Why does rigidity hold only if rank >1?

Question

In simple words, why does Margulis' superrigidity and arithemiticit only hold for lattices in Lie groups of rank $>1$? E.g. what is the reason for it to fail for $SL(2,R)$?

Answer 1

It is easy to exhibit counterexamples for $SL(2,\mathbb R)$. E.g. there are non-constant families of curves of genus $>1$, such as $$y^2 = x(x-1)(x-2)(x-a)(x-b)(x-c)$$ (just to write down an example for genus two).

By the uniformization theorem, these give rise to a non-constant family of conjugacy classes of cocompact lattices in $SL(2,\mathbb R)$. In particular, since there are only countably many conjugacy classes of arithmetic groups in $SL(2,\mathbb R)$, most of these (uncountably many) conjugacy classes must consist of non-arithmetic lattices.

One way to summarize this answer is to say that curves are not rigid. Since curves are not rigid, and they correspond to conjugacy classes of cocompact lattices in $SL(2,\mathbb R)$ by uniformization, it must be that cocompact lattices in $Sl(2,\mathbb R)$ are not rigid either.

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By considering punctured curves, we can similarly show that finite covolume lattices in $SL(2,\mathbb R)$ need not be rigid either.