I'm currently taking a course on the mathematical foundations of QM and we're formalizing tensor products. The definition I'm used to which was taught in my differential geometry course seems to be different than the one I'm seeing now.

Here's the definition I'm used to: Let $V$ be a vector field over a field $F$. Then a $(p,q)$ tensor is a multi linear map $T: V \times \cdots \times V \times V^* \times \cdots \times V^* \rightarrow F$ where we have $p$ copies of $V$ and $q$ copies of $V^*$. We denote the space of all such $T$ as $V^* \otimes \cdots \otimes V^* \otimes V \cdots \otimes V$.

The definition I encounter now is $V_1 \otimes \cdots \otimes V_k = (\text{k-Lin}(V_1, \ldots, V_k; F))^*$ where $\text{k-Lin}(V_1, \ldots, V_k; F)$ is the set of k-linear maps into $F$.

What's going on here? If $V$ is infinite dimensional then it is not necessarily the case that $V \cong (V^*)^*$ so these definitions don't actually match up?


The point I was trying to make was that the first definition implies that elements of $V \otimes \cdots \otimes V$ are multi linear maps on $V^* \times \cdots \times V^*$ whereas the second definition says that the elements are maps on $(V_1 \times \cdots \times V_k)^*$.

An answer suggested that both these definitions apply only to finite dimensional space in which case this could be resolved if $$ V^* \times \cdots \times V^* \cong (V_1 \times \cdots \times V_k)^* $$ is this correct?

  • 1
    $\begingroup$ The first definition seems to be the definition of a tensor on a vector space while the second is a definition of tensor product of several vector spaces. $\endgroup$
    – md2perpe
    Feb 1, 2019 at 17:34
  • $\begingroup$ @md2perpe hello! what I meant was that the first definition defining a tensor leads defining the tensor product space as the space of all such objects $T$, I made an edit to make it more clear what im trying to say $\endgroup$
    – David Feng
    Feb 1, 2019 at 18:13

1 Answer 1


Both of these definitions assume finite-dimensionality of $V$, if you want them to agree with what most mathematicians might say. For example, one might say that $V_1\otimes\dots\otimes V_k$ is an $F$-vector space along with an $F$-multilinear map $\mu: V_1\times\dots\times V_k \to V_1\otimes\dots\otimes V_k$ so that for any $F$-vector space $W$ and $F$-multilinear map $A: V_1\times\dots\times V_k \to W$, there is a unique $F$-linear map $\alpha:V_1\otimes\dots\otimes V_k \to W$ such that $\alpha \circ \mu = A$. (Such $F$-vector-space-map pairs exist, as one can construct such a thing, and they are unique up to unique isomorphism, in a precise way).

One can prove that if $B_1,\dots,B_k$ are bases of $V_1,\dots,V_k$, then the dimension of $V_1\otimes\dots\otimes V_k$ is $|B_1\times\dots\times B_k|$. This will rule your definitions out.

However: I don't know whether these definitions are just extended to infinite-dimensional vector spaces in physics, ignoring that it will no longer match the above definition. That could well be.

In order to see whether the two definitions at least agree for infinite-dimensional spaces, I would want a more precise statement of the first definition. Are you saying that the first definition would say that $V\otimes\dots\otimes V$ (the space of $(k,0)$-tensors) is the space of $F$-multilinear maps $V^*\times\dots\times V^* \to F$, where $V$ is repeated $k$ times in the first and $V^*$ is repeated $k$ times in the second?

  • $\begingroup$ hi thanks for the response. The definition you provided in the infinite dimensional case is not one i've encountered in the case of physics, although in our course we do make it explicit that we only study finite dimensional case, I don't know about how this definition is extended in physics. For the second point, you are right. If we look at the specific case of $(k, 0)$ tensors, the dual is applied to each individual copy of $V$ whereas the second definition is the dual of the products of $V$, I made an edit to clarify. $\endgroup$
    – David Feng
    Feb 1, 2019 at 18:18
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    $\begingroup$ Okay. Then in the infinite-dimensional case, your two definitions do in fact disagree. And they also disagree with the definition I gave. All three are different when $V$ is infinite-dimensional. I should also say that by 'different', we are just talking about having different dimension. That's not really the end-all. The definitions are just different regardless -- there are no natural isomorphisms of the three, when considered as functors on vector spaces, even in the finite-dimensional case. (At least I think there aren't.) $\endgroup$
    – user555203
    Feb 1, 2019 at 18:39
  • $\begingroup$ does your definition for the infinite dimensional case reduces to either of the two definitions I have in the finite dimensional case ? $\endgroup$
    – David Feng
    Feb 1, 2019 at 18:42
  • $\begingroup$ Because it would seem that the two definitions I have do in fact agree in finite dimensions up to isomorphism, if $V \cong V^*$ I think.... $\endgroup$
    – David Feng
    Feb 1, 2019 at 18:43
  • $\begingroup$ What do you mean? They all have the same dimension in the finite-dimensional case, but that's not really satisfying anyway ($V$ and $V^*$ are isomorphic in the finite-dimensional case, but this isomorphism isn't natural, in a precise sense). In the infinite-dimensional case, all three definitions differ even in dimension. $\endgroup$
    – user555203
    Feb 1, 2019 at 18:44

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