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Fix $n\in \mathbb N$ and a partition $\lambda$ with at most $n-1$ parts (of length at most $n-1$). Let $V$ be the irreducible $GL_n \mathbb R$-representation with highest weight $\lambda$ and $D$ the determinant representation.

Is it possible that $$ Res_{GL_n \mathbb Z} V \cong Res_{GL_n \mathbb Z} (V \otimes D)?$$

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  • $\begingroup$ What is the situation if you stay with $GL_n(\mathbb{R})$ and do not restrict down to $GL_n(\mathbb{Z})$? $\endgroup$ – Vincent Aug 8 '16 at 9:55
  • $\begingroup$ $GL_n(\mathbb{Z})$ may be replaced by the algebraic subgroup $H$ of $GL_n(\mathbb{R})$ with determinant $\pm 1$. In which case, whether the determinant on $V(\lambda) $ is $\pm 1$ on $H$ plays a role. $\endgroup$ – Venkataramana Aug 8 '16 at 10:14
  • $\begingroup$ @Venkataramana Are you saying it is not possible? If yes, is there some kind of reference for this? (You could put this in an answer and I'll except it ;) ) $\endgroup$ – Peter Patzt Aug 8 '16 at 10:23
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    $\begingroup$ @anton: $GL_n(\mathbb{Z})$ is NOT Zariski dense in $GL_n$. The Zariski closure is precisely the subgroup of matrices in $GL_n(\mathbb{R})$ whose determinant is $\pm 1$. $\endgroup$ – Venkataramana Aug 8 '16 at 11:50
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    $\begingroup$ Wait no ... that is obvious because the zariski closure of $SL_n \mathbb Z$ is $\SL_n \mathbb R$. It is clear to me, why the isomorphism is not possible if $\dim V$ is odd. I think I can even prove that $Res V \not\cong Res V \otimes D$ if $V$ is the 2-dimensional defining representation of $GL_2 \mathbb R$. But I don't have a general argument. $\endgroup$ – Peter Patzt Aug 8 '16 at 12:34
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First of all, if $V$ is not irreducible, this can happen. Every matrix in $GL_n(\mathbb{Z})$ has determinant $\pm 1$ so $D^{\otimes 2}|_{GL_n(\mathbb{Z})}$ is trivial. Therefore, $(1 \oplus D) \cong (1 \oplus D) \otimes D$ on $GL_n(\mathbb{Z})$.

However, with the stated hypothesis that $V$ is the irrep of $GL_n$ with highest weight $\lambda$, this is true. Note that the Schur polynomial $s_{\lambda}(x_1, x_2, \ldots, x_n)$ is homogenous of degree $|\lambda|$. Therefore, it cannot be divisible by the inhomogoneous polynomial $1+x_1 x_2 \cdots x_n$. Write $s_{\lambda}$ as a polynomial in the elementary symmetric polynomials $e_1$, $e_2$, ..., $e_n$, we have just shown that it is not divisible by the polynomial $1+e_n$. Therefore, we can find integers $(f_1, f_2, \ldots, f_n)$ with $f_n=-1$ so that $s_{\lambda}$ does not vanish when the $f_j$ are plugged in for the $e_j$.

Let $g \in GL_n(\mathbb{Z})$ be a matrix with characteristic polynomial $x^n - f_1 x^{n-1} + f_2 x^{n-2} - \cdots + (-1)^n f_n$ (for example the companion matrix), and let $\alpha_1$, ..., $\alpha_n$ be its roots. Then $\mathrm{Tr} \ \rho_{V}(g) = s_{\lambda}(\alpha_1,\ldots,\alpha_n) \neq 0$. We have $\mathrm{Tr} \ \rho_{V \otimes D}(g) = (\det g) \mathrm{Tr} \ \rho_{V}(g) = -\mathrm{Tr} \ \rho_{V}(g)$. So $V$ and $V \otimes D$ have different characters at $g$ and are not isomorphic.

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  • $\begingroup$ Thanks for the answer David! +1 The only step that I don't understand right away is, why does there for every integer polynomial with constant coefficient $\pm 1$ of degree $n$ exist an integer $n\times n$ matrix that has that polynomial as a characteristic polynomial? $\endgroup$ – Peter Patzt Aug 8 '16 at 13:46
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    $\begingroup$ @PeterPatzt, this is just the companion matrix. $\endgroup$ – LSpice Aug 8 '16 at 13:47
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    $\begingroup$ Companion matrix, that's the word I couldn't remember! Thanks @LSpice, editing that in now. $\endgroup$ – David E Speyer Aug 8 '16 at 14:16

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