Lie Bracket, closure under the Lie Bracket

(If you're just here for the easy way to see closure, skip ahead to Closure under the Lie Bracket)

In the previous article, I introduced Lie Groups and Lie Algebras by talking about Lie Algebras as a parameterisation for the Lie Group -- we said that the elements of the Lie Group could be written as exponentials of these parameters (not uniquely, sure, but they can be written in this way). Some things to note here:

• What we've called "Lie Groups" refers only to connected Lie Groups, as motivation. In general, the theory of Lie groups considers any group that is also a manifold -- for instance, the non-zero real numbers are also a Lie Group (even though their Lie Algebra is identical to that of the positive real numbers -- can you see why?). We will hereby use this more general definition.
• It's not really true that any Lie group can be parameterised in this fashion by writing each element as an exponential of a Lie Algebra element -- even for connected groups. This shouldn't be surprising -- given a term of the form $\exp X$ and a term $\exp Y$, their product $\exp X\exp Y$ is in the group by closure, but it isn't necessarily equivalent to $\exp(X+Y)$ on a non-Abelian group (could it be the exponential of something else? We'll find out later).
• A parameterisation of this form is not the same as a co-ordinate system.

The last point is what we will concentrate on in this article, because not being described fully by the Lie algebra is what makes things interesting, right?

What is a co-ordinate system on a manifold? Well, they key point is that any element of the manifold can be decomposed in terms of its components along the co-ordinates. On a Lie Group, this means that there should exist a "basis" for the Lie Group $\exp(X_1),\ldots\exp(X_n)$ corresponding to the basis $X_1,\ldots X_n$ for the Lie Algebra vector space such that every element of the Lie Group can be written as products of powers of these elements, and any rearrangement of the terms in the product should leave it invariant (i.e. the elements should commute with each other).

Note that it is possible to decompose elements of a connected Lie Group as a product of some exponentials, but this is different from there being specifically $n$ elements that one can write any Lie group element as products of.

But clearly, this can only be possible if the group is Abelian, commutative. This is a special case of the more general fact that only a holonomic basis gives rise to a co-ordinate system on a manifold. The idea is -- a closed loop should produce no overall group action. If you flow $\varepsilon$ in the $X$ direction, then flow $\varepsilon$ in the $Y$ direction, then flow $\varepsilon$ back in the $X$ direction and flow $\varepsilon$ back in the $Y$ direction, you should end up back where you started. If you don't, then the resulting difference is the infinitesimal "group commutator" of the Lie Group:

$$e^{\varepsilon X}e^{\varepsilon Y}e^{-\varepsilon X}e^{\varepsilon Y}$$
One can check via a Taylor expansion that this is equal, to second order, to:

$$1+\varepsilon^2(XY-YX)$$
The first thing to note about this is that the $\varepsilon^1$ term is zero -- this may seem like a surprising coincidence, but perhaps it isn't that surprising (I mean, there's nothing else it could be, right? If the commutator was to first-order $1+\varepsilon z$, $\exp z$ would be equal to 1, and so it would give no characterisation at all of the amount of non-commutativity of the flows $X$ and $Y$) -- it's analogous to vector calculus, where the curl of a vector field is proportional to $\varepsilon^2$ (i.e. a line integral along the curve is proportional to its area, so you divide it by this area in the definition of curl, etc.).

The second-order term, $XY-YX$, is more interesting. This may seem weird because so far, we've been considering the Lie algebra purely as a vector space, with addition and scalar multiplication being the only things going on. But clearly, this cannot be the entire picture, or a connected Lie group would be characterised entirely by the dimension of its Lie algebra. This operation -- the Lie Bracket or Lie Algebra commutator represented by $[X,Y]$ -- as we will see, gives some additional structure to the Lie Algebra, and in fact characterises it (we'll see what this means).

So far, we've obtained no motivation for why this operation $XY-YX$ is actually of any significance. Sure, it appeared in our second-order approximation for the group commutator, but is the group commutator we defined really so great? Surely there could be other ways one could measure the non-commutativity of a group. And the $\varepsilon^2$ business is weird. Things that arise proportional to $\varepsilon$ live in the tangent space, in the Lie Algebra. Where does $[X,Y]$ even live?

Two facts will convince us that the Lie Bracket is indeed the "right" measure of non-commutativity of a Lie Algebra:

• The Lie Algebra is closed under the Lie Bracket -- we will see that in fact, $[X,Y]$ lives in the lie algebra, so it is in fact a binary operation on the Lie Algebra, and really does add structure to the Lie Algebra.
• It characterises the entire Lie Algebra -- not only is it part of the structure of the Lie Algebra, it characterises the entire structure of the Lie Algebra. What this means is that defining the Lie Bracket on the vector space allows a full characterisation of the part of the group connected to the identity (the "connected part" of the group), so we can say that any Lie Algebras with the same dimension and Lie Bracket are isomorphic.

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Closure under the Lie Bracket

If you're like me, you might've thought of several analogous situations to our $1+\varepsilon^2(XY-YX)$ expression -- e.g. in (complex) analysis, at a point where the derivative of a function is zero, the function is characterised by its second derivative (consult Needham's Complex Analysis, p. 205-207 for an explanation). Another example is -- if the first derivative of a function is zero, the second derivative satisfies the product rule (this is actually directly related, in a way we won't go into now).

Here's an idea you might think of: as we discussed earlier, the infinitesimal group commutator is $e^{\varepsilon X}e^{\varepsilon Y}e^{-\varepsilon X}e^{-\varepsilon Y}= 1+\varepsilon^2 (XY - YX) + O(\varepsilon^3)\in G$. But for a moment let $\varepsilon$ not be infinitesimal. So $\varepsilon (XY - YX) + O(\varepsilon^2)\in \mathfrak{g}$, the Lie Algebra corresponding to Lie Group $G$, so by scaling $XY-YX+O(\varepsilon)\in\mathfrak{g}$ and by connectedness of the vector space $XY-YX\in\mathfrak{g}$.

But this argument is incorrect -- this becomes obvious if you try to formally write it down -- In general, $1+\varepsilon T\in G$ does not imply $T\in\mathfrak{g}$ for non-infinitesimal $\varepsilon$. It's close to an element in $\mathfrak{g}$ (for small $\varepsilon$), but how close? You might get the feeling that it is "sufficiently close", in that the limit $\varepsilon\to0$ of the sequence $\left(c_\varepsilon(X,Y)-1\right)/\varepsilon^2$ (where $c_\varepsilon(X,Y)$ is the group commutator) indeed ends up in the Lie Algebra.

To make this feeling formal, consider instead the curve parameterised differently as $\gamma(\varepsilon)=e^{\sqrt\varepsilon X}e^{\sqrt\varepsilon Y}e^{-\sqrt\varepsilon X}e^{-\sqrt\varepsilon Y}$. Then $\gamma'(0)=XY-YX$, and we're done.

think about the Taylor expansion here of this new curve for a while