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I get that the dimension for $\mathscr{H}^q(M)$ is $(2q-1)(g-1)$ for $q > 1$ by the Riemann-Roch theorem. Here $M$ is a compact connected Riemann surface with genus $g > 0$ and $\mathscr{H}^q(M)$ is the vector space consisting all the holomorphic $q$-differential forms, which formally is all the holomorphic sections of the cotangent bundle tensoring itself $q$ times.(The terminologies and notations are from Farkas&Kra's Riemann Surface)

But taking $g=1$, I get that $\mathbb{C}/\Lambda$ for some maximal lattice $\Lambda$ has no nonzero $q$-differential for $q > 1$. I am quite confused about this. Since $\mathrm{d}z$ is kind of holomorphic differential form in $\mathbb{C}/\Lambda$ and $(\mathrm{d}z)^2$ has to be $0$. I cannot get it how the power of a nonzero 1-form can be zero. And it seems kind of a contradiction and makes me hesitate if my proof for the dimension is true because I use an argument that $w^q$ is a nonzero $q$-differential for a non-zero holomorphic 1-form $w$. How can we expalain this result in the $g=1$ cases?

Attach my approach for proving $\mathscr{H}^q(M)=(2q-1)(g-1)$:

Let $w$ be a non-zero holomorphic 1-form on $M$. Then $w^q$ is a non-zero holomorphic $q-$differential.(Is this true?) Denote $D=\mathrm{Div}(w^q)$. Using the fact that $w$ has $2g-2$ zeros(for the proof see: Why does a holomorphic differential has $2g-2$ zeros?), I conclude that $\mathrm{Deg}(D)=2q(g-1)$.

Denote $l(D) = \{f \in \mathcal{M}(M)|\mathrm{div}(f)+D \geq 0\}$, where $\mathcal{M}$ is all the meromorphic functions on $M$. Then by the map

$$ \phi: l(D) \to \mathscr{H}^q(M) \\ f \mapsto f w^q $$

and

$$ \psi: \mathscr{H}^q(M) \to l(D)\\ u \mapsto u/w^q $$

I get a linear isomorphism between $\mathscr{H}^q(M)$ and $l(D)$.

Denote $i(D) = \{w \in \mathcal{R}(M)|\mathrm{div}(w) \geq D\}$. Then $i(D) = \{0\}$ since all holomorphic differential forms has zero number $2g-2 < 2q(g-1)$. Then $ \mathrm{dim}(\mathscr{H}^q(M)) = \mathrm{dim}(l(D)) - \mathrm{dim}(i(D)) = \mathrm{Deg(D)} -g +1=(2q-1)(g-1). $ ∎

onRiv
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    Your inequality $2g-2 < 2q(g-1)$ is not valid when $g=1$. In fact, when $g=1$, $i(D)=1$. – Function Jun 15 '22 at 14:08
  • @Function Ah thanks a lot! You are right. I was not even notice that my problem is in this inequality equation. – onRiv Jun 15 '22 at 14:18
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    You're confusing the tensor power and the exterior power here - when you say $(dz)^2=0$, you're thinking about the exterior power, but your definition of $q$-differential forms involves the tensor power. – KReiser Jun 15 '22 at 14:53
  • @KReiser Thanks. Yeah...at first I was confused for Farkas&Kra's $\mathrm{d}z^q$ notation for $q$-differential forms. But in the question that I wrote $(\mathrm{d}z)^2=0$ is because I thought my proof holds for $g=1$(and then the dimension is $0$, which is wrong). Mr. Function pointed out my mistake. Making such a elementary mistake make me feel some kind of depressed. – onRiv Jun 15 '22 at 17:04

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