# Relationship between real Lie groups/algebras, complex Lie groups/algebras and complexification

Having read posts such as this and the links within, as well as the Wikipedia articles on complex Lie groups, complex Lie algebras and complexification, I am still rather confused about what they are and the relationship between them.

The context of this question is in trying to understand the representation theory of the Lorentz group. My current understanding is as follows:

Every real Lie group has an associated real Lie algebra, and every real Lie algebra $$\mathfrak{g}$$ has a unique associated simply connected real Lie group $$G$$. The basis of $$\mathfrak{g}$$ can be chosen to be some set of $$d$$ "generators" $$\{t_a\}$$, where $$d$$ is the dimensionality of the algebra, and the elements of $$G$$ are obtained from the exponentiation of real linear combinations of $$\{t_a\}$$.

In this language, is the complexification of the algebra $$\mathfrak{g}$$ a complex Lie algebra $$\mathfrak{g}_\mathbb{C}$$, and is the corresponding Lie group, where we instead exponentiate complex linear combinations of the generators, the complexification $$G_\mathbb{C}$$ of the Lie group $$G$$?

Expanding out my comment into a full answer. There are two ideas of complexification of Lie groups that are used. Because of the ambiguity and technical awkwardness we tend to sweep it under the rug and talk about complexified Lie algebras and their Lie groups rather than complexified Lie groups.

The first idea is perhaps the obvious one. A complexification of a real Lie group $$G$$ is a complex Lie group $$G_\mathbb{C}$$ which contains $$G$$ as a real Lie subgroup and whose Lie algebra $$\mathfrak{g}_\mathbb{C}$$ is a complexification of the Lie algebra of $$G$$. As an example of this $$SL_n(\mathbb{C})$$ is the complexification $$SL_n(\mathbb{R})$$. See here for more detail: https://encyclopediaofmath.org/wiki/Complexification_of_a_Lie_group

This works quite well in many cases and nicely interfaces with the Lie group/algebra correspondence. However it has a major flaw. Not every real Lie group has a complexification even among the semisimple groups. As I understand having a complexification in this sense is equivalent to being linear (so if you're into linear algebraic groups you can stop here). In particular the universal cover of $$SL_n(\mathbb{R})$$ has no complexification in this sense

To fill in the gaps we define the universal complexification. This is the definition that Wikipedia uses https://en.wikipedia.org/wiki/Complexification_(Lie_group). This one is defined by a universal property. The complexification of $$G$$ is a complex Lie group $$G_\mathbb{C}$$ together with a continuous homomorphism $$\varphi:G \to G_\mathbb{C}$$ such that for any homomorphism to a complex Lie group $$f:G\to H$$ there is a unique complex analytic homomorphism $$F: G_\mathbb{C}\to H$$ such that $$f= F \circ \varphi$$.

Constructively, we can build this as follows. Take the simply connected group $$\tilde{G}$$ covering $$G$$ and the natural homomorphism $$\Phi:\tilde{G}\to \tilde{G}_\mathbb{C}$$ from this into the simply connected complex group (which we find by complexifying the Lie algebra). The natural projection $$\pi:\tilde{G}\to G$$ has kernel equal to the fundamental group of $$G$$. Now we define the universal complexification as $$\tilde{G}_\mathbb{C}/N$$ where $$N$$ is the smallest normal subgroup which contains $$\Phi(\ker \pi)$$.

As you can see the second is much more complicated and can even produce a group whose Lie algebra isn't even the complexification of the original group's but a quotient of it. One neat thing is the universal complexification of a simply connected group is exactly the corresponding simply connected complex group.

• Thank you for this. I just stumbled upon math.stackexchange.com/q/3183/96384 and its answers which do not talk about complexification, but I think the issue is related. Commented Dec 19, 2021 at 6:23

Partial answer, mostly a too long comment.

The basis of $$\mathfrak g$$ is the set [of] "generators" $$\{t_\alpha\}$$:

Any real Lie algebra $$\mathfrak g \neq 0$$ has infinitely many choices of bases, although some might be handier for computational purposes than others. Often people fix a basis somewhere early in the theory and then never bring up the issue if anything depends on that choice (some things don't, some do).

and the elements of G are obtained from the exponentiation of real linear combinations of $$\{t_\alpha \}$$.

