I am reading the paper of Fulton and Lazarsfeld on the connectivity of degeneracy loci of morphisms of vector bundles, but there is a comment in the article that I don't quite understand.

Let $G$ be a smooth complex projective variety of dimension $r$ and let $Y \subseteq G$ be a closed algebraic subset. How do I prove that there is an injection (I assume it will be an isomorphism?) in singular cohomology $$ H^1(G,Y;\mathbb Q) \overset{\simeq ?}{\longrightarrow} H_{2r-1}(G-Y;\mathbb Q) $$ I actually don't know what they meant by Lefschetz duality since I am using Hatcher to learn algebraic topology, and in that book they mention that the isomorphism $H^k(M;R) \simeq H_{r-k}(M, \partial M;R)$ (where $M$ is compact and orientable) is sometimes called "Lefschetz duality" (page 254, Theorem 3.43) but I don't know what Fulton & Lazarsfeld mean by using Lefschetz duality in this context. The theorem in Hatcher works also with boundaries of smooth compact submanifolds of codimension $1$ but my closed algebraic subset $Y$ is not a smooth $(r-1)$-manifold (in fact, it never has codimension $1$ in $G$ as a real manifold).

Another way to obtain this injection, according to the paper, is to write down the exact sequence of low degree terms of the Zeeman spectral sequence. I couldn't find it anywhere (the paper of Zeeman explains the spectral sequence, but I don't have enough experience with spectral sequences to write it down myself), so it would be nice to at least see it and wonder if I can work out the injection.

  • $\begingroup$ You're misreading Theorem 3.43. It's saying that the case $H^k(M;R)\cong H_{n-k}(M,\partial M;R)$ is called Lefschetz duality $\endgroup$ – Denis Nardin Apr 23 '16 at 16:43
  • 1
    $\begingroup$ @DenisNardin : Right. I actually typed it in wrong, but I read it correctly! Thanks, let me correct that in the question. $\endgroup$ – Patrick Da Silva Apr 23 '16 at 16:43
  • $\begingroup$ Doing algebraic geometry over $\mathbb{C}$, you should not forget that you are speaking about manifold, so all power of algebraic topology applies. A bit of standard shamanism (tubular neighborhoods and deformation retraction) shows that, for your isomorphism, you only need that $G\setminus Y$ should be a manifold (of course, assuming that everything is nice, e.g., triangulizable). This question is more appropriate for MSE. $\endgroup$ – Alex Degtyarev Apr 23 '16 at 16:50
  • $\begingroup$ @Alex Degtyarev : I know, I have been doing singular cohomology forever since I started reading this paper! But I thought I'd post it here because I was also curious about the Zeeman spectral sequence part. Is the triangulizable assumption reasonable for a complex variety? $\endgroup$ – Patrick Da Silva Apr 23 '16 at 16:53
  • $\begingroup$ Of course: in algebraic geometry, everything is triangulizable. A standard (?) ref would be Hironaka [MR 0374131]. $\endgroup$ – Alex Degtyarev Apr 23 '16 at 16:58

A general form of Lefschetz duality is that if $X$ is a compact topological space, $A \subset X$ is a closed subset (which needs some very mild point-set assumptions, maybe a neighborhood retract or something), and $X \setminus A$ is an oriented $n$-manifold, then $$ H^k(X,A) \cong H_{n-k}(X \setminus A).$$ I find it easiest to think of this as the conjunction of two assertions: firstly that $H^k(X,A) \cong H^k_c(X \setminus A)$, which is an isomorphism having nothing to do with manifolds: for any "reasonable" topological space, compact support cohomology can be defined as relative cohomology for an arbitrary compactification. Secondly, Poincaré duality on an oriented manifold, in the form of an isomorphism between homology and compact support cohomology.

The dual form of Lefschetz duality $H_k(X,A) \cong H^{n-k}(X \setminus A)$ can be understood similarly in terms of Borel--Moore homology.

  • $\begingroup$ Do you have a reference for the first isomorphism (relative cohomology with compactly supported cohomology)? I have one for the second and I'd like to read up on that. Thanks for your answer! $\endgroup$ – Patrick Da Silva Apr 23 '16 at 17:15
  • $\begingroup$ @PatrickDaSilva Since you're familiar with Hatcher's book, you can use its proof of Alexander duality (which is the case $X=S^n$ of this statement), it will work in this more general setting. $\endgroup$ – Denis Nardin Apr 23 '16 at 21:55
  • $\begingroup$ @DenisNardin : I'm sorry if I misunderstood you, but I couldn't find the proof of my statement (even in the case $X=S^n$) within Hatcher's proof of Alexander duality. The "closest" (in terms of symbols, I guess) I found was $H_i(X-A) \simeq H_c^{n-i}(X-A) \simeq \tilde H^{n-i-1}(A)$, but nothing related to $H^i(X,A) \simeq H^i_c(X-A)$. Can you explain a bit more? $\endgroup$ – Patrick Da Silva Apr 23 '16 at 22:11
  • $\begingroup$ @PatrickDaSilva I've just checked and I'm sorry, I thought it was done more explicitely. Remember that $H^*_c(X\smallsetminus A)$ is defined as the colimit of $H^*(X\smallsetminus A,U\smallsetminus A)=H^*(X,U)$ where $U$ ranges across all neighbouroods of $A$ and then proceed like in the proof of Alexander duality (i.e. there is a cofinal system of $U$'s such that the inclusion $(X,A)\to (X,U)$ is an equivalence of pairs) $\endgroup$ – Denis Nardin Apr 23 '16 at 22:28
  • $\begingroup$ @DenisNardin : This sounds more reasonable! I'll give it some thought and come back to you. For the moment, +1 :) $\endgroup$ – Patrick Da Silva Apr 23 '16 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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