I am reading some lie groups/lie algebras on my own..

I am using Brian Hall's Lie Groups, Lie Algebras, and Representations: An Elementary Introduction

I was checking for some other references on lie groups and found J. S. Milne's notes Lie Algebras, Algebraic Groups,and Lie Groups

It was written in introductory page of algebraic groups chapter that :

Most of the theory of algebraic groups in characteristic zero is visible already in the theory of Lie algebras

I would like to know if anybody wants to make it more clear..

I am planning to read some algebraic groups also and I was kind of happy to see that lie groups/lie algebras and algebraic groups are related.

I had a very basic course in algebraic geometry and I want to learn algebraic groups as well...

I would be happy if one can give some other references or exposition to comment made by Milne or give some idea of how much algebraic geometry is related to algebraic groups.

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    $\begingroup$ I think, Milne is already an excellent reference. In chpater II, section 3 he explains exactly what you want (the Lie algebra of an algebraic group etc.). His new version of LAG is of May 5, 2013. $\endgroup$ Commented Jun 13, 2014 at 8:30
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    $\begingroup$ @DietrichBurde : I am unable to follow his ideas as i am just a beginner.. so i thought some comments from others would be helpful.... I would be so thankful if you want to say anything about this... :) $\endgroup$
    – user87543
    Commented Jun 13, 2014 at 10:18
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    $\begingroup$ I'm living proof for the claim that you can do a PhD's worth of algebraic groups and rep theory without learning much at all about Lie groups. Characteristic zero algebra is kinda boring anyway :-/ Anyway, I learned the basics from Humphreys' books (GTM series, Intro to Lie Algebras... and Algebraic groups). The former is IMHO great, but I am not equally sold on the latter. Humphreys covers the necessary algebraic geometry, but some people who know more AG than I ever will have told me that it could have been better. $\endgroup$ Commented Jun 21, 2014 at 16:49
  • $\begingroup$ @JyrkiLahtonen : Sir, I am sorry that i could not understand what you are trying to convey :(... $\endgroup$
    – user87543
    Commented Jun 21, 2014 at 16:57
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    $\begingroup$ @JyrkiLahtonen What a coincidence. I did essentially the same (PhD in algebraic groups without learning about Lie groups). $\endgroup$ Commented Jun 21, 2014 at 17:54

4 Answers 4


I will focus on the representation theory of the objects involved, which is a place where the quote mentioned is pretty much precisely what happens.

So let's start with an algebraic group $G$ over $\mathbb{C}$. We would like to know something about the representations of this group (by representation I mean rational representation).

Let's also assume that we are very good at this whole business with representations of Lie algebras, so if we could somehow get a Lie algebra from our algebraic group, that would be good. If this Lie algebra can somehow tell us something about the representations of the algebraic group, then that will be even better.

Fortunately, this is exactly what we can do. Associated to $G$ is a Lie algebra $\operatorname{Lie}(G)$, and the representations of this Lie algebra correspond directly to those of $G$ (I will make this more precise a bit later).

Before I go into further details about this Lie algebra and how the representations relate to each other, I would like to take a brief detour through a more general case, namely that where we work over an algebraically closed field whose characteristic need not be $0$ (to do this properly, one should then have $G$ be a group scheme, but I will not get into that here).
When the characteristic is not $0$, we still get a Lie algebra, but this Lie algebra is "too small" to capture the representations of the group (instead, it is a restricted Lie algebra, and its restricted representations correspond to the representations of the first Frobenius kernel of $G$. Ignore the previous sentence if you are not familiar with schemes).

So is there something we can do that works in all characteristics?
The answer is yes, and it is called the algebra of distributions on $G$ (or sometimes the hyperalgebra of $G$). I will not go into how this is constructed here, but just note that no matter the characteristic, we have an equivalence of categories between the representations of the algebra of distributions on $G$ and the representations of $G$.

