I have this question in my textbook, and I can't seem to solve it on my own:
Let $P \subset GL(n,\mathbb{C})$ be a subgroup as following:
$P$ consists of all matrices in block upper-triangular form ${g}=\left[\begin{array}{*{2}{c}}{a}&b\\{0}&{d}\\\end{array}\right]$ where $a \in GL(p,\mathbb{C})$, $d \in GL(n-p,\mathbb{C})$ and $b \in M_{p,n-p}(\mathbb{C})$
- Prove that the ideal $J_p$ is generated by the matrix entry functions $x_{ij}$ with $p\lt i\le n$ and $1\le j\le p$ .
- Use (a) to prove that $Lie(P)$ consists of all matrices in $2$ x $2$ block upper triangular form (with diagonal blocks of sizes $P$ x $P$ and $(n-P)$ x $(n-P)$.
Let $I_p$ denote the ideal in $\mathbb{C}[x_{11},x_{12},\ldots,x_{nn}]$ generated by all $x_{ij}$ with $i > p \geq j$. Clearly, each $f \in I_p$ vanishes on $P$, as so does each of the defining generators of $I_p$; thus, $I_p \subset J_p$. On the other hand, if all polynomials $f \in I_p$ vanish at some $g \in \operatorname{GL}(n,\mathbb{C})$, then in particular $x_{ij}(g) = 0$ whenever $i>p\geq j$, and thus $$ g = \begin{bmatrix}a & b\\ 0 & d\end{bmatrix} \quad \text{with} \quad a \in \operatorname{M}_{p}(\mathbb{C}), \ b \in \operatorname{M}_{p,n-p}(\mathbb{C}), \ d \in \operatorname{M}_{n-p}(\mathbb{C}). \qquad (\ast)$$ Since in this case, $\det(g) = \det(a)\det(d) \neq 0$, we actually have $a\in\operatorname{GL}(p,\mathbb{C})$ and $d \in \operatorname{GL}(n-p,\mathbb{C})$. Hence, $P = \operatorname{GL}(n,\mathbb{C}) \cap \mathcal{V}(I_p)$, where $\mathcal{V}(I_p)$ denotes the vanishing set of $I_p$ in $\operatorname{M}_{n}(\mathbb{C})$. Since $\operatorname{GL}(n,\mathbb{C})$ is dense in $\operatorname{M}_n(\mathbb{C})$, it follows that $P$ is dense in $\mathcal{V}(I_p)$. Therefore, every polynomial $f \in J_p$ vanishes also on $\mathcal{V}(I_p)$. For the sake of simplicity of notation, put $m= n^2$, $s = p(n-p)$, and let $z_1,\ldots,z_s$ denote the matrix coefficients $x_{ij}$ with $i > p \geq j$ (in some order), while $z_{s+1},\ldots,z_m$ denote the remaining $(m-s)$ matrix coefficients. Now, suppose that $f \in J_p$, and write $$ f(z_1,\ldots,z_m) = \sum_{k_1,\dots,k_m \geq 0} c_{k_1,\ldots,k_m}z_1^{k_1}\cdots z_m^{k_m} \quad \text{with} \quad c_{k_1,\ldots,k_m} \in \mathbb{C}.$$ Putting $z_1 = \cdots = z_s = 0$, so that $(z_1,\ldots,z_m) \in \mathcal{V}(I_p)$, we obtain $$ 0 = f(0,\ldots,0,z_{s+1},\ldots,z_m) = \sum_{k_{s+1},\cdots,k_m \geq0} c_{0,\ldots,0,k_{s+1},\ldots,k_m}z_{s+1}^{k_{s+1}}\cdots z_{m}^{k_m} \qquad (\dagger)$$ for all $(z_{s+1},\ldots,z_m) \in \mathbb{C}^{m-s}$. Taking all possible partial derivatives in $(\dagger)$, we obtain $$ 0 = \frac{\partial^{k_{s+1}+\cdots+k_m}f}{\partial z_{s+1}^{k_{s+1}}\cdots \partial z_m^{k_m}} (0,\ldots,0,z_{s+1},\ldots,z_m) = k_{s+1}!\cdots k_m!\,c_{0,\ldots,0,k_{s+1},\ldots,k_m}$$ Therefore, the only non-zero coefficients $c_{k_1,\ldots,k_m}$ of the Taylor expansion of $f$ at $0 \in \operatorname{M}_n(\mathbb{C})$ can be among those with $k_j >0$ for at least one $j \in \{1,\ldots,s\}$, which shows that $f$ lies in the ideal $I_p$ generated by $z_1,\ldots,z_s$. Thus, $J_p \subset I_p$. As noted at the beginning, $I_p \subset J_p$, as well, whence we conclude that $J_p = I_p$.
As observed in part (1.), we have $P = \operatorname{GL}_n(\mathbb{C}) \cap \mathcal{V}(I_p)$. Since $I_p$ is generated by linear functions, $\mathcal{V}(I_p)$ is a vector subspace of $\operatorname{M}_n(\mathbb{C})$, in fact the space of all $g\in \operatorname{M}_n(\mathbb{C})$ given by $(*)$ above. Since $P$ is open in $\mathcal{V}(I_p)$, the tangent space to $P$ (at any point of $P$, and in particular, at the identity) can be identified with $\mathcal{V}(I_p)$, in the same way as $\mathfrak{gl}(n,\mathbb{C}) = \operatorname{Lie(GL}(n,\mathbb{C}))$ is identified (as a vector space) with $\operatorname{M}_n(\mathbb{C})$. Thus, with this convention, the underlying vector space of the Lie algebra $\operatorname{Lie}(P) \subset \mathfrak{gl}(n,\mathbb{C})$ is $\mathcal{V}(I_p) \subset \operatorname{M}_n(\mathbb{C})$.