Generators of Product Rep: t_{a}^{R_1 \otimes R_2}

[sgd37]

Homework Statement

Write the generators of the product of representation t_{a}^{R_1 \otimes R_2} in terms of t_{a}^{R_1}, t_{a}^{R_2}

Homework Equations

[t_{a}^{R_i} , t_{b}^{R_i}] = i f_{abc} t_{c}^{R_i}

I don't believe that the BCH formula is relevant here since that relates elements within the same representation

The Attempt at a Solution

 

[fzero]

You can figure this out by considering the tensor product of representations as vector spaces. Suppose that \text{dim}~R_1 = n, \text{dim}~R_2 = m. Then an element of R_1 is an n vector (a_1, \cdots a_n)^T, while an element of R_2 is (b_1, \cdots b_m)^T. To form an element of the product R_1\otimes R_2, we construct an nm dimensional vector by replacing each b_i by a vector in R_1. You can determine a basis of generators on R_1\otimes R_2 by figuring out how you have to combine the generators on each space to form something that acts properly on the R_1\otimes R_2 vectors constructed above.

 

[sgd37]

well i guess my main hangup is whether the two sets of generators commute, if they do its quite clear that they add

 

[fzero]

well i guess my main hangup is whether the two sets of generators commute, if they do its quite clear that they add

I'm not sure what you mean here. It doesn't make sense to do ordinary multiplication or addition of matrices of different dimensions.

 

[sgd37]

That is right (an aside what is the difference between a direct product and a tensor product). So then is it just t_{a}^{R_1} \otimes t_{a}^{R_2} = t_{a}^{R_1 \otimes R_2} it might have the right index structure to transform a V_{R_1} \otimes V_{R_2} tensor but I am unsure about the generator basis index 'a'

 

[fzero]

That is right (an aside what is the difference between a direct product and a tensor product). So then is it just t_{a}^{R_1} \otimes t_{a}^{R_2} = t_{a}^{R_1 \otimes R_2} it might have the right index structure to transform a V_{R_1} \otimes V_{R_2} tensor but I am unsure about the generator basis index 'a'

The direct product is the Cartesian product. Since this gives an m+n dimensional representation, it is the same as the direct sum of the representations. The tensor product takes the two vectors from R_1 and R_2 and forms a matrix V_{\mu\alpha} = v_\mu v_\alpha. The construction I gave above maps this matrix into an mn dimensional vector which is in a reducible representation of the algebra.

The generators that act on V_{\mu\alpha} can be written as

(t_a^{(1)})_{\mu\nu} (t_a^{(2)})_{\alpha\beta},~~\text{no sum on}~a,

where we use the indices in the representation spaces to show that this is a higher rank tensor. It's simplest to pair matching generators with the index a, but other choices should just correspond to a change of basis. Alternatively we can map these onto the mn representation to express the generators as ordinary matrices.

 

[sgd37]

cool thanks for conferring with me you've cleared some things up. But can I mix generators with different group index or am I restricted to like indices even if there is no sum

 

[fzero]

cool thanks for conferring with me you've cleared some things up. But can I mix generators with different group index or am I restricted to like indices even if there is no sum

A different choices of pairing correspond to different choices of basis elements for the representation.