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).