This question occurred to me while thinking on another one here, Name for an operation on matrices?

Can one define in an invariant way a binary operation on finite-dimensional vector spaces - let us denote it somehow suggestively by $(V,W)\mapsto V^{\otimes W}$ - with the property$$\dim(V^{\otimes W})=\dim(V)^{\dim(W)}?$$

Added later (and I should do it from the beginning as it is important): the construction from the linked question suggests that this operation seemingly should act on linear operators by assigning to $f:V\to X$ and $g:Y\to W$ certain operator (explicitly given there in terms of chosen bases) $$ f\dagger g:V^{\otimes W}\otimes Y\to X\otimes W, $$ with $\operatorname{rank}(f\dagger g)\geqslant\operatorname{rank}(f)\operatorname{rank}(g)$.

I'm aware that this looks even more impossible, but anyway - with bases it is done in that question.

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    $\begingroup$ I think the answer is morally no. If you additionally require that $V^{\otimes (-)}$ sends direct sums to tensor products and that $V^{\otimes 1} \cong V$, then using Schur-Weyl duality you can conclude that if $\dim W = 2$ then $GL(W)$ can never act faithfully on $V^{\otimes W}$ (in a way compatible with the natural action of $GL(V)$), regardless of the size of $V$. I think the lesson of Schur-Weyl duality here is that unlike the situation with taking iterated direct sums, the exponent when taking iterated tensor products can be upgraded to at best a set but not really a vector space. $\endgroup$ Sep 30 '15 at 4:35
  • $\begingroup$ I liked your answer very much and almost accepted it, but since you deleted it let me add one thing here :D It occurred to me that $\Lambda^*(W_1\oplus W_2)\cong\Lambda^*(W_1)\otimes\Lambda^*(W_2)$ allows one to view $\Lambda^*(W)$ as "$2^W$" of sorts, so there must be more to it. I mean, It might be that there is some more tricky invariance wrt $V$ (whereas action on the $W$ side might be "ordinary"). What do you think? $\endgroup$ Sep 30 '15 at 4:59
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    $\begingroup$ My answer was incorrect; the correct observation (that $GL(W)$ can't act faithfully) is in the comment above. Anyway, also note that if you want functoriality in $W$ with respect to non-invertible morphisms then there are generally no nontrivial $GL(V)$-equivariant maps $V \to V^{\otimes 2}$ or $V^{\otimes 2} \to V$, etc. In that exterior algebra observation the "base" of the exponential is not a vector space but a graded vector space, so my comment above doesn't apply. $\endgroup$ Sep 30 '15 at 5:01
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    $\begingroup$ @Arul Sorry I don't see how algebra structures might enter the picture (except I have some vague association with the formalism of plethories) $\endgroup$ Sep 30 '15 at 6:22
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    $\begingroup$ Over different fields?? $\endgroup$ Sep 30 '15 at 6:26

Okay, so you can get pretty close as follows: I still don't think $V^{\otimes W}$ makes sense, but riffing off of your comment, we can make sense of $(1 \oplus V)^{\otimes W}$ (where $1$ denotes the $1$-dimensional vector space). The guiding intuition is the binomial expansion

$$(1 + V)^W = 1 + {W \choose 1} V + {W \choose 2} V^2 + \dots $$

which can be upgraded to the functor

$$(1 \oplus V)^{\otimes W} = 1 \oplus (W \otimes V) \oplus (\Lambda^2 W \otimes V^{\otimes 2}) \oplus \dots $$

When $V = 1$ we reproduce the exterior algebra functor. Note that there's no hope to upgrade the $GL(V)$ action on this to a $GL(1 \oplus V)$ action: each component of this direct sum contains as a factor a different irrep of $GL(W)$, so any such upgrade must continue to respect this direct sum decomposition, but this is impossible already for the summand $W \otimes V$.

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    $\begingroup$ This is somehow reminiscent of the chain complex structure (I think due to Illusie?) on $\Lambda^*(W)\otimes\operatorname{S}^*(W)$ quasiisomorphic to $1$, with (super-) $\Lambda^*(W)$ being "like $(1-k)^{\otimes W}$" and $\operatorname{S}^*(W)$ "like $(1-k)^{\otimes(-W)}$" - yours looks like a "noncommutative version" of it with tensor algebra in place of the symmetric algebra... $\endgroup$ Sep 30 '15 at 6:50
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    $\begingroup$ ...no, this analogy is not quite correct, as $V$ takes place of $k$ here. Still there is some connection I think $\endgroup$ Sep 30 '15 at 6:57
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    $\begingroup$ So this is natural w.r.t. the $GL(V)$ action. How about either of the parabolics containing it? (I.e. replace $1\oplus V$ by an extension of $V$ by $1$, or of $1$ by $V$). $\endgroup$ Oct 1 '15 at 0:06

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