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When $V$ is a finite dimensional $k$-vector space, I understand that there is the natural bijection between $(p, q)$-tensor product $\bigotimes_{p}V\otimes \bigotimes_{q}V^*$ and the space of multilinear maps $L\left(\prod_p V^*\times\prod_{q} V;K\right)$.

Is there a bijection between them when $V$ is infinite dimensional?

If $\{e_i\}_{I\in I}$ is a basis of $V$, I think a multilinear map $f$ is an infinite linear combination of $e_{i_1}\otimes\dots\otimes e^*_{i_{p+q}}$ and not a element of the tensor product space.

Ponta
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    If $V$ is countably infinite dimensional, then so is $V \otimes V$. However, $L(V \otimes V; K)$ is uncountably infinite dimensional and so there can be no isomorphism. – Joppy Nov 06 '19 at 23:24
  • @Joppy How could it be uncountably infinite-dimensional? (I'm assuming you mean non-separable by that.) If $V$ is separable (ie. there exists a dense countable set ${v_k} \subset V$), then the set of pairs $(v_i, v_j)$ is dense in $V^2$, isn't it? Then $V \otimes V$ should be also separable, as should its dual $L(V\otimes V; K)$. Am I missing something? – m93a Feb 09 '21 at 17:38
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    @m93a I was using $L(-, -)$ to mean the space of all linear maps (not necessarily bounded or continuous, since there was no mention of Banach spaces or norms/topology in the question). If you take $V$ to be a Banach space and $L(-, -)$ to mean the space of continuous linear maps, then I believe you do have $V \otimes V \cong L(V \otimes V, K)$. – Joppy Feb 09 '21 at 21:13

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