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Every vector space $V$ could be embedded into $V^{\ast}$ (see here) after choosing a basis, for a given vector $v \in V$ denote this embedding by $v^{\ast}\in V^{\ast}$. Now for given vector spaces $V_1, \ldots, V_k$ over some field $F$, let $V = \{ \varphi : V_1 \times \ldots \times V_k \to F \mbox{ multilinear } \}$. Why not define the tensor product of $V_1, \ldots, V_k$ simply as $T = \{ \varphi^{\ast} \mid \varphi\in V\}$. Then the universal property is obviously fulfilled, for if we define $\pi : V_1 \times \ldots \times V_k \to T$ by $\pi(v_1, \ldots, v_k) = \Phi \in V^{\ast}$ with $$ \Phi(\varphi) = \varphi(v_1, \ldots, v_k). $$ Then if we have some multilinear $\varphi : V_1 \times \ldots \times V_k \to F$ define the linear map $h_{\varphi} : T \to F$ by $$ h_{\varphi}(\Phi) = \Phi(\varphi) $$ and we have $$ h_{\varphi}(\pi(v_1, \ldots, v_k)) = \varphi(v_1, \ldots, v_k) $$ i.e. it factors through $T$ by $\pi$ and $h_{\varphi}$. Then everything works out quite easily, no nasty "quotient constructions", it even appears too simple for me...

I have nowhere seen this definition? So why not define it that way? Have I overlooked something? Note that we do not rely on reflexivity here, as $T$ does not has to be all of $V^{\ast}$, but just those elements that arise from elements of $V$ (the image of the embedding). Maybe the universal property breaks down because the linear map is not unique, but I do not see other choices for it?

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    $\begingroup$ Note that your proof of the universal property only works for multilinear maps to $F$. You should prove that it true for multilinear maps to any vector space $W$. This can be done for vector spaces once you choose a basis of $W$, but you should prove that it doesn't depend on the chosen basis. $\endgroup$ – Arnaud D. Oct 19 '17 at 12:28
  • $\begingroup$ You say that $T$ does not need to be the whole dual of $V$, but it's not obvious that your construction of $\pi$ actually makes it a multilinear map $V_1\times\dots \times V_k\to V^*$. It's not obvious to me that it really factors through $T$! In fact, I'm not even sure that your $T$ is really well-defined, since you need to pick a basis of $V$... $\endgroup$ – Arnaud D. Oct 19 '17 at 12:45
  • $\begingroup$ You asking if $\pi$ is multilinear? Yes it is, $\pi(v_1 + v_1', \ldots) = \Phi$, in short: $\Phi(\varphi) = \varphi(v_1 + v_1',\ldots) = \varphi(v_1,\ldots) + \varphi(v_1',\ldots)$ by multilinearity of $\varphi$ and the last two summand equal $\pi(v_1,\ldots)$ and $\pi(v_1',\ldots)$. That it factors is also a short computation, actually it is just "reading" the definitions verbatim, it is choosen so that it works out in the end. What you do not understand? $\endgroup$ – StefanH Oct 19 '17 at 12:53
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    $\begingroup$ No, I can see that $\pi$ is multilinear. What I'm asking is why $\Phi\in T$. What is the multilinear map $\mu:V_1\times \dots \times V_k \to F$ for which $\Phi=\mu^*$? $\endgroup$ – Arnaud D. Oct 19 '17 at 12:56
  • $\begingroup$ Oh yes, have not noticed that, but indeed guess this is the point where it breaks down. Would you like to make your comment into an answer so I can give it credit... $\endgroup$ – StefanH Oct 19 '17 at 13:13
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You claim that you only need $T$ to be a subspace of $V^*$; but in your construction, you define $\pi$ as a multilinear map $V_1\times \dots \times V_k\to V^*$, and it is not obvious that $\pi$ can be restricted to have domain $T$. In other words, given $v_i\in V_i$ for $i=1,\dots,k$, it is not obvious $\pi(v_1,\dots,v_k)=\Phi$ can be written as $\mu^*$ for some $\mu\in V$.

This is particularly difficult because the embedding $V\to V^*$ depends on the choice of a basis for $V$.

