# Invariant for group actions

Hello everybody!

Define the action of $SL_4({\mathbb{Z}})$ on alternating 2-forms or simply skew-symmetric matrices of degree 4 according to the following:

For $B \in SL_4({\mathbb{Z}})$ and an alternating matrix like $M$ define: $B.M = BMB^{T}$

Question: Why the single invariant for this action is $Pf(M)$?

Question2: Or why the invariant will be the coefficient of $Pf(Mx-Ny)$ by considering the action of $SL_4({\mathbb{Z}})$ on the pairs of skew-symmetric matrices instead?

Thanks ;)

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What is your motivation for asking this? Where did you read it? – David Roberts Jun 28 '12 at 6:50
In Bhargava's Papers; Higher Composition Laws I – user24766 Jun 28 '12 at 7:08
Thanks. Such detail is what makes a good question, because it gives context in which people can frame their answers. – David Roberts Jun 28 '12 at 9:05

One of the coincidences of simple Lie groups is that (over $\mathbb{C}$) $SL(4)=Spin(6)$. From the point of view of $SL(4)$, the three fundamental representations are three exterior powers of the vector representation. From the point of view of $Spin(6)$ they are the vector representation and the two spin representations.
You can find the invariant theory of the vector representation of $SO(2n)$ in Weyl's seminal book "Classical groups" and presumably in more recent accounts of this material.
I don't know about the passage from $\mathbb{C}$ to $\mathbb{Z}$.
As for your first question, the single invariant is not the Pfaffian. You can "diagonalize" any alternating form by finding a Frobenius basis: in this case $e_1, f_1, e_2, f_2$ with $e_i \wedge f_i = d_i$ (and other wedge products $0$) with non-negative integers $d_1 | d_2$. Then in fact $d_1$ and $d_2$ are the two invariants of $SL_4$ acting on alternating $2$-forms. The correct statement is that the Pfaffian is the only polynomial invariant: here it equals (up to sign?) $d_1 d_2$. To see this I believe you can argue as follows: $SL_4({\mathbb Z})$ is Zariski dense in $SL_4({\mathbb C})$, so the polynomial invariants agree. The point is that the Pfaffian is defined over $\mathbb Z$, so it is in fact a poly invariant over $\mathbb Z$. The answer to the second question is similar. Another place to learn a bit of the classical invariant theory is the book by Goodman and Wallach.