# Tensor operation on a vector space

From the various definitions provided in the article https://en.wikipedia.org/wiki/Tensor, the tensor seems always to be defined, even in the more abstract forms, as a multilinear map, from a product of vector and dual spaces to the underlying field.

However, in applied mathematics, one often come across a tensor when used in the form that maps elements of vector and dual spaces to elements of vectors and dual spaces, like this, for example:

$\theta^j=\mu_l^j dx^l$

No operation is defined in the article between a tensor and a dual vector, which gives a dual vector. But since other operations are defined on tensors, should the previous notation actually be read in two steps:

1. a tensor product: $\mu_l^j dx^k$
2. followed by a contraction of the indices $k$ and $l$

You seem to be confused by notationnal issues. With some luck I can deconfuse you, with less luck your confusion may increase ;-)

Usually - in classical differential geometry - one would write something like $\mu^j_l dx^l\otimes \frac{\partial}{\partial x^j}$ to define locally a $(1,1)$ tensor or tensor field. The $\mu_l^j$ are just the coefficients, the $dx^l$ form a base of the cotangent space and the $\frac{\partial}{\partial x^j}$ a base of the tangent space. The tensor products $dx^k \otimes \frac{\partial}{\partial x^j}$ form a base of the space of $(1,1)$ tensors. The $\theta^j$ from your example are just coefficients of a vector field, the base vectors are omitted for some reason.

If $v= v^j \frac{\partial}{\partial x^j}$ is, in this formalism, a vector field and $\omega = \omega_k dx^k$ a one-form, then you may look at the $(1,1)$ tensor field $v\otimes \omega$ with coefficients $v^j\omega_k$. Written out more explicitly, the $(1,1)$ tensor field is

$$v\otimes \omega= \omega_k v^j dx^k \otimes \frac{\partial}{\partial x^j}$$

This is just writing down a tensor field in a local base representation. Applying a contraction to that tensor field means kind of inserting the vector field $v^j\frac{\partial}{\partial x^j}$ into the one form $\omega_kdx^k$ resulting in the sum $v^j\omega_j$, since $dx^j(\frac{\partial}{\partial x^k})=\delta^j_k$ is just a scalar. In a coordinate free notation you'd write $$C(v\otimes \omega) = \omega(v)$$ ($C$ denoting contraction) that is, you apply the linear functional to the vector.

A general $(1,1)$ tensor is, locally, written as $\mu^j_k\frac{\partial}{\partial x^j}\otimes dx^k$ 'Applying' this to a one form $\sigma_k dx^k$ results in a one form $$\mu^j_k \sigma_j\frac{\partial}{\partial x^j}\otimes dx^k \otimes dx^j = \mu^j_k\sigma_j dx^k$$

• I see what you mean, but my question is more about the signification of "applying" a (1,1) tensor to a 1-form. I do not see how this operation is defined, and I guessed it might have something to do with tensor product and contraction. I think I used en.wikipedia.org/wiki/Abstract_index_notation . – vkubicki Jun 7 '12 at 20:46
• @JohnDeas I edited my answer in order to cover that aspect of your question. – user20266 Jun 7 '12 at 20:58
• OK, so in that case "applying" means contracting the result of the tensor product. Thanks you for the detailed answer. – vkubicki Jun 8 '12 at 7:08

Yes. This is what is called the Einstein summation convention.

(I happen to think it is ultimately less confusing to define tensors as elements of tensor products of a vector space and its dual, but whatever floats your boat, I suppose.)