α : π1(Y )→ U(n) a representation. De�ne

V αY := ker{Hk(Σ;Cnα)

i∗→ Hk(Y ;Cnα)}PD∼= im{Hk(Y ;Cα)

i∗→ Hk(Σ;Cnα)}.

Proposition 55. V αY is Lagrangian with respect to IαΣ.

Proof (Sketch). Consider the commutative diagram below:

Hk(Y ;Cnα) Hk(Σ;Cnα) Hk+1(Y,Σ;Cnα)

Hk(Σ;Cnα) Hk(Y ;Cnα).

i∗

PD PD

We show the inclusion V αΣ ⊆ (V α

Σ )⊥. Given a, b ∈ V αΣ , we have

IαΣ(a, b) = 〈b,PD(a)〉Σ = 〈i∗(b′),PD(a)〉Σ = 〈b′, i∗PD(a)〉Y = 0.

The other inclusion can be proved in a similar way.

Example 56. Let N be a compact oriented 3-manifold. Consider the map

induced by inclusion H1(∂N ;Z)i∗→ H1(N ;Z). Then ker(i∗) is half-rank

("half lives, half dies principle").

Corollary 57. If Σ = ∂Y , then σα(Σ) = 0 for all representations α : π1(Σ)→U(n).

De�nition 58. V αY is called the natural Lagrangian of Hk(Σ,Cnα) associated

to Y .

22

De�nition 59. A complex symplectic space is a pair (H,β) where H isa �nite dimensional vector space over C and β is a non-degenerate skew-Hermitian form.

Example 60. 1. C with β = (i).

2. C2 with

β =

(0 −11 0

).

De�nition 61. (H,β) is called balanced if sign(iβ) = 0.

Example 62. Let Σ2k be closed with Σ = ∂Y and α : π1(Y ) → U(n) arepresentation. Then Hk(Σ;Cα) is a balanced complex symplectic spacewith

β =

{IαΣ, k odd

iIαΣ, k even.

6.2 The Maslov Triple Index

Let (H,β) be a balanced complex symplectic space with L1, L2, L2 ⊆ HLagrangians. De�ne

hL1,L2,L3 : ((L1 + L2) ∩ L3)×((L1 + L2) ∩ L3)→ C(a1 + a2, b) 7→ β(a1, b).

Exercise. Show that hL1,L2,L3 is well-de�ned and Hermitian.

De�nition 63 (Maslov triple index). The Maslov triple index is de�nedas

ζ(L1, L2, L3) := sgn(hL1,L2,L3).

Remark. ζ(L1, L2, L3) depends on the position of the Lagrangians in H,that is, if you move one Lagrangian around with a symplectomorphism,ζ(L1, L2, L3) might change (it stays the same however if you move all La-grangians Li around with the same symplectomorphism).

Proposition 64. Let σ ∈ S3. Then

ζ(Lσ(1), Lσ(2), Lσ(3)) = sign(σ)ζ(L1, L2, L3).

Remark. In particular ζ(L,L,L′) = 0 etc.

23

Example 65. Let C2 with

β =

(0 −11 0

).

The Lagrangians of (C2, β) are lines L = span{z1e1 + z2e2 | z1z2 ∈ R}.Restricting to R2, we see that the Maslov triple index depends on whetherthe Lagrangian L1 can be turned to L2 and then to L3 via a clockwise orcounterclockwise rotation (see Figure 5 below).

Figure 5: Lagrangians sitting in R2

Therefore

ζ =

0, if at least two Lagrangians coincide,

+1, if L1 turns into L2 and then L3 counterclockwise,

−1, if L1 turns into L2 and then L3 clockwise.

Recall Novikov additivity: if M = M1 ∪∂ M2, then σ(M) = σ(M1) +σ(M2). Problem: what ifM1 andM2 have additional boundary componentsthat are not a�ected by the gluing? (See Figure 6 below).

24

Figure 6: A schematic drawing of M

Observe that Σ = ∂Y1 = ∂Y2 = ∂Y0. If M is oriented, then M1 and M2

are oriented as well (since they are codimension 0 manifolds). In particular,∂M = Y1∪ΣY2 is oriented. Convention: we pick on Σ the orientation comingfrom the relation Σ = ∂Y1.

Theorem 66 (Wall). Let M,Mi, Yi,Σ be as above and let α : π1(M) →U(n) be a representation. Then

σα(M) = σα(M1) + σα(M2)− ζ(V αY0 , V

αY1 , V

αY2),

where V αYi

are Lagrangians of Hk−1(Σ,Cnα).

