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

  • 1
    What is the definition of $\frac{\delta}{\delta x}$? – T. Eskin May 29 '15 at 3:45
  • The link doesn't show the page for me. Could you reproduce the definitions and exercise? – user7530 May 29 '15 at 3:51
  • @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 3:51
  • @user7530 I added what the book says – Ana Galois May 29 '15 at 4:01
  • 2
    $\partial/\partial x$ 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 4:07
up vote 1 down vote accepted

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.

  • 1
    @Ana Galois I wrote 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 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 4:33
  • Thank you, this is very helpful and interesting – Ana Galois May 29 '15 at 4:38
  • 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 5:59

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.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.