As the question suggests, what is the intuition behind the Kodaira Vanishing Theorem? The Kodaira Vanishing Theorem says that the cohomology groups $H^q(M, L \otimes K_M)$ vanish for $q \ge 1$ when $L$ is a holomoprhic line bundle over a compact complex manifold $M$ and $L$ admits a Hermitian metric whose curvature form is positive definite. Here, $K_M$ denotes the canonical line bundle.

7$\begingroup$ There should not be any exponent "$1$" on the dualizing invertible sheaf. $\endgroup$ – Jason Starr Jul 23 '15 at 17:43

1$\begingroup$ I'm not sure what you are asking. There are many proofs available. I think the easiest way to see that a theorem is true is to read a proof. I personally like Kollár's proof: goo.gl/bzE5Ho $\endgroup$ – Sándor Kovács Jul 23 '15 at 18:32

$\begingroup$ If you want an intuition for a theorem about "$H^q$" of a "line bundle with positivedefinite curvature form", you probably need an intuition for what $H^q$ and that condition on $L$ mean. Is there a standard intuitive/heuristic interpretation of these notions? $\endgroup$ – Noam D. Elkies Jul 23 '15 at 20:12

$\begingroup$ Perhaps stupidly, for curves the statement is clear. So one could think of it along the lines of what can one generalize from the curve case. RR and vanishing allows one to calculate global sections and hence provides motivation for why one would try and prove such a result. $\endgroup$ – meh Jul 23 '15 at 20:22

5$\begingroup$ I think somebody changed the statement of the question. Before, if memory serves, the question asked for a proof. Now the question asks for intuition. If you are going to change the statement, you should correct that exponent $1$. $\endgroup$ – Jason Starr Jul 23 '15 at 20:55
Here are some thoughts. For simplicity I will call a line bundle positive if it admits a metric as you described.
According to Serre's vanishing theorem for any (fixed) coherent sheaf $F$, the sheaf $F\otimes L^m$ has no higher cohomology for $m\gg 0$. So as a first approximation you could consider Kodaira's vanishing theorem as a more precise version of that in the case when $F$ is the canonical line bundle.
It might also help to look at examples.
As Adam points out, the statement is trivial for curves: Using Serre duality the (correct) statement is equivalent to $$ H^q(M, L^{1}) =0 \text{ for } q<\dim M. $$ So if $\dim M=1$, then this only says that $H^0(M, L^{1}) =0$ (which is obvious).
One can also look at the cohomology of line bundles on a projective space. In that case the canonical line bundle is just $\mathscr O_{\mathbb P^n}(n1)$ and that's exactly so to speak the "breaking point" where the cohomology groups start behaving differently.
As an alternative, one could say that Kodaira's vanishing theorem is a vast generalization of the behaviour of the cohomology of line bundles on projective spaces.
Next, as an exercise, you should do the same comparison for complete intersections. You will find two things:
 The pattern is similar, the behaviour of cohomology groups changes when you hit the canonical line bundle and what matters whether your line bundle is more or less positive than the canonical.
 You will discover that the canonical line bundle itself tends to be positive. A simple form of Serre's vanishing says that a high power of a positive line bundle has no positive cohomology. Kodaira's vanishing is again a more precise version of that; measuring by the canonical line bundle it tells how high a power you need. In particular, if the canonical line bundle itself is positive, then Kodaira says that Serre vanishing hold for this particular line bundle already from the second power (and by Serre duality you know that it cannot hold for the first power). Given the importance of the canonical line bundle this is pretty useful.
It is well known that a cohomology class can be uniquely expressed by a harmonic form. To show that a certain cohomology group vanishes, it is equivalent to show that every harmonic form form that cohomology group is identically zero.
How does one show that a harmonic form vanishes? Let us look at the case of a realvalued harmonic function. We know that on a compact manifold, every realvalued harmonic function is constant. There are two ways of showing this. One is to use the maximum principle, and another way is to use integration by parts. Instead of considering a harmonic function, we consider a realvalued function $f$ which satisfies the equation$$\Delta f = cf,$$for some positive constant $c$. Such a function $f$ must vanish identically. Here, one can use either the maximum principle or integration by parts.
For the maximum principle, we consider a point where the maximum of $f$ is achieved. At that point, $\Delta f$ is nonpositive. Because of the positivity of $c$, it forces $f$ to be nonpositive. So $f$ is nonpositive everywhere. Applying the same argument to $f$ instead of $f$, we conclude that $f$ is nonnegative everywhere. Hence, $f$ is identically zero.
For integration by parts, we simply multiply$$\Delta f = cf$$ by $f$, integrate over $M$, and get$$(df, df)_M = c(f, f)_M,$$ which forces $f$ to be identically zero by the positivity of $c$. The argument of integration by parts is the same as saying the operator $\Delta + c$ is positive definite.
Let us go back to the consideration of harmonic forms $\varphi$. Instead of $\Delta$, we have the operator$$\square = \overline{\partial}^* \overline{\partial} + \overline{\partial}\overline{\partial}^*.$$When we compute $\square \varphi$, we will get$$\square\varphi = (\text{Tr}\, \nabla \overline{\nabla} \varphi) + A\varphi,$$ where $A$ is a zeroorder linear operator. The term $(\text{Tr}\,\nabla \overline{\nabla}\varphi)$ corresponds to $\Delta f$, and the term $A\varphi$ corresponds to $c \varphi$. When the curvature form of $L$ is positive, the operator $A$ is a positive operator, it corresponds to the positivity of $c$, and we get the vanishing of the harmonic form $\varphi$.
In order to rigorously prove the Kodaira Vanishing Theorem, we would derive the formula for $\square\varphi$.

