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

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 $$so^+(1,3)\sim su(2)\times su(2)$$. 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?

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 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 $$\mathfrak{su}(2)\oplus \mathfrak{su}(2)$$.

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.

• 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? Jan 8 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. Jan 8 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. Jan 8 at 13:08
• Does the last fact derive from the exponential matrix form that links the group to its algebra? Thanks again! Jan 8 at 13:20
• 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. Jan 8 at 14:05

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.

It does make sense to talk about the irreps of $$\mathfrak{so}^+(1,3)$$, and therefore of $$SO^+(1,3)$$: