First of all, assume all finite-dimensional vector spaces.I have two other questions about this subject that maybe the one would like to read to understand my kind of doubts and some self progress about the subject.

Silly Question about tensor products and universal property

Difficulties about Tensor Products

$$ * * *$$

$\textbf{Definition (Tensor Product):}$ A $Tensor$ $Product$ constructed from vector spaces $V$ and $W$ is a pair $(T, \vec{\otimes})$, where $T$ is a vector space and a bilinear map:

$$\vec{\otimes}: V\times W \to T \\ (v,w)\to v\vec{\otimes}w$$

that satisfies the $Universal$ $Property$: given any other vector space $Z$ and any other Bilinear map $B:V\times W \to Z$ always exists just one linear map $L:T\to Z$ such that $B = L\circ \vec{\otimes}$

In other words,the whole construction means that you earn the abillity to change multilinear studies by linear ones, via the linear map $L$. Another way to say that is by figure 1, that shows the commutative diagram.

figure 1

This is the definition and intuition about this machinery. In despite of say whats is a Tensor product, we want to show explicitly an example of the space $(T,\vec{\otimes})$. Commonly we introduce this construction with Quotient spaces, but there's another identification that is the core of the tensors of physics: when we show some isomorphisms using spaces like $Lin(V\times W;R)$ or $Lin(V^{*}\times W^{*};R)$. Here is the point of my doubts!

$$ * * * $$

$\textbf{Proposition:}$ The map: $$B: V\times W \to Lin(V^{*}\times W^{*};R) \\ (v,w) \to [B(v,w)](f,g):=f(v)g(w)$$

induces an isomorfism:

$$(T,\vec{\otimes}) \cong Lin(V^{*}\times W^{*};R) $$

So, cleary $B$ is bilinear. By Universal Property we know that exists a linear map $L: T\to Lin(V^{*}\times W^{*};R)$ such that:

$$[L(v\vec{\otimes}w)](f,g) = [B(v,w)](f,g):=f(v)g(w)$$

in the next figure we have the commutative diagram. enter image description here

Well, here the things start to get really obscure. The author concludes the proof defining another bilinear function called $tensor$ $product$: $$ \otimes: V\times W\to Lin(V^{*}\times W^{*};R)\\ [v\otimes w](f,g):= f(v)g(w)$$

in order to prove the isomorphic relation. He said that this is just for the sake of convenience, but I already read that trick in other books too. It seems that you must to define another bilinear function (being the same of your $B$) to throw up the Quotient space construction and prove the isomorphism above. Here is another "commutative diagram" just to illustrate that:

enter image description here


$$[(T,\vec{\otimes}) \equiv V\vec{\otimes}W] \cong Lin(V^{*}\times W^{*};R)$$

and if,

$$[v\otimes w](f,g):= f(v)g(w)$$ and $$[L(v\vec{\otimes}w)](f,g) = [B(v,w)](f,g):=f(v)g(w)$$

Then: $$[v\otimes w](f,g) = [L(v\vec{\otimes}w)](f,g)$$

And finally we have:

$$V\vec{\otimes}W \cong [V\otimes W\equiv (Lin(V^{*}\times W^{*};R),\otimes)]$$

$$ * * * $$

Well, I don't understand the need of the new bilinear map $\otimes$, and why we can throw up quotient spaces constructions!Also, the author concludes that we can stablish other identifications like: $V\otimes V^{*}$,$V^{*}\otimes V$,etc... And actually I don't be able to see this whole machinery acting when I read things like:

$$g = g_{ij}\textbf{d}x^{i}\otimes\textbf{d}x^{j}$$

It's a serious problem because I know how to calculate with tensors but I don't know what they really are.

  • $\begingroup$ The identification of $V\otimes W$ with $(V^*\times W^*)^*$ is only valid when $V$ and $W$ are finite-dimensional vector spaces. An argument that attempts to prove this without using finite-dimensionality cannot be sound. $\endgroup$ Mar 3, 2018 at 6:12
  • $\begingroup$ All the vector spaces here are finite-dimensional. $\endgroup$
    – M.N.Raia
    Mar 3, 2018 at 6:55
  • $\begingroup$ But in the proofs one must use finite-dimensionality not just assert it. $\endgroup$ Mar 3, 2018 at 6:56
  • $\begingroup$ Ok, I didn't follow you. Clearly my problem is about other thing, and not whether the vector spaces here are finite-dimensional vector spaces. $\endgroup$
    – M.N.Raia
    Mar 3, 2018 at 12:57
  • 1
    $\begingroup$ Please, next time you ask a question (if possible, also in this one), consider this advice. Specifically, the part concerning "Make your actual question stand out". $\endgroup$
    – Aloizio Macedo
    Mar 8, 2018 at 23:36