Note also that your proof of the universal property is incomplete : you only consider it for maps to $F$, but the universal property should hold for multilinear map to any vector space. The extension can be done for vector spaces if you choose a basis, but that would fail for modules over $\Bbb Z$, for example; and moreover, you would have to prove that it is independent on the chosen basis (otherwise you wouldn't have uniqueness).

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I haven't read all the details of your construction, but if you look at Halmos' classic text on linear algebra, Finite dimensional vector spaces, you will see he defines $V\otimes W$ as the linear dual of bilinear forms $V\times W\to k$, where $k$ is the base field, which seems to be what you suggest. This is Chapter 1, $\S$25.

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The problem with this is extending it to infinite dimensional vector spaces and to arbitrary modules over arbitrary rings, of course, as Halmos explains:

enter image description here

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  • $\begingroup$ (Halmos' book is, albeit not modern, a beautiful text to read!) $\endgroup$ – Pedro Tamaroff Oct 19 '17 at 12:28
  • $\begingroup$ He says this relies on reflexivity, but by just considering the embedding of $V$ into $V^{\ast}$ we do not need reflexivity! So what I suggest should also works for infinite dimensional spaces, and hence should be totally equivalent to the usual definition! $\endgroup$ – StefanH Oct 19 '17 at 12:28
  • $\begingroup$ Surely this is in line with what Halmos is doing, but just do not consider the entire dual. I am not sure why Halmos, as a true master, has not made this small step to even include the infinite dimensional one (and in essence that is what I am asking) by just considering the image of the embedding instead of the entire dual space. $\endgroup$ – StefanH Oct 19 '17 at 12:31
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    $\begingroup$ @StefanH The clue is in the title of the book: "Finite dimensional vector spaces." It might be a "small step" to extend this detail to infinite dimensional spaces, but it would have been a "giant leap" to add a general treatment of infinite dimensional spaces to the whole book! Introductions to linear algebra often ignore infinite dimension spaces completely, except possibly to mention that they exist. $\endgroup$ – alephzero Oct 19 '17 at 20:43
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Let me mention something which has not been explicitly stated in any of the other answers: your construction is emphatically wrong for infinite-dimensional vector spaces (not just, you have to tweak it or do more work prove it works, but it just doesn't give the right answer at all).

This is easiest to see by considering the case $k=1$. The tensor product of just a single vector space $V_1$ should, of course, be (isomorphic to) $V_1$ itself, since a multilinear map on the one-fold product $V_1$ is the same thing as a linear map on $V_1$. However, your definition of $T$ would make it isomorphic to $V$, which is just the dual space $V_1^*$ in this case. If $V_1$ is infinite-dimensional, then $V_1^*$ is not isomorphic to $V_1$!

(The basic thing that goes wrong with your argument in this case is that you haven't checked uniqueness of your $h_\varphi$. You have only constrained the value of $h_\varphi$ on elements $\Phi$ of the form $\pi(v_1)$, but not every element of $T$ has this form, since $V=V_1^*$ is bigger than $V_1$.)

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  • $\begingroup$ Thanks for pointing out. But I have defined $h_{\varphi}$ just for $T$ (not the entire dual space, $T$ is just the image of the embedding map, and for me its not obviously non-unique). Your other remark, yes its quite natural to demand that the $1$-fold tensor product is isomorphic to $V$. But let me add that in Spivaks text the $1$-fold tensor product is also $V^{\ast}$. $\endgroup$ – StefanH Oct 20 '17 at 14:54
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    $\begingroup$ But in this case $T$ is (isomorphic to) the entire dual space $V_1^*$, since $T$ is by definition isomorphic to $V$ and $V$ is the dual space of $V_1$. The elements of $T$ of the form $\pi(v_1)$ form a subspace isomorphic to $V_1$, which is therefore not the entirety of $T$ since $T$ has larger dimension than $V_1$ does. So a linear map on $V$ is not uniquely determined by its restriction to this subspace, and so $h_\varphi$ is not unique. $\endgroup$ – Eric Wofsey Oct 20 '17 at 15:09

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