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Problem:

Let $V$ be a finite dimensional vector space. $T^{1}_{1}(V)\cong End(V)$ canonically.

Attempt:

Define $\varphi: T^1_1(V)\rightarrow End(V)$ by

$\varphi(f)(v)(\omega)=f(\omega,v)$ for $\omega \in V^{*}$ and $v\in V$

$\varphi$ is easily to be seen linear. We thus only establish $\varphi$ is an isomorphism.

$\varphi$ is injective: Assume $\varphi(f)=0$. We must show for $\omega \in V^{*}$ and $v\in V$, $f(\omega,v)=0$. Indeed, $0=\omega(f)(v)(\omega)=f(\omega,v)$. $\varphi$ is surjective: Let $g\in End(V)$. Define $f:V^{*}\times V\rightarrow \mathbb{R}$ by $f(\omega,v)=g(v)(\omega)$. Then, $\varphi(f)=g$

Is the proof for showing it is an isomorphism correct?

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    This is not true if $V$ is not finite dimensional, because the image of $\varphi$ is the subset of linear endomorphisms with finite rank. Injectivity is still true, and in finite dimension, injectivity and surjectivity are equivalent when the dimensions of the LHS and RHS are the same. – Didier Nov 23 '21 at 12:10
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    @Didier where does the proof go wrong? –  Nov 23 '21 at 12:17
  • First, you should look at $V^\otimes V$ instead of $V^\times V$. – Didier Nov 23 '21 at 12:21
  • @Didier we haven't defined $V^{*}\otimes V$ in class –  Nov 23 '21 at 12:23
  • Then how did you define tensors in $T^1_1$ without tensors products? – Didier Nov 23 '21 at 12:28
  • @Didier multilinear maps $V^{*}\times V\rightarrow \mathbb{R}$ –  Nov 23 '21 at 12:34
  • Well, then it is not really $V^\times V$. If $V$ is $n$ dimensional, so is $V^$, and $V^*\times V$ is $2n$ dimensional, which has no big chance to be equal to the dimension of $End(V)$ (which is $n^2$)! – Didier Nov 23 '21 at 16:57
  • @Didier I don't think there's anything wrong with saying $V^\otimes V$ is identified with multilinear maps $V^ \times V\to \mathbb{R}$ though? Note that in the proof attempt, $f$ is not the isomorphism $T_1^1(V)\to\operatorname{End}(V)$, but $\phi$ is, and $f$ is just an element of the former space. – Elliot Yu Nov 23 '21 at 18:25
  • @ElliotYu You are totally right, there is nothing wrong with the identification $Bil(V^,V;\Bbb R) \simeq V^\otimes V$, but it seems the question does not refer to any "multilinear $(V^*,V) \to \Bbb R$" in the question, and I somehow missed it in the comment. The result it is not surjective in infinite dimension stays true however – Didier Nov 23 '21 at 18:45
  • @Didier Though your point still stands, namely that when $V$ is infinite-dimensional, $\phi$ is not an isomorphism. In that case $\operatorname{End}(V)$ has larger cardinality than $T_1^1(V)$, so I suspect it's the surjectivity part of the proof that went wrong (as you pointed out in the first comment). In particular, I notice that the bilinearity of $f$ defined by $f(\omega, v) = g(v)(\omega)$ is not verified. – Elliot Yu Nov 23 '21 at 19:09
  • @ElliotYu notice that in my proof, i'm canonically identifying $V$ with $V^{**}$, this identifiation is canonical provided $V$ is finite dimensional –  Nov 23 '21 at 19:41
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    In that case you should put "finite dimensional" in the statement, and your proof would probably be fine. Right now it only says that "Let $V$ be a vector space" without any mention of finite-dimensionality. – Elliot Yu Nov 23 '21 at 19:46

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