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Let a finite-dimensional vector space $V$ over $\mathbb R$ or $\mathbb C$ with dual $V^*$ and a group $G$ be given. Let $\rho:G\to\mathrm{GL}(V)$ be a representation, and let $T_kV$ and $V^{\otimes k}$ denote the vector space of $k$-tensors on $V^*$ and the $k$-fold tensor product of $V$ with itself respectively. The $k$-fold tensor product representation $\rho^{\otimes k} :G\to\mathrm{GL}(V^{\otimes k})$ is defined in the standard way as \begin{align} \rho^{\otimes k}(g)v_1\otimes\cdots\otimes v_k = (\rho(g)v_1)\otimes\cdots\otimes(\rho(g)v_k) \end{align} Now, I'm not sure this is standard, but let's now define a dual representation $\rho^*:G\to\mathrm{GL}(V^*)$ as follows \begin{align} (\rho^*(g)f)(v) = f(\rho(g^{-1})v) \end{align} a rank $k$ tensor representation $\rho^k:G\to\mathrm{GL}(T_kV)$ as \begin{align} (\rho^k(g)T)(f_1, \dots,f_k) = T(\rho^*(g^{-1})f_1,\dots,\rho^*(g^{-1})f_k) \end{align} Then

Is $\rho^k$ equivalent to $\rho^{\otimes k}$?

I'm relatively certain that the answer is yes, as I think I can prove it. My proof, however, proceeds by choosing a basis for $V$ and corresponding dual bases for $V^*, V^{**}$, exhibiting an explicit isomorphism $V^{**}\to V$, using this to induce an isomorphism $T_k(V)\to V^{\otimes k}$, and then showing that this isomorphism is an equivalence of representations between $\rho^k$ and $\rho^{\otimes k}$.

If the answer to the above question is yes, then is there a slicker way to prove it that what I outlined? Perhaps one that does not appeal to bases?

Note. This is all inspired by theoretical physics in which one encounters objects like the tensor representation I defined above all the time, and statements are made about the decomposition of such tensors using results about tensor product representations. I'm trying to make mathematical sense out of this stuff.

Thanks for the help.

joshphysics
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  • How are you defining $k$-tensors if it isn't in terms of the tensor product? – Qiaochu Yuan Aug 13 '13 at 19:57
  • $T_k(V) = \text{multilinear maps on $(V^*)^k$}$ – joshphysics Aug 13 '13 at 19:58
  • Hello. Perhaps I am misundertanding your notation, but it seems like the character of the one is $\chi_\rho^k$ and the other is $\overline{\chi_\rho}^k$. No? – Alex Youcis Aug 13 '13 at 20:19
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    You can define an explicit map $V \to V^{\ast\ast}$ without choosing bases. You need to choose a basis in order to prove that it is an isomorphism, but that's only a proof of a lemma. Once you know that it is isomorphism, you can take its inverse, which is an isomorphism $V^{\ast \ast} \to V$, and proceed as you do. There is no slicker argument like this. If you could completely avoid choosing bases and similar things, then the result would hold for infinite-dimensional vector spaces and for non-free modules over commutative rings; but it does not, and thus this is the best you can do. – darij grinberg Aug 13 '13 at 20:20
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    You can construct an isomorphism $V \to V^{*}$, independent of any basis: given $v \in V$ we get a linear map $V^ \to \mathbf R$ by $f \mapsto f(v)$. You can use a basis to verify this is an isomorphism but the definition of the map itself is natural. – Eric O. Korman Aug 13 '13 at 20:21
  • @AlexYoucis: I don't think there is any complex conjugation involved in his definition of $V^{\ast}$. – darij grinberg Aug 13 '13 at 20:21
  • @darijgrinberg Hmm, then I must be misinterpreting something. Because, it seems to me like that $\rho^\ast$ is the dual representation to $\rho$, and has character $\overline{\chi_\rho}$ for a complex representation. – Alex Youcis Aug 13 '13 at 20:23
  • @darijgrinberg Thanks that clears things up a bit. So that I can sleep well, is it implicit in your answer that the equivalence holds? Thanks again. – joshphysics Aug 13 '13 at 20:27
  • @EricO.Korman Oh right; thanks! – joshphysics Aug 13 '13 at 20:28
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    @joshphysics: yes, it is. – darij grinberg Aug 13 '13 at 20:44
  • @AlexYoucis I know next to nothing about characters, but I can say that I think I had an error in the original post; the dual rep needs to be defined with $g^{-1}$ on the RHS instead of $g$, and similarly for the tensor rep. Does this relate to your observation? – joshphysics Aug 13 '13 at 22:10

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Why hasn't any mathematician said anything...: there is no canonical map from a vector space to its dual!

But given a basis (which is also possible in infinite dimension if one uses the axiom of choice) then one can build a linear map V to its dual. But given another basis than one has another map. Moreover in infinite dimension this is not an isomorphism.

However when one has a scalar product then one has a canonical isomorphism to the dual even in infinite dimension.

Also there is no need to consider k fold tensor product. The dual representation also know as contragredient representation is of course not always equivalent to the original representation otherwise there won't be two different words for it. That is very much related to "covariant" and "contravariant" tensors in general relativity. The two are equivalent if one has a unitary representation of your group.

Noix07
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