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Is there some way of giving a lower bound on the dimension of the Mumford-Tate group of a hypersurface? Let's say it's of general type, say, of degree $10$ inside $ \mathbb{P}^3$.

(Edited from here onward, because I forgot about Fermat hypersurfaces):

I would expect small Mumford-Tate groups to be rare. For example, a Fermat hypersurface has Mumford-Tate group a torus. Is it possible to say, for example, that points with toral Mumford-Tate group are not Zariski-dense?

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For hypersurfaces in $\mathbb P^2$, i.e. curves, this follows from the Andre-Oort conjecture for $\mathcal A_g$, $g=(d-1)(d-2)/2$, as long as $d>4$. The moduli space of hyper surfaces is a sub variety of the moduli space of abelian varieties. By Andre-Oort, if the CM points are dense then it must be a special sub variety, that is, a Shimura variety. But the monodromy group of $H^1$ is full symplectic, so if it is a Shimura variety it is all of $\mathcal A_g$, which only happens for $d=3,4$ by dimension counting.

For higher dimension hyper surfaces I would imagine this follows from an Andre-Oort-like statement, although the moduli space of Hodge structures would not usually be a Shimura variety.

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    $\begingroup$ You probably want $d>4$, because it's still dense in ${\cal A}_g$ for $d=4$. $\endgroup$
    – Noam D. Elkies
    Commented Jul 6, 2015 at 3:31
  • $\begingroup$ Thanks for your answer. I was actually primarily thinking of the case of surfaces and higher dimension precisely because (as you said) it's usually not a Shimura variety, and I'm trying to understand how different that case is to the A-O story. $\endgroup$
    – FriendlyWendy
    Commented Jul 7, 2015 at 4:45
  • $\begingroup$ @FriendlyWendy Yeah, I suspected that the higher-dimensional case was much more interesting but I answered the case I knew. In general, it seems unlikely to me that the CM points can be dense when the degree is large enough, and even more unlikely that one can prove that they are dense. If one instead wants to prove that they are not dense, that naively seems even more difficult than Andre-Oort, which is already quite hard. So you might hope only for a conjectural answer. If this question is as tricky as I think and no one on MO answers it, you might want to ask some experts on Andre-Oort. $\endgroup$
    – Will Sawin
    Commented Jul 7, 2015 at 13:35

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