# Definition of a tensor in a given basis

To understand my question you can directly go to the last section. The rest is juste here to give the context.

I don't understand why we can say that a tensor can be written as :

$$T=T^{\mu_1\cdots \mu_q}_{\nu_1\cdots\nu_r} \frac \partial {\partial x^{\mu_1}} \cdots\frac \partial {\partial x^{\mu_q}} dx^{\mu_1}\cdots dx^{\mu_r}$$

For example : what is the product in $\dfrac \partial {\partial x^{\mu_q}} *dx^{\mu_1}$ ?

Indeed, for me a tensor is a multilinear object which takes $q$ elements from $T^*_pM$ and $r$ elements from $T_pM$ and associates to them a number.

So the only thing we can say about $T$ is :

$$T(\omega_1,\ldots,\omega_q,V_1,\ldots,V_r)=\\ T\left(dx^{\mu_1},\ldots,dx^{\mu_q},\frac{\partial}{\partial x^{\nu_1}},\ldots,\frac \partial {\partial x^{\nu_r}}\right) dx^{\mu_1}(\omega_1)\cdots dx^{\mu_q} (\omega_q) \frac \partial {\partial x^{\nu_1}}(V_1)\cdots \frac \partial {\partial x^{\nu_r}}(V_r)$$

And in this last expression the product has a sense because it is a product of real numbers. But I don't get the product between $\frac{\partial}{\partial x^{\mu_q}} *dx^{\mu_1}$ because we would have a product of functions that "don't eat the same thing".

In fact, according to the comment from Ted Shifrin it seems to be a tensorial product but I would like to understand how we can see this.

Indeed the starting point is the multilinearity of my object which gives us this equation :

$$T(\omega_1,\ldots,\omega_q,V_1,\ldots,V_r)=\\ T\left(dx^{\mu_1},\ldots,dx^{\mu_q},\frac{\partial}{\partial x^{\nu_1}},\ldots,\frac \partial {\partial x^{\nu_r}}\right) dx^{\mu_1}(\omega_1)\cdots dx^{\mu_q} (\omega_q) \frac \partial {\partial x^{\nu_1}}(V_1)\cdots \frac \partial {\partial x^{\nu_r}}(V_r)$$

Now, how can we see from this equation that we have in fact a tensorial product behind ?

I understand tensorial products as the following :

Let $V$ and $W$ vectorials space with basis $e_i$ and $f_j$. We define $V \otimes W$ as the vectorial space that has a basis $(e_i,fj)$ that we note : $e_i \otimes fj$.

We then define the tensorial product of vectors as :

$$u \otimes w = \sum_{i,j} v_i w_i ~ e_i \otimes f_j$$

(Where $u=\sum_i u_i e_i$ and $v=\sum_j v_j f_j$).

More precisely :

How do we know that $\frac \partial {\partial x^{\mu_q}}\otimes dx^{\mu_1}(\omega_q \otimes V_1)=\frac \partial {\partial x^{\mu_q}}(\omega_q)*dx^{\mu_1}(V_1)$

Where we finally have a product in $\mathbb{R}$. Is that a property ? Because in Wikipedia, they say that for $S$ and $V$ linear maps, $S\otimes V(u\otimes v)=S(u)\otimes V(v)$ and we see that we still have a tensor product at the end.

• The tensor product symbols are all missing. The way you apply $S\otimes T$ to $(v,w)$ is to feed the first vector, $v$ to $S$ and the second vector, $w$, to $T$, and then multiply $S(v)T(w)$. [The same applies if $v$ is a covector, etc.] Apr 3, 2017 at 16:23
• How can I prove that from my last equation (with the $dx^{\mu_i}(\omega_i)$), that I have a tensorial product in the first equation ? What I mean is that the multilinearity which is the starting hypothesis implicates my last equation. But how from this I can say "Oh well, in fact it is just a tensor product !". Apr 4, 2017 at 10:56
• More specifically you can read my last paragraph beginning with How do we know that $\frac \partial {\partial x^{\mu_q}}\otimes dx^{\mu_1}(\omega_q \otimes V_1)=\frac \partial {\partial x^{\mu_q}}(\omega_q)*dx^{\mu_1}(V_1)$ to understand precisely what I don't totally get. Apr 4, 2017 at 11:11

