# The lie bracket.

1. Mar 13, 2010

### i_emanuel

If we have vect (u) which denotes an infinite-dimensional vector space of all vector fields on u. As infinitesimal elements of the continuous group of Diff(u) they form a Lie Algebra. We then can define the bracket of two vector fields in v and w. If in coordinates:

v = $$\sum_{i}$$V i $$\partial$$$$/$$$$\partial$$X$$^{i}$$

w = $$\sum_{j}$$Wj $$\partial$$$$/$$ $$\partial$$X$$^{j}$$

the components of [v,w]

[v,w]:=$$(\sum_{i,j}($$ $$V^{i}$$ $$\frac{d}{dx}$$$$X^{i}$$ $$W^{j}$$ - $$W^{i}$$ $$\frac{d}{dx}$$$$X^{i}$$ $$V^{j} )$$ $$\frac{d}{dx}$$

if the definition is independent of the choice of coordinates is it bilinear by nature? if so it must be antisymmetric [v,w] = -[w,v];

therefore the jacobi identity would yield [v,[u,w]] = [[v,u],w] + [u, [v,w]]

how can i go about verifying this for a lie bracket?

Last edited: Mar 13, 2010
2. Mar 14, 2010

### Fredrik

Staff Emeritus
You know that

$$[X,Y]_pf=X_p(Yf)-Y_p(Xf)$$

where Xf is defined as the function that takes p to $X_pf$, right? So why not just use this on $[U,[V,W]]$? When you have your result, just substitute U→V→W→U to get a new equation, and then do it again to get a third. Then add the three equations together.

You might be interested in this thread too, if not for anything else, just to see how to LaTeX these things. Click on the math expression or quote the post to see the LaTeX code.

Last edited: Mar 14, 2010
3. Mar 14, 2010

### i_emanuel

oh, thank you, I believe I got incredibly confused by trying to relate the 0-forms to any p-form. e.g: let's say i have a linear map

D: $$\Lambda^{p}U \rightarrow \Lambda^{p+d}U$$

if I were to take a graded derivation of $$\Lambda\U$$ of degree d $$\in Z$$

if it satisfies,

$$D(\varphi \wedge \Psi ) = ( D \varphi ) \wedge \Psi + ( -1)^{d deg \varphi} \varphi\wedge D\varphi$$

the set of all graded derivations of $$\Lambda U$$ is an infinite-dimensional graded lie algebra with a bracket:

$$[D_{1}, D_{2}]: = D_{1}D_{2} - ( -1) ^{d_{1} d_{2}$$ $$D_{2} D_{1} }$$

and then I can evaluate the Inner Lie, and exterior derivative togheter to form a graded subalgebra, then probably use Cartan's method of commutation relation (where U is equipped with an orientation and a metric.)

Last edited: Mar 14, 2010
4. Mar 16, 2010

### i_emanuel

I have another question though:

How can I verify the lie bracket of two vector fields on a manifold using the method you transcribed?

Last edited: Mar 16, 2010
5. Mar 16, 2010

### Fredrik

Staff Emeritus
I don't understand. What do you want to verify? That the commutator is a Lie bracket? Are you asking for the details of the derivation I sketched? Have you tried and got stuck somewhere?

6. Mar 16, 2010

### i_emanuel

What I am trying to do is prove mathematically the existence of vector fields on open subsets of $$\textbf{R}^{n}$$. Assuming the tangent space and vector fields lie on differentiable manifold M. Identifying these vector fields would allow me to start defining the tangent vectors on the manifold.

Let $$(U, \alpha)$$ be a chart for M and denote the corresponding coordinates by $$X^{1}, X^{2}, ... X^{n}$$

let

$$f : U \rightarrow \textbf{R}$$

be differentiable at $$x \in U$$.

$$f \circ \alpha ^{-1} : \alpha (U) \rightarrow \textbf{R}$$

which is differentiable at $$\alpha (x)$$.

$$\frac{\partial}{\partial_x^{i}} |_{x} f: =\frac{\partial}{\partial^{i}}( f \circ\alpha^{-1}|_{\alpha (x)} = \frac{\partial}{\partial \(x)^{i}} f ( x^{1}, \x^{2}, \....\x^{n}.$$

satisfying liebniz:

$$\frac{\partial}{\partial^{i}}|_{x} (f\g) = (\frac{\partial}{\partial^{i}}|_{x} f ) g(x) + f(x) \frac{\partial}{\partial^{i}} |_{x} \\\\\\g,$$

where f and g are differentiable at x. Is my reasoning correct?

7. Mar 17, 2010

### Fredrik

Staff Emeritus
Looks like you need to work on your LaTeX skills. Click the quote buttons or the math in my posts here to see how I wrote similar expressions. I also suggest that you preview before you post.

If you just want to show that local vector fields exist, by showing that a partial derivative operator $\partial_i$ is a local vector field, all you need to show is that $p\mapsto\partial_i|_pf$ is smooth for every smooth f. (Edit: This is assuming that we already know that $\partial_i|_p$ is a tangent vector at p. See Isham's book if you don't). The thing on the right there is defined as $(f\circ\alpha^{-1})_{,i}(\alpha(p))$, where I'm using the notation ",i" for the ith partial derivative. Now what do we mean when we say that f is smooth? It means precisely that $f\circ\alpha^{-1}$ is smooth for every coordinate system (chart) $\alpha$. And when we have realized that, we're already done with the proof.

Isham's book is a really good place to read about tangent vectors. Lee is good too.

Last edited: Mar 17, 2010