Importance of Vanishing Cohomology

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.

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 $i$.