So, I've been trying to get some grasps on Tensors, and now I've come to look at one particular classical definition, that goes around Tensor Products.

So, I think I was able to follow the conventional way of defining that, which is through its universal product. That really clears up how those are much (of course) closely related to multilinear maps. Explicitly, for a $k\in\mathbb{Z}_{\geq0}$, a tensor product for a family of vector spaces $V_1,\ldots,V_k$ is the pair $(\varphi,V_1\otimes\cdots\otimes V_k)$, for a k-linear map $\varphi:X\to V$ and a vector space $V_1\otimes\cdots\otimes V_k$ which is universal over $X=V_1,\ldots,V_k$ under the properties of every map with domain $X$, say, $\psi:X\to Z$ is $k$-linear (call that $P_1$), and a respective unique map $\tilde{\psi}:V_1\otimes\cdots\otimes V_k\to Z$ is $\text{linear}$ (call that $P_2$). That is, $(\varphi,V_1\otimes\cdots\otimes V_k)$ is a tensor product for the family of vector spaces defined if \begin{equation}{\rm for \ all \ } \psi\in\text{Hom}_{\mathbb{F}}^{k}(V_1,\ldots,V_k,Z) \ {\rm there \ is \ unique \ } \tilde\psi\in\text{Hom}_{\mathbb F}(V_1\otimes\cdots\otimes V_k,Z)\ \text{s.t}\ \psi=\tilde\psi\ \circ\ \varphi\end{equation} For a vector space $Z$. Equivalently, it's a tensor product if the following diagram commutes:


That really says the exact same thing, in an illustrative manner. The original definition even uses directly $V$ instead of the notation $V_1\otimes\cdots\otimes V_k$. Of course, understanding that would be almost sufficient to see why tensors are elements of tensor products, but I don't really see it clearly. Why is $V_1\otimes\cdots\otimes V_k$ actually an arbitrary vector space $V$?

Goes a little fuzzier when we define the tensor itself: \begin{equation}T\in\bigotimes_{p\in P}V\otimes\bigotimes_{q\in Q}V^{*},\end{equation} which is not quite different from simply defining it via a simple multilinear mapping from $V^p\times \prod_{q\in Q}V^*$ to the field of scalars$^{[1]}$.

Why do we take the product within the dual as well?

Summarizing, two basic questions: How does it follow from those definitions that (i) a tensor product of copies of vector spaces is itself a vector space, implying that any vector space is a tensor product and (ii) why do we have to include the dual to properly define the tensor?

Since the dual traces all the linear maps to the scalars, it's not completely obscure why it would be interesting to include it, but I think a little push is yet needed for me to completely accept the thing.

(1): I've alternated the notations for the cartesian product $V^k=\prod_{k\in K}V$ of the copies of the vector space $V$, just to avoid the not so nice notation for the dual $V^{*k}$ or $V^{k*}$. I've also freely used index sets without expliciting it, guess that's okay.

  • $\begingroup$ If $V$ is finite dimensional, a $(p,q)$ tensor over $V$ is an element of $V^{\otimes p}\otimes (V^*)^{\otimes q}$, or equivalently a multilinear map $(V^*)^p\times V^q \to F$. Notice the powers and the duals. You have them wrongly placed in your post. $\endgroup$ Commented Sep 28, 2022 at 1:47
  • $\begingroup$ Note the LaTeX commands \cdots (centered dots) and \ldots (lower dots). Use \cdots instead of typing \cdot three times, and use \ldots for lists rather than \cdots. $\endgroup$
    – KCd
    Commented Sep 28, 2022 at 2:15
  • $\begingroup$ @JackozeeHakkiuz Sorry, I don't know how does that affect anything. I'm using a slightly (yet quite the same) different notation than you, since I took the actual operators $\prod, \bigotimes$, instead of the exponents. $\endgroup$ Commented Sep 29, 2022 at 0:01
  • $\begingroup$ @Adrien-MarieDeschamps it's not the notation. It's the amount of $p$s and $q$s. $\endgroup$ Commented Sep 29, 2022 at 2:12

1 Answer 1


Let me try to answer question (ii). If $V$ is a vector space over a field $F$, then the space of $(p,q)$-tensors over $V$ is the space $$T^p_q(V)=V^{\otimes p} \otimes (V^*)^{\otimes q}.$$ A $(p,q)$-tensor over $V$ is an element of $T^p_q(V)$. Now, if $V$ has finite dimension, then you have \begin{align} T^p_q(V) &\simeq (T^p_q(V))^{**} \\ &= [V^{\otimes p} \otimes (V^*)^{\otimes q}]^{**} \\ &\simeq [(V^*)^{\otimes p} \otimes V^{\otimes q}]^{*} \\ &= \hom_F((V^*)^{\otimes p} \otimes V^{\otimes q},F) \\ &\simeq \mathrm{mult}((V^*)^{\times p} \times V^{\times q},F). \end{align} The first isomorphism is because every $T^p_q(V)$ also has finite dimension and so it is isomorphic to its double dual, the second isomorphism is because the tensor product commutes with the dual and the third isomorphism is the defining property (universal property) of the tensor product.

So $(p,q)$-tensors over $V$ are in correspondence with multilinear maps $$(V^*)^{\times p} \times V^{\times q} \to F.$$ Notice that the $p$ now is on $V^*$ and the $q$ is on the $V$. This is the reason of my comments about the position of $p$ and $q$: you have them the other way around in your post.

Now, this said, let me address your question (i). Any vector space $V$ is isomorphic to its space of $(1,0)$-tensors: you have $T^1_0=V^{\otimes 1}\otimes V^{\otimes 0}=V$.

Please let me know if my answer was helpful.

  • $\begingroup$ To add on to this a bit, linear maps $(V^*)^{\otimes p}\otimes V^{\otimes q} \to F$ are equivalent (in the sense that they are completely determined by) bilinear maps $(V^*)^{\otimes p}\times V^{\otimes q} \to F$; so since $(V^{\otimes p})^*$ is naturally isomorphic to $(V^*)^{\otimes p}$, the tensors $T^p_q(V)$ are equivalent to all linear maps $V^{\otimes q} \to V^{\otimes p}$. That is to say that tensors $T^p_q(V)$ are just the linear maps between tensor powers. $\endgroup$ Commented Sep 30, 2022 at 19:09
  • $\begingroup$ It is helpful, thank you. Question (i) was silly. I do see how using duals work, but my question remains: Why do we use them? The correspondence would follow as well if we defined $T_{q}^{p}(V)=V^{\otimes p}$. There's surely something deeper I'm missing. $\endgroup$ Commented Sep 30, 2022 at 19:48
  • $\begingroup$ @Adrien-MarieDeschamps what correspondence are you talking about? the correspondences I mentioned in my post do not work for $V^{\otimes p}$ alone. $\endgroup$ Commented Oct 3, 2022 at 17:40

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