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I have just started learning about Lie algebras and got confused with this:

Some sources say the generators are $J_0,J_1$ and $J_2$ and some use $J_0,J_+$ and $J_-$. Which set is correct?

Or if both are correct what key concept am I missing here?

My understanding is that if have certain commutation relations then we know that Lie Algebra is such and such.

But if we have two such choices then this understanding falls apart?

How do I figure out what is $\mathfrak{su}(2)$ Lie Algebra in general?

Saurabh Shringarpure
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  • When I learned quantum mechanics, $J_x, J_y and J_z$ were introduced as operators and $J_+ and J_-$ were introduced as raising and lowering operators for $J_z$. This allows one to derive all the possible eigenvalues of $J_z$, and of $J^2$ etc. Although $L_+ and L_-$ are linear combinations of generators, they are not Hermitian, and so they do not generate infinitesimal group transformations. – tippy2tina Aug 14 '20 at 01:39
  • So, are $J_+$ and $J_-$, not generators? – Saurabh Shringarpure Aug 14 '20 at 01:50
  • The literature is not entirely clear to me, but I think they are sometimes included as generators of the Lie Algebra, but they do not generate infinitesimal group transformations. – tippy2tina Aug 14 '20 at 03:01
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    I wonder what are $J_0,J_1,J_2,J_{+},J_{-}$ here, can you make them explicit? – Alexey Do Aug 14 '20 at 05:17
  • $\begin{equation} J_1=\frac{1}{2} \begin{pmatrix} 0 & 1\ 1 & 0 \end{pmatrix} \end{equation}$, $\begin{equation} J_2=\frac{1}{2} \begin{pmatrix} 0 & -i\ i & 0 \end{pmatrix} \end{equation}$, $\begin{equation} J_0=\frac{1}{2} \begin{pmatrix} 1 & 0\ 0 & -1 \end{pmatrix} \end{equation}$, $J_+=J_1+iJ_2$ and $J_-=J_1-iJ_2$. – Saurabh Shringarpure Aug 14 '20 at 10:40

1 Answers1

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Remember, there is the Lie group $SU(2)$ and its Lie algebra $\mathfrak{su}(2)$ which are two entirely different concepts. We are talking about the latter.

Anyway, a Lie algebra is in particular a vector space, and since $\mathfrak{su}(2)$ has dimension 3 (why?), it follows that the above matrix sets will generate it as a vector space, since they are linearly independent sets.

The Lie bracket is bilinear, so if we know what it does to a basis, we know the whole Lie algebra. This is why the commutation relations are practical to write down, but remember that it is basis dependent, so different generator sets will give different commutation relations.

Edit: Another notation thing. Don't use the word "representation" as in the title, representations of Lie algebras is a huge subject in itself. Presentation would be a better word, though I'm not sure if it is that fitting in the context of Lie algebras.

Richard Jensen
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  • I read in related questions in the physics StackExchange that $J_\pm$ are related to something called "complexification" and so it is not valid. I'm really confused about what this term is why it is not valid? – Saurabh Shringarpure Aug 14 '20 at 11:22
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    Can't quite remember the exact relations, let me get back to you when I get back home :). – Richard Jensen Aug 14 '20 at 11:32