If you take a standard representation of a symmetric group, take an alternating tensor power of it, what groups appear as stabilizers of vectors? - MathOverflow

If you take a standard representation of a symmetric group, take an alternating tensor power of it, what groups appear as stabilizers of vectors?

2010-06-30T20:44:31Z

I'm particularly interested in the case $\Lambda^3 \mathbb{F}_3^n$, and specifically, just stabilizers of vectors that satisfy the two conditions (i) there are no zero coordinates (in the basis induced from the standard basis of $\mathbb{F}_3^n$) and (ii) they are in the image of a map $(\mathbb{R}^n)^3 \to \Lambda^3 \mathbb{F}_3^n$ that I will now describe.

We start with the obvious map $(\mathbb{R}^n)^3 \to \Lambda^3 \mathbb{R}^n$. Then write our vector in the standard coordinates $\sum_{i < j < k} a_{ijk} e_i \wedge e_j \wedge e_k \in \Lambda^3 \mathbb{R}^n$ and then replace $a_{ijk}$ with $0 \in \mathbb{F}_3$ if it is 0, $1 \in \mathbb{F}_3$ if it is positive, and $-1 \in \mathbb{F}_3$ if it is negative.

I can calculate with GAP all stabilizers for n = 4, 5, and stabilizers for given vectors for n = 6, 7.

For n = 4 I get $\mathbb{Z} / 4, \mathbb{Z} / 3$, and $Alt(4)$, for n = 5 I get $1, \mathbb{Z} / 3$ and $\mathbb{Z} / 5$, and for n = 6 and 7 I can find cyclic groups of orders 1, 2, 3, 5, 6, and 1, 3 respectively.

It seems like the sort of problem that should have a solution....

Answer by zzl for If you take a standard representation of a symmetric group, take an alternating tensor power of it, what groups appear as stabilizers of vectors?

2010-07-02T04:41:29Z

I assume that the field has characteristic $\neq 2 $.

(1) Assume the vector is the standard basis vectors in wedge product with the standard basis in the standard permutation representation $V$, i.e., $v=v_{1}\wedge v_2\wedge \cdots \wedge v_r\in \Lambda^r V$. The any element in the stabilizer stabilizes the subset ${ 1,\dots, r}\subseteq {1,\dots, n}$ and its restriction to the subset ${1,\dots, r}$ is an even permutation and thus is an element in $A_r$. Thus the stabilizer is $A_r\times S_{n-r}$. For other basis elements, simply replace ${1, \dots, r}$ by any subset ${ i_1 < \dots < i_r}$.

(2) If the vector is sum of a set of basis elements. Let $A\subseteq {1, \dots, n}$ is a subset of $r$ elements and $v_A$ be the basis elements discussed in (1). Hence $v=\sum_{s=1}^t v_{A_s}$. The the stabilizer consists of elements that stabilizers each $A_1, \dots, A_t$ and restricts to each is even, and multiplied by $\sigma \in S_n$ which defines a permutation of the above sets $A_1, \dots, A_r$. Now you know how to determine this group.

(3) For a general vector, write it as linear combination of the standard basis and group them so that they is a linear combination of elements of type (2) with distinct non-zero coefficients so that two different terms have common basis terms. The stabilizer is the intersection of the stabilizers described in (2). For $\mathbb{F}_3$. there are at most two such subgroups to intersect.

I hope this helps.

Answer by zzl for If you take a standard representation of a symmetric group, take an alternating tensor power of it, what groups appear as stabilizers of vectors?

2010-07-06T21:32:26Z

I take back (3) (and partial (2) as well)! Let $C(r,n)$ be the set of all $r$-combinations of {$ 1,2, \dots, n$}, which has the natural order. $\wedge ^r(V)$ is a signed permutation representation of $ S_n$ with a basis {${ e_{A}| A\in C(r,n)}$} such that $\sigma e_A=sign(\sigma(A))e_{\sigma(A)}$. I missed the sign in my (3).