Week 3, Day 1 or Properties of Lie Brackets

Today is the first day of the third week of the semester. I know that counting down like this is going to make it seem longer than it already seems, but it seems so long that I can’t seem to help remaining conscious of the precise time frame I’m dealing with.

Such is life, I suppose.

I’ve noticed an amusing trend in my page views involving solutions from Dr. Hatcher’s book, namely that I’ve been receiving an abnormally-high level of page views lately, almost all of which seem to center on those solutions. I guess that means that the semester has started elsewhere too and that people find topology as difficult and frustrating as I do.

For those of you who fit this bill and who are reading this right now: My plan is to start doing more problems ASAP, so that page might get its first update in quite a while.

Today, though, I want (read: need) to talk about differential geometry. In particular, we spent some time in class last week discussing the Lie bracket and its properties, and because we have a derivation of one particular property, I wanted to take the time to put that here for my own benefit.

Given a differentiable manifold M^n=M of dimension n with coordinate chart (x,U), two differentiable vector fields X and Y on M can be written in terms of the related local coordinates (x_1,x_2,\ldots,x_n) as follows:

X=\sum_{i=1}^n a_i\dfrac{\partial}{\partial x_i} and Y=\sum_{i=1}^n b_i\dfrac{\partial}{\partial x_i}

where a_k,b_k:M\to\mathbb{R} are smooth functions. For brevity, let \partial_i denote \partial/\partial x_i. Note, then, that one can always write down a formula for, say, XY:

XY=\sum_{i,j}a_ib_j\partial_i\partial_j=\sum_{i,j}a_ib_j\dfrac{\partial^2}{\partial x_i\partial x_j}.

On the other hand, the resulting operator may not be a vector field due (roughly) to the existence of the second-order derivatives in the resulting expression. Therefore, the Lie bracket is a mechanism for remedying that.

Definition. Given two differentiable vector fields X and Y on M as above, the Lie bracket is an operator which assigns to this pair the operator [X,Y]=XY-YX which is guaranteed to be a vector field on M.

The Lie bracket has a whole slew of nice, useful, or otherwise worthwhile properties. The one I’m here to talk about today is as follows: Given differentiable functions f,g:M\to\mathbb{R},

[fX, gY]=fg[X,Y]+fX(g)Y-gY(f)X.

Now one of the things I’ve always struggled with in this field is notation, so here’s a brief rundown: Because f is a function and X is a vector field (both on M), the quantity fX acts on an arbitrary point p\in M as (fX)(p)=f(p)X(p). This is a vector. Similarly, X(f) is a function which assigns to each point p\in M the point (Xf)(p)\in\mathbb{R}. So then, we can now begin:

Proof. By definition of the Lie bracket,

[fX,gY]=fX(gY)-gY(fX)          (1).

Therefore, it suffices to determine what each of these summands does, and perhaps the easiest way is to determine directly how the operator X(gY) acts on functions. To that end, let h:M\to\mathbb{R} be a differentiable function. One can easily show (based on the properties of derivations, say) that for arbitrary functions \ell, m, X(\ell m)=\ell X(m)+mX(\ell) (i.e., the Leibniz product rule). Therefore, it follows that (X(gY))(h) = X(gY(h))=gXY(h)+X(g)Y(h). Substituting into (1), it follows that

\begin{array}{rcl} [fX,gY](h) &= &(fX(gY))(h)-(gY(fX))(h)\\ &= &fX(g)Y(h)+fgXY(h)-gY(f)X(h)-gfYX(h)\,\,\,\,\,(2)\\ &= &fg(XY(h)-YX(h))+fX(g)Y(h)-gY(f)X(h)\\ &= &fg[X,Y](h) + fX(g)Y(h)-gY(f)X(h)\\ &= &\left( fg[X,Y]+fX(g)Y-gY(f)X\right)(h)\end{array}.

Because this identity holds for h arbitrary, the identity holds for all differentiable functions from M to \mathbb{R} and hence the result is proved. Note that the step in (2) stems from the Leibniz rule as noted above. \square

I can say without a doubt that this topic is absurdly notation-heavy, but I haven’t not-loved a single day of the class so far.

Excitement: I haz it.

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