Exercise. Use Theorem 66 to compute the twisted signature σα of a pair ofpants again, where α : π1 → U(1).

Example 67. Let K ⊂ S3 be a knot with Seifert surface F ⊆ S3, andlet WF = D4\νF . Recall that a representation α : π1(WF ) → U(1) factors

25

through Z =< µ >, where µ is a meridian. We have seen that

σα(WF ) = σK(α(µ)).

Consider the decomposition of the boundary of WF and the construction ofa new manifold W as shown below.

Figure 7: A decomposition of ∂WF and construction of W

Figure 8: The construction of W

Observe that ∂W = XK ∪D2×S1 =: MK is the 0-framed surgery (here,XK denotes the knot exterior of K in S3).

Proposition 68. 1. σα(W ) = σα(WF ).

2. σ(W ) = σ(WF ).

Corollary 69. ρα(MK) = −σK(α(µ)).

26

7 Lecture 7: Cut-and-Paste & Applications

Let K ⊂ S3 be a knot with F ⊂ D4 a pushed-in Seifert surface of K. LetXK := S3\νK denote the knot complement and WF := D4\νF . Observethat ∂WF = XK ∪ϕ (−F × S1), where ϕ identi�es the meridian µ with S1

and the longitude λ with ∂F . Let MK := XK ∪ −(D2 × S1) denote the0-framed surgery of K. Observe that H1(MK ;Z) = Z (0-framed surgery hasno e�ect on homology compared to XK since it kills the null-homologouslongitude in XK). Recall that σK(ω) = σα(WF ) and σ(WF ) = 0, where αmaps µ to ω ∈ S1. By Atiyah-Patodi-Singer, we have

ρ(∂WF ) = σ(WF )− σα(WF ) = −σK(ω).

By Wall σ(WF ) = σ(W ) and σα(WF ) = σα(W ) where W = WF ∪ (H × S1)and H is a handlebody (see previous lecture), so

ρ(∂WF ) = ρ(MK) = σ(W )− σα(W ).

We recall the situation in Wall's non-additivity result in Figure 9 below.

Figure 9: The situatuion in Wall's non-additivity result

We have

σ(W1 ∪x0 W2) = σ(W1) + σ(W2)− ζ(Vx0 , Vx1 , Vx2)

σα(W1 ∪x0 W2) = σα(W1) + σα(W2)− ζα(V αx0 , V

αx1 , V

αx2),

soρ(X1 ∪Σ X2) = ρ(X1 ∪X0) + ρ(−X0 ∪X2) + ζα − ζ.

Theorem 70 (To�oli). In the setting above,

ρα(X1 ∪Σ X2) = ρα(X1 ∪X0) + ρα(−X0 ∪X2) + ζα − ζ.

27

Proof (Sketch). De�ne W1 := I × (X1 ∪Σ X0) and W2 := I × (−X0 ∪X2).By Wall,

σ(W1 ∪W2) = σ(W1) + σ(W2)− ζσα(W1 ∪W2) = σα(W1) + σα(W2)− ζα.

Since σ(Wi) = σα(Wi) = 0, we have

σ(W1 ∪W2)− σα(W1 ∪W2) = ζα − ζ.

But σ(W1∪W2)−σα(W1∪W2) = ρ(X1∪ΣX2)−ρ(X1∪X0)−ρ(−X0∪X2).Therefore

ρα(X1 ∪Σ X2) = ρα(X1 ∪X0) + ρα(−X0 ∪X2) + ζα − ζ

as claimed.

The procedure where we have X1 ∪∂ X2 and change it to X1 ∪∂ X0 t−X0 ∪X2 is called cut-and-paste.

Example 71. Let Y1 = X1 ∪ ball and Y2 = X2 ∪ ball. Then cut-and-pasteyields the ordinary connected sum Y1#Y2.

Example 72. Consider the cut-and-paste where we change XK ∪ (−F ×S1)to XK ∪ (−D2 × S1) tD2 × S1 ∪ (−F × S1). Then

ρα(XK ∪ (−F × S1)) = ρ(XK ∪ (−D2 × S1)) + ρ(D2 × S1 ∪ (−F × S1))

This is the same result as in the previous lecture, but without using handle-bodies.

Remark. Let T 2 be the torus and α : π1(T 2)→ U(1) a non-trivial represen-tation. Then H∗(T

2;Cα) = 0 (exercise).

