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Jan 11, 2018 at 10:50 comment added Asaf Shachar Thanks. This was also my natural guess, but that doesn't seem to be well-defined (consider what happens when you switch $v_1^*,v_2^*$). In fact, there is a possibility that there is no injection at all: see here.
Jan 11, 2018 at 9:54 comment added Peter Michor Link corrected. I did not work out the injection. The most natural one is: $(v_1^*\wedge\dots\wedge v_k^*)\otimes (w_1^*\wedge \dots\wedge w_k^*) \mapsto \pm (v_1^*\otimes w_1^*)\wedge \dots \wedge (v_k^*\otimes w_k^*)$. One should look at the corresponding Young tableaus and identify the irreducible components as $GL(V)$-representations.
Jan 11, 2018 at 9:45 history edited Peter Michor CC BY-SA 3.0
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Jan 11, 2018 at 8:08 comment added Asaf Shachar Sorry, one more question: In my formulation $g:V \to V^*$, and $\Lambda^kg:\Lambda^{k} V \to \Lambda^{k} (V^*) \cong (\Lambda^{k} V)^*$ is the induced map on exterior powers, that is an element of $\Lambda^{k} V^* \otimes \Lambda^{k} V^*$. You seem to consider it as an element in $\Lambda^{k}(V^* \otimes V^*)$. Can you please say how exactly do you identify $\Lambda^{k} V^* \otimes \Lambda^{k} V^*$ as a subset of $\Lambda^{k}(V^* \otimes V^*)$? Thanks.
Jan 11, 2018 at 7:39 comment added Asaf Shachar Thanks, though I think the link you provided doesn't work (at least for me).
Jan 11, 2018 at 7:28 history answered Peter Michor CC BY-SA 3.0