Finally, how does this algebra of distributions on $G$ relate to the Lie algebra of $G$?
Here, the answer is very nice in characteristic $0$, since the algebra of distributions on $G$ is in fact isomorphic to the universal enveloping algebra of $\operatorname{Lie}(G)$, and hence representations of this algebra correspond precisely to representations of $\operatorname{Lie}(G)$.

The above of course lacks any sort of detail, but often it is good to see the big picture first, so that is what I decided to provide here.
For more details, the books called "Linear Algebraic Groups" are a common reference (there is one each by Borel, Humphreys and Springer, though I am not familiar with the Borel one myself). I am not sure if any of those actually discuss the algebra of distributions (as the main focus there is on characteristic $0$ where it is not really needed), so if you want a more advanced treatment, I recommend Representations of Algebraic Groups by Jantzen (but beware that this is a very advanced book that assumes a good familiarity with at least one of the previously mentioned books as well as some further algebraic geometry).

  • $\begingroup$ i can not say i understand what you have said.. it is not of my level... may be i need some more time to understand.... $\endgroup$
    – user87543
    Commented Jun 19, 2014 at 14:07
  • $\begingroup$ @PraphullaKoushik Which part of it? Note that some of the parts are on purpose at a more advanced level (but without any actual details). $\endgroup$ Commented Jun 19, 2014 at 19:45
  • $\begingroup$ As i am not familiar with algebraic groups so much i am getting confused with the way it was explained(almost all)... :) $\endgroup$
    – user87543
    Commented Jun 20, 2014 at 9:11
  • $\begingroup$ @PraphullaKoushik Are you familiar with representations of Lie algebras and similar objects? $\endgroup$ Commented Jun 20, 2014 at 9:16
  • $\begingroup$ I know representation theory of finite groups and now i am studying representations of lie algebras on my own.... I can not say confidently that i know representations of lie algebras.. I am just a beginner.. $\endgroup$
    – user87543
    Commented Jun 20, 2014 at 9:19

I state without a complete proof the following interesting theorem.

Theorem 1. Let $G$ be a connected, complex, semisimple Lie group. $G$ admits a unique algebraic group structure.

Proof. In [OV], one can find the proof that for any connected, complex Lie group $G$ such that it is equal to $G^{\prime}$ (the derived subgroup) the holomorphic representations in an algebraic group are polynomial; in particular, this statement holds for $G$ semisimple.

By definition of adjoint representation and by the hypothesis, $\ker\mathrm{ad}=Z(\mathfrak{g})=\{0\}$ and therefore $\mathfrak{g}$ has a faithful representation in $\mathfrak{gl}(\mathfrak{g})$; where $\mathfrak{g}$ is the Lie algebra of $G$ and $\mathfrak{gl}(\mathfrak{g})$ is the general linear group over the vector space $\mathfrak{g}$.

Considered the diagram: \begin{matrix} \mathfrak{g} & \stackrel{\mathrm{ad}} {\longrightarrow} & \mathfrak{gl}(\mathfrak{g})\simeq\mathfrak{gl}(\mathbb{V})\\ {\exp}{\downarrow} & & {\downarrow}{\exp}\\ G & \stackrel{\rho} {\cdots>} & \mathrm{GL}(\mathbb{V}) \end{matrix} as $G$ is connected then $\mathrm{ad}$ determines a unique holomorphic representation $\rho$ of $G$ on $\mathbb{V}$.

By the previous statement, $\rho$ is a polynomial representation of $G$ on $\mathbb{V}$; as $G$ is a connected group then $\rho$ is a faithful representation.

Summing up, $G$ is a closed subgroup of $\mathrm{GL}(\mathbb{V})$ with respect to the Zariski topology, in particular $G$ is an algebraic group. (q.e.d.) $\Box$

By the proof of previous theorem, we can state the following corollary.

Corollary. Let $G$ be a connected, complex, semisimple Lie group; any closed Lie subgroup of $G$ is an algebraic subgroup of $G$ regarded as an algebraic group.

