# Group representations on tensor basis.

1. Jul 19, 2011

### rkrsnan

I am a physicist, so my apologies if haven't framed the question in the proper mathematical sense.

Matrices are used as group representations. Matrices act on vectors. So in physics we use matrices to transform vectors and also to denote the symmetries of the vector space.
v_i = Sum M_ij v_j
Is there an equivalent group representation that act on tensors instead of vectors?
T_ij = Sum M_iajb T_ab
I guess then we need some group multiplication law...
may be
M_abcd = Sum M_axcy M_xbyd

Does this make any sense? Please provide some references where I can find more information.

2. Jul 20, 2011

### kdbnlin78

Tensor representations are given by the direct (tensor) product of copies of the vector representation.

They act on the set of tensors of a given rank, which is indeed a linear vector space. These tensors are called Lorentz 4-tensors. For example, consider (2, 0) tensors, that is, tensors with two contravariant (upper) indices T^{/mu /nu).

A representation of the Lorentz group will be a 16 x 16 matrix /Lambda^{/mu /nu}_{/rho /sigma} = /Lambda^{/mu /nu}/Lambda_{/rho /sigma}.

Search for "Lorentz spinors" and the documents you find will give both the matirx version and the tensorial version of the transformations too.

3. Jul 20, 2011

### Fredrik

Staff Emeritus
I think the natural way to do something like this is to consider a representation into the group of invertible linear operators on the tangent space at a point p of a manifold M. (Do you have a way to associate a representation with each point?) When an invertible linear operator acts on the members of a basis, the result is another basis. Now you can just use the standard stuff about how tensors at p transform under a change of basis of the tangent space at p.

4. Jul 20, 2011

### rkrsnan

Thanks guys for the replies.

What I meant was not a tensor product. Because then it will be
T_ij = Sum M1_ia M2_jb T_ab
What I wrote in my question was a general linear tranformation acting on all components of a two component thingy(need not even be a tensor. I guess if it is a tensor, then I will have to associate a representation at each point etc.)

So the bottomline is that my T_ij is just like a vector with a single index, but the index runs from 1 to n where n is the sum of the individual ranges of i and j. So the question was rather stupid.

5. Jul 20, 2011

### Fredrik

Staff Emeritus
With my approach, you get transformation rules like $$T'^i{}_{jk}=(M^{-1})^i{}_l M^m{}_j M^n{}_k T^{\,l}{}_{mn}$$ Not sure if that matters. It seems that what you had in mind was something different.

6. Jul 21, 2011

### rkrsnan

Yeah, I had something else in my mind.
Let me explain why I asked the question.
Consider Z3 group. We may visualize it as a 120 degree rotation of a vector. So the three elements are $\left( \begin{matrix}1 \\ 0 \end{matrix}\right )$, $\left( \begin{matrix}-1/2 \\ \sqrt{3}/2 \end{matrix}\right )$and$\left( \begin{matrix}-1/2 \\ -\sqrt{3}/2 \end{matrix}\right )$. So in physics if I see a vector quantity which is proportional to $\left( \begin{matrix}-1/2 \\ \sqrt{3}/2 \end{matrix}\right )$, then I may make a guess that the underlying symmetry is Z3.
Now in my case I am encountering a matrix of the form
T = $\left( \begin{matrix}1 && 1/\sqrt{2} \\ 1/\sqrt{2} && 1 \end{matrix}\right )$ and I need to use some mathematics to 'explain' this particular matrix. I am not sure how to proceed. I mean what is special about this matrix mathematically? What gives you a simple ratio between the diagonal and offdiagonal parts?

Last edited: Jul 21, 2011
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook