[Math] Intutive interpretation about Linking forms

3-manifoldsat.algebraic-topologyhomotopy-theoryknot-theory

Let $M^3$ be a rational homology 3-sphere. (i,e, $M^3$ is closed 3-manifold with
$H_{*}(M;Q)=H_{*}(S^3;Q)$

As beautifully explained in Ranicki's Algebraic and Geometry surgery book and Davis-Kirk's Lecture notes in Algebraic toplogy book, we have a $Q/Z$ valued linking form, $\lambda\colon H_{1}(M;Z)\times H_{1}(M;Z)\to Q/Z$ defined by the adjoint to following isomorphism.

(Actually we can define linking form in more general setting, e.g.) odd dimensional manifold without the restriction such as rational homology sphere condition. Because I want to just intuitive idea about linking form, I restricted the case)

$H_1(M;Z)\cong H^2(M;Z)\cong H^1(M;Q/Z)=Hom(H_1(M;Z),Q/Z)$

First isomorphism : Poincare duality,

Second isomorphism : Inverse of Bockstein homomorphism $\delta$ induced form $0\rightarrow Z \rightarrow Q\rightarrow Q/Z\rightarrow 0$. More precisely, induced long exact sequence is that $\ldots\rightarrow H^1(M;Q)\rightarrow H^1(M;Q/Z)\rightarrow H^2(M;Z)\rightarrow H^2(M;Q)$ and here $H^1(M;Q)$ and $H^2(M;Q)$ vanishes. Therefore, the long exact sequence shows that $H^1(M;Q/Z)$ and $H^2(M;Z)$ are naturally isomorphic (if $M$ is rational homology 3-sphere) and we call that homomorphism as Bockstein homomorphism.

Third isomorphism : Universal Coefficient theorem

In short, $\lambda\colon H_{1}(M;Z)\times H_{1}(M;Z)\to Q/Z$ is defined by $\lambda(x,y)=<\tilde{x}\cup\delta^{-1}(\tilde{y}),[M]>$, where $\tilde{x},\tilde{y}$ are Poincare dual to $x,y$ and $\delta\colon H^1(M;Q/Z)\to H^2(M;Z)$ is a Bockstein homomorphism as defined above. Cup products are defined in $H^1(M;Z)\times H^2(M;Q/Z) \to H^3(M;Q/Z)$ induced from the Bilinear map $Z \times Q/Z \to Q/Z$.

I understand this linking form only algebraic viewpoint. Therefore, I can play with this form only by using dilluminating algebraic topology.

I'm struggle to find geometric interpretation but I have no idea to express the Q/Z valued in terems of geometric language which seems to be highly algebraic. (Feeling like Injective, divisible, Ext or something linke that )

Are there any intuitve and geometric (clean) interpretation about this linking form? (e.g.)such as Alexander duality, like the argument that ………we can find a dual basis which represents meridian……)

Best Answer

This may be a good place to explain the well-known principle $$\text{intersection in the interior = linking in the boundary}$$ in an oriented $m$-dimensional manifold with boundary $(M,\partial M)$. Let $$f~:~(K,\partial K)\subset (M,\partial M)~,~g~:~(L,\partial L) \subset (M,\partial M)$$ be embeddings of oriented manifolds with boundary, such that $${\rm dim}(K)~=~k~,~{\rm dim}(L)~=~\ell~,~k+\ell~=~m~,~ f(\partial K) \cap g(\partial L)~=~\emptyset \subset \partial M~.$$ Assume there exists an isotopy (= homotopy through embeddings) rel $\partial K$ $$f_t~:~K \to M~~(0 \leqslant t \leqslant 1)$$ such that $f_0=f$ and $f_1(K)\subset \partial M$, with each $f_t(K), g(L)$ intersecting transversely in $M$, so that $f_t(K) \cap g(L)\subset M$ is a finite set with an intersection index $I(x)\in \{\pm 1\}$ at each point $x \in f_t(K) \cap g(L)$ according to the orientations. A continuity argument shows that the function $$\lambda~:~[0,1] \to {\mathbb Z}~:~t\mapsto \lambda(t)= \sum\limits_{x \in f_t(K) \cap g(L)}I(x)$$ is constant, so that $${\rm intersection}(f(K),g(L)\subset M)~=~\lambda(0)~=~\lambda(1)$$ $$=~{\rm linking}(f(\partial K),g(\partial L) \subset \partial M) \in {\mathbb Z}~.$$ This is best seen by drawing pictures for $(M,\partial M)=(D^2,S^1)$.

The localization exact sequence in algebraic $L$-theory is based on an abstract homological version of this principle.

