5
$\begingroup$

As part of my masters project I have been working through Serre's FAC. Below are three closely related results I will be presenting as part of my defense. These results are from n$^{°}$ 52, page 63 from the English translation here.

The results are as follows:

  1. Let $X = \mathbb{P}_r(K)$ and let $\mathscr{F}$ be a coherent algebraic sheaf on $X$. Then $H^{n}(X, \mathscr{F}) = 0$ for $n > r$.

  2. Result 1. can be generalized as follows: Let $V$ be an algebraic variety isomorphic to a locally closed subvariety of the projective space $X$. Let $\mathscr{F}$ be an algebraic coherent sheaf on $V$ and let $W$ be the subvariety of $V$ such that such that $\mathscr{F}$ is zero outside $W$. We then have $H^n(V,\mathscr{F}) = 0$ for $n > \mathrm{dim} \, W$.

  3. In particular taking $W = V$ in result 2. we see that $H^n(V, \mathscr{F}) = 0$, for $n > \mathrm{dim} V$.

These results are, more or less, supposed to be a big culmination of my project. I have worked through the material in FAC pretty rigorously up to this point, but I am not enough of an expert yet to see the connections and implications to geometry. Also, my intuition is still fairly poor. I have to write out everything and work through it slowly. What I have learned from this paper so far is a lot of elementary general sheaf theory (I say elementary, but it was by no means easy for me), a lot of general Cech cohomology, and now I am in the chapters where he ties it into the geometry (chapters II and III on varieties and protective varieties respectively). In the interest of time I have skipped some sections, but not too many. I am sure if I keep going and finish the paper, and keep studying AG for years to come, the importance of such results will be obvious to me. But, I don't have that much time before I defend my masters project and I will almost surely be asked to explain the importance of these results.

Right now, the best answer I would have is that I do not know why, but I trust that they are important. Ha! Some reasons I trust that they are important are 1) Serre chose to label it as a proposition / corollary which means he thinks it is important. 2) Results like this seem to be ubiquitous, Serre gives versions and analogues of this result in the chapter on varieties and the chapter on projective varieties. In the introduction to this paper he mentions it is one of the main results in the non-projective varieties chapter. There are also analogue assertions like this made in other big results like the Weil conjectures (not exactly the same, but the same flavor).

Can anyone offer me some insight as to why these vanishing cohomology results are relevant? Why are they important for geometry? Why are they important from an abstract homological algebra view? Why are they natural questions to try to answer?

In response to the obvious question "why would you set out to study this if you don't know why it is important?" - Because my adviser recommended it and I trust him. In response to "well then why don't you ask your adviser why it is important?" - I already did. Now I am asking you, too.

$\endgroup$
3
$\begingroup$

Here are three concise remarks on a question which would take hundreds of pages to answer in a reasonably complete way.

1) Many theorems, like Riemann-Roch, concern themselves with the calculation of the Euler characteristic $\chi (X,\mathcal F)=\sum_{i=0}^{\infty}\dim _KH^i(X,\mathcal F)$ of a coherent sheaf $\mathcal F$ on the projective variety $X$.
The sum wouldn't even make sense if we didn't know that $\dim _KH^i(X,\mathcal F)=0$ for $i\gt \dim X$.
Then the Hilbert polynomial of the coherent sheaf $\mathcal F$ on$X$ is the polynomial $H_\mathcal F(t)=\chi(X,\mathcal F\otimes\mathcal O(t))\in \mathbb Z[t]$.
This polynomial yields several invariants for $(X,\mathcal F)$ (like the dimension of X for example) and is the prerequisite for Grothendieck's path-breaking construction of Hilbert schemes.

2) Grothendieck has proved the amazing generalization that for a noetherian topological space of Krull dimension $n$ and an arbitrary sheaf of abelian groups $\mathcal A$ on $X$ we have $H^i(X,\mathcal A)=0$ as soon as $i\gt n$.
Let me insist that $X$ has nothing to do with algebraic varieties !
So for example the circle $S^1$ with its usual metric topology has Krull dimension $0$ and we can thus deduce $H^1(S^1,\mathbb R)=0$, right?
Wrong! Because an infinite Hausdorff topological space is never noetherian (since it has infinitely many irreducible components: its points).
So, Grothendieck's theorem apparently doesn't require one to work in algebraic geometry, but in practice it is not so useful in other domains of mathematics :-)

3) For context and contrast note that Barratt-Milnor have proved that there exists a compact $n$-dimensional infinite bouquet $B$ of spheres whose singular cohomology groups $H^i(B,\mathbb Q)$ are $\neq 0$ for infinitely many values of $i$.

$\endgroup$
2
$\begingroup$

It can help to look up some of the primary sources that start dealing with cohomology groups to see what problems they were trying to solve. Today's treatment of them is very streamlined and nice, but tends to leave some of the motivation or intuition behind. There are two examples that can be good to keep in mind:

The first is basically where Cech cohomology comes from. We have a variety $X$, and want to construct a holomorphic function $f$ on it. To do this, we pick a covering $(U_j)$ of $X$ by open sets, and construct a function $f_j$ on each $U_j$. To be able to "glue" these together and define a function $f$ on the whole of $X$, we need the different parts to agree on the intersections; that is, we need that $f_j = f_k$ on $U_j \cap U_k$ for all $j, k$. This is just saying that the Cech cocycle defined by $f_{jk} = f_j - f_k$ is zero in $H^1(X, \mathbb C)$.

If we know that $H^1(X, \mathbb C) = 0$, then this condition is automatically satisfied, and we can construct whatever holomorphic function we need.

In practice, this problem often comes up when we have a function $f$ on a subvariety $Y$ of a variety $X$, and we ask whether we can extend $f$ to all of $X$.

The second example is when we want to solve differential equations on some variety. The Cauchy-Riemann equations being PDEs helps explain why we'd want to do that. Suppose, for example, that we want to construct a smooth $p$-form $u$ on a variety $X$ such that $$ d u = v. $$ For this to work, we need to have $d v = d^2 u = 0$. That is, the element $[dv]$ of the de Rham cohomology group $H^{p+1}(X,\mathbb C)$ must be zero. Again, if that group itself is zero, we win by default and can solve whatever PDE we want. This problem can again arise if we're trying to extend a differential form from a subvariety to the whole ambient variety.

Very similar holomorphic examples happen when we consider the Dolbeault cohomlogy groups $H^p(X, \mathcal F)$ with values in some holomorphic sheaf $\mathcal F$, where we try to solve PDEs of the form $\bar\partial u = v$ where $u$ is a smooth $(p,0)$-form.

For a more concrete example, you can consider the projective space $\mathbb P^{n+1}$ and a hypersurface $X$ therein defined by a homogeneous polynomial of degree $d$. Look up the Euler short exact sequence associated to the hypersurface, and find the Kodaira vanishing theorems. Combine the two to see that almost all of the cohomology groups of the hypersurface $X$ vanish, which we can interpret as saying as we can extend any interesting object of certain degree on it to the whole of the projective space, except in one degree where some interesting things finally happen.

$\endgroup$
  • $\begingroup$ You both have great answers I don’t know what one to choose $\endgroup$ – Prince M Jan 29 at 21:08
  • $\begingroup$ @PrinceM I vote for Georges. :) $\endgroup$ – Gunnar Þór Magnússon Jan 30 at 9:51
  • $\begingroup$ A humble genius! $\endgroup$ – Prince M Jan 30 at 18:35

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.