# representation of Lie groups-complexification

Suppose that $G$ is a (real) Lie group with Lie algebra $\mathfrak{g}$. The complexification of Lie groups and Lie algebras is part of the Weyl’s trick which I will talk about in the following.

Note that $\mathfrak{g}$ is only a real vector space, not a complex one. One naive way of giving it a complex structure is to get it tensor with $\mathbb{C}$. That is, $\mathfrak{g}_c=\mathfrak{g}\bigotimes_{\mathbb{R}}\mathbb{C}$. Then the problem is to give this new vector space a $\mathbb{C}$-linear Lie bracket. One natural way is this, suppose that $v\bigotimes a,u\bigotimes b\in \mathbb{C}$, we set $[v\bigotimes a,u\bigotimes b]=[v,u]\bigotimes ab$ and then expand this Lie bracket linearly to the whole space.

Complexifying a vector space is always easy, yet the problem is to complexify a Lie group. Here we give one definition of the complexification of a Lie group(note here I use the word ‘the’ to indicate the uniqueness of complexification, which can be seen from the definition):

Definition: a complex Lie group $G'$ is called complexification of $G$, if there is a morphism of (real) Lie groups, $i:G\rightarrow G'$, such that, for any complex Lie group $H$ and any morphism of (real) Lie groups $h:G\rightarrow H$, there is a unique morphism of complex Lie groups $h':G'\rightarrow H$ such that $h'\circ i=h$.

(We will often note the complexification of a Lie group $G$ as $G_c$)A priori, there is no reason that the complexification of a Lie group exists. Indeed, this can happen. For example, the universal cover of $SL_2(\mathbb{R})$(often written as $\tilde{SL_2(\mathbb{R})}$) does not have a complexification(see here).

Yet as for compact Lie groups, there always exist complexifications. This is really an important result in the representation theory, which will lead to the Weyl’s trick:

Theorem: $G$ is a compact Lie group, then there exists a complexification $(G_c,i)$ such that (1)$i:G\rightarrow G_c$ is injective;(2)$i(G)$ is a maximal compact subgroup of $G_c$;(3)$\mathfrak{g}(G_c)=\mathfrak{g}(G)\bigotimes \mathbb{C}$.

Note in the second phrase, we use the word ‘subgroup’, which is a very strong result in Lie group theory, as is known that a subgroup of a Lie group is by definition closed.

The essential ingredient in the proof of this result is the fact that, the complexification of the unitary group $U_n$ is GL_n(\mathbb{C}). If this is established, then by Peter-Weyl theorem we know that every compact Lie group $G$ can be injected into some $GL_n(\mathbb{C})$(write it as $f:G\rightarrow GL_n(\mathbb{C})$). What is more, $U_n$ is a maximal compact subgroup of $GL_n(\mathbb{C})$, so $f(G)$ is included in some $gU_ng^{-1}$ for some $g\in GL_n(\mathbb{C})$. Then we construct $G_c$ as $G_c=\{g e^{iX}|g\in G,X\in \mathfrak{g}\}$. Then we should verify that $G_c$ is indeed a complex Lie group and is a complexification of $G$.

Remark: the construction of $G_c$ ressembles very much to the polar decomposition of a matrix.

As for why the complexification of $U_n$ is $GL_n(\mathbb{C})$, this can be better seen using their Lie algebras. It is not hard to construct an isomorphism between $\mathfrak{u}_{n c}$ and $\mathfrak{gl}_n(\mathbb{C})$. For example, for $v\bigotimes a\in \mathfrak{u}_{n c}$, we send it to $av$ in the latter space. For any $M\in \mathfrak{gl}_n(\mathbb{C})$, we send it to $(M-M^{h})/2\bigotimes 1+(-i)(M+M^h)/2\bigotimes i$ where $M^h$ means the complex transpose of $M$. This in some sense means that their corresponding complex Lie groups are isomorphic.

Now we can say something about the Weyl’s trick. For example, we want to study the representations of some Lie group, say, $SL_n(\mathbb{R})$. We do not know much about this. Then we can study the representations of its Lie algebra, $\mathfrak{sl}_n(\mathbb{R})$. Perhaps it is still a bit difficult. Then we lift to the complex representations of its complexification $\mathfrak{sl}_n(\mathbb{C})$(this can be easily verified). Here we can easily construct an isomorphism between $\mathfrak{sl}_n(\mathbb{C})$ and $\mathfrak{su}_n$. So the complex representations of $\mathfrak{sl}_n(\mathbb{C})$ are equivalent to the representations of $\mathfrak{su}_n$, which are again equivalent to the representations of $SU_n$. Since $SL_n(\mathbb{R})$ is connected, this means that the representations of $SL_n(\mathbb{R})$ are equivalent to the complex representations of $SL_n(\mathbb{C})$. The latter are equivalent to the complex representations of its Lie algebra, $\mathfrak{sl}_n(\mathbb{C})$. So, in this way we see that the representations of $SL_n(\mathbb{R})$ are equivalent to the representations of $SU_n$. In other words,

$Rep(SL_n(\mathbb{R}))\approx Rep(\mathfrak{sl}_n(\mathbb{R}))\approx Rep_c(\mathfrak{sl}_n(\mathbb{C})) \approx Rep_c(\mathfrak{su}_n(\mathbb{C}))\approx Rep(\mathfrak{su}_n)\approx Rep(SU_n)$

where the subscript $c$ means complex representations. This equivalence also means that, any irreducible representation in one category corresponds to an irreducible representation in another category. So, since $SU_n$ is compact, which means that all its finite dimensional representations are semi-simple. This implies that all the finite dimensional representations of $SL_n(\mathbb{R})$ are also semi-simple. This result is not so easily obtained if we don’t use Weyl’s trick.