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Do all the homomorphisms $\phi: SL(2,\mathbb{Z})\ltimes \mathbb{Z}^2 \to GL(2,\mathbb{R})$ always have that $\phi_{|\mathbb{Z}^2}$ is trivial, i.e. $\phi(\mathbb{Z}^2)=I_2$?

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What kind of maps? Group homomorphisms, I guess? – Marc Palm Mar 20 '14 at 16:39
@MarcPalm Yes, it is the group homomorphism. – user119197 Mar 20 '14 at 16:52
A few comments: (1) It's best to edit the opening to read "Do all homomorphisms ..." and change the subscript $n$ at the end to 2. (2) I assume your semidirect product is the usual one, but keep in mind that there are others for which the answer to your question will be different. (3) The question is offbeat but also a little artificial, so it would be worthwhile to indicate some of the motivation. – Jim Humphreys Mar 20 '14 at 23:12
up vote 12 down vote accepted

Yes. Consider $\mathbb R^2$ as a representation of the semidirect product, hence as a representation of $\mathbb Z^2$, and write it as an extension of irreducible characters $\chi_1$, $\chi_2$. If any nontrivial character $\chi$ appears, then all conjugates of that character by automorphisms of $\mathbb Z^2$ in $SL_2(\mathbb Z)$ must appear as well. But each nontrivial character has at least three distinct conjugates (with the minimum achieved by the three nontrivial quadratic characters), so none can appear. Hence the representation of $\mathbb Z^2$ is unipotent, but this just gives a character $\mathbb Z^2 \to \mathbb R^{+}$. Again all conjugates must appear, but there is just one conjugate, so it must be trivial.

So $\mathbb Z^2$ acts trivially.

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