Intutive interpretation about Linking forms - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:18:36Z http://mathoverflow.net/feeds/question/36987 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/36987/intutive-interpretation-about-linking-forms Intutive interpretation about Linking forms Topologieee 2010-08-28T18:41:42Z 2010-08-31T05:22:08Z <p>Let $M^3$ be a rational homology 3-sphere. (i,e, $M^3$ is closed 3-manifold with $H_{*}(M;Q)=H_{*}(S^3;Q)$</p> <p>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. </p> <p>(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)</p> <p>$H_1(M;Z)\cong H^2(M;Z)\cong H^1(M;Q/Z)=Hom(H_1(M;Z),Q/Z)$ </p> <p>First isomorphism : Poincare duality, </p> <p>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.</p> <p>Third isomorphism : Universal Coefficient theorem</p> <p>In short, $\lambda\colon H_{1}(M;Z)\times H_{1}(M;Z)\to Q/Z$ is defined by $\lambda(x,y)=&lt;\tilde{x}\cup\delta^{-1}(\tilde{y}),[M]>$, 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 $Z \times Q/Z \to Q/Z$.</p> <p>I understand this linking form only algebraic viewpoint. Therefore, I can play with this form only by using dilluminating algebraic topology. </p> <p>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 )</p> <p>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......) </p> http://mathoverflow.net/questions/36987/intutive-interpretation-about-linking-forms/36993#36993 Answer by Ryan Budney for Intutive interpretation about Linking forms Ryan Budney 2010-08-28T20:10:39Z 2010-08-28T23:11:08Z <p>You get the geometric interpretation of the linking form by tracing through exactly what the isomorphisms are and looking at what they do. </p> <p>The upshot is the answer is this. Let $[a]$ and $[b]$ be torsion classes in $H_1(M;\mathbb Z)$. Let $n[a]=\partial [A], m[b]=\partial [B]$ for $n,m \neq 0$. $A$ and $B$ are $2$-cycles in $M$ is a $3$-manifold.</p> <p>Then </p> <p>$$\langle [a],[b]\rangle = \frac{1}{m} a\pitchfork B = \frac{1}{n} A \pitchfork b \in \mathbb Q / \mathbb Z$$</p> <p>here $\pitchfork$ means the transverse intersection, meaning you have to find chain representatives $a$ and $B$ (or $A$ and $b$) such that they intersect transversely -- the representatives do not need to be manifolds, as you can make sense of this in the PL-category. This is the algebraic intersection number, where every point of intersection is given a weight $\pm 1$, according to how the relative orientations add up to either the local orientation of the manifold or its reverse. </p> <p>Perhaps the simplest context where a proof of this is relatively easy would be if your $3$-manifold is triangulated, then your Poincare duality isomorphism is naturally between the simplicial homology and the CW-cohomology of the dual polyhedral decomposition, so transversality is for free in this context.</p> http://mathoverflow.net/questions/36987/intutive-interpretation-about-linking-forms/37217#37217 Answer by Andrew Ranicki for Intutive interpretation about Linking forms Andrew Ranicki 2010-08-31T05:22:08Z 2010-08-31T05:22:08Z <p>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 $(M,\partial M)=(D^2,S^1)$.</p> <p>The localization exact sequence in algebraic $L$-theory is based on an abstract homological version of this principle.</p>