What interpretation of the Lie braket is this? 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
 A: 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.
A: 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.
