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). $ ∎