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I was reading exercise 1.96 of Gadea's Analysis and Algebra on Differentiable Manifolds. I don't know what definition of Lie braket the author used, but I'm confused, as far as I know $[\frac{\partial}{\partial x},X]$ is the Lie braket, isn't it? But I thought that was defined for vector fields, and I don't see how $\frac{\partial}{\partial x}$ is one. I don't understand the answer given in the book.

The problem is:

Find the general expression for $X\in \mathscr X(\mathbb R^2)$ in the following cases:
(i) $[\frac{\partial}{\partial x},X]=X$ and $[\frac{\partial}{\partial y},X]=X$;
(ii) $[\frac{\partial}{\partial x}+\frac{\partial}{\partial y},X]=X$.

Where I suppose that $[\frac{\partial}{\partial x},X]$ is the Lie braket.
Now the answer: see here

Ana Galois
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    What is the definition of $\frac{\delta}{\delta x}$? – T. Eskin May 29 '15 at 03:45
  • The link doesn't show the page for me. Could you reproduce the definitions and exercise? – user7530 May 29 '15 at 03:51
  • @ThomasE. If I'm not mistaken or understanding wrong what the author wanted to say, it would be an element of the coordinate basis in the tangent space, obtained with the parametrization of the coordinate basis in $\mathbb R ^2$. – Ana Galois May 29 '15 at 03:51
  • @user7530 I added what the book says – Ana Galois May 29 '15 at 04:01
  • @ThomasE. Oh! I just noticed, maybe what it's confusing is that I messed up the notation, I meant $\frac{\partial}{\partial x}$ – Ana Galois May 29 '15 at 04:04
  • What is $\partial/\partial x$? What is curly-$X$? I also doubt that $[\ ,\ ]$ is the Lie bracket since otherwise bilinearity gives you $X=2X$. – user7530 May 29 '15 at 04:05
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    $\partial/\partial x$ is a vector field on $\mathbb{R}^2$. In the non-derivative notation, it is just $\hat{i}$ or $\langle 1,0 \rangle$ if you prefer. – James S. Cook May 29 '15 at 04:07
  • @user7530 $\mathscr X(\mathbb R^2)$ would be the set of all the vector fields of $\mathbb R^2$. – Ana Galois May 29 '15 at 04:07
  • @JamesS.Cook I'm sorry, but why $\partial / \partial x$ is a vector field? I'm sorry if it is too obvious, is just for now I don't see it – Ana Galois May 29 '15 at 04:09
  • There are several equivalent definitions of "vector field". What one does your book use? In some books, the definition is essentially "a first-order differential operator", in which case $\partial/\partial x$ certainly is one. – Nate Eldredge May 29 '15 at 04:22
  • @AnaGalois the custom of differential geometry is to use derivations to represent tangent vectors. So, the $\partial/\partial x$ at a point represents the unit-vector in the $x$-direction, but, since it is defined on all of $\mathbb{R}^2$ it can be viewed as a vector field. I can't find a nice link for this at the moment... I'm searching. – James S. Cook May 29 '15 at 04:24
  • @JamesS.Cook Thank you! Hope you find it – Ana Galois May 29 '15 at 04:27

2 Answers2

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Let $X = a\partial_x +b \partial_y$. Then, $$ [\partial_x, X]f = \partial_x(X(f))-X(\partial_x f)$$ But, $X(f) = a\partial_xf +b \partial_yf$ and likewise for the second term, hence: $$ [\partial_x, X]f = \partial_x(a\partial_xf +b \partial_yf)-a\partial_{xx}f +b \partial_{xy}f$$ Of course, $a,b$ are functions thus there are product rules to consider in the first expression. Half of the terms cancel and we derive: $$ [\partial_x, X]f = (\partial_x a)\partial_x f +(\partial_x b) \partial_yf \ \ \star. $$ We wish to select functions $a,b$ for which $[\partial_x, X]=X = a\partial_x +b \partial_y$. Comparing our calculation $\star$ to the desired outcome and using the linear independence of the coordinate basis yields: $$ \partial_x a = a \qquad \& \qquad \partial_x b = b$$ There are many solutions, but, one I like is $a=e^x$ and $b=e^x$. Thus $X =e^x(\partial_x+\partial_y)$. Your other problems can be solved by similar calculations.

James S. Cook
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    @Ana Galois I wrote http://math.stackexchange.com/a/189529/36530 to contrast the different views of vectors in geometry. The wikipedia article is a bit abstract for our current discussion. Maybe also see http://math.stackexchange.com/q/58084/36530 to dig in further. But, this really ought to be in your textbook... – James S. Cook May 29 '15 at 04:33
  • Thank you, this is very helpful and interesting – Ana Galois May 29 '15 at 04:38
  • Glad it was helpful, btw, you might notice we can multiply my current answer $X$ by a function of $y$ and it still solves $\partial_x a = a$ and $\partial_x b = b$. – James S. Cook May 29 '15 at 05:59
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Given a $C^{\infty}$ manifold $M$, a vector field $X$ is a derivation on the algebra $C^{\infty}(M)$ of $C^{\infty}$ real functions on $M$. That means:

$X$ is a map $X: C^{\infty}(M) \rightarrow C^{\infty}(M)$ that respects the following properties:

(i) $X$ is linear: $X(\alpha f + \beta g)=\alpha Xf+\beta X g; \quad \alpha, \beta \in \mathbb{R}, \quad f,g, \in C^\infty(M)$

(ii) $X$ satisfies: $X(fg)=fXg+gXf ; \quad f,g \in C^\infty(M)$

Now, consider a local chart $(\phi,U)$. Define the following map from $C^\infty(M)$ to itself:

$\displaystyle g \mapsto \left( \frac{\partial (g\circ \phi^{-1})}{\partial x_1}\right)\circ \phi$

Note that this is a conventional partial derivative. It is easily verified (by the properties of partial derivatives) to be a derivation. This derivation is your $\displaystyle \frac{\partial }{\partial x}$, hence a vector field*.

*Note: It is not a vector field ipsis litteris: it is a local vector field, but that is no problem. If you want a global one, you can extend it to a global one, cutting down a bit your $U$ and using a bump function.

Aloizio Macedo
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