# Lie Bracket of Pushforwards

Suppose $$f(x,y)=(f_1,f_2,...,f_n)(x,y):\mathbb R^2\to\mathbb R^n$$ is a smooth parameterization of a smooth surface $$S$$ in $$\mathbb R^n$$. The pushforwards $$\mathrm{d}f\frac{\partial}{\partial x}$$ and $$\mathrm{d}f\frac{\partial}{\partial y}$$ are vector fields in $$\mathbb R^n$$ tangent to $$S$$, so their Lie bracket $$X=\left[\mathrm{d}f\frac{\partial}{\partial x},\mathrm{d}f\frac{\partial}{\partial y}\right]$$ is too. Therefore $$X$$ should admit a representation in local coordinates, as the pushforward of some $$\left(\alpha\frac{\partial}{\partial x}+\beta\frac{\partial}{\partial y}\right)\in\mathfrak{X}(\mathbb{R}^2)$$ for some $$\alpha,\beta:\mathbb{R}^2\to\mathbb{R}$$.

Is my reasoning above correct? If so, how may we find $$\alpha$$ and $$\beta$$? If not, what can we say about $$X$$? An example would be extremely helpful.

I have seen the formula $$[Y,Z]=\left(Y^i\frac{\partial Z^j}{\partial x^i}-Z^i\frac{\partial Y^j}{\partial x^i}\right)\frac{\partial}{\partial x^j}$$, and I think this is what I need, but I do not see how to apply it in this case. I have also seen $$\mathrm{d}f\left[\frac{\partial}{\partial x},\frac{\partial}{\partial y}\right]=\left[\mathrm{d}f\frac{\partial}{\partial x},\mathrm{d}f\frac{\partial}{\partial y}\right]$$ when $$f$$ is a local diffeomorphism, but that does not appear to be the case here.

Thank you!

P.S. this is not homework. I am working through Lee's Introduction to Smooth Manifolds and I cannot find any examples where $$S$$ is not simply $$\mathbb{R}^n$$ to answer my question.

On the other hand, if $$f : M \to N$$ is a smooth bijection, then one can push forward $$X \in \mathcal{X}(M)$$ to a smooth vector field $$df \cdot X \in \mathcal{X}(N)$$ defined by pushing forward at each point, that is, setting $$(df \cdot X)_{f(p)} = df_p \cdot X_p$$ for all $$p \in M$$. In particular, for a smooth parameterization $$f : \Bbb R^2 \to \Bbb R^n$$, $$n \geq 3$$, a vector field $$X$$ on $$\Bbb R^2$$ determines a vector field on the surface $$f(\Bbb R^2)$$ but not on all of $$\Bbb R^n$$. (After all, how would you define the pushforward of $$X$$ at a point not on the surface?)
On the other hand, the Lie bracket is compatible with pushforwards in the sense that if $$f : M \to N$$ is a smooth map, $$X$$ is $$f$$-related to $$X'$$, and $$Y$$ is $$f$$-related to $$Y'$$, then $$df_p \cdot [X, Y]_p = [X', Y']_{f(p)}$$ for all $$p \in M$$. In particular, if $$M = \Bbb R^m$$ and $$X, Y$$ are coordinate vector fields, $$[X, Y] = 0$$ and so $$[X', Y']_{f(p)} = df_p \cdot 0_p = 0_{f(p)} .$$
• Ah ok, that makes sense. Perhaps I should have said the pushforwards are vector fields along $S$ in $\mathbb{R}^n$. I take it the short answer is that $\alpha=\beta=0$ because $(x,y)$ are local coordinates? Feb 27 '19 at 21:15
• I believe that the pushforwards should at the very least extend locally to vector fields on an open $n$-submanifold of $\mathbb{R}^n$ containing $S$, which are tangent to $S$ along $S$. Feb 27 '19 at 21:25
• You can certainly think of the pushforwards as being vector fields along the surface $f(\Bbb R^2)$, but the statement about $f$-related maps---in particular the case where $f$ is a coordinate chart map---says exactly that there's no loss of information when computing Lie brackets in coordinates, and hence why working on $\Bbb R^n$ is actually totally general (at least for local computations). And yes, in short the fact that the pushed-forward vector fields are coordinate vector fields implies that $\alpha = \beta = 0$. Feb 27 '19 at 21:26
• That's correct, and in fact using partitions of unity we can show that the pushforwards extend to global vector fields on $\Bbb R^n$. In general, however, there's no canonical way to construct such an extension, i.e., doing so involves an arbitrary choice. For $n > 2$ there are infinitely many vector fields on $\Bbb R^n$ $f$-related to any vector field on $\Bbb R^2$. Feb 27 '19 at 21:29