Intuition behind Alexander duality - MathOverflow

Intuition behind Alexander duality

2010-10-13T05:17:28Z

I was wondering if anyone could offer some intuition for why Alexander duality holds. Of course, the proof is easy enough to check, and it is also easy to work out many examples by hand. However, I don't have any feeling for why it is true.

To give you an example of what I am looking for, when I think of Poincare duality I think of the picture in terms of triangulations and dual triangulations. Is there any picture like that for Alexander duality? Is there at least maybe some kind of obvious bilinear pairing between the two sides of it or something?

Answer by Emerton for Intuition behind Alexander duality

2010-10-13T06:45:13Z

Let $M$ be a closed orientable $n$-manifold containing the compact set $X$. Given an $n-q-1$-cocyle on $X$ (I am choosing this degree just to match with the notation of the Wikipedia article to which you linked), we extend it to some small open neighbourhood $U$ of $X$. By Lefschetz--Poincare duality on the open manifold $U$, we can convert this $n-q-1$-cocylce into a Borel--Moore cycle (i.e. a locally-finite cycle made up of infinitely many simplices) on $U$ of degree $q+1$. Throwing away those simplices lying in $U \setminus X$, we obtain a usual (i.e. finitely supported) cycle giving a class in $H_{q+1}(U,U\setminus X) = H_{q+1}(M,M\setminus X)$ (the isomorphism holding via excision). Alexander duality for an arbitrary manifold then states that the map $H^{n-q-1}(X) \to H_{q+1}(M,M \setminus X)$ is an isomorphism. (If $X$ is very pathological, then we should be careful in how define the left-hand side, to be sure that every cochain actually extends to some neighbourhood of $X$.)

Now if $M = S^{n+1}$, then $H^i(S^{n+1})$ is almost always zero, and so we may use the boundary map for the long exact sequence of a pair to identify $H_{q+1}(S^{n+1}, S^{n+1}\setminus X)$ with $H_{q}(S^{n+1}\setminus X)$ modulo worrying about reduced vs. usual homology/cohomology (to deal with the fact that $H^i(S^{n+1})$ is non-zero at the extremal points $i = 0$ or $n$).

So, in short: we take a cocycle on $X$, expand it slightly to a cocyle on $U$, represent this by a Borel--Moore cycle of the appropriate degree, throw away those simplices lying entirely outside $X$, so that it is now a chain with boundary lying outside $X$, and finally take this boundary, which is now a cycle in $S^{n+1} \setminus X$.

(I found these notes of Jesper Moller helpful in understanding the general structure of Alexander duality.)

One last thing: it might help to think this through in the case of a circle embedded in $S^2$. We should thicken the circle up slightly to an embedded strip. If we then take our cohomology class to be the generator of $H^1(S^1)$, the corresponding Borel--Moore cycle is just a longitudinal ray of the strip (i.e. if the strip is $S^1 \times I$, where $I$ is an open interval, then the Borel--More cycle is just $\{\text{point}\} \times I$).

If we cut $I$ down to a closed subinterval $I'$ and then take its boundary, we get a pair of points, which you can see intuitively will lie one in each of the components of the complement of the $S^1$ in $S^2$.

More rigorously, Alexander duality will show that these two points generate the reduced $H^0$ of the complement of the $S^1$, and this is how Alexander duality proves the Jordan curve theorem. Hopefully the above sketch supplies some geometric intuition to this argument.

Answer by Greg Kuperberg for Intuition behind Alexander duality

2010-10-13T09:15:35Z

I read Alexander duality as saying "If you have any type of doughnut in a sphere, then the outside must have handles or islands that fill the doughnut's holes." The proof that Ryan outlines exactly matches this intuition. If $D \subseteq S^n$ is the doughnut, then you start with the exact sequence of a pair: $$\cdots \to H_{k+1}(S^n) \to H_{k+1}(S^n,D) \to H_k(D) \to H_k(S^n) \to \cdots,$$ and assume some middle value of $k$. Then you can read this diagram fairly directly as $$\cdots \to 0 \to \text{handles of $S^n\setminus D$} \to \text{holes of $D$} \to 0 \to \cdots.$$ Clearly $H_k(D)$ measures the holes of $D$, and the only question is why $H_{k+1}(S^n,D)$ can be interpreted as the handles/holes of $S^n \setminus D$. As Ryan says, if $D$ and $E = \overline{S^n \setminus D}$ are manifolds that meet at a common boundary, then $$H_{k+1}(S^n,D) \cong H_{k+1}(E,\partial E) \cong H^{n-k-1}(E),$$ where the first isomorphism is excision and the second one is Poincaré duality.

You can also extend this from submanifolds of $S^n$ to general closed subsets by taking limits. On one side you have a decreasing sequence of compact sets, and the answer is Cech cohomology, by the old-fashioned definition of Cech cohomology as a direct limit of cohomology groups. On the other side, you have an increasing sequence of compact sets whose union is open, and the answer is homology because it just is — the homology functor commutes with this type of direct limit of spaces.

In response to an anonymous question just now, I ought to improve my answer, partly also for my own sake. It is a theorem of Steenrod (see Spanier 1948 and Steenrod 1936) that Cech cohomology satisfies the "continuity axiom" in the category of compact Hausdorff spaces, i.e., it converges to itself when you take an inverse limit of spaces. Looking now at the historical record, I see no evidence that the continuity axiom was ever used as a definition of Cech cohomology; and if not I shouldn't have called it "the old-fashioned definition". Spanier does explain that you can prove fairly quickly that the continuity axiom plus the Eilenberg-Steenrod axioms uniquely determines Cech cohomology for compact Hausdorff spaces. Maybe if this was said in 1948 and isn't part of the standard script now, then you could call it an old-fashioned characterization of Cech cohomology. I would say instead that this was forward-thinking whether or not textbooks now mention it.

For one thing, Borsuk's shape theory is an important generalization of this interpretation of Cech cohomology theory. If you express a compact Hausdorff space $X$ as an inverse limit of polyhedra (which you can always do), then the homotopy type of the inverse system of those polyhedra is a topological invariant, the shape of $X$. Instead of considering only inverse and direct limits, you keep the inverse system as a categorical object; then you can take the entire homotopy type of the inverse system instead of just looking at cohomology.

Answer by Jim Conant for Intuition behind Alexander duality

2010-10-13T13:14:19Z

I like to think of Alexander duality in terms of linking numbers of submanifolds (or, in general, k cycles). This is one way to define the pairing you are looking for. In general, consider a $k$-cycle $z$ in the space $X$, and an $(n-k-1)$-cycle $w$ in the complement, then $w=\partial v$ in $\mathbb R^n$. Now take the algebraic intersection of $z$ and $w$. This defines a bilinear pairing $H_k(X)\otimes H_{n-k-1}(\mathbb R^n\setminus X)\to \mathbb Z$ as desired.

Answer by John Klein for Intuition behind Alexander duality

2011-10-16T02:21:05Z

Yet another way to get it. Alexander duality is closely related to Poincare duality: suppose we can write $$ S^n = K \cup_A C $$ where $K$ and $C$ are codimension zero compact manifolds with common boundary $A$ (typically, $K$ is a regular neigbhorhood of a nicely embedded finite polyhedron). Then $$ K/A \cong S^n/C $$ or in homology: $H_{\ast}(K,A) \cong H_{\ast}(S^n,C) \cong H_{\ast-1}(C)$ where the last isomorphism holds when $\ast < n$ using the boundary operator in the long exact homology sequence.

Then Alexander duality can be viewed as this isomorphism coupled with with the Poincare duality isomorphism $H_\ast(K,A) \cong H^{n-\ast}(K)$.

Another way to understand this is to realize these isomorphisms using the composite $$ S^n \overset{pt}{\to} K/A \overset{diagonal}{\longrightarrow} K_+ \wedge K/A \cong K_+ \wedge S^n/C $$ where pt is the Pontryagin-Thom map which collapses $C$ to a point, and the diagonal map is the map of quotients coming from the evident map of pairs $(K,A) \to (K\times K,K\times A)$. The image of the fundamental class of $S^n$ gives rise to a slant product which in turn gives the Alexander duality isomorphism.

Answer by Andrew Ranicki for Intuition behind Alexander duality

2011-10-16T06:10:44Z

Of course Alexander duality is most exciting for arbitrary compact subspaces $X \subset S^n$, Cech cohomology etc. But there is a very direct combinatorial proof for subcomplexes $X \subset \partial \Delta^{n+1}$, due to Blakers and Massey "The homotopy groups of a triad II" (Ann. of Maths. 55, 192-201 (1952)), which harks back to Poincare's proof of his duality using dual cells. For a simplicial complex $K$ let $K'$ denote the barycentric subdivision, and let $\sigma^* \subset K'$ denote the dual cell of a simplex $\sigma \in K$. Define the supplement of the subcomplex $X\subset \partial \Delta^{n+1}$ to be the subcomplex $\overline{X} \subset \Sigma^n=(\partial \Delta^{n+1})'$ consisting of the dual cells $\sigma^* \subset \Sigma^n$ of the simplices $\sigma \in \partial \Delta^{n+1} \backslash X$. The relative cellular chain complex $C^{cell}(\Sigma^n,\bar{X})$ is isomorphic to the $n$-dual ${C^{simp}(X)}^{n-*}$ of the simplicial chain complex $C^{simp}(X)$, inducing the Alexander duality isomorphisms

$H_* (\Sigma^n,\overline{X}) \cong H^{n-*}(X)$ .

I used this combinatorial approach to Alexander duality in my book "Algebraic L-theory and topological manifolds" (CUP (1992)) to construct a combinatorial model for the generalized homology theory with algebraic surgery spectrum coefficients.