Let $M$ be a manifold and let $ X , Y $ two vector fields on $M$. I know ths definition of the Lie bracket between $X $ and $Y$ $[X,Y]: C^\infty(M)\rightarrow C^\infty(M)$ is an $\mathbb R$-linear map such that $[X,Y]f:=X(Yf)-Y(Xf)$.
However, if $p \in M $ I'm not sure what is the explicit expression of $[X,Y](p)$?
Best Answer
Suppose $(x^1,\ldots,x^n)$ is a local coordinate patch and $\left(\partial_1,\ldots,\partial_n\right)$ are the associated tangent vector fields. Then locally, there exist functions $X^i$ and $Y^j$ such that: \begin{align} X &= \sum_{i=1}^nX^i\partial_i, & Y &=\sum_{j=1}^nY^j\partial_j. \end{align} If $f$ is a function, then: \begin{align} Y\cdot f = \mathrm{d}f(Y) = \sum_{i=1}^n\left(\frac{\partial f}{\partial x^i}\mathrm{d}x^i\right) Y = \sum_{i=1}^n Y^i\frac{\partial f}{\partial x^i}. \end{align} Thus: \begin{align} X\cdot(Y\cdot f) &= \sum_{j=1}^n \left(\frac{\partial}{\partial x^j}\left(Y\cdot f \right)\mathrm{d}x^j\right)X \\ &= \sum_{j=1}^nX^j\left(\frac{\partial}{\partial x^j}\left(\sum_{i=1}^n Y^i \frac{\partial f}{\partial x^i} \right) \right) \\ &=\sum_{i,j=1}^nX^j \left(\frac{\partial Y^i}{\partial x^j}\frac{\partial f}{\partial x^i} + Y^i \frac{\partial^2 f}{\partial x^j\partial x^i} \right). \end{align} Therefore, it follows that: \begin{align} [X,Y]f &= X\cdot(Y\cdot f) - Y\cdot(X \cdot f)\\ &=\sum_{i,j=1}^n X^j\frac{\partial Y^i}{\partial x^j}\frac{\partial f}{\partial x^i} - Y^j\frac{\partial X^i}{\partial x^j}\frac{\partial f}{\partial x^i} + X^jY^i\frac{\partial^2 f}{\partial x^j\partial x^i} - Y^jX^i\frac{\partial^2 f}{\partial x^j\partial x^i}\\ &= \sum_{i=1}^n\left(\sum_{j=1}^n X^j\frac{\partial Y^i}{\partial x^j}- Y^j\frac{\partial X^i}{\partial x^j} \right) \frac{\partial f}{\partial x^i} + 0 \tag{$\star$}\\ &= \mathrm{d}f \left(\sum_{i=1}^n \left(\sum_{j=1}^n X^j\frac{\partial Y^i}{\partial x^j}- Y^j\frac{\partial X^i}{\partial x^j} \right) \partial_i \right), \end{align} where ($\star$) is a consequence of Swcharz's lemma, that is $\frac{\partial^2f}{\partial x^j \partial x^i} = \frac{\partial^2f}{\partial x^i\partial x^j}$. Hence, in local coordinates: $$ [X,Y] = \sum_{i=1}^n\left(\sum_{j=1}^n X^j \frac{\partial Y^i}{\partial x^j} - Y^j\frac{\partial X ^i}{\partial x^j}\right) \partial_i. $$ Note that thanks to Schwarz's lemma, it is canonically defined. Otherwise, as the second partial derivative of functions depends on the choosen chart, it would not have been a canonical expression.
To answer your comment: if $X(m)=0$, the above expression shows that $$ [X,\xi](m) = - \sum_{i=1}^n\left(\sum_{j=1}^n \xi^j(m)\frac{X^i}{\partial x^j}(m)\right) \partial_i, $$ which does not depend on the local behaviour of $\xi$ around $m$ but just on its value $\xi(m)$.