1 Answer 1


The map which the author defined as the tensor map $ \ \otimes: V \times W \to Lin(V^* \times W^*; R) \ $ is exactly the map $B$ previously defined. That is just unnecessary to prove the theorem and a bad notation. Forget about that. I will drop the arrow notation for your original tensor map $ \ \vec{\otimes} \ $ and denote it simply by "$\otimes$" as usual. Moreover, let us denote the set of real numbers by "$\mathbb{R}$".

Let $ \ n=dim_{\mathbb{R}} (V) \, $, $ \, m=dim_{\mathbb{R}} (W) \, $, $ \, C_V = \{ v_1,...,v_n \} \ $ a basis for $V$, $ \, C_W = \{ w_1, ..., w_m \} \ $ a basis for $W$, $ \, C_V^* = \{ f^1,...,f^n \} \ $ the dual basis of $ \, C_V \, $, $ \, C_W^* = \{ g^1,...,g^m \} \ $ the dual basis of $ \, C_W \, $ and $ \ \mu \in \mathbb{R}^{\mathbb{R} \times \mathbb{R}} \ $ the multiplication function of $\mathbb{R}$, ie, $ \mu(r,s)=rs \, $, $ \, \forall r,s \in \mathbb{R}$.

For each $ \ i \in \{ 1,...,n \} \ $ let $ \ \pi_i : V^* \to \mathbb{R} \ $ be the projection onto de $i$-th coordinate, that is, for all $ \ f \in V^*$, $$\pi_i (f) = \pi_i \left( \sum_{k=1}^{n} f(v_k) \cdot f^k \right) = f(v_i) \ \ .$$ Analogously, for each $ \ j \in \{ 1,...,m \} \ $ let $ \ \varpi_j : W^* \to \mathbb{R} \ $ be the projection onto de $j$-th coordinate, that is, for all $ \ g \in W^*$, $$\varpi_j (g) = \varpi_j \left( \sum_{ \ell =1}^{m} g(w_\ell) \cdot g^\ell \right) = g(w_j) \ \ .$$

For each $ \ i \in \{ 1,...,n \} \ $ and each $ \ j \in \{ 1,...,m \} \ $ let $ \ B_{ij} = \mu \circ (\pi_i \times \varpi_j) \, $. Hence, $\forall (f,g) \in V^* \times W^*$, $\forall i \in \{ 1,...,n \}$, $\forall j \in \{ 1,...,m \}$, $$B_{ij} (f,g) = [\mu \circ (\pi_i \times \varpi_j)](f,g) = f(v_i) g(w_j) = [B(v_i,w_j)](f,g) \ \text{ .}$$ Ergo $ \ B_{ij} = B(v_i,w_j) \, $, $ \, \forall i \in \{ 1,...,n \}$, $\forall j \in \{ 1,...,m \}$.

Exercise: Show that $$B[C_V \times C_W] = \Big\{ B_{ij} \in \mathbb{R}^{V^* \times W^*} \, : \ i \in \{1,...,n \} \ \text{ and } \ j \in \{1,...,m \} \Big\}$$ is a basis for $ \, Lin(V^* \times W^* ; \mathbb{R}) \, $. Therefore, $dim_{\mathbb{R}} \big( Lin(V^* \times W^* ; \mathbb{R}) \big) = nm \, $.

We also know that $$\otimes [C_V \times C_W] = \Big\{ v_i \otimes w_j \in V \otimes W \, : \ i \in \{1,...,n \} \ \text{ and } \ j \in \{1,...,m \} \Big\}$$ is a basis for $ \ V \otimes W$. Thus $ \ dim_{\mathbb{R}} (V \otimes W) = nm \, $.

Now it is enough to note that, $\forall i \in \{ 1,...,n \}$, $\forall j \in \{ 1,...,m \}$, $$B_{ij} = B(v_i,w_j) = (L \circ \otimes)(v_i,w_j) = L(v_i \otimes w_j) \ . $$ Then $L$ is an isomorphism of $\mathbb{R}$-vector spaces.

The quotient space construction do not play an important role here because all the information needed to manipulate tensor products lies in the universal property.

For the isomorphisms $ \ V \cong V^* \, $, $ \ V \otimes V^* \cong V^* \otimes V \ $ etc., my suggestion to you is that you ask specific questions in another posts.


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