Let $X$ be the sort of topological space for which it makes sense to talk about the intersection homology. Fix a perversity $p$, or just take $p= 1/2$ if you like.
Is there some naturally defined $X'$ such that ${}^p IH_* (X) = H_* (X')$ ?
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asked Dec 2, 2010 at 21:53
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3 Answers
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I'm not sure this is the kind of answer you want, but if $X$ has a small resolution $f:X' \rightarrow X$ (so $X'$ is a manifold and the dimension of fibers is sufficiently small), then there is an induced isomorphism $IH_{\ast}(X) = IH_{\ast}(X')$, and because $X'$ is smooth, the latter group is $H_{\ast}(X')$.
More precisely, a proper (birational) map is small if the set $${x \in X | \dim f^{-1}(x) \geq r }$$ has codimension more than $2r$; such maps induce isomorphisms on IH.
Of course, there is nothing natural about $X'$, nor do small resolutions necessarily exist (see the comment below by Mike Skirvin for an easy example). Hopefully someone more knowledgeable about IH will have something to say.
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4$\begingroup$ It's probably worth mentioning that small resolutions do not always exist. The simplest example is given by the quadric cone in $\mathbb{C}^3.$ More generally, the Springer resolution is semismall, but not small. This, of course, has led to a lot of interesting mathematics. $\endgroup$ Commented Dec 2, 2010 at 23:23
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1$\begingroup$ @Dave: could you include the definition of "small resolution" in your answer please? $\endgroup$ Commented Dec 3, 2010 at 10:52
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$\begingroup$ Although probably obvious, we implicitly assume something like complex dimensions, for uniqueness reasons: The wedge of two circles has two resolutions (a circle, versus a disjoint union of two circles) where cohomology is thus different. $\endgroup$ Commented Sep 8, 2016 at 22:40
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Such a space $X'$ cannot possibly depend functorially on $X$: suppose that there is a functor $F$ from the category of topological spaces (that have reasonable stratifications) and continuous maps to itself such that IH(X)=H(F(X)). Then IH would be functorial, but considering the example $$(S^1 \hookrightarrow X \rightarrow S^1)=id_{S^1}$$ where $X$ is the pinched torus and $S^1$ is the non-collapsed circle (see pages 55 and 56 of Kirwan-Woolf "Introduction to Intersection Homology") would imply that the intersection homology of $S^1$ (=ordinary homology of $S^1$) imbeds in the intersection homology of $X$. But the first IH group of $X$ (middle perversity) is $0$.
answered Dec 3, 2010 at 16:14
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$\begingroup$ @Vivek: So, given that by "naturally defined" you mean "functorial", I guess the answer is no (maybe it would help the people trying to answer the question if the meaning of "naturally defined" is pinned down in the question?). I don't know of a reasonable class of spaces strictly containing manifolds such that IH is functorial on that class. $\endgroup$ Commented Dec 3, 2010 at 16:36
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$\begingroup$ @Srrb: Thanks for this very helpful example! I didn't want to say functorial in the question, because I don't know with respect to what class of maps the construction should be functorial. E.g., maybe only "stratified smooth" maps, whatever that means... $\endgroup$ Commented Dec 3, 2010 at 18:02
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The short answer is "no". But you might be interested in some work Markus Banagl is doing along these lines. He has a functor that assigns to a space X an "intersection space" $I^{\bar p}X$. Then rather than studying $I^{\bar p}H_\ast(X)$, he looks at $H_\ast(I^{\bar p}X)$. However, this is not generally the same thing as $I^{\bar p}H_\ast(X)$.
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6$\begingroup$ "no" in the sense that no such construction is known, or "no" in the sense than no such construction could possibly exist (eg. if $X'$ is required to be a smooth manifold and the construction should be say functorial for smooth maps) ? $\endgroup$ Commented Dec 3, 2010 at 13:35