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EDIT again: I've now also done it for $\operatorname{Sym}^3V$, and it followed the same pattern. I did it (as much as I could) in the unitary case, using the representation $V \otimes V$ of the group $\operatorname{GL}_2 \times \operatorname{GL}_2$, where the first copy of $\operatorname{GL}_2$ acts on the first copy of its standard representation $V$, and the same for the second copies. Since I didn't know what to plug in for the derivatives of the inputs, I left them as $a^\prime, b^\prime$ and so on. In the end the terms paired up in a way that implies the pattern is present here as well, but it is difficult to see the actual final answer since I don't know how the coordinates vary with each $z_{ij}$. (In fact that was what I was hoping to get from this specific exercise, but the fact that I had to deal with the Lie algebra of $\operatorname{GL}_2 \times \operatorname{GL}_2$ gave enough of a wrinkle that I couldn't recover that from what I found.)

EDIT: Maybe I should include the fact that I know the answer to the question in the case described above, but I'd like to know how to determine the answer in more generality. The Lie algebra for $\operatorname{GL}_n(\mathbb{C})$ is $M_n(\mathbb{C})$, with a $\mathbb{C}$-basis $\{\varepsilon_{ij}\}$ where $\varepsilon_{ij}$ has a $1$ in its $ij$ component, and $0$ in all others. Then it should be the case for $i \neq j$ that $\rho(s)\frac{\partial}{\partial z_{ij}}\left[ \rho(s)^{-1} \right] = \sum_{\ell=1}^n s_{\ell i} \varepsilon_{\ell j} + s_{\ell j} \varepsilon_{\ell i}$ (at least in terms of its action on $V$). If $i=j$, it's $\sum_{\ell=1}^n s_{\ell i}\varepsilon_{\ell i}$. This reduces for $n=1$ to $s\varepsilon \cdot$, and $\varepsilon$ acts on $V$ by multiplication by $k$. So it gives the right answer in the case I can do explicitly.

It feels like we're using $\frac{\partial Z}{\partial z_{ij}}$ is $\varepsilon_{ij} + \varepsilon_{ji}$ when $i \neq j$ and $\varepsilon_{ii}$ when $i=j$, and we're picking out these components of the matrix $\begin{pmatrix} \varepsilon_{11} & \dots & \varepsilon_{n1} \\ \vdots & \ddots & \vdots \\ \varepsilon_{1n} & \dots & \varepsilon_{nn} \end{pmatrix}\begin{pmatrix} s_{11} & \dots & s_{1n} \\ \vdots & \ddots & \vdots \\ s_{n1} & \dots & s_{nn} \end{pmatrix}$, but I'm not sure why this would give the right answer.

My last comment for the edit: I tried to do the calculation explicitly for $n=2$ and the representation $V \otimes V$, where $V$ is the standard 2-dimensional representation of $\operatorname{GL}_2$. I think I ended up with the two matrices (the matrix I get from $\sum_{\ell} Y_{\ell i}\varepsilon_{\ell i}$ and the matrix I get from differentiating $\rho(s^{-1})$ directly) as transposes of each other, so I did something wrong. Actually, I did the calculation for $\frac{\partial}{\partial z_{11}}$ and $\frac{\partial}{\partial z_{22}}$ and got transposes, but the calculation for $\frac{\partial}{\partial z_{12}}$ was just wrong.

EDIT: I should include the fact that I know what the answer should be, from the paper I'm reading. It writes $\rho(s)\frac{\partial \rho(s^{-1})}{\partial z_{ij}} = \sum_{1 \leq \ell \leq n} s_{\ell i} \varepsilon_{\ell j} + s_{\ell j} \varepsilon_{\ell i}$ for $i \neq j$, and $\sum_{1 \leq \ell \leq n} s_{\ell i} \varepsilon_{\ell i}$ for $i=j$. Here $\varepsilon_{pq}$ denotes the matrix with a $1$ in its $pq$ component and a $0$ in all other components, and represents the matrix it uses to act on $V$. When $n=1$, this reduces to $sk$ as we wanted since $\varepsilon$ acts as $k$.

I did an example for $n=2$ and the representation $V \otimes V$ of $\operatorname{GL}_2$, where $V$ is its standard representation. From this, I think the pattern may be $\rho(s)\frac{\partial \rho(s^{-1})}{\partial z_{ij}} = \sum_{1 \leq \ell \leq n} s_{\ell i} \varepsilon_{j\ell} + s_{\ell i}\varepsilon_{j \ell}$ for $i = j$, and $\sum_{1 \leq \ell \leq n} s_{\ell i} \varepsilon_{i\ell}$ for $i=j$. The fact that one of the indices is switched is not worrying, since $s_{ij} = s_{ji}$.

This possible answer feels like picking out a component (or two components) of the matrix $$ \begin{pmatrix} \varepsilon_{11} & \dots & \varepsilon_{1n} \\ \vdots & \ddots & \vdots \\ \varepsilon_{n1} & \dots & \varepsilon_{nn} \end{pmatrix}\begin{pmatrix} s_{11} & \dots & s_{1n} \\ \vdots & \ddots & \vdots \\ s_{n1} & \dots & s_{nn} \end{pmatrix} $$ For $i=j$, we're picking out the $ii=jj$ component, and for $i\neq j$ we're summing the $ij$ component and the $ji$ component. I'm hoping to do another example (hopefully in the unitary case, so that $s_{ij} \neq s_{ji}$), but I don't think I know enough of the Lie group/Lie algebra theory to prove it even if the pattern continues.

I'm learning about Maass--Shimura operators, and there's a term that I'm not sure how to generalize nicely.

Let $\mathfrak{h}$ be the upper half-plane with parameter $z= x + iy$, and write $s = \frac{1}{z - \overline{z}} = \frac{1}{2iy}$. For a modular form $f \colon \mathfrak{h} \to \mathbb{C}$ of weight $k$, the Maass--Shimura operator is $D_k(f) = s^k \frac{\partial}{\partial z}\left[ s^{-k}f \right] = s^k \frac{\partial}{\partial z}[s^{-k}] f + s^ks^{-k}\frac{\partial f}{\partial z} = ksf + \frac{\partial f}{\partial z}$. The $k$ can be viewed as coming from the action of an element of the Lie algebra of $\mathbb{G}_m$. The $s$ pops out if you do the calculation directly, since you lose a power of $y$. This is a pretty nice, simple formula.

On the other hand, let $\mathfrak{h}_n = \{Z \in M_n(\mathbb{C}) \mid \, ^t\!\!Z = Z, Z = X + iY \text{ with }Y \text{ positive definite}\}$ and write $s = (Z - \overline{Z})^{-1}$. Fix a representation $\rho \colon \operatorname{GL}_n(\mathbb{C}) \to \operatorname{GL}(V)$. For a Siegel modular form of genus $n$ and weight $\rho$, $f \colon \mathfrak{h}_n \to \mathbb{C}$, the Maass-Shimura operator is $D_\rho(f) = \rho(s)\operatorname{d}\left[ \rho(s^{-1})f \right]$, where $\operatorname{d}$ is the usual exterior derivative. Using the product rule, we should get a term which is $\operatorname{d}(f)$, and a term which is $\rho(s) \operatorname{d}(\rho(s^{-1}))$ times $f$. I don't know how to calculate this, at least to the point of getting a nice, simple formula like the $skf$ above.

In fact, I am interested in the directional derivative $\rho(Z - \overline{Z})^{-1}\frac{\partial \rho(Z - \overline{Z})}{\partial z_{ij}}$, where the $z_{ij}$ means the partial derivative with respect to the $ij$ entry of the matrix $Z$. It should be able to be given in terms of the action of the Lie algebra of $\operatorname{GL}_n(\mathbb{C})$.

