5
$\begingroup$

I want to find the irreducible representations of the Lie algebra $\mathrm{so}(3,1)$. I know the standard procedure goes on like

  1. Complexify the Lie algebra, obtaining the complex Lie algebra $A_1\oplus A_1$;
  2. Obtain all the irreducible representations of $A_1$
  3. Build a representation of $A_1\oplus A_1$ from the product of two irreducible representations of $A_1$.

Now I should go back to the real algebra $\mathrm{so}(3,1)$, but I don't know how. I know that if we have a (complex-linear) representation of a complex Lie algebra $L$ we can use its Weyl canonical basis (constructed using the Cartan subalgebra and the root system) to create a real-linear representation of the compact real section of $L$. In my case, however, the compact real section is $\mathrm{so}(4)$ (or $\mathrm{su}(2)\oplus\mathrm{su}(2)$ if you like), not $\mathrm{so}(3,1)$.

In its article on the representation theory of the Lorentz group, Wikipedia says that

[...] all irreducible representations of $\mathrm{so}(3,1)_{\mathbb{C} }$, and, by restriction, those of $\mathrm{so}(3,1)$ are obtained.

What "restriction" is it about? Would someone explain how can we construct a representation of a real Lie algebra from the representations of its complexification?

$\endgroup$
1

1 Answer 1

Reset to default
4
$\begingroup$

Let $\mathfrak{g}$ be a Lie algebra, then $\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$ is a complex Lie algebra where we extend the Lie bracket by complex linearity. The map $i : \mathfrak{g} \to \mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$ given by $X \mapsto X\otimes 1$ is a Lie algebra homomorphism.

Recall that a representation of a Lie algebra is just a Lie algebra homomorphism to $\mathfrak{gl}(V)$. Therefore, if $\rho : \mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C} \to \mathfrak{gl}(V)$ is a representation of $\mathfrak{g}\otimes_{\mathbb{R}}\mathbb{C}$, then $\rho\circ i : \mathfrak{g} \to \mathfrak{gl}(V)$ is a representation of $\mathfrak{g}$ (because the composition of Lie algebra homomorphisms is a Lie algebra homomorphism).

$\endgroup$
5
  • 1
    $\begingroup$ So can we say that given a Lie algebra $g_0$ with basis $\{e_k\}$, its complexification $g$ and a representation $\rho$ of $g$, we can obtain a representation of $g_0$ by restricting to real-linear combinations of the elements $\rho(e_k)$, that is $\rho(i(e_k))$? $\endgroup$
    – yellon
    Jun 15, 2018 at 12:37
  • 1
    $\begingroup$ Moreover, can every representation of the real algebra be found this way? $\endgroup$
    – yellon
    Jun 15, 2018 at 12:42
  • $\begingroup$ In general, I don't think every representation of $\mathfrak{g}$ arises this way. $\endgroup$
    – Michael Albanese
    Jun 19, 2018 at 19:11
  • 1
    $\begingroup$ If we are talking about representations on complex vector spaces, complexification does induce a bijection between complex rep's of a real Lie algebra and complex rep's of its complexification, cf. math.stackexchange.com/q/1408894/96384 (and that's straightforward: every rep of the real $g$ is the restriction to $g$ of the complexified rep. of $g_{\mathbb C}$). $\endgroup$
    – Torsten Schoeneberg
    Jul 5, 2019 at 21:47
  • $\begingroup$ If, on the other hand, we start with rep's of $g$ on a real vector space $V$ (which is rarely studied actually), then one can first extend that to one on the complex vector space $\mathbb C \otimes_{\mathbb R}V$, and then that to a rep of $g_{\mathbb C}$. Here however the induced map on isoclasses of irreducibles is definitely not surjective, i.e. not every complex representation comes from a real representation, cf. math.stackexchange.com/a/1026919/96384; and I think (as Qiaochu seems to suggest in his comment there) it is not injective either. $\endgroup$
    – Torsten Schoeneberg
    Jul 5, 2019 at 21:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .