1
$\begingroup$

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

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.