# Tensor representation of the Lorentz Group

• I
chingel
I've been trying to understand representations of the Lorentz group. So as far as I understand, when an object is in an (m,n) representation, then it has two indices (let's say the object is ##\phi^{ij}##), where one index ##i## transforms as ##\exp(i(\theta_k-i\beta_k)A_k)## and the other index as ##\exp(i(\theta_k+i\beta_k)B_k)##, where A and B are commuting su(2) generators of dimension (2m+1) and (2n+1) respectively and ##\theta## are the rotation angles and ##\beta## the rapidities.

It is often said (for example I am reading the Schwartz QFT book, where it is mentioned), that the (n/2,n/2) representation corresponds to a tensor with n indices.

How is this the case? How can I see it?

Is the meaning of the tensor representation that each index transforms using the usual 4x4 Lorentz transformation matrix? How is it the same as a two index object where each index transforms under a (n+1) dimensional su(2) representation?

## Answers and Replies

chingel
At least for the (1/2, 1/2) representation I figured it out by surfing on the internet. I can choose for A representation the ##\sigma/2## matrices and for B take ##-\sigma^*/2##, then the two index object ##\phi^{ij}## transforms as $$\phi \rightarrow \exp(i(\theta_k-i\beta_k)\sigma_k/2)\, \phi\, \exp(i(\theta_k-i\beta_k)\sigma_k/2)^\dagger,$$ where we have just matrix multiplication. Note that the element ##\exp(i(\theta_k-i\beta_k)\sigma_k/2)## belongs to SL(2) with complex entries. So the determinant of ##\phi## is constant under the transformation, and the transformation keeps the ##\phi## Hermitian if it was Hermitian to begin with (it has to be Hermitian to be connected to the identity). Now we can identify the matrix ##\phi## as (you might want to double check the signs) $$\phi = \begin{bmatrix} p^0+p^3 & p^1-ip^2 \\ p^1+ip^2 & p^0-p^3 \end{bmatrix},$$ where the determinant gives just ##p_\mu p^\mu=m^2##, which is good because it is constant, and also then we can check that under the ##\phi## transformation the ##p^\mu## transform as we would expect under Lorentz transformations. So in the end we can say that the (1/2,1/2) representation indeed can be written using four parameters that transform as a four vector.

But now how do I do it for the (1,1) case to show that it has 4x4 components that transform as a two index Lorentz tensor ##T_{\mu\nu}##?

Gold Member
2021 Award
The point is that ##(j_1,j_2)## describe irreducible representations. It's ##(2j_1+1) (2j_2+1)## dimensional, i.e., for ##(1,1)## you have 9, not 16 components. This is easy to understand, because you can decompose the general 2nd-rank Tensor into a scalar part (1 component), given by its trace, the antisymmetric piece (6 components) and the traceless symmetric piece (9 components).

For a very good and didactical introduction to the group- and representation-theoretical analysis of the Lorentz and Poincare groups, see

Sexl, Urbandtke, Relativity, Groups, Particles, Springer (2001)

chingel and dextercioby
chingel
Thanks, I managed to take a quick look at the book and chapter 8 seems to cover exactly what I was asking about (and that I didn't find in my QFT books).

On a quick skim apparently if we had for a four vector ##p_\mu\sigma^\mu## and transforming this under (1/2,1/2) gave the proper transformation for ##p_\mu##, then more generally there is this quantity ##T^{ij\dots}\sigma_i\sigma_j\dots##, and using this I think they can show the equivalence to tensors. However I'll have to read the chapter in more detail to figure out the details how exactly this works out.

chingel
So for anybody who is interested, the argument is quite simple in my opinion.

First we know from angular momentum addition that by adding m copies of spin 1/2 (I mean by doing a tensor product of them) and symmetrizing, we get a m/2 representation of SU(2).

So a (m/2,m/2) representation of the Lorentz group can be written as ##\phi_{i_1\dots i_m j_1\dots j_m}##, where it is symmetric over the i indices, and also symmetric over the j indices, and each index transforms as (1/2) representation of SU(2) either LH for i or RH for j.

Now before we showed that (1/2,1/2) is a four vector, where we used ##\phi_{ij} = p^\mu \sigma_{\mu,ij}## (or the other way around it is ##\frac12 \sigma_{\mu,ji}\phi_{ij} = p^\mu ##) where ##\sigma_\mu## is ##(1,\vec{\sigma})##. So now we can just consider each pair of indices ##(i_1, j_1)## etc and each of those is a four vector, so we just have a tensor product of m pairs each of which transforms as a four vector, so the 4 tensor is like this $$T^{\mu_1\dots\mu_m} = \phi_{i_1\dots i_m j_1\dots j_m} \sigma^{\mu_1}_{i_1j_1}\dots\sigma^{\mu_m}_{i_mj_m} .$$ From this we see that the 4 tensor T has to be symmetric, because the right hand side is due to ##\phi## being symmetric.

dextercioby