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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. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 30, 2015 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? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 30, 2015 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. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 30, 2015 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) $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 30, 2015 at 6:22
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    $\begingroup$ Over different fields?? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 30, 2015 at 6:26

2 Answers 2

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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... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 30, 2015 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 $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Sep 30, 2015 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 $V$). $\endgroup$
    – Allen Knutson
    Commented Oct 1, 2015 at 0:06
  • $\begingroup$ @AllenKnutson, re, doesn't the argument about different $\operatorname{GL}(W)$ representations already rule that out? For example, if $T - I$ is $0$ on $V$ and maps $1$ non-$0$-ly into $V$, then $T^{\otimes W} - I$ has nowhere to $\operatorname{GL}(W)$-equivariantly map the $1$ summand; whereas, if $T - I$ is $0$ on $1$ and maps $V$ non-$0$-ly to $1$, then $T^{\otimes W} - I$ has nowhere to map the $\bigwedge^\text{top}W \otimes V^{\otimes\text{top}}$ summand, at least if $\dim(\text{top}) \ne 0$ as a scalar. $\endgroup$
    – LSpice
    Commented Jun 24, 2023 at 19:00
  • $\begingroup$ This construction still makes perfectly good sense over any ring, in which generality it (more or less formally) is monoidal in $W$ (converting $\oplus$ canonically to $\otimes$); but I can't seem to show that it is monoidal in $V$ (converting $\otimes$ canonically to $\otimes$) unless $W$ is free of finite rank—which is, of course, no problem in the generality in which the problem was proposed, but makes me wonder whether it's not true in general, or I'm just not clever enough. $\endgroup$
    – LSpice
    Commented Jun 27, 2023 at 17:19
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I know it's not the question, but, anent now-user76479's / I guess-then Arul's comment and your response, one way to make use of an algebra structure on $W$, if it is present, is to define $V^{\otimes W} = (V^{\otimes\operatorname{Hom}_\text{$k^\text{sep}$-alg}(W, k^\text{sep})})^{\operatorname{Gal}(k^\text{sep}/k)}$ (i.e., one ‘factor’ in the tensor product for each element of the set of algebra homomorphisms), where $k$ is the underlying field and $k^\text{sep}/k$ is a separable closure. Of course this can fail to exhibit the desired dimension behaviour for a randomly chosen finite-dimensional $k$-algebra $W$, but, if $W$ is an étale algebra, then it works as desired.

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