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This is my attempt at an answer. I think I might be off in justifying the second diagram. It's based on the proof in Bredon, but I think it might be a little simpler. Corrections are welcome, of course!

For convenience, we introduce some notation and conventions. We define subsets of $$S^n= \{ (c,t):c\in S^{n-1},t\in I \} /\sim$$ by $X=\{(c,t):t\leq 1/2\}$, $Y=\{(c,t):t\geq 1/2\}$, and $Z=\{(c,1/2)\}$. We define subsets of $$S^{2n-1}=\{ (a,t,b):a,b\in S^{n-1},t\in I\}$$ by $A=\{(a,t,b)\in S^{2n-1}:t\leq 1/2\}$ and $B=\{(a,t,b)\in S^{2n-1}:t\geq 1/2\}$. We define subsets of $A\cap B=\{(a,1/2,b)\}$ by $S^{n-1}_A = \{(a,1/2,b_0)\}$ and $S^{n-1}_B = \{(a_0,1/2,b)\}$ for fixed $a_0,b_0\in S^{n-1}$. We consider $S^{2n-1}$ as the boundary of $e^{2n}$; in this cell, $S^{n-1}_A$ is the boundary of a cell $e^n_A$ and $S^{n-1}_B$ is the boundary of a cell $e^n_B$. We write $K=S^n\cup_{h(g)}e^{2n}$ for the complex in which we must compute $H(h(g))$. Throughout, all cohomology is integral.

To compute cup products in $K$, we use the commutative diagram

H^n(K) x H^n(K) ---------> H^{2n}(K)
       ^             cup       ^
       |                       |
       |             cup       |
H^n(K,X) x H^n(K,Y) -----> H^{2n}(K,S^n)

which follows from the map $(K,\emptyset,\emptyset)\rightarrow (K,X,Y)$ and naturality; the vertical maps are isomorphisms by the long exact sequence in cohomology for pairs. Let the generator $x$ of $H^n(K)$ correspond to the generator $x_Y$ of $H^n(K,X)$ and the generator $x_X$ of $H^n(K,Y)$, so that $x_Y\cup x_X$ in $H^{2n}(K,S^n)$ corresponds to $x^2$ in $H^{2n}(K)$.

To understand the image of the cup product, we use the evident map $j:(e^{2n},A,B)\rightarrow (K,X,Y)$. We have the diagram

H^n(K,X)  =  H^n(S^n,X)  =  H^n(Y,Z)  =  H^{n-1}(Z)
   |                                       |
   | j^*                                   | j^*
   |                                       |
   V                                       V
H^n(e^{2n},A) = H^n(e^n_A,S^{n-1}_A) = H^{n-1}(S^{n-1}_A)

which commutes because of the naturality of the isomorphisms involved (namely, along the top: cellular cohomology, excision and homotopy invariance, lexseq for pairs; along the bottom: homotopy invariance, lexseq for pairs). By definition of degree, the map $j^\ast:H^{n-1}(Z)\rightarrow H^{n-1}(S^{n-1}_A)$ is multiplication by $\alpha$. Hence, so is the map $j^\ast : H^n(K,X)\rightarrow H^n(e^{2n},A)$. Similarly, the map $j^\ast:H^n(K,Y)\rightarrow H^n(e^{2n},B)$ is multiplication by $\beta$. Now, $j^\ast:H^{2n}(K,S^n)\rightarrow H^{2n}(e^{2n},S^{2n-1})$ is an isomorphism, and $j^\ast(x_Y\cup x_X)$ is $\alpha\beta$ times a generator of $H^{2n}(e^{2n},S^{2n-1})$. Thus $x_Y\cup x_X$ is $\alpha\beta$ times a generator of $H^{2n}(K,S^n)$. So by the first diagram, $H(h(g))=\alpha\beta$.

[Math] A question about the topological proofs of Bott periodicity

There is a beautiful observation of my advisor John Moore that to my mind ought to be part of the focus of any such argument: the Hopf algebra $H_{\ast}(BU;Z)$, or equivalently $H^*(BU;Z)$ is self-dual. This is pure algebra and streamlines the presentation. My preferred presentation is in Section 21.6 of May and Ponto, More Concise Algebraic Topology, available at Amazon or University of Chicago Press. Incidentally, there is a proof of real Bott periodicity along the same lines due to Moore (and written up by Cartan if memory serves) in one of the 1950's Cartan seminars and another such proof by Dyer and Lashof. The funny thing is that each of these proofs finds one of the six equivalences one has to prove more difficult than all of the others, but they find difficulty with different ones: combining the easier of the respective arguments gives a clean and instructive homological proof.

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