It is well known that the Mumford-Tate group of a polarizable pure $Q$-Hodge structure is reductive. (this is proved for instance in Deligne et al LNM 900, Voisin's books on Algebraic Geometry,..)

Question: Is there an example of a non-polarizable $Q$-Hodge structure whose Mumford-Tate group is reductive?



1 Answer 1


Indeed. Look at Hodge structures as finite dimensional vector spaces $V$ over $\mathbb Q$ together with an action of the Deligne torus $\mathbb S$ on $V \otimes \mathbb C$, i.e. a morphism of algebraic groups $\rho: \mathbb S \to \mathrm{GL}(V \otimes \mathbb C)$. The Mumford--Tate group of $V$ is then just the Zariski closure over $\mathbb Q$ (!) of the imagr of $\rho$.

Now take $V$ of dimension $\geq 4$ and a representation $\rho: \mathbb S \to \mathrm{GL}(V\otimes \mathbb C)$ of pure weight $k$ say, such that the Zariski closure over $\mathbb Q$ of the image of $\rho$ is $\mathrm{GL}(V)$. So the resulting Mumford--Tate group is $\mathrm{GL}(V)$ which is reductive. But the Hodge--structure $V$ is not polarisable, indeed, any polarisation $\psi: V \otimes V \to \mathbb Q(k)$ would confine the Mumford--Tate group to the Goup of symplectic similitudes $\mathrm{GSp}(V,\psi)$ which is strictly smaller than $\mathrm{GL}(V)$.

  • $\begingroup$ Cool! I understand the idea, but could you still add some details on how to arrange for the Zariski closure of the image of $\rho$ to be all of GL(V)? Thanks! $\endgroup$
    – SGP
    Apr 13, 2011 at 23:36

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