Trouble seeing tensors as multilinear maps because of the quotient.

Question

Good evening,

I saw on this page ( https://fr.wikipedia.org/wiki/Tenseur : for those who understand French ) a definition of a tensor as a multilinear map ( and we see it more like this in my physics course ). We can take, for instance the example of a 1-order tensor. Following this definition, it will be a map $\phi : E \rightarrow K $, where $ K $ is a field such that $\phi$ is linear and will follow some basic transformation rule ( that differs following it's a covariant or contravariant tensor ).

But I learned from my multilinear algebre course that a tensor is an element of the free vector space that we quotient in order to obtain a bilinear map $ \otimes $ between $E$ and $F(E)/V$.

My question is the following : How do we keep the linear map after the quotient ? Because i know $\psi : E \rightarrow K $ is linear but how do we define the linear map after quotienting ? Because $\chi_{\lambda \cdot v} \neq \lambda \cdot \chi_{v}$ where $\chi_v$ is the function that is equal to $1$ if $x = v$ and $0$ otherwise. Do we just say $[\chi_{\lambda \cdot v}](x) = \lambda \cdot \chi_{v}(x) $ ? Because other defintions would be possible depending of the representative.

Excuse me if similar questions were already asked but i didn't an answer who satisfies me.

Thanks a lot.

Answer 1

Based on the word 'bilinear map', it seems you refer here to the tensor product of two vector spaces $E_1$ and $E_2$, which is commonly defined as a quotient of the free vector space $F(E_1\times E_2)$, such that it induces a universal bilinear map $\beta:E_1\times E_2\to F(E_1\times E_2)/V$, meaning that, every bilinear map $E_1\times E_2\to U$ uniquely factors through it.

Now, this tensor product $F(E_1\times E_2)/V$ is denoted by $E_1\otimes E_2$, and its elements are also called tensors, and the image of a pair of vectors $(e_1,e_2)\ \in E_1\times E_2$ under the natural map $E_1\times E_2\,\to\, E_1\otimes E_2$ (which is just $\beta$) is also denoted by $e_1\otimes e_2$.

An order one contravariant tensor refers to an element of $E^*=\hom(E, K)$, and an order one covariant tensor is just a vector, i.e. element of $E$, which indeed can be seen as an element of $E^{**}$, especially in the finite dimensional case.

A tensor of order 2 (element of a tensor product) is basically a matrix, since if $b_1, \dots, b_n\in E_1$ and $c_1,\dots, c_m\in E_2$ are bases, then $(b_i\otimes c_j)_{i,j}$ is a basis for $E_1\otimes E_2$.

Answer 2

Let $U$ be a linear space and $N \subseteq U$ be a linear subspace. Define an equivalence relation by $x \sim y$ if $x - y \in N.$ Then we will define $U/N := U/\sim$ to be a linear space taking

1. scalar multiplication be defined as $c[u] := [cu]$,
2. addition be defined as $[u]+[v] := [u+v]$.

First we should check that these operations are well-defined. For scalar multiplication we take $u_1, u_2 \in [u]$ and need to show that $[c u_1] = [c u_2]$. This is easy: $c u_1 - c u_2 = c (u_1 - u_2) \in N$ since $u_1 - u_2 \in N,$ which is a linear space. For addition we take $u_1, u_2 \in [u]$ and $v_1, v_2 \in [v]$ and need to show that $[u_1 + v_1] = [u_2 + v_2].$ I leave this to you.

Then we need to show that $U/N$ is in fact a linear space by checking the axioms. I'll show a couple of them.

1. There is a zero element; it is given by $[0] = N$ since $[0] + [u] = [0+u] = [u] = [u+0] = [u] + [0]$.
2. Every element has an additive inverse; we have $-[u] = [-u]$ since $[u] + [-u] = [u+(-u)] = [0].$
3. Scalar multiplication is distributive over addition of vectors; $c([u]+[v]) = c[u+v] = [c(u+v)] = [cu+cv] = [cu]+[cv] = c[u]+c[v]$.