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Suppose $V_1,\dots,V_k$ are vector spaces of finite dimension. Then I could prove easily that $(V_1\otimes\cdots\otimes V_k)^\ast\simeq V_1^\ast\otimes\cdots\otimes V_k^\ast$. My proof was like that: first of all, I've shown that if for each $i$ we have $W_i$ another vector space such that $V_i\simeq W_i$ then $V_1\otimes\cdots\otimes V_k \simeq W_1\otimes\cdots\otimes W_k$.

Then, since I'm supposing each $V_i$ finite dimensional, each $V_i\simeq V_i^\ast$ and also, we have $V_1\otimes\cdots\otimes V_k$ also finite dimensional, so that

$$(V_1\otimes\cdots\otimes V_k)^\ast \simeq V_1\otimes\cdots\otimes V_k\simeq V_1^\ast\otimes \cdots \otimes V_k^\ast$$

and so it is proved. Now, if the spaces are not finite dimensional this proof cannot be used. In that case, the property still holds? Is it possible to prove for infinite dimensional spaces?

I've tried to prove it directly, constructing an isomorphism. I've picked first the mapping $\psi : V_1^\ast\times\cdots\times V_k^\ast \to \mathcal{L}(V_1,\dots,V_k;\mathbb{K})$ given by

$$\psi(f_1,\dots,f_k)(v_1,\dots,v_k) = f_1(v_1)\cdots f_k(v_k)$$

this map is multilinear and hence by the universal property there corresponds a unique linear mapping $\phi : V_1^\ast\otimes \cdots \otimes V_k^\ast \to \mathcal{L}(V_1,\dots,V_k;\mathbb{K})$ such that:

$$\phi(f_1\otimes\cdots\otimes f_k)(v_1,\dots,v_k) = f_1(v_1)\cdots f_k(v_k)$$

To show that this $\phi$ is isomorphis I would need to find an inverse, but I didn't have any idea. Is it possible to complete this proof?

Thanks very much in advance.

Gold
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    1. The infinite-dimensional case is false. A counterexample is the element $e_1^{\ast} \otimes e_1^{\ast} + e_2^{\ast} \otimes e_2^{\ast} + e_3^{\ast} \otimes e_3^{\ast} + ...$ in $\left(V\otimes V\right)^{\ast}$, where $V$ is the free vector space with basis $\left(e_1,e_2,e_3,...\right)$, and where $\left(e_1^{\ast}, e_2^{\ast}, e_3^{\ast}, ...\right)$ denotes the corresponding dual basis of $V^{\ast}$. This is well-defined (check why! it's not like every infinite sum automatically makes sense, but this one does) but manifestly not in $V^{\ast} \otimes V^{\ast}$. – darij grinberg Feb 04 '14 at 20:04
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    2. Your proof of the finite-dimensional case is correct, but it proves the letter of the statement, not the spirit. What you should be proving is that the canonical injection $V_1^{\ast} \otimes V_2^{\ast} \otimes ... \otimes V_k^{\ast} \to \left(V_1 \otimes V_2 \otimes ... \otimes V_k\right)^{\ast}$ (which sends $f_1 \otimes f_2 \otimes ... \otimes f_k$ to $f_1 \otimes f_2 \otimes ... \otimes f_k$ with the understanding that the tensor signs in these two terms have completely different meanings!) is an isomorphism. This is a stronger claim! – darij grinberg Feb 04 '14 at 20:07

1 Answers1

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The situation here is a little subtle.

The canonical inclusion $(V^*_1\otimes\dots\otimes V_k^*)\hookrightarrow(V_1\otimes\dots\otimes V_k)^*$ isn't an isomorphism (it's a strict injection), but (assuming the axiom of choice) the spaces are nevertheless isomorphic by some other (non-canonical) map.

The proof that the canonical map isn't an isomorphism is explained in detail in this answer to another question.

The proof that they are isomorphic works by just calculating the dimensions on both sides and finding that they are the same. The dimensions are infinite cardinals. That proof is given here.

Ken
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Oscar Cunningham
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  • I don't think your second link gives a proof that the dimensions are equal. I think it just shows that the dimensions can be equal. It assumes that both dimensions are $\geq$ to the size of the ground field. – darij grinberg Feb 03 '19 at 06:39
  • @darijgrinberg You're right. I think the statement is still true. There's a proof here, but it involves Choice. Do you know of a proof or counterexample not needing Choice? – Oscar Cunningham Feb 03 '19 at 09:20
  • I think it's worth pointing out that this non-canonical isomorphism is pretty much useless in virtually all situations. The results is theoretical, even in theory. – shuhalo Nov 24 '22 at 21:22