Definition of a tensor in a given basis

Question

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".

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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 $v=\sum_j v_j f_j$).

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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.

Answer 1

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.