1

Compute the partial derivative of $f(x,y) = 2x + y^3$ at $a = (x_0,y_0)$.

I know this looks easy but the purpose of me asking this question is to see how this question is worked as if it we were just working on a general manifold & needed to use all the formalism required for this purpose. Thus on a manifold you say $f$ is differentiable at $a$ in $U \cap V$ on a manifold $M$ if it's coordinate representation $F$ on $V$ is differentiable, where $ f = F \circ \phi$ such that $\phi$ is a chart on the manifold with domain $U$. I'd just like to see if how other people do it is the way I would do it, where you do it pedantically invoking the notation - thanks!!

derive
  • 11

1 Answers1

2

If you're just beginning to work with manifolds, then I think pedantically invoking the notation as you suggest is natural. Once you get comfortable with them, you'll find yourself thinking several steps at a time and the formalism will be well-absorbed. You'll skip right to the intuitive thinking. That's just from my personal experience.

Anyway, if $f:M\to N$ is a differentiable map between manifolds, then we can pick $p\in M$ and coordinate charts $(U,\phi)$ and $(V,\psi)$ containing $p$ and $f(p)$ respectively such that $f(U)\subseteq V$. The differentiability of $f$ implies that $\psi\circ f\circ \phi^{-1}:\phi^{-1}(U)\to \psi(V)$ is differentiable. Now the partial derivatives of this function between open subsets of Euclidean space will depend on the coordinate charts chosen. But you can still compute it, of course.

I hope this helps!

Amitesh Datta
  • 20,779
  • Yes thank you, I learned to be pedantic from painful experience. But you haven't shown me how to compute it, I see ambiguity as to whether you can even write $f(x,y) = 2x + y^3$ or whether you call your function just $f$ & then say that in specific coordinates it becomes $2x + y^3$ or whether you can use that notation from the start. Further this ambiguity highlights itself when you try to actually compute the partial derivative, when you write $\frac{\partial f}{\partial x} \ = \ \frac{\partial F}{\partial x} \cdot \phi$ - it tells me writing $f(x,y) = 2x + y^3$ encodes tons of assumptions! – derive Jul 10 '13 at 18:43
  • I'm trying to unwire these assumptions & since you seem to see the way forwards I would greatly appreciate you pointing out your ideas on how to do this! – derive Jul 10 '13 at 18:45