How to write action on vector space as a matrix?

Question

This question is motivated by a similar but more complicated one that came up in my research. My first step is understanding a simpler example.

Say I have a $\mathbb{C}$-vector space $V$ on basis $\{e_1,...,e_4 \}$. $\operatorname{GL}_2(\mathbb{C})$ acts on $V$ in the following way: given $M \in \operatorname{GL}_2(\mathbb{C})$, create the matrix in $\operatorname{GL}_4(\mathbb{C})$: $$\begin{pmatrix} M & 0\\ 0 & (M^t)^{-1} \end{pmatrix} $$ and this matrix acts on our basis for $V$ in the standard way. As a $\operatorname{GL}_2(\mathbb{C})$ representation, $V$ is the direct sum of the standard representation (on basis $\{e_1, e_2 \}$) and its dual (on basis $\{e_3, e_4 \}$).

I'm interested $\bigwedge^3 V$, which has a basis $\{e_2 \wedge e_3 \wedge e_4, e_1 \wedge e_3 \wedge e_4, e_1 \wedge e_2 \wedge e_4, e_1 \wedge e_2 \wedge e_3 \}$. We have an action of $\operatorname{GL}_2(\mathbb{C})$ on $\bigwedge^3 V$, inherited from the action on $V$. Given some $M \in \operatorname{GL}_2(\mathbb{C})$ acting on $V$, how can I write this inherited action on $\bigwedge^3 V$ as a matrix in $\operatorname{GL}_{\text{dim}(\bigwedge^3 V)}(\mathbb{C})$?

(Note: $\text{dim}(V) = \text{dim}(\bigwedge^3 V)$ in this example is a coincidence and will not be true in other dimensions.)

This example is small enough that I assume it can be brute forced, but I want to be able to do this for more complicated examples.

Bonus: for calculations like this, is there a computer program (Mathematica, Matlab?) that works best? Ultimately, I'll want a computer to help me solve a problem similar to: given $M \in \operatorname{GL}_2(\mathbb{C})$, find the fixed set of $M$ in $\bigwedge^3 V$.

What you're looking for is the exterior power of this $4 \times 4$ matrix, see this post for more details on computation. Each entry will ultimately be a $3 \times 3$ minor of your $4 \times 4$ matrix — Ben Grossmann, Sep 27 '24 at 14:55
That said, because $3 = 4 - 1$, the exterior power that you're calculating corresponds in some way to the adjugate of that $4 \times 4$ matrix. With a suitable basis, you should be able to obtain the matrix representation $$ \pmatrix{M^{-1} & 0\ 0 & M^t} $$ — Ben Grossmann, Sep 27 '24 at 14:57
Actually, the basis that you've presented is the basis relative to which we obtain this matrix — Ben Grossmann, Sep 27 '24 at 14:59
@BenGrossmann This is exactly what I needed, thank you! (If you wanted to post your comments as an answer, I'd accept it) — Chase, Sep 27 '24 at 16:22

score 3 · Accepted Answer · answered Sep 27 '24 at 18:30

As is explained here, it is a general fact that the action induced by $A \in GL_n(\Bbb C)$ on $\bigwedge^{n-1}\Bbb C^n$ has $\text{adj}(A)$ (the adjugate of $A$) as its matrix representation relative to the basis $$ \{\mathbf{e}_1 \wedge \dots \wedge \hat{\mathbf{e}}_k \wedge \dots \wedge \mathbf{e}_n\}_{k=1}^n, $$ where $\hat v$ indicates that $v$ is omitted from the wedge product. Consequently, the action you describe should have the matrix $$ \text{adj}\pmatrix{M & 0\\0 & M^{-t}} = \pmatrix{M^{-1} & 0\\0 & M^t} $$ relative to the basis $$ \{e_2 \wedge e_3 \wedge e_4, e_1 \wedge e_3 \wedge e_4, e_1 \wedge e_2 \wedge e_4, e_1 \wedge e_2 \wedge e_3 \}. $$

How to write action on vector space as a matrix?

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