3
$\begingroup$

Consider the following chain $\{A_1,A_2,A_3,\cdots,A_{n}\}$ of orbit spaces of even-rank anti-symmetric tensors, where $$A_k:=\frac{\Lambda^{2k}(\mathbb{R}^{2n})}{e_{i_1}\wedge \cdots \wedge e_{i_{2k}}\mapsto Re_{i_1}\wedge \cdots \wedge Re_{i_{2k}}},~~~~~~~~~R\in O(2n),$$ The base case is well-known, and corresponds to the canonical form for a real, antisymmetric matrix: $$A_1=\frac{\Lambda^2(\mathbb{R}^{2n})}{e_i\wedge e_j\mapsto Re_i\wedge Re_j}=\left\{\bigg[\sum_{j}\lambda_{j} e_{2j-1}\wedge e_{2j}\bigg],~~\lambda_j\in\mathbb{R}\right\}\simeq \mathbb{R}^n$$ Are there any results for $A_2$?

$\endgroup$

1 Answer 1

4
$\begingroup$

Actually, there is a fair amount known in the first nontrivial case: $(k,n) = (2,4)$. For example, see Calibrations on $R^8$ by J. Dadok, R. Harvey and F. Morgan Transactions of the American Mathematical Society Vol. 307, No. 1 (May, 1988), pp. 1-40.

Also, see Antonyan, L. V. Classification of four-vectors of an eight-dimensional space. (Russian) Trudy Sem. Vektor. Tenzor. Anal. No. 20 (1981), 144–161.

Already for higher values $(k,n) = (2,n)$ for $n>4$, I think very little is known.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .