# Confusion in group theory at the example of the Lie algebra $su(2)$

Please excuse me for this unstructured question: I have some problems with understanding group theoretical aspects relevant to physics and would like to make one of my confusions clear by discussing the $$su(2)$$ algebra, given by:

$$[T_i,T_j] = i \epsilon_{ijk} T_k.$$

As far as I understood, if I now choose matrices that fulfill these relations denoting them $$T_1, T_2$$ and $$T_3$$ they render a representation.

First question: Can they be understood as a basis for the algebra and wouldn't then linear combinations of them yield new representation? Or would they still describe the same representation? I.e. are the generators of a specific representation unique? In the $$su(2)$$ case I am specifically confused about the step that a lot of authors perform by defining new generators as:

$$J_{\pm} = \frac{1}{\sqrt{2}}(J_1 \pm i J_2)$$

satisfying:

$$[J_{+},J_{-}] = J_3$$ and $$[J_3,J_{\pm}]= \pm J_{\pm}.$$

What does this specifically "do", as we clearly changed the generators?

1. Yes, the $$T_i$$ are supposed to be a basis of the algebra as a vector space. Therefore, choosing n-by-n matrices $$\rho(T_i)$$ that represent each of the $$T_i$$ defines a linear map $$\rho : \mathfrak{su}(2) \to \mathfrak{gl}(n)$$ ($$\mathfrak{gl}(n)$$ is just notation for the n-by-n matrices here; linear maps are fully specified by giving their values on a basis), hence a representation of $$\mathfrak{su}(2)$$ in the proper sense.

2. Choosing a new basis $$T'_i$$ does not change the map $$\rho$$ - if you don't suddenly claim that $$\rho(T'_i)$$ has different values (and hence define a different representation map $$\rho'$$ by that), what set of generators you're looking at doesn't affect any representations.

3. The $$J_\pm = J_1\pm\mathrm{i} J_2$$ does something often glossed over in physics texts: Tt complexifies the Lie algebra to be $$\mathfrak{su}(2)_\mathbb{C}$$ instead of $$\mathfrak{su}(2)$$ in addition to choosing a basis that makes it particularly easy to evaluate what the possible representations are.

When you are looking at complex representations (i.e. representations of the algebra by potentially complex-valued matrices), then the representations of a Lie algebra and its complexification are the same. It might sounds strange to stress the difference nevertheless, but it will come back when you look at representations of the Lorentz algebra, where we have $$\mathfrak{so}(1,3)_\mathbb{C} \cong \mathfrak{su}(2)_\mathbb{C} \oplus \mathfrak{su}(2)_\mathbb{C}$$ and leaving out the complexification can lead to confusion, see e.g. this question and my answer there.

...and wouldn't then linear combinations of them yield new representation?

Not necessarily. Suppose you have a representation in matrix form like: $$[M_i, M_j] = i\epsilon_{ijk}M_k$$

A linear combination of the $$M_i$$ looks like $$\tilde M_i = \Lambda_{i\ell}M_\ell\;,$$ where the $$\Lambda_{i\ell}$$ are numbers.

Then you have: $$[\tilde M_i, \tilde M_j] = \Lambda_{i\ell}\Lambda_{jm}[M_\ell, M_m]$$ $$=\Lambda_{i\ell}\Lambda_{jm}i\epsilon_{\ell m n}M_n$$ $$=\Lambda_{i\ell}\Lambda_{jm}i\epsilon_{\ell m n}\Lambda^{-1}_{np}\tilde M_p$$

So, you need $$\Lambda_{i\ell}\Lambda_{jm}\epsilon_{\ell m n}\Lambda^{-1}_{np} = \epsilon_{ijp}$$

For example, in the case where you specialize to $$\Lambda^{-1} = \Lambda^T$$ (e.g., rotations of the $$M_i$$ into each other) then the condition above can be rewritten as: $$\det(\Lambda) = 1\;.$$

...I am specifically confused about the step that a lot of authors perform by defining new generators as:

$$J_{\pm} = \frac{1}{\sqrt{2}}(J_1 \pm i J_2)$$

satisfying:

$$[J_{+},J_{-}] = J_3$$ and $$[J_3,J_{\pm}]= \pm J_{\pm}.$$

These are not the same commutation relations. Here, if you let $$i$$ range over $$+,-,3$$ you have $$[J_i,J_j] = \epsilon_{ijk} J_k\;,$$ without the factor of $$i$$.