Now let VX := ker(H1(∂X;C)→ H1(X;C)). Since

V αXK

, V αF×S1 , V

αD2×S1 ⊆ H1(T 2 = ∂XK ;Cα) = 0,

we have ζα = ζα(V αXK

, V αF×S1 , V

αD2×S1) = 0. In the non-twisted case,

VXK︸︷︷︸=<λ>

, VF×S1︸ ︷︷ ︸=<λ>

, VD2×S1︸ ︷︷ ︸=<λ>

⊆ H1(T ;C) =< µ, λ >,

so ζ = ζ(VXK , VF×S1 , VD2×S1) = 0 as well.

28

Now let MK(pq ) := XK ∪ϕ (−D2 × S1) where (p, q) = 1 and ϕ identi�es

∂D2 with pµ+ qλ. Consider a representation

α : H1(Mk(p

q)) ∼= Zp → U(1), [1] 7→ ω,

where ω is a pth-root of unity. Then by cut-and-paste,

ρα(MK

(pq

)) = ρα(MK) + ρα(D2 × S1 ∪ (−D2 × S1))

= −σK(ω)− ρα(L(p, q), α),

where L(p, q) is a lens space (there is an explicit formula for computingρ-invariants of lens spaces).

Let L ⊆ S3 be an oriented link and F a Seifert surface. As for knots, wehave a Levine-Tristram signature σL(ω), and σL(ω) = σα(WF ), σ(WF ) = 0.Hence

−σL(ω) = ρα(∂WF ),

where ∂WF = XL ∪∂ (−F × S1).

Example 73 (Hopf link). Consider the Hopf link with linking number +1and Seifert surface as shown below.

��� � ��� ����������������������������

������������������������ ����� ������

�����������������������������

�������� ���� �������� ���������� ����

Figure 10: Hopf link with Seifert surface

The Seifert matrix is V = (−1), so we get

σL(ω) =

{−1, ω 6= 1

0, ω = 1.

29

Observe that λ1 = µ2 and λ2 = µ1 in H1(XL;Z), so ML = S3. Problem:cannot de�ne rho invariants in terms of L1 and L2. Trick: use a di�erentframed surgery. Let α : H1(XL;Z)→ U(1) be a representation with µ1, µ2 7→ω. De�ne

λ′1 := λ1 − µ1, λ′2 := λ2 − µ2.

Then α(λ′1) = ωω = 1, and α(λ′2) = 1. De�ne

ML(GL) := XL ∪ϕ (D21 × S1 tD2

2 × S2),

where ϕ identi�es ∂D2i with λ

′i = λi − lk(L1, L2)µ1.

Remark. In general, for a k-component link L = L1 ∪ · · · ∪Lk, the framing

λ′i := λi −∑j 6=i

lk(Li, Lj)µi

is called the Seifert framing.

Proposition 74. ρ(ML(GL)) = −σL(ω) + sign(λL), where sign(λL) is aninteger only depending on the linking numbers of L.

7.2 Multivariable Signatures

Let L = L1∪· · ·∪Lk be a k-component link and let F := F1∪· · ·∪Fk where

1. Fi is a Seifert surface for Li;

2. Fi ∩ Fj is a union of clasps;

3. there are no triple intersection points.

Such an F is then called a C-complex for the link L. The de�nition is due toCooper and revived by Cimasoni. Depending on the direction of the push-o�of curves on the individual Seifert surfaces, one obtains a Seifert matrix V ε

for every ε ∈ {−1, 1}k. Fix such an ε and write V = V ε. Given a k-tupleω = (ω1, . . . , ωk) ∈ (U(1)\1)k, one can de�ne a generalization of the matrixH(ω) de�ned for knots as

(1− ω)V + (1− ω)V t =: H(ω)

(see [CF08] for details).

De�nition 75 (Multivariable Signature (Cimasoni-Florens)). Themultivariable signature of a link L = L1 ∪ · · · ∪ Lk is de�ned as

σL(ω) := sign(H(ω)),

where ω ∈ (U(1)\1)k.

30

References

[CF08] D. Cimasoni and V. Florens. Generalized Seifert surfaces and sig-natures of colored links. Trans. Amer. Math. Soc., 360(3):1223�1264 (electronic), 2008.

[FK06] Stefan Friedl and Taehee Kim. The Thurston norm, �bered man-ifolds and twisted Alexander polynomials. Topology, 45(6):929 �953, 2006.

[FNOP19] Stefan Friedl, Matthias Nagel, Patrick Orson, and Mark Pow-ell. A survey of the foundations of four-manifold theory in thetopological category, 2019. https://arxiv.org/abs/1910.07372.

[Her19] Gerrit Herrmann. Sutured manifolds, L2-betti numbers and anupper bound on the leading coe�cient, 2019. https://epub.uni-regensburg.de/40578/.

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