Let $X$ be a projective algebraic variety over a algebraic closed field $k$. Let $\mathcal{F}$ be a coherent sheaf on $X$. We know that $H^0(X, \mathcal{F})$ is the vector space of global sections of $\mathcal{F}$. This gives us a geometric illustration of $H^0$. For example, let $I\_D$ be the ideal sheaf of a hypersurface $D$ of degree $>1$ in a projective space $\mathbb{P}^n$, then it is easy to see that
$$H^0(\mathbb{P}^n,I\_D\otimes\mathcal{O}\_{\mathbb{P}^n}(1))=0.$$ 


In fact, there is no hyperplane containing $D$, which means that there is no global section of $\mathcal{O}\_{\mathbb{P}^n}(1)$, which are hyperplanes, containing $D$. Hence $H^0(\mathbb{P}^n, I\_D\otimes\mathcal{O}\_{\mathbb{P}^n}(1))=0$. However, my first question is how to understand higher cohomologies of sheaves in geometric ways. The following questions then come out:


1) How to understand Serre's vanishing theorem, i.e., is there a geometric way to think about the vanishing of $H^q(X, \mathcal{F}\otimes A^n)$ for $n\gg1$, where $\mathcal{F}$ is coherent and $A$ is ample.


2) How to understan Kodaira's vanishing theorem geometrically.


Maybe a concrete question will help, say $D$ a subvariety of $\mathbb{P}^n$, how to determine geometrically whether $H^0(\mathbb{P}^n, I\_D\otimes\mathcal{O}\_{\mathbb{P}^n}(1))$ vanishes or not.



Geometry meaning of higher cohomology of sheaves?
Let's start way back. The invention of schemes moved algebraic geometry away from thinking about varieties as embedded objects. However, embedding an abstract scheme into projective space has a lot of advantages, so if we can do that, it's useful. And even if we cannot embed our scheme into projective space, but we can find a non-trivial map, that gives us some way to understand our abstract scheme. In order to find a non-trivial map we need a line bundle with sections.


So we are interested in finding sections of various sheaves, but primarily line bundles (and of course for this sometimes we need to deal with other kind of sheaves). Thus we are interested in $H^0$. 


On the other hand, computing $H^0$ is non-trivial. There are no good general methods. One reason for this is that, for instance, $H^0$ is not constant in families, or put it another way it is not deformation invariant. On the other hand, $\chi(X,\mathscr F)$ behaves much better. It is constant in flat families and if $\mathscr F$ is a line bundle, then it is computable using Riemann-Roch. 


Then, if we know that $H^i=0$ for $i>0$, then $H^0=\chi$ and we're good.




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Here is an explicit example for a typical use of Serre vanishing:


**Example 1** Suppose $X$ is a smooth projective variety and $\mathscr L$ is an ample line bundle on $X$. 
Then we know that $\mathscr L^{\otimes n}$ is very ample and $H^i(X, \mathscr L^{\otimes n})=0$ for $i>0$ and $n\gg 0$. Then $\mathscr L^{\otimes n}$ induces an embedding $X\hookrightarrow \mathbb P^N$
where $N=\dim H^0(X,\mathscr L^{\otimes n})-1=\chi(X,\mathscr L^{\otimes n})-1$ by Serre's vanishing and hence $N$ is now computable by Riemann-Roch. 


The only shortcoming of the above is that in general there is no way to tell what $n\gg0$ really means and so it is hard to get any explicit numerical estimates out of this. This is where Kodaira vanishing can help.


**Example 2** In addition to the above assume that $\mathscr L=\omega\_X$, or in other words assume that $X$ is a smooth canonically polarized projective variety. There are many of these, for instance all smooth projective curves of genus at least $2$ or all hypersurfaces satisfying $\deg > \dim +2$. 
In particular, these are those of which we like to have a moduli space.
Anyway, the way Kodaira vanishing changes the above computation is that now we know that already $H^i(X,\omega\_X^{\otimes n})=0$ for $i>0$ and $n>1$! In other words, as soon as we know that $\omega\_X^{\otimes n}$ is very ample and $n>1$, then we can compute the dimension of the projective space into which we can embed our canonically polarized varieties. In fact, perhaps more importantly than that we can compute it, we know (from the above) without computation that this value is constant in families.
So, once we have a boundedness result that says that this happens for any $n\geq n\_0$ for a given $n\_0$, and Matsusaka's Big Theorem says exactly that, then we know that all such canonically polarized smooth projective varieties (with a fixed Hilbert polynomial) can be embedded into $\mathbb P^N$, that is, into the **same** space. 


This implies that then all of these varieties show up in the appropriate Hilbert scheme of $\mathbb P^N$ and we're on our way to construct our moduli space. 


Of course, there is a lot more to do to finish the whole construction and also this method works in other situations, so this is just an example.




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There is one more thing one might think regarding your question, that is, ask the more abstract question:


"What does higher cohomology of sheaves mean (e.g., geometrically)?"


This is arguable, but I think that the essence of higher cohomology is that it measures the failure of something we wish were true all the time, but isn't. More specifically, if you're given a short exact sequence of sheaves on $X$
$$
0\to \mathscr F' \to \mathscr F \to \mathscr F'' \to 0
$$
then we know that even though $\mathscr F \to \mathscr F''$ is surjective, the induced map on the global sections $H^0(X,\mathscr F) \to H^0(X,\mathscr F'')$ is not. However, the vanishing of $H^1(X,\mathscr F')$ implies that for any surjective map of sheaves with kernel $\mathscr F'$ as above the induced map on global sections is also surjective. Since you already have a geometric interpretation of $H^0$, this gives one for $H^1$: it measures (or more precisel


For a more detailed explanation of the same idea see [this MO answer](https://mathoverflow.net/questions/125200/coboundaries-and-gluing-in-cech-cohomology-intuition/125236#125236). 


In my opinion the best way to understand higher ($>1$) cohomology is that it is the lower cohomology of syzygies. In other words, consider a sheaf $\mathscr F$ and embed it into an acyclic (e.g., flasque or injective or flabby or soft) sheaf. So you get a short exact sequence:
$$
0\to \mathscr F\to \mathscr A\to \mathscr G \to 0 
$$
Since $\mathscr A$ is acyclic, we have that for $i>0$
$$
H^{i+1}(X,\mathscr F)\simeq H^i(X,\mathscr G),
$$
so if you understand what $H^1$ means, then $H^2$ of $\mathscr F$ is just $H^1$ of $\mathscr G$, 
$H^3$ of $\mathscr F$ is just $H^2$ of $\mathscr G$ and so on. 


