# How to View Tensors

1. Jan 20, 2012

### blashmet

Is it correct to view tensors as multi-variable functions? For example, it seems the permutation tensor is a function of three variables and the metric tensor is a function of two variables. Of course, these "functions" turn into constants when i,j, and k (the indices) are known, but it seems like these constants can be viewed as the output of the function.

So, am I viewing tensors correctly? Thanks!

2. Jan 21, 2012

### jeo23

Well, a tensor is a multi-linear functional on vector and dual spaces, and it can take multiple arguments so I suppose in that sense it can be tought of as a multi-variable function as long as one realizes that each of the "variables" are vectors or dual vectors.

3. Jan 21, 2012

### blashmet

Are all tensors multi-linear functionals? Do the arguments of all tensors have to be vectors? That is, are these necessary conditions of being a tensor? Thanks for the help!

4. Jan 21, 2012

### wisvuze

a k-tensor is a function of k vector variables ( a multi linear one, that is ). So yes, each of the arguments must be vectors

5. Jan 21, 2012

### blashmet

Can you have a tensor that isn't multi-linear? Thanks!

6. Jan 21, 2012

### jeo23

I just would like to add that the arguments don't have to be all vectors (contravariant). They can be dual vectors (covariant) as well.

As to your last question, you can have a rank (0,1) tensor that takes a single vector as input. Similarly, you can have a rank (1,0) tensor that takes a single dual vector as input. However, all tensors must be linear in their arguments.

7. Jan 21, 2012

### tiny-tim

hi blashmet!
it's easiest to define a tensor in terms of its effect on "inputting" individual vectors,

but the "input" can be any tensor …

for example the metric gij can have any tensor Aijklm as "input"
no!

but although it's multi-linear, it isn't necessarily linear in each vector, see the pf library on tensor …

"Non-linear" tensors:

The equation $a^i\ =\ T^i_{\ jk}b^jc^k$ is linear in both b and c (meaning that a linear transformation on b or c induces a linear transformation on a).

But the equation $a^i\ =\ T^i_{\ jk}b^jb^k$ is not linear in b.

For example, in "non-linear optics", there are "higher-order" susceptibility tensors, each with the electric field vector as the only input (exactly as in the linear case), but with that input repeated one or more times:

$\frac{1}{\varepsilon_0}P^i\ =\ \chi^i_{\ j}E^j\ +\ \chi^{(2)i}_{\ \ \ jk}E^jE^k\ +\ \chi^{(3)i}_{\ \ \ jkl}E^jE^kE^l\ +\ \cdots$​

8. Jan 21, 2012

### Alesak

I'm real amateur in this, but I think only covariant tensors can be viewed as multilinear functionals, that is, only (0, n)-tensor. See wiki for more. You put n vectors in it and it spits out real number, that's how I understand it.

In the book that I'm reading, Lee's Introduction to smooth manifolds, when he says for example 2-tensor he means (0, 2)-tensor, so I guess covariant tensors are more important than contravariant(of type (n, 0)).

9. Jan 21, 2012

### mathwonk

in finite dimensions, V** = V where * means dual space. so you can define the other kind of tensor as a dual of the linear function kind.

i.e. linear functions on a tensor space of V are the same as multilinear functions on V. Hence the dual of the multilinear functions on V are the tensor space of V. and the multilinear functions themselves are the tensor space of V*.

I.e. the (0,2) tensors on V are the (2,0) tensors on V*, so if you know about V and V*, you only need to know about one type of tensors.

there is also another more formal way to define tensors on V as elements

written as xoyoz, where x,y,z are in V, and satisfying rules like

(u+v)oyoz = uoyoz + voyoz, but when you get through the thing you have

is dual to the multilinear (trilinear in this case) functions on V.

10. Jan 22, 2012

### Fredrik

Staff Emeritus
This is how I define "tensor" and "tensor field": (People often say "tensor" when they mean "tensor field").

This post
and the ones it link to explains the basics of manifolds, tangent and cotangent spaces, and the relationship between coordinate systems and bases.

Excellent choice.

The metric g is a tensor field that assigns a tensor gp to each point p. The domain of gp is $T_pM\times T_pM$, so the "input" of gp is always two tangent vectors at p. The input of g is a point p in the manifold.

If X and Y are vector fields, the notation g(X,Y) can be used for the map $p\mapsto g_p(X_p,Y_p)$. So it would make sense to say that g takes two vector fields to a scalar field (i.e. a real-valued function defined on a subset of the manifold). But neither g nor gp can have "any tensor" as input.

$g_{ij}A^{ij}{}_{lm}$ isn't the result of g taking A as input. It can denote either the tensor field $g(\partial_i,\partial_j) A(\mathrm{d}x^i,\mathrm{d}x^j,\cdot,\cdot)$ (abstract index notation) or its lm component $g(\partial_i,\partial_j) A(\mathrm{d}x^i,\mathrm{d}x^j,\partial_l,\partial_m)$.

Last edited: Jan 22, 2012