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

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....

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Can you correct the typo in (i)? As is, the only vector is the zero vector. It seems like if "nonzero" is replaced by "zero", then vectors satisfying both (i) and (ii) exist only for n=3 and n=4. However, for n=4 my answer agrees with yours. I'm assuming F3 is the field with 3 elements and the standard representation is by permuting coordinates. – Jack Schmidt Jul 1 2010 at 0:46
Hey a response! Sorry. about the typo. Yes, standard representation is permuting the coordinates, Yes $\mathbb{F}_3$ is the field with three elements. – rt-ist Jul 1 2010 at 18:49
Huh. I didn't expect that for n > 4. My bad. I thought it was an equivalent problem. I've fixed it now. The original motivation is actually configurations of $n$ vectors in $\mathbb{R}^3$ such that each triple spans $\mathbb{R}^3$. Then we want to know which permutations of the vectors preserve the orientation of each triple. – rt-ist Jul 1 2010 at 19:46
Thanks. I'll look at this later today. – Jack Schmidt Jul 1 2010 at 20:05
Two observations: The standard representation does not permute coordinates (this is the permutation representation which is given by the standard plus the trivial). Secondly, alternate powers of the standard representation are exactly the $L-$shaped representations, at least in characteristic zero. I do not know what happens to them in characteristic $3$. – Roland Bacher May 11 2011 at 11:27

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).

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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.

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Thanks for your reply. I agree with (1) but for (3) the situation is a bit more complicated. Write $e_{ijk} = e_i \wedge e_j \wedge e_k$. In $\Lambda^3 F^4$ (if char of $F$ is not 2) the stabilizer of $e_{123} + e_{124}$ is generated by $(3,4)$ and the stabilizer of $e_{234} + e_{134}$ is generated by $(1,2)$ but the stabilizer of $e_{123} + e_{124} - e_{234} - e_{134}$ is generated by $(1,3,4,2)$ which has nothing to do with the other two groups. – rt-ist Jul 2 2010 at 12:09