I've recently become curious about Lie groups/algebras. In particular, I'm quite interested in Lie's methods for solving differential equations.

Unfortunately, there's a bit of a disconnect between how I think about groups, and how Lie groups are described in the relevant literature. I tend to think of groups in terms of a set and a binary operation. However, I've come to notice that a lot of mathematicians talk about Lie groups in terms of geometry (e.g. Lie groups are smooth manifolds where multiplication and inversion are smooth maps). I'm trying to bridge that gap.

I can see how that works for some simple case. (The real number under addition are geometrically equivalent to a line. The circle group is, unsurprisingly, like a circle.) I'm not sure how you would formalize this, though, or how you would do this for a general Lie group.

Here's how I currently understand things:

-Every point on the manifold corresponds to an element of the group, and vice-versa (does this have to be one-to-one?)

-If a the group is parametrized by parameters $\theta_1,\theta_2,\theta_3...$, then small changes to the parameters correspond to a new point on the manifold which is geometrically close to the starting point (assuming we don't violate any constraints).

-If we have two group elements $G(\theta_1,\theta_2,...)$ and $G(\gamma_1,\gamma_2,...)$, then $$ G(\theta_1,\theta_2,...)*G(\gamma_1,\gamma_2,...) = G(f_1(\theta_1,\gamma_1,\theta_2,...), f_2(\theta_1,\gamma_1,\theta_2,...),...)$$ Where each $f_i$ is smooth. (Similarly for inverses)

Is this correct?

How do we then use this set+operation description to come to a description of the manifold? In particular, is there an easy way to go from a representation of a Lie group to a manifold description? (this is my main question.)

side question: I'm an undergrad in applied mathematics with little experience in differential geometry. Can you recommend any good books on Lie theory for someone with my background? Alternatively, can you recommend any books on geometry that would help me understand Lie theory?

  • $\begingroup$ I do think your description of the multiplication is accurate. Notice Baker Campbell Hausdorff formula I mentioned in my answer actually gives you the mechanics of the multiplication for the exponential chart near the identity. If a different chart was used then the rule could be described in some other way. $\endgroup$ Aug 9, 2013 at 21:27

1 Answer 1


A Lie group $G$ is a manifold with smooth group operations $m: G \times G$ and $inv: G \rightarrow G$. This means $G$ is a set with a binary operation. It just happens to have infinitely many elements, but, it still is a set with an associative, unital operation which is closed under multiplication and inverse. Smoothness of these maps means that $m$ and $G$ have smooth local coordinate representatives for all charts in the atlas for $G$. Furthermore, the use of the word "smooth" in the previous sentence refers to a smooth mapping on $\mathbb{R}^n$-type spaces, this reduces to requiring the Jacobian matrices have smooth component functions in the sense they have arbitrarily many continuous partial derivatives. All of that is wrapped up in saying "smooth group operations". That said, your idea that a small change in parameters of the group should produce near-by group elements is basically the idea.

As I read your post, it seems to me that you are making a distinction between the manifold and the group for a Lie group. If we're just talking about a Lie group then this is not necessary. It's better than 1-1, they're actually the same set.

On the other hand, if we are considering a manifold on which some group of transformations acts then the distinction is warranted. Both structures are of interest. For example, on $\mathbb{R}^3$ we can consider the group of rotations and reflections which is $O(3) = \{ R \in \mathbb{R}^{3 \times 3} \ | \ R^TR=I \}$. For each $R \in O(3)$ the point $p \in \mathbb{R}^3$ is transformed to $Rp$. Euclidean three dimensional space is a manifold. $SO(3)$ is a Lie group.

Of course, $\mathbb{R}^3$ is also an additive Lie group with respect to the usual vector addition so, in that sense, it is a Lie group and the addition of vectors gives you the group of translations acting on three dimensional space.

Finally, as to how we find coordinates on a Lie group verses how we describe the group multiplication. I think it depends on the example. One standard chart to consider is given by the exponential at the origin. It turns our that Lie algebra is isomorphic to the tangent space to the identity of the group. The Lie algebra is a vector space and hence permits identification (non-cannonically) with $\mathbb{R}^n$. Then, exponentiation produces a group element from a given algebra element. Turns out this is not one-one, but if you stay close enough to the origin it works.

For matrix Lie groups, the beautiful formula below shows that the group multiplication is fixed by knowledge of the commutators of the matrix Lie algebra: the commutator The notation $[A,B]=AB-BA$ in what follows: $$ exp(A)exp(B) = exp(A+B+\frac{1}{2}[A,B] + \cdots ) $$ here the $+ c\dots$ contains terms like $\frac{1}{12}[A,[A,B]]$ and $\frac{1}{12}[B,[B,A]]$ at next order ( I forget the $\pm$ here) and beyond that you have three, four, five...-fold commutators. This is the Baker-Campbell-Hausdorff formula.

Where to start reading? Well, it depends on your background. Certainly you want a mastery of multivariate calculus. Beyond that, you'll need to study differential geometry or manifold theory a bit before attempting Lie groups proper. But, there might be a book out there with a bit of both written with you in mind. I think your question is interesting, I hope someone more knowledgeable brings added insight tomorrow.

  • $\begingroup$ Thanks for the explanation. It doesn't quite answer all my questions to my satisfaction, but it did help confirm/clear up a lot of things. $\endgroup$
    – hasnohat
    Aug 9, 2013 at 22:42
  • $\begingroup$ @Julien Here's a nice exercise to do when one first starts learning about Lie groups. Prove that the condition that the inversion map $i: G\to G$ be smooth is superfluous, namely that smoothness of $i$ comes from smoothness of the multiplication map. Hint 1) Calculate the tangent space of $G \times G$ at the identity. 2) Use the inverse function theorem. $\endgroup$
    – user38268
    Aug 10, 2013 at 0:08

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