# Mobius: Representations of SU(2,1)=U(1,1)

#### TriTertButoxy

I'm studying the representations of SU(2,1) [or U(1,1)], and since they are non-compact, their representations are necessarily infinite dimensional.

I have a couple questions.

In the literature, they say the algebra satisfied by the three generators, $T_1, T_2, T_3$ is

$$[T_1,\,T_2]=-i T_3$$
$$[T_2,\,T_3]=i T_1$$
$$[T_3,\,T_1]=i T_2$$​

Is this correct?

And, secondly, given that the group is non-compact, I should expect the generators to be represented by infinite-dimensional matrices only. But, in the literature, I found the following:

$$T_1 = \frac{1}{2}\begin{pmatrix}0&1\\-1&0\end{pmatrix}\qquad T_2=\frac{1}{2}\begin{pmatrix}0&-i\\-i&0\end{pmatrix}\qquad T_3=\frac{1}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$​

and these appear to satisfy the algebra above. So why aren't they infinite-dimensional? What's the deal?

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#### chrispb

So, su(2,1) can't be u(1,1). I would expect to be (2+1)^2 - 1 = 8-dimensional, and u(1,1) to be (1+1)^2 = 4-dimensional. The algebra you listed is 3d. It looks like it may very well be su(1,1) = so(2,1) = sl(2,R), I think. I expect one would get to u(1,1) by adding a central charge (a generator that commutes with everything), at which point the commutation relations wouldn't change. I did not verify that that algebra is consistent with the definition of su(1,1), but it certainly appears to be at first glance.

To answer your question, non-compact groups have UNITARY representations that are necessarily infinite-dimensional. They can have finite-dimensional representations that are not unitary. What you listed was an example of such a representation. The way to see whether a representation of a group is unitary by means of looking at the representation of the algebra alone is to ask whether or not the representation is hermitian. In this case, T3 is indeed hermitian, but T1 and T2 are not. The hermitian representations of this would be infinite-dimensional; for example, the action of this algebra on functions/fields on 2+1d spacetime. (This algebra would act as the lorentz algebra on those functions.)

#### TriTertButoxy

Awesome! Thank you for your very clear answer.

*And I totally botched the names, I really mean to say SO(2,1) [SU(1,1)]. My brain was apparently dead when I was typing this out.

But also, you mentioned sl(2,R). What does that mean? and how is it different from SU(1,1) and SO(2,1)?

#### chrispb

Just so I'm clear, I'm using capital letters (e.g SU(2)) to mean the group generated by the algebra denoted by the same lowercase letters (su(2)). I don't know if that's common practice or not.

SL(2,C) is the group of 2x2 complex matrices of determinant one. It can be generated by three matrices (including the three you listed above) over the field C. What I mean by that is if you took e^ix_a T_a for x_a in C and T_a one of your three matrices, you'd get an element of the group. To form the group SL(2,R), one would demand x_a be in R instead of C. Strictly speaking, this doesn't change the commutation relations of sl(2), the algebra, so perhaps it was a poor choice of notation on my part to say sl(2,R). (It DOES change at the group level, though.) They are all isomorphic structures... what I mean by that is the algebras are identical. You may want to read literature regarding how SO(3) and SU(2) (the groups) are not the same (though the latter is the double cover of the former), but so(3) and su(2) (the algebras) are identical. This is the sense in which I meant the equality above. Hopefully this helps!

#### TriTertButoxy

Yes yes, That was very clear. Thank you.

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