I'm reading this. The construction is as follows:
Let $E$ be an elliptic curve, $P$ be a positive degree invertible sheaf , and $N$ be a non-torsion degree 0 invertible sheaf.
First claim : $H^0(E, N^m \otimes P^n)$ is nonzero iff $n > 0$ or $m=n=0$. Is this direct sum or tensor product? That is, does $N^m$ mean $N \oplus N … \oplus N$ or $N \otimes N … \otimes N$? I'll assume it's the tensor product for what comes below:
I also want to know why the first claim is true. By Riemann Roch, and that the sheaf of differentials is isomorphic to the structure sheaf on an elliptic curve, we have $h^0(L) – h^0(L^\vee) = deg(L) + 1 – 1$. If $L$ has positive degree then $h^0(L) = deg(L)$.
But if $n = 0$, but $m \neq 0$, then why is $h^0(N^m) = 0$ ? I think that if it's not zero, then a global section can be used to trivialize $N^m$ (using that $N^m$ is degree 0), contradicting that $N$ is not torsion. But I'm not sure how to make this precise.
Second claim : $R = \oplus_{m,n \geq 0} H^0(E, N^m \otimes P^n)$ is a ring. How is multiplication defined on this ring?
Third claim : Let $X$ be the total space of $N \oplus P$. I think this means $\underline{Spec}_E(Sym(N \oplus P))$. Then the global sections of $X$ is $R$. Why? I don't know how to compute the global sections of a relative spec. The definition is so abstract : define a functor and then show that it's representable.
Best Answer
The $ m $ and $ n $ in the index mean tensor product obviously, for the reason outlined by Mariano Suarez-Alvarez.
(1) If $ n > 0 $ then $ L = N^m \otimes P^n $ has positive degree and therefore its dual has no sections so by Riemann-Roch and the fact that the canonical bundle of $ E $ is trivial, we get $ h^1(E,L) = h^0(E, L^{\vee}) = 0 $ and $ h^0(E, L) = \deg L - 1 + 1 = n > 0 $, proving one half. For the second half, if $ n = 0 $ and $ m \neq 0 $ we get by Riemann-Roch again that $ h^0(E, N^m) = h^0(E, (N^m)^{\vee} ) $ and if both sides are positive, the bundle $ N^m $ is trivial i.e. $ N $ is torsion, a contradiction.
(2) There is obviously a map on global sections $$ H^0(E, N^m \otimes P^n ) \times H^0(E, N^a \otimes P^b) \rightarrow H^0(E, N^{m+a} \otimes P^{n+b} ) $$ just by tensor product of the underlying quasicoherent sheaves, in this case line bundles.
(3) The total space $ T $ of $ N \oplus P $ is $ \operatorname{Spec}_E( \operatorname{Sym}^{\bullet} (N \oplus P)^{\vee} ) $ for functorial reasons. However I think it is your space $ X $ that we want for us to get the correct ring, with no duals. Then giving a global section of $ X $ just amounts to giving a morphism $ X \rightarrow \mathbb{A}^1 \times E = \operatorname{Spec}_E ( \mathcal{O}_E[x] )$ over $ E $, which is just to give a morphism of the associated sheaves of $ \mathcal{O}_E $-algebras $ \mathcal{O}_E[x] \rightarrow \operatorname{Sym}^{\bullet} (N \oplus P) $ and such a morphism is completely determined by wherever $ x $ maps, so the morphisms are in one-to-one correspondence with global sections $$ H^0(E, \operatorname{Sym}^{\bullet} (N \oplus P) ) = \oplus_{m,n \ge 0} H^0(E, N^m \otimes P^n ) = R $$ as desired.