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
$$ Res\_{GL\_n \mathbb Z} V \cong Res\_{GL\_n \mathbb Z} (V \otimes D)?$$



Can these two irreducible $GL_n \mathbb Z$-representations be isomorphic?
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](https://en.wikipedia.org/wiki/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.


