Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 ;)

share|improve this question
2  
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
4  
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

2 Answers 2

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 $\mathbb{Z}$.

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.