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Let $V$ be an $n$-dimensional real vector space, and let $2 \le k \le n-2$.

Definitions

  1. We say an element $\omega \in \Lambda^k V$ is decomposable if $\omega=\alpha_1 \wedge \dots \wedge \alpha_k$, for some $\alpha_i \in V$. We say $\omega$ is a power if $\omega=\alpha \wedge \dots \wedge \alpha$ for some $\alpha \in V$.

  2. We say an element $h \in \Lambda^k V^* \otimes \Lambda^k V^* \cong \operatorname{Hom}(\Lambda^k V,\Lambda^k V^*)$ is a power if $h=\Lambda^k g$ for some linear map $g:V \to V^*$. Here, $\Lambda^k g:\Lambda^k V \to \Lambda^k V^*$ is the induced map on exterior powers, that is $$ \Lambda^k g(v_1 \wedge \dots \wedge v_k)=g(v_1) \wedge \dots \wedge g(v_k).$$

Question: Is there a linear injection $$ \Lambda^k V^* \otimes \Lambda^k V^* \to \Lambda^k (V^* \otimes V^*)$$ which maps power elements to decomposable elements?

Even better, is there an injection which maps power elements to power elements?

Note: For dimensional reasons, there is always some linear embedding of $\Lambda^k V^* \otimes \Lambda^k V^*$ in $ \Lambda^k (V^* \otimes V^*)$.

Motivation:

This question arose in the context of applying the Plucker relations, to characterise which metrics on exterior powers are induced by metrics at the base. See here.

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    $\begingroup$ I think this is equivalent to $e_k\left[e_1^2\right] - e_k^2$ being Schur-positive, where the $e_i$ are the elementary symmetric functions and the brackets mean plethysm. But this is false for $k = 2$: In fact, $e_2\left[e_1^2\right] - e_2^2 = -s_{1,1,1,1} + s_{2,1,1} - s_{2,2} + 2s_{3,1}$. (Though I have ignored your $2 \leq k \leq n-2$ condition. Maybe it changes things -- although I don't expect it to.) $\endgroup$ – darij grinberg Jan 11 '18 at 9:40
  • $\begingroup$ "Schur-positive" means "an $\mathbb{N}$-linear combination of Schur functions". Plethysm... that's a longer story. See Chapter 7 (and its Appendix 2) in Stanley's Enumerative Combinatorics (volume 2). Anyway, the main thrust of the argument is that $\Lambda^2 \left(V \otimes V\right)$, as an $\operatorname{GL}\left(V\right)$-representation, contains the irreducible subrepresentation corresponding to the partition $\left(2,2\right)$ one less time than $\Lambda^2 V \otimes \Lambda^2 V$ does (when $\dim V \geq 2$), and ... $\endgroup$ – darij grinberg Jan 11 '18 at 10:10
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    $\begingroup$ ... thus there is no injective $\operatorname{GL}\left(V\right)$-representation homomorphism from the latter to the former. You can probably check this by hand, painful as it will be. Start by proving that the character of $\Lambda^2 \left(V\otimes V\right)$ is $2s_{2,1,1} + 2s_{3,1}$. $\endgroup$ – darij grinberg Jan 11 '18 at 10:11
  • $\begingroup$ By the way: I am more used to seeing $\bigwedge^k V$ (\bigwedge^k V) than $\Lambda^k V$ (\Lambda^k V), but perhaps notations differ. $\endgroup$ – Omnomnomnom Jan 11 '18 at 15:01
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    $\begingroup$ darij’s argument rules out an embedding which does not depend on a choice of coordinates, or equivalently which is equivariant with respect to the automorphism group of $V$. $\endgroup$ – Qiaochu Yuan Jan 14 '18 at 8:20

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