Lie bracket of canonical vectors on tangent space to a point on a manifold is zero.

Let M be a manifold and $$T_p(M)$$ be the tangent space at $$p$$, and $$\phi$$ a local chart around $$p$$. Let $$\left.\frac{\partial}{\partial\phi^1}\right|_{_p},\ \cdots\ ,\left.\frac{\partial}{\partial\phi^n}\right|_{_p}$$ be a basis of this vector space. Apparently the lie bracket of these basis vector is $$0$$, i.e. for instance, $$\left[\frac{\partial}{\partial\phi^1},\frac{\partial}{\partial\phi^2}\right]_pf=0, \quad\forall p, \forall f$$ smooth, i.e.
$$\frac{\partial}{\partial x^1}\left(\phi^{-1} \circ \frac{\partial f}{\partial \phi^2}\right)(\phi(p))=\frac{\partial}{\partial x^2}\left(\phi^{-1} \circ \frac{\partial f}{\partial \phi^1}\right)(\phi(p)).$$

This seems an easy exercise but I am not sure how to prove it properly... Note that here $$\frac{\partial f}{\partial \phi^2} :M \rightarrow \mathbb{R},\quad p \mapsto \left.\frac{\partial}{\partial \phi^2}\right|_{p}(f)$$

• I just noticed that in the last equation the order of the composition should be interchanged. Mar 18 '19 at 2:20

By definition of the Lie bracket

$$\left[\frac{\partial}{\partial\phi^i},\frac{\partial}{\partial\phi^j}\right]_p(f)=\frac{\partial}{\partial\phi^i} |_p(\frac{\partial}{\partial\phi^j} |_{(\cdot)}f)-\frac{\partial}{\partial\phi^j} |_p(\frac{\partial}{\partial\phi^i} |_{(\cdot)}f)$$

where $$\frac{\partial}{\partial\phi^k} |_{(\cdot)}f$$ denotes the smooth function $$q\mapsto \frac{\partial}{\partial\phi^k} |_{q}f$$.

Now since $$\frac{\partial}{\partial\phi^k}|_{(\cdot)}f=\frac{\partial}{\partial x^k}|_{\phi(\cdot)}(f\circ\phi^{-1})$$

$$\frac{\partial}{\partial\phi^i} |_p(\frac{\partial}{\partial\phi^j} |_{(\cdot)}f) =\frac{\partial}{\partial x^i} |_{\phi (p)}((\frac{\partial}{\partial\phi^j} |_{(\cdot)}f)\circ\phi^{-1}) \\=\frac{\partial}{\partial x^i} |_{\phi(p)}(\frac{\partial}{\partial\phi^j} |_{\phi^{-1}(\cdot)}f) =\frac{\partial}{\partial x^i} |_{\phi(p)}(\frac{\partial}{\partial x^j} |_{(\cdot)}(f\circ\phi^{-1})) =\frac{\partial^2}{\partial x^i\partial x^j}(f\circ\phi^{-1})_{|\phi(p)}$$

Similar

$$\frac{\partial}{\partial\phi^j} |_p(\frac{\partial}{\partial\phi^i} |_{(\cdot)}f) =\frac{\partial^2}{\partial x^j\partial x^i}(f\circ\phi^{-1})_{|\phi(p)}$$

so by Schwarz's theorem we have $$\left[\frac{\partial}{\partial\phi^i},\frac{\partial}{\partial\phi^j}\right]_p(f)=0$$.