Computation of modulus character of the mirabolic subgroup of $\operatorname{GL}_n$ using roots

Question

$\DeclareMathOperator\GL{GL}$For parabolic subgroup of a general linear group or classical group $G$, we can compute its modulus character using the positive roots associated to them.

But it seems that there is no such formula for arbitrary closed subgroup of $G$ and it is very hard to compute it from the very definition of the modulus character.

In the meanwhile, I am wondering whether we can describe the modulus character of mirabolic subgroup of $\GL_n$ restricted to $\GL_{n-1}$ using the roots associated to its unipotent radical. Because mirabolic subgroup is similar to parabolic, I strongly guess it should be!

Answer 1

For a general Lie group $G$, the modulus character is the absolute value of the determinant of the adjoint action of $G$ on $\DeclareMathOperator\Lie{Lie}\Lie(G)$. For convenience, I'll refer to the determinant of the adjoint action itself, not its absolute value, as the modulus character.

If we are in the algebraic setting, as I think one usually assumes when discussing reductivity (and anyway it's the setting of your question!), then this determinant is algebraic—that is, not just an abstract homomorphism on rational points, but one deduced from a homomorphism of algebraic groups. In particular, I may freely replace the base field by a larger one. (Actually I no longer even care that we started off over the real numbers, although I do care, for formation of the unipotent radical, that I am working over a perfect field.) I will once more refer to this homomorphism of algebraic groups as the modulus character.

If $\newcommand\Ru{\operatorname R_u}\Ru(G)$ is the unipotent radical of $G$, then the obvious action of $G$ on $\Lie(G/{\Ru}(G))$ factors through the adjoint action of the reductive group $G/{\Ru(G)}$, and so is unimodular (i.e., has determinant $1$). That is, we may as well compute the determinant of the adjoint action of $G$ on $\Lie(\Ru(G))$, as you hoped.

Now suppose that $G$ is reductive and $H$ is a closed regular subgroup of $G$, in the sense that it is normalised by some maximal torus $T$ in $G$ (assumed split after base change, if needed). Then the restriction to the maximal torus $T_H$ in $T \cap H$ of the modulus character of $H$ is the sum of the restrictions to $T_H$ of the roots of $T$ that occur in $\Lie(H)$ (or, equivalently, $\Lie(\Ru(H))$). Since the modulus character is trivial on the unipotent radical of $H$, this completely determines the modulus character. Notice that this requires only that $G$ be reductive, not that it be general linear or classical.

In your case, if we choose coördinates so that the mirabolic subgroup $H$ sits inside $\operatorname{GL}_n$ as the subgroup of matrices of the form $\begin{pmatrix} A & b \\ 0 & 1 \end{pmatrix}$, then the recipe above shows that the modulus character agrees with the restriction to the group of diagonal matrices in $H$ of, and therefore equals, the usual determinant character $\begin{pmatrix} A & b \\ 0 & 1 \end{pmatrix} \mapsto \det(A)$.