Is the Lorentz group $O(1,3)$ irreducible, i.e. simple?

Question

I'm studying group theory and in particular the Lorentz group $O(1,3)$. The book I'm studying from talks about the proper orthochronus Lorentz group. I've studied that I can rearrange the generators of its algebra (that I'll denote as $so^+(1,3)$) in such a way that the six generator that I've formed are two distinct sets of $su(2)$'s generators, showing that $so^+(1,3)\cong su(2)\oplus su(2)$. The book goes on discussing about the irreps of $so^+(1,3)$, and so my question: why does it make sense to talk about $so^+(1,3)$ irreps when I know that there are two distinct invariant subalgebras? Shouldn't I say that $so^+(1,3)$ is indeed completely reducible? Am I missing something?

score 12 · Accepted Answer · edited Mar 30 '22 at 19:59

12

Groups are not reducible or irreducible, representations are.

You are thinking of the notion of simplicity vs. semi-simplicity.

Let me discuss it at the level of Lie algebras so we don't have any global issues: the complexification of the Lie algebra $\mathfrak{so}(1,3)$ is semisimple but it is not simple because, as you noted, it is the direct sum of two algebras $\mathfrak{su}(2)_\mathbb C\oplus \mathfrak{su}(2)_\mathbb C$.

A consequence of this is that the adjoint representation is reducible, indeed it is $\mathbf{adj} = (1,0)\oplus(0,1)$. In general $$ \mbox{$\mathfrak{g}$ is simple} \;\Longleftrightarrow\; \mbox{$\mathbf{adj}(\mathfrak{g})$ is irreducible} $$

Nevertheless, non-simple groups can and do have irreducible representations.

edited Mar 30 '22 at 19:59

Torsten Schoeneberg

103

answered Jan 08 '21 at 12:29

MannyC

6,706

1

I messed it up, thanks for the correction, that was clear. Just to know, where does the representation should take place in this reasoning? Could you suggest me a book where I can read more about this argument? – Matteo Brini Jan 08 '21 at 12:41
Just one more question. I just noticed that $so(1,3)\sim su(2)\otimes su(2)$: how can I see that $so(1,3)$ is a direct sum of two $so(2)$? Thanks. – Matteo Brini Jan 08 '21 at 13:00
There are many books of representation theory. Try to ask a resource recommendation question and many people will reply.
As for the second comment: at the level of groups you do a direct product and at the level of algebras you do a direct sum. I meant to write $su$, not $so$, I'll fix it.
– MannyC Jan 08 '21 at 13:08
Does the last fact derive from the exponential matrix form that links the group to its algebra? Thanks again! – Matteo Brini Jan 08 '21 at 13:20
Yes I'd say so. – MannyC Jan 08 '21 at 13:29
1

It is adjoint is irreducible over the complex field, but not over the reals, i.e. your argument holds for the complexification but not for the real form. – ZeroTheHero Jan 08 '21 at 14:05
@ZeroTheHero Thanks for the clarification – MannyC Jan 08 '21 at 14:14
1

I think it's exactly the other way around, @ZeroTheHero. The adjoint representation of the complexification is reducible (because the complexification is semisimple but not simple), whereas the adjoint representation of this real form is irreducible (because it is a simple real Lie algebra). – Torsten Schoeneberg Mar 30 '22 at 04:33
1

@TorstenSchoeneberg oups. Good catch. you are correct I got this the wrong way about. – ZeroTheHero Mar 30 '22 at 08:51

score 8 · Answer 2 · answered Jan 08 '21 at 14:44

For what it's worth:

The real Lorentz Lie algebra $so(1,3;\mathbb{R})\cong sl(2,\mathbb{C})$ is simple.
Its complexification $so(1,3;\mathbb{C})\cong sl(2,\mathbb{C}\oplus sl(2,\mathbb{C})$ is semisimple but not simple.

Is the Lorentz group $O(1,3)$ irreducible, i.e. simple?

3 Answers3