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.


In general one cannot coherently define the pushforward of a smooth vector field by a smooth function. (Lee's book discusses why this is the case.)

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 $$[X', Y']_{f(p)} = df_p \cdot 0_p = 0_{f(p)} .$$

  • $\begingroup$ 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? $\endgroup$ Feb 27 '19 at 21:15
  • $\begingroup$ 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 $S$. $\endgroup$ Feb 27 '19 at 21:25
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    $\begingroup$ 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 $\alpha = \beta = 0$. $\endgroup$ Feb 27 '19 at 21:26
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    $\begingroup$ 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 $\Bbb R^2$. $\endgroup$ Feb 27 '19 at 21:29

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