Remark 1. [BA] proves that an algebraic group (over an algebraically closed field) is affine if and only if it is isomorphic to a closed subgroup of some general linear group (of finite dimension over the same field) thus, any affine algebraic groups can also be called linear algebraic groups.

In particular, any algebraic group is a smooth algebraic variety (for a proof, one can consult [BS]) so one proves that any complex algebraic group has a structure of complex Lie group. $\Diamond$

Remark 2. The tangent space $T_PM$ of a (differential) manifold $M$ at a point $P$ is isomorphic to the space of (real) linear derivation of the ring of germs of $M$ at $P$ (see [Spk]).

The same construction and the same result hold for the complex manifolds, that is the tangent space $T_PM$ of a complex manifold $M$ at a point $P$ is isomorphic to the space of (complex) linear derivation of $M$ at $P$.

On the other hand, the Zariski tangent space $T_PX$ of an arbitrary algebraic variety $X$ over a field $\mathbb{K}$ at a closed $\mathbb{K}$-valued point $P$ is isomorphic to the space of $\mathbb{K}$-linear derivation of $X$ at $P$ (see [FOAG]).

From all this, the Lie algebra $\mathfrak{g}$ of a connected, complex, semisimple or reductive Lie group $G$ is the Lie algebra of $G$ as algebraic complex group.

Moreover, the dimension of $G$ is the dimension of $\mathfrak{g}$, by the previous statement, the dimension of $\mathfrak{g}$ is the dimension of the Zariski tangent space to $G$ at $1_G$; as proved in [FOAG], the local dimension of $G$ as algebraic group at $1_G$ is the dimension of the Zariski tangent space to $G$ at $1_G$. $\Diamond$

From all this, I can state the following theorem.

Theorem 2. Let $G$ be a connected, complex, semisimple Lie group. The Lie algebra $\mathfrak{g}$ of $G$ as complex Lie group is the Lie algebra of $G$ regarding as complex algebriac group; moreover, the categories of holomorphic representations $\mathbf{Rep}_{hol}(G)$ of $G$ is equivalent to the category of algebraic representations $\mathbf{Rep}_{alg}(G)$ of $G$, regarding $G$ as algebraic complex group.

Is it all clear?


  1. [BA] A. Borel (1991) Linear Algebraic Groups, II Edition, Springer-Verlag
  2. [BS] S. Bosch (2011) Algebraic Geometry and Commutative Algebra, Springer-Verlag
  3. [OV] A. L. Onishchik - E. B. Vinberg (1988) Lie Groups and Algebraic Groups, Nauka
  4. [Spk] M. Spivak (1999) A Comprehensive Introduction to Differential Geometry; Volume 1, III Edition, Publish or Perish Inc.
  5. [FOAG] R. D. Vakil (2015) Foundations of Algebraic Geometry, http://math.stanford.edu/~vakil/216blog/FOAGdec2915public.pdf, Stanford University

This would not be very clear until you started to read algebraic groups (like T.A.Springer's book) and see how the structure theorems of algebraic groups are very similar to these of the Lie groups. So to really understand what he/she means you need to learn basics algebraic groups, like how Jordan decomposition carries out, when is it semisimple/unipotent/borel, etc.

It is like someone told you elements of Galois theory is already presented in various reciprocity laws. This is certainly true; but unless you started to learn Galois theory you would not really appreciate what it means. So you have to learn it.

  • $\begingroup$ i do not really know what to comment on this.... Thank you so much for your advice.. $\endgroup$
    – user87543
    Commented Jun 19, 2014 at 6:29
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    $\begingroup$ A good reference book is Humphrey's book, though it might be dense. Springer's book is hard to read. I have not read Milne's book so I cannot comment on it. $\endgroup$ Commented Jun 19, 2014 at 6:34

[Remark 2.11][1]

I haven't checked this very carefully, but it seems Remark 2.11 in Milne's notes "Lie algebras, algebraic groups, and Lie groups" indicates what we want. [1]: https://i.sstatic.net/gCjhL.png


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