As a physics student, I’ve come across mathematical objects called

tensorsin several different contexts. Perhaps confusingly, I’ve also been given both the mathematician’s and physicist’s definition, which I believe are slightly different.I currently think of them in the following ways, but have a tough time reconciling the different views:

- An extension/abstraction of scalars, vectors, and matrices in mathematics.
- A multi-dimensional array of elements.
- A mapping between vector spaces that represents a co-ordinate independent transformation.
In fact, I’m not even sure how correct these three definitions are. Is there a particularly relevant (rigorous, even) definition of tensors and their uses, that might be suitable for a mathematical physicist?

Direct answers/explanations, as well as links to good introductory articles, would be much appreciated.

**Answer**

At least to me, it is helpful to think in terms of bases.

(I’ll only be talking about tensor products of finite-dimensional vector

spaces here.)

This makes the universal mapping property that Zach Conn talks about

a bit less abstract (in fact, almost trivial).

First recall that if $L: V \to U$ is a linear map, then $L$ is completely determined

by what it does to a basis $\{ e_i \}$ for $V$:

$$L(x)=L\left( \sum_i x_i e_i \right) = \sum_i x_i L(e_i).$$

(The coefficients of $L(e_i)$ in a basis for $U$ give the $i$th column

in the matrix for $L$ with respect to the given bases.)

Tensors come into the picture when one studies *multilinear* maps.

If $B: V \times W \to U$ is a bilinear map, then $B$

is completely determined by the values $B(e_i,f_j)$ where

$\{ e_i \}$ is a basis for $V$

and $\{ f_j \}$ is a basis for $W$:

$$B(x,y) = B\left( \sum_i x_i e_i,\sum_j y_j f_j \right) = \sum_i \sum_j x_i y_j B(e_i,f_j).$$

For simplicity, consider the particular case when $U=\mathbf{R}$;

then the values $B(e_i,f_j)$

make up a set of $N=mn$ real numbers (where $m$ and $n$ are the

dimensions of $V$ and $W$), and these numbers are all that we need to keep

track of in order to know everything about the bilinear map $B:V \times W \to \mathbf{R}$.

Notice that in order to compute $B(x,y)$ we don’t really need to know the

individual vectors $x$ and $y$, but rather the $N=mn$ numbers $\{ x_i y_j \}$.

Another pair of vectors $v$ and $w$ with $v_i w_j = x_i y_j$ for all $i$ and $j$

will satisfy $B(v,w)=B(x,y)$.

This leads to the idea of splitting the computation of $B(x,y)$ into two stages.

Take an $N$-dimensional vector space $T$ (they’re all isomorphic so it doesn’t matter

which one we take) with a basis $(g_1,\dots,g_N)$.

Given $x=\sum x_i e_i$ and $y=\sum y_j f_j$,

first form the vector in $T$

whose coordinates with respect to the basis $\{ g_k \}$ are given by the column vector

$$(x_1 y_1,\dots,x_1 y_m,x_2 y_1,\dots,x_2 y_m,\dots,x_n y_1,\dots,x_n y_m)^T.$$

Then run this vector through the *linear* map $\tilde{B}:T\to\mathbf{R}$ whose matrix

is the row vector

$$(B_{11},\dots,B_{1m},B_{21},\dots,B_{2m},\dots,B_{n1},\dots,B_{nm}),$$

where $B_{ij}=B(e_i,f_j)$.

This gives, by construction, $\sum\sum B_{ij} x_i y_j=B(x,y)$.

We’ll call the space $T$ the **tensor product** of the vector spaces $V$ and $W$

and denote it by $T=V \otimes W$;

it is “uniquely defined up to isomorphism”,

and its elements are called **tensors**.

The vector in $T$ that we formed from $x\in V$ and $y\in W$ in the first stage above

will be denoted $x \otimes y$;

it’s a “bilinear mixture” of $x$ and $y$ which doesn’t allow us to

reconstruct $x$ and $y$ individually,

but still contains exactly all the information needed

in order to compute $B(x,y)$ for any bilinear map $B$;

we have $B(x,y)=\tilde{B}(x \otimes y)$.

This is the “universal property”; any bilinear map $B$ from $V \times W$

can be computed by taking a “detour” through $T$, and this detour

is unique, since the map $\tilde{B}$ is constructed uniquely from

the values $B(e_i,f_j)$.

To tidy this up, one would like to make sure that the definition is

basis-independent. One way is to check that everything transforms

properly under changes of bases. Another way is to do the construction

by forming a much bigger space and taking a quotient with respect to

suitable relations (without ever mentioning bases).

Then, by untangling definitions, one can for

example show that a bilinear map $B:V \times W \to \mathbf{R}$ can be

canonically identified with an element of the space $V^* \otimes W^*$,

and dually an element of $V \otimes W$ can be identified with a

bilinear map $V^* \times W^* \to \mathbf{R}$.

Yet other authors find this a convenient *starting* point, so that they

instead *define* $V \otimes W$ to be the space of bilinear maps $V^*

\times W^* \to \mathbf{R}$.

So it’s no wonder that one can become a little confused when trying

to compare different definitions…

**Attribution***Source : Link , Question Author : Noldorin , Answer Author : Hans Lundmark*