TITLE: Riemann-Roch for vector bundles
QUESTION [6 upvotes]: The Riemann-Roch theorem is one of the most essential theorems on Riemann Surfaces, or so I am told. I have encountered two formulations for vector bundles (and clearly there are many more), and I am trying to understand why it is that they are equivalent. I would greately appreciate some explanation of this issue.
In either case, we are given a compact Riemann surface $\Sigma$, and and a holomorphic line bundle $E$ over $\Sigma$. The genus of $\Sigma$ is $g$, and degree of $E$ is $d$. We consider the space of holomorphic functions by $H^0(E)$, and take $h^0(E) = \dim H^0(E)$.
Now, the first formulation says that:
$$
h^0(E) - h^0(E^{-1} \otimes K) = d + 1 - g
$$
where $K$ is the canonical bundle, and (I think) in this context the symbol $E^{-1}$ can be taken to mean just $E^*$ (the dual bundle) for reasons having to do with divisors.
Now, for the second formulation, we take $H^1(E)$ to be the quotient space of (the closed $1$-forms with coefficients in $E$) divided by (the exact $1$-forms with coefficients in $E$). We define $h^1(E) = \dim H^1(E)$ and the second version of the theorem says:
$$
h^0(E) - h^1(E) = d + 1 - g
$$
Now, it is far from clear for me why it is that these two formulations are equivalent. Unless I'm getting something seriously wrong here (rather plausible), it should hold that $ h^0(E^{-1} \otimes K) = h^1(E) $, but I would very much appreciate an explanation why it should be so. I would be also grateful for ideas, references and examples concerning computation of the quantities occurring in the Riemann-Roch formula (i.e. starting from a given Riemann surface and ending with (all but one of) $h^0(E), h^1(E), d, g$) [Note: this last question is realted to a homework I have].
Also, remarks on any blunders that I have written above are more than welcome (the only way to get rid of one's misconceptions, I guess).

REPLY [2 votes]: Here is a quite original proof without using Serre's duality. It's left to prove that the dualizing sheaf is the differentials if we work on smooth curves:
Given an algebraic curve $(C,O_C)$ (complete, over $k$) , the idea is to consider the functor $H^1(C, \_)^*$ over the category of quasi-coherent $O_C$-modules. This functor is left exact (because $H^i(C,M)=0$ for $i>1$) and transforms inductive limits in projective limits, so it is representable by a pair $(\omega_C, res)$ where we call $\omega_C$ the dualizing sheaf and $res\in H^1(X, \_)^*$ will be the classical residue when working over $\mathbb{C}$ (and considering the differentials). In other words, for all  quasi-coherent module $M$ we have: $$H^1(X,M)=Hom_{O_X}(M,\omega_X)$$ (where $Hom_{O_X}(\_,\_)$ is the global sections of the homomorphism sheaf).
Then we can prove Riemann-Roch theorem (for singular curves):
Let $C$ be as before, $D=\sum_{i=1}^{k} n_i x_i$ a divisor of non-singular points and $L_D$ the associated line bundle. Then:
$$h^0(L_D)-h^0(\omega_C\otimes_{O_C}L_{-D})= \chi(C,O_C)+ deg(D)$$
where $h^i$ stands for the dimension of the $i$-th cohomology group.
Proof:
By definition we have $$H^1(C, L_D)^*=Hom_{O_C}(L_D,\omega_{C})=Hom_{O_C}(O_C, L_{-D}\otimes_{O_C}\omega_{C})=H^0(X,L_D\otimes_{O_C}\omega_{C})$$ so $h^1(L_D)=h^0(\omega_C\otimes_{O_C}L_{-D})$ and we conclude by weak Riemann-Roch.
Weak Rieman-Roch:(see Hartshorne for instance)
With same notations as in the preceding theorem:
$$  \chi(C,L_D)=\chi(C,O_C)+ deg(D)$$