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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?

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3 Answers 3

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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.

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    $\begingroup$ 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? $\endgroup$ Commented Jan 8, 2021 at 12:41
  • $\begingroup$ 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. $\endgroup$ Commented Jan 8, 2021 at 13:00
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    $\begingroup$ 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. $\endgroup$ Commented Jan 8, 2021 at 14:05
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    $\begingroup$ 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). $\endgroup$ Commented Mar 30, 2022 at 4:33
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    $\begingroup$ @TorstenSchoeneberg oups. Good catch. you are correct I got this the wrong way about. $\endgroup$ Commented Mar 30, 2022 at 8:51
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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.

See also this related Phys.SE post.

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

  • the presence of subgroups doesn't imply the reducibility/irreducibility of representations. The subgroup/subalgebra is a feature of the group (or algebra), while reducibility is a feature of the representation and the representation vector space. Even in the case of the adjoint representation, for which the algebra and the representation space coincide, you can say something about the adjoint representation only.
  • even if a representation of the group was reducible, this doesn't mean that there aren't any irreducible ones! If you find a reducible representation, it means it can be reduced... into irreducible representations.
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