# Cohomology--What Is Going on with the Geometry?

This question is pretty open-ended, so I'm expecting that some intuitions and potentially even straight up philosophies of (Cech) cohomology will be popping up here. We've started doing cohomology in my alg. geo. class (and I've been studying the stuff solo for a little while now), and we computed $H^0(X,\mathcal{F}) = \mathcal{F}(X)$, which is all well and good. So the $0^{\text{th}}$ cohomology group of an arbitrary sheaf of Abelian groups on a scheme (or, I imagine, a topological space) $X$ is just the group of global sections of that sheaf. Neat.

I have also seen (and proved for myself) the familiar fact that $H^1(X,\mathscr{O}_X^*)\cong \text{Pic}(X)$ where $\text{Pic}(X)$ denotes the set of isomorphism classes of line bundles on $X$. (Again, $X$ here is an arbitrary scheme.) So we can identify isomorphism classes of line bundles, isomorphism classes of divisors, and the first cohomology group of the sheaf of nowhere vanishing regular functions on $X$. Also cool; there is some geometry happening there with line bundles and such.

My question, then, is what other (classical) geometric information is contained in cohomology?

For instance, my prof is having us compute $H^1(\mathbb{P}_A^1,\mathscr{O}_{\mathbb{P}_A^1}(n))$ for an arbitrary commutative ring with unity, $A$. Again, I can understand how this cohomology group contains the idea of "how easy it is to pick local sections on an (the canonical) affine cover of $\mathbb{P}_A^1$ which don't glue to a global section on the entire space." Is there anything else latent in this computation which I'm not seeing because of just trying to do all the algebra?

Update Jul, 2018: I'm still hoping for more answers and ideas about this question, maybe some things from category theory/homological algebra, or the relevance of cohomology particularly in arithmetic geometry and maybe some relationship with Galois groups, etc.

• For your specific example you could in principle notice Serre duality and try to make sense of it. I vaguely recall that there is a pretty geometric way to go about Serre duality in this example, but I don't remember. I can look it up if you are curious. If you take a variety over $C$ with the euclidean topology, and take the constant sheaf, then sheaf cohomology is usual singular cohomology. You (using GAGA) can connect this to your example for $Pic$ using the exponential short exact sequence. This leads to Chern classes, which measure twistyness of line bundles. So that's pretty geometric. – Lorenzo Apr 23 '17 at 7:07

General philosophy about $H^1$ : $\newcommand{\O}{\mathcal O}$ These cohomology groups are fully characterized by the usual abstract nonsense as "obstruction space", (i.e the long exact sequence) and I think this is a good way of interpret them. As an example, if $X$ is a differentiable manifold, then $H^1(X, \Bbb Z/2 \Bbb Z) = 0$ if and only if $X$ is orientable.

In general, $H^1$ can be seen as a way of classify isomorphisms class of "fiber bundle" which are "locally trivial". In the previous example, the bundle was the orientation cover, which is trivial if and only if the manifold is orientable. Notice that manifolds are locally orientable as they are locally diffeomorphic to $\Bbb R^n$. So being non-orientable is a global property and measured by cohomology.

Another example is from deformation theory : fix $\mathscr{F,G}$ line bundles on a variety $X$. Then, isomorphismes classes of exact sequence on the form $0 \to \mathscr F \to \mathscr H \to \mathscr G \to 0$ are classified by $H^1(\O_X, \mathscr F \otimes \mathscr G^{\vee})$. For example $H^1(\O_X, \O(-2)) = \Bbb C$ on $X = \Bbb P^1$. This means that there should be a non-split exact sequence $$0 \to \O(-1) \to \mathscr H \to \O(1) \to 0$$ and indeed by chance the Euler sequence is non-split : $$0 \to \O(-1) \to \O^2 \to \O(1) \to 0$$ Notice that locally sections always exists so one can locally write a splitting, and again we find the principle that $H^1$ classify "isomorphism class of object which are locally isomorphic to something". More about this principle can be found in the last chapter of the book by Miranda, including this example.

A more concrete example : Consider $X = p_1 \cup \dots \cup p_k$ be a collection of points in $\Bbb P^2$. What is the dimension of the space of conics passing through these points ? I claim that this is given by $r = \dim H^0(I_X(2)) - 1$, since more or less by definition $I_X(2)$ is the vector space of polynomials of degree $2$ which vanishes on $X$, and I added $-1$ because two proportional polynomials define the same curve. How to compute $H^0(I_X(2))$ ? We have the exact sequence of sheaves $0 \to I_X \to \O \to \O_X \to 0$. Tensoring by $\O(2)$ and taking the long exact sequence gives us the following exact sequence $$0 \to I_X(2) \to \O(2) \to \O_X(2) \to H^1(I_X(2)) \to 0$$ which means precisely that $r = h^0(I_X(2)) - 1= h^1(I_X(2)) - k + 5$.

How to interpret this ? $5$ is the dimension of the space of plane conics, and $-k$ means that every point will reduce this space by $1$ which seems clear. Where does appears this mysterious $h^1(I_X(2))$ ? The map $\O(2) \to \O_X(2)$ simply evaluate a polynomial of degree $2$ simultaneously in $p_1, \dots, p_k$. When is it surjective ? We want a quadric polynomial $f \in \O(2)$ such that $f(p_i) = \lambda_i$, since $\O_X(2) \cong \Bbb C^k$, as $\O_{pt}(d) \cong \Bbb C$ for any $d \geq 0$. These equations $f(p_i) = \lambda_i$ will determine equations in the space of conics $\Bbb P^5$. If the space of solutions of these equations has at least dimension $0$ we'll have a solution. As an example, if $p_1, \dots, p_6$ lies on the same conic, then there is no polynomial $f$ of degree $2$ with $f(p_1, \dots, p_6) = (1,0,0, \dots, 0)$. But what object is here for measure the failure of surjectivity $\O(2) \to \O_X(2)$ ? Well, this is exactly $H^1(I_X(2))$ ! Here we are : generically, we expect that this map is surjective for a small number of points, but of course "accident" like linear dependance or quadratic dependance can happen, and there are exactly detected by the first cohomology group $H^1(I_X(2))$.

In the same spirit, if $k=3$, we have $H^1(I_X(1)) = \Bbb C$ if $p_1,p_2,p_3$ are collinear and $0$ else.

• N.H. are you just going through all my posts? :P – Tanner Strunk May 12 '17 at 15:44
• I'm upvoting this but not going to check it, because I want to see what other people say. With a somewhat open-ended, philosophical question like this, I feel it's good to get all the answers possible and just see what people think and experience in their own work! :D – Tanner Strunk May 12 '17 at 16:02
• @TannerStrunk : I just looked to this question and though it was interesting to put my answer (and my class was a bit boring this morning :D ). I understand, you better have to wait, there are many things to say. – user171326 May 12 '17 at 17:44