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Let $E$ be a smooth elliptic curve over algebraically closed field $k$ of characteristic zero, $\mathcal{L}$ is a line bundle over $E$, $\operatorname{deg}(\mathcal{L})=n \geq 1$. Then I define the group of translations of $\mathcal{L}$ as $H(\mathcal{L})=\{x \in E: T_x \mathcal{L} \cong \mathcal{L}\}$ and theta group (or Heisenberg group) $G(\mathcal{L})$ of line bundle $\mathcal{L}$ as central extension $$ 0 \to k^* \to G(\mathcal{L}) \to H(\mathcal{L}) \to 0. $$ Group $G(\mathcal{L})$ acts on $H^0(E, \mathcal{L})$ in an obvious way and provide an irreducible representation.

Sometimes another version of Heisenberg group is used for elliptic curves. I denote this group $\Gamma_n$. As abstract group $\Gamma_n$ has generators $\epsilon_1$, $\epsilon_2$ and $\delta$ with relations $\epsilon_1^n=\epsilon_2^n=\delta^n=1$, $\epsilon_1 \epsilon_2 =\delta \epsilon_2 \epsilon_1$. In other words it is a central extension of the group $H(\mathcal{L})$ (generated by $\epsilon_1$ and $\epsilon_2$) by group of roots of unity in $k$ (generated by $\delta$). First of all, in order to get an action on $H^0(E, \mathcal{L})$ I need to see that translation by $x \in H(\mathcal{L})$ induce an automorphism of $\mathcal{L}$ given by some root of unity. How to show this?

Where I can find a proof that $\Gamma_n$ acts irreducibly on $H^0(E, \mathcal{L})$? Is it true for higher rank vector bundles over $E$?

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    $\begingroup$ Look at Th. 2 in Mumford's 'On the equations defining abelian varieties. I' (Invent. Math. 1, 287--354 (1966)). See also Mumford's book 'Tata lectures on theta I'. These references are for abelian varieties in general - I don't know of a reference where the elliptic curve case is treated directly. I am not sure that I understand your question well but I have never heard of an extension of the theory of the Heisenberg and theta groups for vector bundles of higher rank on abelian varieties. $\endgroup$ – Damian Rössler Jan 4 '14 at 23:01
  • $\begingroup$ +1, but it seems better to rephrase a question as: how to describe geometrically the action of the Heisenberg group on section of line bundel ? Otherwise, for the question like it is phrased now - the answer is trivial - finite Heisenberg group has n - dimensional irrep, space of sections is n-dimensional - so just choose any isomorphism of vector spaces and get the irrep. $\endgroup$ – Alexander Chervov Jan 5 '14 at 13:00
  • $\begingroup$ The n-dim irrep of Heisenberg is easy to construct: choose a basis v_1...v_n, let e_1 act as a cyclic SHIFT: v_i->v_{i+1}, and e_2 acts on v_l as multiplication by exp(2pi i l /n), and delta is scalar operator acting as multiplication on exp(2pi i /n). $\endgroup$ – Alexander Chervov Jan 5 '14 at 13:05
  • $\begingroup$ @Alexander Chervov Yes, I reformulated my question in order to stress what I really want to understand. $\endgroup$ – Sasha Pavlov Jan 6 '14 at 21:23
  • $\begingroup$ Any element of $H(\mathcal L)$ with order $n$ can be lifted to an element of $G(\mathcal L)$ with order $n$. because it can be lifted to an arbitrary element, whose $n$th power is in $k^*$, and then we divide by an $n$th root of that element of $k^*$. Then if we lift the two generators to have order $n$, their commutator must have order $n$, because $[g,h]^n=[g,h^n]=[g,1]=1$. So this is why you get an action of the smaller Heisenberg group. $\endgroup$ – Will Sawin Jan 6 '14 at 21:40

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