What you are claiming here seems to be that the exponential map $$exp: \mathfrak g \rightarrow G$$ is surjective. In the general case, deciding whether this is true is a notoriously difficult question, cf. for example the answer https://math.stackexchange.com/a/3859991/96384 with further links, as well as On surjectivity of exponential map for Lie groups and https://mathoverflow.net/q/41644/27465 and https://en.wikipedia.org/wiki/Exponential_map_(Lie_theory)#Surjectivity_of_the_exponential.

To be fair, you restrict to the case that the Lie group $$G$$ is connected and simply connected. It is certainly true according to the above that then $$G$$ is generated by the image of the exponential, i.e. you can write any element of $$G$$ as a a product of exponentials of real linear combinations of your basis $$\{t_\alpha\}$$. Whether $$exp$$ is surjective, i.e. you can actually write each element as such an exponential (as opposed to a product of exponentials), I am not sure right now, but I doubt it. The standard example for non-surjectivity of $$exp$$ is the Lie group $$SL_2(\mathbb R)$$; now, that one is not simply-connected; what I guess is that if the exponential to its universal cover were surjective, then so would this one be; I am just not sure if this slick argument overlooks something.

What it does point out though is that the Lie-theoretic "exponential map" might be not just the matrix exponential, for which the above is an example: The matrix exponential of $$\mathfrak{sl}_2(\mathbb R)$$ lands in $$SL_2(\mathbb R)$$, but not in its universal cover (which is the simply connected Lie group you are looking for), which is "much bigger".

In this language, is the complexification of the vector space spanned by $$\{t_\alpha\}$$ a complex Lie algebra $$\mathfrak g_{\mathbb C}$$

Yes. The formal mathematical definition is with a tensor product, $$\mathfrak g_{\mathbb C} := \mathbb C \otimes_{\mathbb R} \mathfrak g$$. If $$\{t_\alpha\}$$ is a basis of $$\mathfrak g$$ (as a real vector space), then one can think of $$\mathfrak g_{\mathbb C}$$ as a complex vector space with "the same" basis, just that now complex linear combinations are allowed (formally, $$\{1 \otimes t_\alpha\}$$ is a basis of the complex space $$\mathfrak g_\mathbb C$$). Often it is useful to use different bases though.

and is the corresponding Lie group (where we instead use complex linear combinations of $$\{t_\alpha\}$$) a complex Lie group $$G_{\mathbb C}$$?

Too many things asked at once. There is a notion of complexification of Lie Groups, taking a real Lie group $$G$$ to a complex Lie group $$G_\mathbb C$$, which of course is made to commute in some way, via the Lie algebra - Lie group correspondence, with the complexification of Lie algebras. What you are asking might be,

if the complexification of the connected simply connected (real) Lie group corresponding to a real Lie algebra $$\mathfrak g$$, is the conneted simply connected (complex) Lie group corresponding to the complexified Lie algebra $$\mathfrak g_\mathbb C$$.

To be honest, again I am not sure. However, I am relatively sure that in the case $$\mathfrak g = \mathfrak{sl}_2(\mathbb R)$$, on the one hand, the simply connected connected complex Lie group corresponding to $$\mathfrak g_\mathbb C = \mathfrak{sl}_2(\mathbb C)$$ is just $$SL_2(\mathbb C)$$, whereas on the other hand, as said before, the simply connected connected real $$G$$ is huge (it is not a matrix gorup), and I very much doubt that its Lie-theoretic complexification would be $$SL_2(\mathbb C)$$.

• Thanks very much! The details of your answer are really useful for someone such as myself who still feels a bit lost on this subject. Your interpretation of what I was asking is correct - I should edit the question a bit to make it clearer :) Commented Dec 13, 2021 at 18:42
• Complexification of Lie groups is a thorny topic and much less clear than the Lie algebra picture. Depending on how you define complexification, the simply connected group covering $SL_2(\mathbb{R})$ either has no complexification or it is equal to the complexification of $SL_2(\mathbb{R})$ which is just $SL_2(\mathbb{C})$. Commented Dec 14, 2021 at 16:58
• A lot of sources make great effort to steer clear of these issues as much as possible Commented Dec 14, 2021 at 17:01
• @Callum: As did I ... -- If you expanded your comment and, in a separate answer, gave a little intro to the at least two definitions/constructions of group complexification you have in mind, and what they (not) give out in our toy example $\mathfrak g =\mathfrak{sl}_2(\mathbb R)$ (or even better, general $\mathfrak{sl}_n(\mathbb R)$), I and probably others would be much obliged. Commented Dec 14, 2021 at 22:42