# Is the Alexander-Pontryagin duality applicable to stratified spaces

If $D$ is the discriminant of the space of all planar curves of a fixed degree, and $D'$ is the subspace whose only singularities are nodes or cusps, then is it possible to apply Alexander-Pontryagin duality to the pair $(D,D\setminus D')$ to conclude that $\mathrm{H}_N(D,D\setminus D';\mathbb{Z}/2)=\mathrm{H}^0(D';\mathbb{Z}/2)$? Here $N$ is the dimension of $D$.

I have seen this being used, but I am not sure why one is permitted to use the duality here. After all, $D$ is not a manifold, but does it form a nice enough space for which the Alexander-Pontryagin duality is applicable? Can the duality be applied to stratified spaces?

Thank you!

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Alexander duality does not need manifolds as input. Have you looked at the statement? en.wikipedia.org/wiki/Alexander_duality It's quite commonly used for singular spaces and discriminants. –  Ryan Budney Jul 11 '13 at 22:13
The wikipedia article only talks about the duality concerning the complement of a subspace of a sphere. From this, is it possible to derive the version mentioned in the question, which applies to the pair of the discriminant and a subspace? –  user1289492 Jul 12 '13 at 6:14
I don't know what $D$ is so it's hard for me to say whether or not the theorem applies, as "planar curves of a fixed degree" means nothing to me. Could you say what $D$ means in more elementary terms, for example, discriminant in what sense? Is $D$ contractible? –  Ryan Budney Jul 12 '13 at 6:18

Set $X=D,Y=D\setminus D',n=\dim D$. I presume you are interested in real algebraic curves (in the complex case everything simplifies). The problem is that $X\setminus Y$ is not smooth. But there is an ad hoc way to handle this. Let us triangulate $X$ so that $Y$ is a subcomplex. Assume that $H_n(X,\mathbb{Z}/2)=\mathbb{Z}/2$ and that $X$ is compact. (Actually, in the real case there are two possible definitions of smoothness: given a plane real curve, one may require that the corresponding complex curve is smooth or that it has no real singularities so depending on the definition of $D$ these assumptions may or may not be true.) The "fundamental class" of $X$ mod 2 is represented in the simplicial chain group by the sum of all the simplices: if we take a combination of all simplices but one, it will come from a subcomplex homotopy equivalent to something of dimension $< n$. So each $n-1$-simplex is in the boundary of an even number of $n$-simplices.

So the relative homology group $H_n(X,Y)$ is freely generated by the sums $\sum_{Int(\sigma)\subset C}\sigma$ where $C$ is a component of $X\setminus Y$ and $Int$ denotes the interior. There is one such element for each component, which proves the isomorphism in the original posting.

Remarks:

1. This does not generalize to higher (co)homology or to other coefficients. The general Alexander duality theorem is the following statement: suppose $X\supset Y$ is a pair of nice spaces (say, the one point compactification of $X$ can be made into a CW-complex so that the closure of $Y$ is a subcomplex) such that $X\setminus Y$ is a manifold. Let's assume it orientable. Then $$H^{BM}_*(X,Y)\cong H^{BM}_*(X\setminus Y)\cong H^{n-*}(X\setminus Y)$$ where $d$ is the real dimension of $X$, $H^{BM}$ denotes the Borel-Moore homology (it coincides with the usual homology for compact spaces) and the second isomorphism is the Poincar\'e-Lefschetz duality. So $H^0(X\setminus Y)\cong H_n^{BM}(X,Y)$. Assuming that in addition to the above $H^{BM}_{n-1-i}(X)=H_{n-i}^{BM}(X)=0$ we get $H^i(X\setminus Y)\cong H_{n-1-i}^{BM}(Y)$ from the long exact sequence for the Borel-Moore homology. In the case $X=\mathbb{R}^n$ we get the usual Alexander duality; for $i=0$ these assumptions are not quite satisfied: $H^{BM}_n(\mathbb{R}^n)=\mathbb{Z}$, so we get an exact sequence $$0\to \mathbb{Z}\to H_n^{BM}(X,Y)\to H^{BM}_{n-1}(Y)\to 0$$ instead.

2. There is a duality theory for singular spaces called the Verdier duality. Informally, it says the following. Let $X$ be a nice space (e.g. a finite CW complex) and $A$ a commutative ring. Let $D^b(X)$ be the derived category of the category of complexes of sheaves of $A$-modules on $X$ whose cohomology sheaves are constructible with respect to some stratification. Then there is an anti-autoequivalence $F\mapsto F^\vee$ of $D^b(X)$ such that $H^*(X,F)\cong H^{-*}(X,F^\vee)$ where $H^i(X,F)$ denotes $\mathop{\mathrm{Hom}}\nolimits_{D^b(X)}(\underline{A},F[i])$, $\underline{A}$ being the constant sheaf with stalk $A$. If $X$ is an orientable manifold (in the usual sense), $\underline{A}^\vee$ is again constant (up to a shift), which gives one the usual Poincar\'e duality, but in general $\underline{A}^\vee$ needn't even be a sheaf (it may have "components" in different degrees).

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