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Let $(X,L)$ be a polarized abelian variety over $k=\overline{k}$, and let $K(L)$ be the kernel of the isogeny $X\to X^\vee$ that sends $x$ to $t_x^*L\otimes L^{-1}$. The theta group $\mathscr{G}(L)$ of $(X,L)$ is a central extension $$0\to k^\times\to\mathscr{G}(L)\to K(L)\to0.$$ Now, $\mathscr{G}(L)$ acts on $H^0(X,L)$, and we get an irreducible representation $$\mathscr{G}(L)\to \mbox{GL}(H^0(X,L)),$$ which can be seen to descend to a projective representation $K(L)\to\mbox{PGL}(h^0(L)-1,k)$. The nice thing about this projective representation is that, if $\varphi:X\to\mathbb{P}^{h^0(L)-1}$ is the map associated to $|L|$, then for all $z\in K(L)$ and $x\in X$, we have that $z\cdot\varphi(x)=\varphi(x+z)$. Using this, it is easily verified that $\mbox{span}\{\varphi(K(L))\}$ is all $\mathbb{P}^{h^0(L)-1}$.

We have that $K(L)=K_1\oplus K_2$, where $K_2$ is the symplectic complement to $K_1$ using the Weil pairing.

My question is the following: Is it true that the projective representation restricted to $K_1$ (or $K_2$) is also irreducible? Put into other terms, is the linear span of $\varphi(K_1)$ all of $\mathbb{P}^{h^0(L)-1}$?

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up vote 3 down vote accepted

I think the answer is no due to the following counterexample (there might be a mistake somewhere though):

Take an elliptic curve $X$ with $L = O_X(3P_0)$ (characteristic $\neq 2, 3$), then $K$ will be the $3$-torsion subgroup and $K_1$ will be generated by a $3$-torsion point $P$. Then in the embedding by $|L|$ of $X$ in $\mathbb{P}^2$, the points $P$, $-P$ and $P_0$ are collinear, that is, $K_1$ is mapped to a lower-dimensional linear subspace.

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Yes, you're right. Great example. Thanks! –  Robert Auffarth Feb 8 '13 at 4:09
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If I understand correctly, $K_1$ and $K_2$ are (mutually orthogonal maximal) isotropic subgroups of $K(L)$. Therefore they both can be lifted (non-canonically) to (finite) commutative (sic!) subgroups of the theta group. In particular, the restriction of the projective representation of $K(L)$ to $K_i$ is actually a projectivization of a certain linear representation of finite commutative $K_i$ in $H^0(X,L)$. Since a linear representation of such groups is irreducible if and only if it is one-dimensional, we get a negative answer to your question (unless $H^0(X,L)$ has dimension 1, i.e. $L$ defines a principal polarization).

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Now, I'm not asking necessarily when the linear representation of $K_i$ es irreducible (since by what you said, $L$ would have to be a principal polarization), but when the projective representation is irreducible; i.e., the group doesn't fix any projective linear space of $\mathbb{P}H^0(L)$... –  Robert Auffarth Feb 10 '13 at 4:04
    
The answer is the same: iff the polarization is principal. If it is not principal then one may take the projectivization $P(W)$ of any proper nonzero $K_i$-invariant vector subspace $W$ of $H^0(X,L)$; this $P(W)$ is $K_i$-stable. –  Yuri Zarhin Feb 15 '13 at 17:07
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