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게시글 주소: https://cuttingedge.orbi.kr/00070074391

Q. Can a boundary map on a long exact sequence of homology on manifold be interpreted as an actual topological boundary of a manifold representing the homology class?

A. True if the class is representable by a manifold with boundary. If $M$ is a compact $n$-manifold with boundary, it has a fundamental class $[M]\in H_n(M,\partial M)$ (coefficients being whatever as long as you're orientable w.r.t. them) and its image under the connecting homomorphism of the pair $(M,\partial M)$ is the fundamental class $[\partial M]\in H_{n-1}(\partial M)$ of the closed $(n-1)$-manifold $\partial M$ with the induced orientation. So, if $f\colon(M,\partial M)\rightarrow(X,A)$ is some map of pairs (the representing manifold of a class), naturality of the pair sequence yields $\partial(f_{\ast}[M,\partial M])=f_{\ast}[\partial M]$ and if $M$ is closed, this is zero, but that's not surprising cause the element then factors through $H_n(X)$ and the composite $H_n(X)\rightarrow H_n(X,A)\rightarrow H_{n-1}(A)$ is zero.

Intuitively, If $[\sigma]\in H_n(X,A)$, then $\sigma$ is some chain in $X$ with boundary inside of $A$. Since it represents a homology class, it should be a cycle, but it need not boundary anything entirely in $A$, so it could be a nonzero representative in $H_{n-1}(A)$. In other words, if $\sigma\mapsto X$ is a chain so that its topological boundary $\partial\sigma$ be mapped entirely into $A$. This boundary represents an element of $H_{n-1}(A)$. Although this is a more or less intuitive argument, this is exactly what's happening on topology. Algebraic machinery is just make this rigorous in algebraic language.

Q. How do you see the Alexander duality?

Rmk. Alexander duality: Let $X\subset S^n$ be a submanifold. Then $H_{p}(S^n\setminus X)\simeq H^{q}(X)$ where $p+q = n-1$. Or, $H_p(\Bbb R^n\setminus X)\simeq H^q(X)$ where $p+q = n-1$.

A. One of the most important interpretation of Alexander duality is via linking numbers of submanifolds, or more generally $k$ cycles. Consider $k$-cycle $z$ in the space $X$ of dimension $k$, and an $(n-k-1)$-cycle $w$ in the complement of $\Bbb R^n$. Then $w = \partial v$ in $\Bbb R^n$ for some cycle $v$. Now take the algebraic intersection (cup product) of $z$ and $v$. This defines a bilinear pairing $H_k(X)\otimes H_{n-k-1}(\Bbb R^n\setminus X)\to\Bbb Z$, called the linking number and gives an Alexander duality. Note that the linking number here is compatible with the linking number in the classical links in $S^3$. This is just a high dimensional analog. See this answer for more geometrical interpretation of high dimensional linking number https://mathoverflow.net/a/332250/323920

Under this interpretation, in case of knot $K$ not link in $S^3$, $S^3\setminus K$ can be thought as a "dual knot" which has linking number 1 with $K$. In particular, every knot complement has $\Bbb Z$ in the first homology, generated by a single "dual unknot" (meridian) of $K$.

One can actually define linking number from Alexander duality as follows: This time we let $M^p,N^q\subset\Bbb R^n$ be closed connected oriented manifolds with dimension $p$ and $q$ and $p+q = n-1$. Then by Alexander duality, we have $\Bbb Z\simeq H^p(M)\simeq H_{q}(\Bbb R^n\setminus Z)$. Now we consider the induced map $i_*:H_q(N)\to H_q(\Bbb R^n\setminus M)$ via inclusion $N\hookrightarrow \Bbb R^n\setminus M$. This map sends the fundamental class of $N$ to some integer times the fundamental class of $H_q(\Bbb R^n\setminus M)$, obtained by the isomorphism from Alexander duality. This integer is exactly the linking number of $M$ and $N$. You will see without much difficulty that these two back and forth are compatible.