I am looking for an explicit isomorphism $Hom(V,V^*)\rightarrow V^*\otimes V^*$ where $V$ is a vector space.
I thought of:
$\phi\rightarrow ((u,v)\rightarrow \phi(u)(v))$
But I'm not sure this works.
Does anyone have a suggestion?
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1What is your definition of $V\otimes V$, and how is the bilinear map$$(u,v)\mapsto \phi(u)(v)$$an element of it? – Zev Chonoles Jun 24 '13 at 10:44
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I see $V\otimes V$ as the space of all bilinear forms on $V\times V$ – Ian Jun 24 '13 at 10:48
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1I think that is what most people would consider $(V\otimes V)^*$. – Zev Chonoles Jun 24 '13 at 10:50
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Isn't it the one also denoted $T^2(V)$? – Ian Jun 24 '13 at 10:56
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For the map in the direction "Hom---> Tensor" you need to work with a basis on $V$. The other way round, as shown by @Andreas Caranti is easier and more natural. – Avitus Jun 24 '13 at 11:18
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I have edited the text accordingly – Ian Jun 24 '13 at 11:19
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I normally think of it the other way around, that is, $$ V^{\star} \otimes W \to \hom(V, W), \qquad \varphi \otimes w \mapsto (v \mapsto \varphi(v) w). $$
PS I am assuming $V, W$ to be finite-dimensional, see the comments below.
Andreas Caranti
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2@Ian: Because vector space homomorphisms can be represented by matrices. The elementary matrix $E_{i,j}$ on a given bases $e,f$ of $V,W$, repspectively, is the image of $e_j^\otimes f_i$. This even works in infinite dimension if you directly give a basis $e^$ of $V^*$. – Marc van Leeuwen Jun 24 '13 at 11:25
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4I retract the part about infinite dimension. Indeed since tensor products contain only finite sums of elementary tensors, any homomorphism in the image of this correspondence has as image a finite dimensional subspace of $W$. Taking $V=W$ of infinite dimension, the identity of $V$ is not in the image of the correspondence. – Marc van Leeuwen Jun 24 '13 at 11:35