Since your question is pure linear algebra, let me answer it in that context. Let $v \in V$ and $\varphi \in V^{*}$. We have a natural identification of $V$ with $\left( V^{*} \right)^{*}$ so we can think of $v$ as defining a linear map $v \colon V^{*} \rightarrow \mathbb{R}$ and $\varphi$ is a linear map $\varphi \colon V \rightarrow \mathbb{R}$. Hence, we can take their tensor product and get a linear map

$$v \otimes \varphi \colon V^{*} \otimes V \rightarrow \mathbb{R} \otimes \mathbb{R}$$

which acts as

$$(v \otimes \varphi)(\psi \otimes w) := v(\psi) \otimes \varphi(w) = \psi(v) \otimes \varphi(w).$$

The multiplication $m \colon \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ of real numbers is $\mathbb{R}$-bilinear and so gives us a map $m \colon \mathbb{R} \otimes \mathbb{R} \rightarrow \mathbb{R}$ which is an isomorphism and so by composing $v \otimes \varphi$ with $m$ we get a map $m \circ (v \otimes \varphi) \colon V^{*} \otimes V \rightarrow \mathbb{R}$ which acts as

$$(m \circ (v \otimes \varphi))(\psi \otimes w) = m(\psi(v) \otimes \varphi(w)) = \psi(v) \varphi(w).$$

Very often the identification of $\mathbb{R} \otimes \mathbb{R}$ with $\mathbb{R}$ via $m$ is not explicit and this is what happens in your case.

• Textbooks that don't bother with abstract nonsense just define $v \otimes \varphi$ (the tensor product of a $(1,0)$ and $(0,1)$ tensor) by the last formula. Apr 4, 2017 at 14:35
• Ok, so to summarize, theoretically, I have : $(S \otimes V)(u \otimes v)=S(u)\otimes S(v)$. And in practice we consider that $S(u)\otimes S(v)=S(u)*S(v)$ as we generally identify $\mathbb{R} \otimes \mathbb{R}$ with $\mathbb{R}$ (which is possible because there is only one vector of basis in $\mathbb{R} \otimes \mathbb{R}$). And more generally, I will have $\otimes_i S_i(u_i)=\prod_{i}S_i(u_i)$ where the last product is the usual product in $\mathbb{R}$. That's right ? Apr 4, 2017 at 17:49
• @user3183950: Yes, although I would write $\left( \bigotimes_{i=1}^n S_i \right) \left( \bigotimes_{i=1}^n u_i \right) = \prod_{i=1}^n S_i(u_i)$ where $S_i \colon U_i \rightarrow \mathbb{R}$. That is, the tensor product of the linear functionals $S_1,\dots,S_n \colon U_i \rightarrow \mathbb{R}$ acts on (elementary) tensors $u_1 \otimes \dots \otimes u_n$ in $U_1 \otimes \dots \otimes U_n$ as the product $\prod_{i=1}^n S_i(u_i)$. Apr 4, 2017 at 18:00
• @user3183950: Also, it's not only there is one vector of basis in $\mathbb{R} \otimes \mathbb{R}$, the point is that you have a distinguished basis in $\mathbb{R}$ given by $(1)$. If $V$ is a one dimensional vector space then $V \otimes V \cong V$ but they are not naturally isomorphic which is important in various places. Apr 4, 2017 at 18:03
• @user3183950: By naturally isomorphic I indeed mean an isomorphism which is independent of the basis. There is also a more technical definition but that is the spirit of what it conveys. Apr 4, 2017 at 18:11