# The lie bracket.

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?

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Staff Emeritus
Gold Member
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.

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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.)

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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?

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Staff Emeritus
Gold Member
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?

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?

Staff Emeritus
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.