$\begingroup$ Thank you for explaining this! I read through that proof pretty carefully in order to teach it a few years ago, and didn't come up with nearly as good a way to summarize it. (See math.lsa.umich.edu/~speyer/632Old/apr12.pdf if you are curious.) One thing I do find helpful is to point out the stronger result $H^q(X, L \otimes \Omega^p)=0$, for $p+q>n$. That makes the $(p,q)$ symmetry much more clear. $\endgroup$ – David E Speyer Jul 24 '15 at 13:18
Let me address the modified question (as corrected by Jason). There are many intuitions depending on which proofstyle you want to use. Here are a few remarks.
The original proof of Kodaira uses the curvature properties directly to show that there are no nonzero harmonic forms representing elements of $H^q(X, K_X\otimes L)$, when $q>0$ and $L$ is positive. In some sense, this is an adaptation of an argument of Bochner that shows that certain Betti numbers vanish for positively curved Riemannian manifolds (like the sphere). There is a slightly slicker proof  again using harmonic forms  of a somewhat stronger result due to Akizuki and Nakano. This is the one you find in many textbooks such as Griffiths and Harris. I think this is quite readable, but too long to explain here. One consequence of Kodaira's vanishing, as originally stated, is to prove that a line bundle is positive iff it is ample. Most algebraic geometers prefer to make this substitution in the statement, and I will do it as well below.
The AkizukiKodairaNakano vanishing implies the Lefschetz hyperplane theorem. Ramanujam turned this around to show that Lefschetz theorem together with the Hodge decomposition actually implies AKNvanising. There are a bunch of other "topological" proofs to due EsnaultViehweg and Kollár. Personally, I find these more insightful than the original arguments, but it is really a matter of taste.
There is now a purely algebraic proof due to DeligneIllusieRaynaud. The idea is to use a bootstrapping argument. If you can show that $\dim H^q(X, K\otimes L)\le \dim H^q(X, K\otimes L^N)$ for $N\gg 0$. You get Kodaira from Serre vanishing. In order to actually get such an inequality, you have to use certain tricks with the Frobenius (i.e. you work in characteristic $p>0$). For example if it happens that after reducing mod $p$, the map $O_X\to F_*O_X$ splits, then the above inequality holds for all $N=p^n$ by the projection formula. Unfortunately, this strategy usually fails for general $X$, but Deligne and Illusie use a more subtle splitting argument which works for all smooth $X$.

$\begingroup$ One nice variant of the argument you talk about in the last paragraph works for toric varieties in characteristic zero. They have something analogous to a Frobenius (indeed, a whole family of Frobenii) for which the map $O_X \rightarrow F_* O_X$ does indeed split. $\endgroup$ – potentially dense Jul 24 '15 at 12:04