Let $f : X \to C$ be a surjective map between projective varieties ($C$ is a curve). Let $C^* = C - \{\text{critical values of $f$}\}$, $X^* = f^{-1}(C^*)$. Fix $t \in C^*$ and let $X_t = f^{-1}(t)$.

There are inclusions $i : X_t \to X^*$ and $j : X^* \to X$ which induces map in singular cohomology : $i^* : H^m(X^*) \to H^m(X_t), j : H^m(X) \to H^m(X^*)$.

1) I saw that "obviously" $H^m(X_t)^{\pi_1(C_t)} = i^*H^m(X^*)$. Why is this true ?

2) The invariant cycle theorem states that $i^*j^* H^m(X) = H^m(X_t)^{\pi_1(C_t)}$. What is the geometric intepretation of this statement ? What are the interesting corollaries ?

  • $\begingroup$ Is there a mistake in 1) ? It seems incompatible with 2) the invariant cycle theorem. $\endgroup$ – Roland Mar 12 '18 at 22:58
  • $\begingroup$ @Roland : thanks for your interest. I couldn't say if there is a typo/mistake since I don't understand why this statement should be true. This is claimed in "A naive guide to mixed Hodge structure" by Durfee, section 7. $\endgroup$ – student Mar 13 '18 at 10:17
  • $\begingroup$ @Roland : ok thanks, it seems to make more sense ! Do you know where I could read a proof of this ? And what about 2) ? $\endgroup$ – student Mar 13 '18 at 14:48
  • $\begingroup$ I deleted my comment because there is indeed a claimed equality (which I find weird). The point is the action of $\pi_1$ is natural, so if a class comes from $X^*$, it is certainly invariant by $\pi_1$. I am thinking about the inverse inclusion. $\endgroup$ – Roland Mar 13 '18 at 14:50
  • $\begingroup$ @Roland : this is what I though too. Also, I know that class in $H^m(X_t)$ might be killed when we go in $H^m(X)$ because of the singular fibers, but normally it should survive in $H^m(X^*)$ since we removed the critical values. But I'm not sure why it's "obvious". Also I made a mistake when copying, it's of course $i^*H^m(X^*)$, sorry about it. $\endgroup$ – student Mar 13 '18 at 14:54

This is just an answer to 1) and some comments.

As I said in the comment, the main point is Leray spectral sequence and its functoriality. So Leray spectral sequence is $$ E_2^{pq}=H^p(C^*,R^qf_*\mathbb{C})\Rightarrow H^{p+q}(X^*) $$ Concretely this means that we can recover the cohomology of the total space if we know the cohomology of the base $C^*$, the sheaves $R^qf_*\mathbb{C}$ and a lot a differentials between the different groups. These differential may be very hard to track, but the functoriality maybe of tremendous help here.

In the present situation, $f:X^*\rightarrow C^*$ is a proper submersion. This implies that $R^qf_*\mathbb{C}$ are local system whose fibers are $H^q(X_t)$. From the general theory of local system, we have $$H^0(C^*,R^qf_*\mathbb{C})=H^q(X_t)^{\pi_1(C^*,t)}$$ What about other values of $p$ ? This is where we will use functoriality : the map $i^*:H^{p+q}(X^*)\rightarrow H^{p+q}(X_t)$ is compatible with a morphism of spectral sequence $$ i^*:H^p(C^*,R^qf_*\mathbb{C})\rightarrow H^p(\{t\},R^qf_*\mathbb{C}|_t)$$ But the last group is zero unless $p=0$ in which case it is $H^q(X_t)$. It follows that the image of $i^*:H^q(X^*)\rightarrow H^q(X_t)$ is exactly the image of $i^*:H^0(C^*, R^qf_*\mathbb{C})=H^q(X_t)^{\pi_1(C^*,t)}\rightarrow H^q(X_t)$.

A word about 2). The Leray spectral sequence is a very powerful to understand the cohomology of a space. But it is often very hard to compute explicitly the differentials, so you can use tricks as above to have more informations. Now from Hodge theory, there is huge restriction on the the differentials : classes in the spectral sequence have weights and the differential preserve them. It turns out that most of the time, they will be zero because they map spaces of different weights. For example we have the very important and powerful theorem that if $f$ is projective smooth between algebraic varieties, then the Leray spectral sequence degenerates at $E_2$ : every single classes survives the spectral sequence and $H^{n}(X^*)$ is actually isomorphic to $\bigoplus_{p+q=n}H^p(C^*,R^qf_*\mathbb{C})$.

Weights behave this way : if $X$ is projective then $H^m(X)$ has weights $\leq m$ whereas if $X$ is smooth $H^m(X)$ has weights $\geq m$. Thus if $X$ is projective and smooth $H^m(X)$ is pure of weight $m$.

But $i^*:H^m(X^*)\rightarrow H^m(X_t)$ preserves weights, so every classes of weight $>m$ is mapped to $0$. We also have $j^*:H^m(X)\rightarrow H^m(X^*)$ and this morphism preserve weights, so every classes of weights $<m$ is mapped to $0$. With a bit more work, both $H^m(X)$ and $H^m(X^*)$ have the same part of weight $m$, and as we said before, this is the part which have a non trivial image in $H^q(X_t)$. Hence the global invariant cycle theorem.

  • $\begingroup$ Perfect. Many thanks for the detailled explanations ! $\endgroup$ – student Mar 13 '18 at 21:57
  • $\begingroup$ For your answer to (1), don't you already need the degeneration of the spectral sequence at $E_2$? (which is a nontrivial theorem of Deligne) Without knowing the degeneration, it seems to me that you can only say that the image of $i^* : H^q(X^*)\rightarrow H^q(X_t)$ is the image of $i^* : E_\infty^{0,q}\rightarrow H^q(X_t)$, and a priori this $E_\infty^{0,q}$ is only a subobject of $H^0(C^*,R^qf_*\mathbb{C}) = H^q(X_t)^{\pi_1}$... $\endgroup$ – oxeimon May 3 '18 at 19:27
  • $\begingroup$ @oxeimon mmh, you are probably right. I don't have the paper I used to work on this question anymore and it seems that I made a mistake. Let me think a little bit more about it, and I will edit my answer. $\endgroup$ – Roland May 4 '18 at 20:21
  • $\begingroup$ Dear @Roland, thanks for your illuminating explanation, for both functoriality in part $(1)$ and weights in part $(2)$. But I have a question regarding your answer to part $(2)$: I know that Deligne's theorem degenerating at $E_2$ says successive quotients of Leray's filtration are isomorphic to $E_2^{p,q}$ (e.g., $H^n(X^*)/L^1H^n(X^*)\cong E_2^{0,n}$, $L^1H^n(X^*)/L^2H^n(X^*)\cong E_2^{1,n-1}$). How far are these from the splitting $H^n(X^*)\cong\oplus_{p+q=n} E_2^{p,q}$ that you mentioned in the first paragraph of $(2)$? $\endgroup$ – Yilong Zhang Apr 29 at 15:04

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