Maybe a general formulation of the question: Let $s \colon \mathfrak{H} \to G(\mathbb{C})$ be a map from a complex manifold $\mathfrak{H}$ to the complex points of an algebraic group $G$. Then let $\rho \colon G(\mathbb{C}) \to \operatorname{GL}(V)$ be a representation of $G$ on the complex vector space $V$. Let $\frac{\partial}{\partial z_{ij}} \in T_{\mathfrak{H}}(U)$ be a vector field on $\mathfrak{H}$. How do I find $\rho(s(z))\frac{\partial \rho(s(z)^{-1})}{\partial z_{ij}}$ in terms of the action of the Lie algebra?

EDIT: Maybe I should include the fact that I know the answer to the question in the case described above, but I'd like to know how to determine the answer in more generality. The Lie algebra for $\operatorname{GL}_n(\mathbb{C})$ is $M_n(\mathbb{C})$, with a $\mathbb{C}$-basis $\{\varepsilon_{ij}\}$ where $\varepsilon_{ij}$ has a $1$ in its $ij$ component, and $0$ in all others. Then it should be the case for $i \neq j$ that $\rho(s)\frac{\partial}{\partial z_{ij}}\left[ \rho(s)^{-1} \right] = \sum_{\ell=1}^n s_{\ell i} \varepsilon_{\ell j} + s_{\ell j} \varepsilon_{\ell i}$ (at least in terms of its action on $V$). If $i=j$, it's $\sum_{\ell=1}^n s_{\ell i}\varepsilon_{\ell i}$. This reduces for $n=1$ to $s\varepsilon \cdot$, and $\varepsilon$ acts on $V$ by multiplication by $k$. So it gives the right answer in the case I can do explicitly.

It feels like we're using $\frac{\partial Z}{\partial z_{ij}}$ is $\varepsilon_{ij} + \varepsilon_{ji}$ when $i \neq j$ and $\varepsilon_{ii}$ when $i=j$, and we're picking out these components of the matrix $\begin{pmatrix} \varepsilon_{11} & \dots & \varepsilon_{n1} \\ \vdots & \ddots & \vdots \\ \varepsilon_{1n} & \dots & \varepsilon_{nn} \end{pmatrix}\begin{pmatrix} s_{11} & \dots & s_{1n} \\ \vdots & \ddots & \vdots \\ s_{n1} & \dots & s_{nn} \end{pmatrix}$, but I'm not sure why this would give the right answer.

My last comment for the edit: I tried to do the calculation explicitly for $n=2$ and the representation $V \otimes V$, where $V$ is the standard 2-dimensional representation of $\operatorname{GL}_2$. I think I ended up with the two matrices (the matrix I get from $\sum_{\ell} Y_{\ell i}\varepsilon_{\ell i}$ and the matrix I get from differentiating $\rho(s^{-1})$ directly) as transposes of each other, so I did something wrong.

I'm learning about Maass--Shimura operators, and there's a term that I'm not sure how to generalize nicely.

Let $\mathfrak{h}$ be the upper half-plane with parameter $z= x + iy$, and write $s = \frac{1}{z - \overline{z}} = \frac{1}{2iy}$. For a modular form $f \colon \mathfrak{h} \to \mathbb{C}$ of weight $k$, the Maass--Shimura operator is $D_k(f) = s^k \frac{\partial}{\partial z}\left[ s^{-k}f \right] = s^k \frac{\partial}{\partial z}[s^{-k}] f + s^ks^{-k}\frac{\partial f}{\partial z} = ksf + \frac{\partial f}{\partial z}$. The $k$ can be viewed as coming from the action of an element of the Lie algebra of $\mathbb{G}_m$. The $s$ pops out if you do the calculation directly, since you lose a power of $y$. This is a pretty nice, simple formula.

