Modular function of the unipotent radical of a parabolic subgroup of a reductive group

Question

Let $G = \text{GL}_n(\mathbb{R})$. For a partition $\underline{n} = (n_1,\ldots,n_t)$ of $n$, let $P = P_{\underline{n}}$ denote the standard, block-upper-triangular parabolic subgroup of $G$ associated to $P_{\underline{n}}$. Let $U_P$ denote the unipotent radical of $P$, i.e., the group $$ \begin{pmatrix} U_{1,1} & X_{1,2} & X_{1,3} & \cdots & X_{1,t} \\ & U_{2,2} & X_{2,3} & \cdots & X_{2,t} \\ & & \vdots & \vdots & \vdots \\ & & & & U_{t,t} \end{pmatrix} $$ where $U_{i,i}$ is the $n_i\times n_i$ identity matrix for each $i$, and the $X_{i,j}$ are $n_i \times n_j$ blocks.

I'm working through a set of notes on automorphic representation theory and am currently reading a section in which the author utilizes the modular function $\delta_P$ of $U_P$ (i.e., the modular function for a Haar measure on $U_P$). (He's actually working adelically (as opposed to working over $\mathbb{R}$, but that shouldn't really matter here.) If $M$ denotes the Levi subgroup of $P$ ($M \cong \text{GL}_{n_1}\times\cdot\times\text{GL}_{n_t}$), then for $m \in M$ the derivation $$ \delta_P(m) = |\det(\text{Ad}(m)|_{\mathfrak{u}_P})| = \prod_{s=1,\ldots,t}|\det(m_s)|^{-\sum_{i < s}n_i + \sum_{j > s}n_j} $$ appears at one point in the text where $\mathfrak{u}_P = \text{Lie}(U_P)$, and for whatever reason I'm not seeing why the second equality holds. (I'm aware of how to derive the first.)

So my question is: How does one show that either $\delta_P(m)$ or $|\det(\text{Ad}(m)|_{U_P})|$ equals the product on the RHS? The problem for me seems to be my lack of knowledge about the algebraic structure of $\mathfrak{u}_P$ here, which limits my options regarding how to compute the determinant of $\text{Ad}(m)|_{\mathfrak{u}_P}$.

I have a feeling that I'm missing something obvious here. Feel free to just provide references, since I couldn't find one!

Edit: Corrected a couple errors.

Answer 1

First, there should be 2 corrections in your post: (1) the $U_{i,i}$ should be the identity, not just unipotent; (2) the $\det(m)$ in the right side of the displayed formula should be $\det(m_s)$, where $m_s$ is the $s$th block of the Levi subgroup.

The matrices of the Lie algebra $\mathfrak{u}_P$ look just like those of the Lie group $U_P$ except that the $U_{i,i}$ are 0 instead of the identity.

First let's do the case where all the $n_i = 1$. Then $U_P$ consist of the upper triangular unipotent matrices, the Lie algebra $\mathfrak{u}_P$ consists of the strictly upper triangular matrices (diagonal=0), and the Levi subgroup $M$ is the set of all nonsingular diagonal matrices. We are looking for the determimant of the linear transformation $u \mapsto mum^{-1}$ where $u \in \mathfrak{u}_P$. If $m$ has diagonal entries $m_1, \ldots m_n$, then each unit matrix $e_{i,j} \in \mathfrak{u}_P$ with $i<j$ is an eigenvector with eigenvalue $m_i m_j^{-1}$. Multiplying all these eigenvalues together gives you the desired formula (with the correction I mentioned above).

In the general case, each $m_i$ will be an $n_i \times n_i$ matrix instead of just a scalar. If we restrict $Ad(m)$ to the subspace $E_{i,j}$ of $\mathfrak{u}_P$ where all the $X$ blocks other than $X_{i,j}$ are zero, then we are looking at the linear transformation from $M_{n_i \times n_j}(\mathbb{R})$ to itself given by $T_{i,j} : x \mapsto m_i x m_j^{-1}$. For each pair of eigenvalues $\lambda$ for $m_i$ and $\mu$ for $m_j$, we get an eigenvalue $\lambda \mu^{-1}$ for $T_{i,j}$. Thus the determinant of $T_{i,j}$ is $\det (m_i)^{n_j} \det(m_j)^{-n_i}$. Multiplying these together (and remembering that $i<j$), we get the desired result.