Mumford-Tate groups of products of Hodge structures Let $V_1$, $V_2$ be two polarised simple $Q$-Hodge structures which are non-isomorphic.
I am assuming that the Mumford-Tate groups of $V_1$ and $V_2$ are semi-simple adjoint.
Is it true in this case that $MT(V_1 \times V_2) = MT(V_1) \times MT(V_2)$?
(I can easily see that this is not true when $MT(V_1)$ and $MT(V_2)$ are tori !)
 A: The answer is no. Namely, let $E$ be an elliptic curve without complex multiplication and $V=H^1(E,Q)$ its first rational cohomology group, which is a 2-dim'l rational Hodge structure of weight 1. Its Hodge (resp. Mumford-Tate) group is $SL(V) \cong SL(2)$ (resp. $GL(V)$). Let $m$ be a positive  integer and $V_m=Sym^{2m}(V)$ be the $2m$-th symmetric power of $V$, which is the (absolutely) irreducible rational Hodge structure of weight $2m$ with Hodge group $PSL(V)\cong PSL_2$. (The irreducibility follows from the representation theory of $SL_2$.) Now its twist $\tilde{V}_m=V_m(-m)$ is the irreducible rational Hodge structure of weight 0 with ``the same"  (adjoint) Hodge/Mumford-Tate group  $PSL(V)$; the Hodge structures  $\tilde{V}_m$ are not isomorphic for different $m$, because they have different dimension, namely, $1+2m$.  Now take two distinct positive integers $m$ and $n$. Then $MT(\tilde{V}_m \times \tilde{V_n})$ is still $PSL(V)$, which is strictly less than $PSL(V)\times PSL(V)$. This provides a desired counterexample.
