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
 A: 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.
