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I am beginner in Lie group theory, and I can't find the answer a question I am asking myself : I know that the Lie algebra $\mathfrak g$ of a Lie group $G$ is more or less the tangent vector of $G$ at the identity, so that $\mathfrak g$ have a very interesting property : linearity.

However $\mathfrak g$ has another property : it is stable under Lie brackets $[.,.]$.

For me when I study Lie groups I always find linearity of Lie algebras really important, and I don't see and I didn't find why the stability under Lie brackets is important. What is the main result/property of Lie groups using this property?

That would be great if you could light me!

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What do you mean by "linearity" of Lie algebras, and "stability under Lie brackets"? –  Santiago Canez Aug 14 '13 at 22:56
@SantiagoCanez I guess "linearity" means that left invariant vector fields on a Lie group form a (finite dimensional) vector space. "Stability" means that the bracket of two left invariant vector fields is again an invariant vector field. –  Sasha Patotski Aug 14 '13 at 23:46
@SashaPatotski, possibly, but the OP also refers to the Lie algebra as the tangent space at the identity, in which case "linearity" and "stability" are unclear. –  Santiago Canez Aug 15 '13 at 0:11
@SantiagoCanez I agree. But in the question it was said that Lie algebra is "more or less" the tangent space at the identity. So it wasn't probably the original definition. –  Sasha Patotski Aug 15 '13 at 0:29
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3 Answers 3

up vote 12 down vote accepted

I think about that this way.

In some sence geometry is "difficult" and algebra is "easy". So you want to obtain as much information as possible from studying Lie algebras instead of Lie groups, and then transering your results from algebras back to groups. So your bracket is the most natural operation on the tangent space that sort of allows you to do that. You can reinterpret a lot of properties of your Lie group (commutativity, solvability, (semi-)simplicity et.c.) into properties of the bracket on the Lie algebra. For example, simplicity for groups encodes into the property that Lie algebra does not have non-trivial ideals, and so on. You also have analogs between subgroups of different kind of your Lie group $G$ and subalgebras of $g$. For example, tangent space to the center of Lie group $G$ is the center of Lie algebra $g$, i.e. $Z(g)=\{x\in g|[x,y]=0,\forall y\in g\}$.

I hope I've helped a bit.

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As a take-home message, I would say that the Lie bracket encodes the structure of group multiplication. The linear structure alone is very uninformative; it really just tells you about the dimension of the manifold.

More precisely, the Baker-Campbell-Hausdorff formula tells you that the group multiplication (at least for elements near the identity, and hence in the image of the exponential map $\exp$) can be entirely obtained from Lie brackets. Therefore, as pointed out in Sasha Patotski's post, (almost) all group properties have Lie bracket analogies.

In fact, the Lie bracket is particularly nice, giving very concise expressions of Lie group properties. Thinking about the adjoint representation, Killing form and so on is a good way in.

Remark: The caveats above reflect the problems like $\exp$ mapping the Lie algebra of $O(3)$ back to just $SO(3)$, and a more subtle problem where the Lie algebra of $SL_2(\mathbb R)$ is mapped to a strict subset of that group (even though the group is connected).

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I think you need a bit more than that. After all, we know there are non-isomorphic Lie groups which share the same Lie algebra. For example, $GL(n)$ the group of non-singular matrices and $GL(n)_+$ the group of positive determinant matrices. These both take $gl(n)$ as their Lie algebra. –  James S. Cook Aug 14 '13 at 23:26
@JamesS.Cook If you prefer, it tells you about the continuous part of group multiplication, or the part close to the identity, or whatever. This is a 'moral' answer, not a rigorous one, in the spirit of the question (in my view, of course). –  Sharkos Aug 14 '13 at 23:34
@JamesS.Cook I've tried to clarify this in an edit. I hope that this makes things clearer. –  Sharkos Aug 14 '13 at 23:40
but what is morality in mathematics if not rigor? Of course, your answer is nearly correct, the points in the connected component of the identity which are not reached by the exponential map directly are still reached by products elements which are reached by the exponential map. But, one has to ask, how do we define those products? In any event, I think your post is clear now +1 –  James S. Cook Aug 15 '13 at 0:17
@JamesS.Cook (I didn't actually even claim that the entire connected component of the identity was addressed, AFAIK. Hence the $SL_2(\mathbb R)$ example quoted.) OED: "moral, n. 3) (a) A moral maxim or practical lesson to be drawn from a story, event, etc.", my emphasis. I think that the "practical" fact is that if you want to know about group multiplication, you want to know about the Lie algebra structure. The point of this answer (and IMHO question) is mathematical pedagogy, not rigour :) –  Sharkos Aug 15 '13 at 0:32
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A good question. There are many aspects of the situation... At least one fundamental structure can be understood in the following way. First, imagining that $t$ is an "infinitesimal", so that $t^3=0$ (not $t^2=0$!) (or equivalent...), and imagining that elements of the Lie group near the identity are $g=1+tx$ and $h=1+ty$ (with $x,y$ in the Lie algebra) observe that $(1+tx)(1+ty)(1-tx)(1-ty)=1+t^2(xy-yx)$. Thus, we care about $xy-yx=[x,y]$.

E.g., for matrix Lie groups, so that $x,y$ are matrices, this makes sense, where $xy-yx$ is in the matrix algebra.

Yes, several issues are left hanging after this walk-through, but the symbol-pattern proves to be excellent, in essentially all incarnations.

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