# Definition of a tensor and tensor product

i was reading Carroll's introduction to general relativity and on page 21 starts talking about tensors. My previous understanding of it was something with indexes that transform in a certian way but now i want to understand them in a more formal way. The definition is a multilinear application of colllections of vectors ($$l$$) and dual vectors ($$k$$) that brings you to a real number.

$$T: T_{p_{1}}^{*} \times T_{p_{2}}^{*}\times ... \times T_{p_{k}}^{*}\times T_{p_{1}} ... T_{p_{l}} \longrightarrow \Re .$$

$$\times$$ indicates cartesian product that gives you a space of ordered pair of vectors or dual-vectors.

I understand that vectors can be represented by linear functions (of the double dual space) and dual space vectors are linear functions, so the Tensor is acting on scalars to bring a another one. But i don't understand what is really happening. The dual vector $$\omega$$ can be represented by $${\omega}_{\mu}$$ and the vector $$V$$ like $$V^{\nu}$$. So your tensor (1,1) would be something like $$T_{\mu}^{\nu}$$ , or not, you tell me.

Well having said that now comes another issue. What is a tensor product? If you have a tensor $$T$$ of rank $$(l,k)$$ and another tensor $$S$$ of rank $$(m,n)$$ then with the tensor product you can obtain a new tensor $$T \otimes S$$ of rank $$(k+m,l+n)$$. But i do not understand how this prodcut is acting , Carroll's gives this definition:

$$T \otimes S ({\omega}^{(1)},...,{\omega}^{(k)},...,{\omega}^{k+m},V^{(1)},...,V^{(l)},...,V^{(l+n)})= T({\omega}^{(1)},...,{\omega}^{(k)},V^{(1)},...,V^{(l)}) \times S({\omega}^{(k+1)},...,{\omega}^{k+m},V^{(l+1),...,V^{(l+n)}}).$$

Does the $$\times$$ mean the same cartesian product? Why isn't it conmutative? Please help

The definition about indexes transforming in a certain way is very much about the tensors built from vectors in the tangent space of a manifold. Not all vectors need to be defined in such a way.

Let there be some vectors space $$\mathcal{U}$$ with basis vectors $$\mathbf{u}_{i=1\dots n}$$ and another vector space $$\mathcal{V}$$ with basis $$\mathbf{v}_{j=1\dots m}$$. Consider a space of all homogeneous bilinear functionals that map a pair of vectors $$\left(a^i\mathbf{u}_i,\,b^j\mathbf{v}_j\right)$$ to real numbers $$\mathcal{L}:\mathcal{U}\times\mathcal{V}\to\mathbb{R}$$. This space of maps can be spanned by the following functionals:

$$\mathbf{l}^{ij}\left(\mathbf{u}_p,\,\mathbf{v}_r\right)=\begin{cases}\begin{array}\\ 1,\quad i=p\,and\,j=r \\ 0,\quad otherwise\end{array}\end{cases}$$

Then every functional can be represented as $$\omega_{ij}\mathbf{l}^{ij}$$ and the application of the functional onto the pair of vectors will lead to:

$$\left(\omega_{ij}\mathbf{l}^{ij}\right)\left(a^p\mathbf{u}_p,\,b^r\mathbf{v}_r\right)=\omega_{ij}a^i b^j$$

There is clearly an isomorphism between the vector space of bilinear functionals (I will skip proof that this is a vector space) and the Cartesian product of two vector spaces. Call it:

$$\phi:\mathcal{U}\times\mathcal{V}\to\mathcal{L}$$

And define it as $$\phi\left(a^p\mathbf{u}_p,\,b^r\mathbf{v}_r\right)=\sum_{p,r}a^p b^r \mathbf{l}^{pr}$$ (yes this breaks the upstairs-downstairs convention, but this is temporary).

Next, since $$\mathcal{L}$$ is a vector space we can consider a vector space dual to it. Let this vector space $$\mathcal{T}$$ be spanned by basis $$\mathbf{t}_{ij}$$. By definition of the dual space:

$$\left(w^{ij}\mathbf{t}_{ij}\right)\left(\omega_{pr}\mathbf{l}^{pr}\right)=w^{ij}\omega_{ij}$$

We can define another isomorphism: $$\psi:\mathcal{L}\to\mathcal{T}$$, where $$\psi\left(\omega_{pr}\mathbf{l}^{pr}\right)=\sum_{pr}\omega_{pr}\mathbf{t}_{lr}$$.

Finally, define the tensor product as:

$$\otimes=\psi\circ\phi:\,\mathcal{U}\times\mathcal{V}\to\mathcal{L}\to\mathcal{T}$$. In particular, by definition, the basis for $$\mathcal{T}$$ can be denoted by: $$\psi\circ \phi\left(\mathbf{u}_i,\,\mathbf{v}_j\right)=\mathbf{u}_i\otimes\mathbf{v}_j$$

It readily follows that $$\psi\circ \phi\left(a^i\mathbf{u}_i,\,b^j\mathbf{v}_j\right)=a^i b^j\mathbf{u}_i\otimes\mathbf{v}_j$$.

The tensor product is not commutative because of the functional space you used in-between ($$\mathcal{L}$$). It was defined specifically for the pair $$\mathcal{U}\times\mathcal{V}$$ and not the other way round.

Note that above procedure can be repeated again and can be combined. For example you can consider bilinear functionals from $$\mathcal{U}\times\mathcal{V}^*$$ and create a tensor with upstairs-downstairs vectors. You can also chain tensor products together, i.e. $$\mathcal{U}$$ could itself be a tensor product space.

The difference between a Cartesian product $$\mathcal{U}\times\mathcal{V}$$ and tensor product $$\mathcal{U}\otimes\mathcal{V}$$ is that the latter is a vector space itself. In particular, you can meaningfully add members of $$\mathcal{U}\otimes\mathcal{V}$$ (thanks to space of bilinear functionals), whereas for Cartesian product such operation is not defined.