MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ${}^p IH_* (X) = H_* (X')$ ?

share|cite|improve this question
up vote 14 down vote accepted

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.

share|cite|improve this answer
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. – Mike Skirvin Dec 2 '10 at 23:23
@Dave: could you include the definition of "small resolution" in your answer please? – André Henriques Dec 3 '10 at 10:52

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 $I^{\bar p}H_\ast(X)$.

share|cite|improve this answer
"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) ? – Vivek Shende Dec 3 '10 at 13:35

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 $0$.

share|cite|improve this answer
@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. – Sheikraisinrollbank Dec 3 '10 at 16:36
@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... – Vivek Shende Dec 3 '10 at 18:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.