On the other hand, let $\mathfrak{h}_n = \{Z \in M_n(\mathbb{C}) \mid \, ^t\!\!Z = Z, Z = X + iY \text{ with }Y \text{ positive definite}\}$ and write $s = (Z - \overline{Z})^{-1}$. Fix a representation $\rho \colon \operatorname{GL}_n(\mathbb{C}) \to \operatorname{GL}(V)$. For a Siegel modular form of genus $n$ and weight $\rho$, $f \colon \mathfrak{h}_n \to \mathbb{C}$, the Maass-Shimura operator is $D_\rho(f) = \rho(s)\operatorname{d}\left[ \rho(s^{-1})f \right]$, where $\operatorname{d}$ is the usual exterior derivative. Using the product rule, we should get a term which is $\operatorname{d}(f)$, and a term which is $\rho(s) \operatorname{d}(\rho(s^{-1}))$ times $f$. I don't know how to calculate this, at least to the point of getting a nice, simple formula like the $skf$ above.

In fact, I am interested in the directional derivative $\rho(Z - \overline{Z})^{-1}\frac{\partial \rho(Z - \overline{Z})}{\partial z_{ij}}$, where the $z_{ij}$ means the partial derivative with respect to the $ij$ entry of the matrix $Z$. It should be able to be given in terms of the action of the Lie algebra of $\operatorname{GL}_n(\mathbb{C})$.

Maybe a general formulation of the question: Let $s \colon \mathfrak{H} \to G(\mathbb{C})$ be a map from a complex manifold $\mathfrak{H}$ to the complex points of an algebraic group $G$. Then let $\rho \colon G(\mathbb{C}) \to \operatorname{GL}(V)$ be a representation of $G$ on the complex vector space $V$. Let $\frac{\partial}{\partial z_{ij}} \in T_{\mathfrak{H}}(U)$ be a vector field on $\mathfrak{H}$. How do I find $\rho(s(z))\frac{\partial \rho(s(z)^{-1})}{\partial z_{ij}}$ in terms of the action of the Lie algebra?

EDIT: Maybe I should include the fact that I know the answer to the question in the case described above, but I'd like to know how to determine the answer in more generality. The Lie algebra for $\operatorname{GL}_n(\mathbb{C})$ is $M_n(\mathbb{C})$, with a $\mathbb{C}$-basis $\{\varepsilon_{ij}\}$ where $\varepsilon_{ij}$ has a $1$ in its $ij$ component, and $0$ in all others. Then it should be the case for $i \neq j$ that $\rho(s)\frac{\partial}{\partial z_{ij}}\left[ \rho(s)^{-1} \right] = \sum_{\ell=1}^n s_{\ell i} \varepsilon_{\ell j} + s_{\ell j} \varepsilon_{\ell i}$ (at least in terms of its action on $V$). If $i=j$, it's $\sum_{\ell=1}^n s_{\ell i}\varepsilon_{\ell i}$. This reduces for $n=1$ to $s\varepsilon \cdot$, and $\varepsilon$ acts on $V$ by multiplication by $k$. So it gives the right answer in the case I can do explicitly.

It feels like we're using $\frac{\partial Z}{\partial z_{ij}}$ is $\varepsilon_{ij} + \varepsilon_{ji}$ when $i \neq j$ and $\varepsilon_{ii}$ when $i=j$, and we're picking out these components of the matrix $\begin{pmatrix} \varepsilon_{11} & \dots & \varepsilon_{n1} \\ \vdots & \ddots & \vdots \\ \varepsilon_{1n} & \dots & \varepsilon_{nn} \end{pmatrix}\begin{pmatrix} s_{11} & \dots & s_{1n} \\ \vdots & \ddots & \vdots \\ s_{n1} & \dots & s_{nn} \end{pmatrix}$, but I'm not sure why this would give the right answer.

My last comment for the edit: I tried to do the calculation explicitly for $n=2$ and the representation $V \otimes V$, where $V$ is the standard 2-dimensional representation of $\operatorname{GL}_2$. I think I ended up with the two matrices (the matrix I get from $\sum_{\ell} Y_{\ell i}\varepsilon_{\ell i}$ and the matrix I get from differentiating $\rho(s^{-1})$ directly) as transposes of each other, so I did something wrong. Actually, I did the calculation for $\frac{\partial}{\partial z_{11}}$ and $\frac{\partial}{\partial z_{22}}$ and got transposes, but the calculation for $\frac{\partial}{\partial z_{12}}$ was just wrong.