Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $U$ be a Riemann surface and let $z:U\longrightarrow B(0,1)$ be a diffeomorphism, where $B(0,1)$ is the open unit disc in $\mathbf{C}$. So $z$ is a coordinate around $P=z^{-1}(0)$.

Let $Q\in U$ be another point and let $V \subset U$ be an open subset. (We choose $V$ such that its image under $z$ is an open disc around $z(Q)$ contained in $B(0,1)$. A picture would help alot here.) Consider the morphism $z^\prime: V\longrightarrow B(0,1)$ defined by $z^\prime(x) = z(x) - z(Q)$. This is a coordinate around $Q$. Essentially, it is the coordinate at $P$ translated by $z(Q)$.

Question. What is the relation between $dz$ and $dz^\prime$?

My other question The chain rule for a function to $\mathbf{C}$ suggests that they are $\textbf{not}$ the same. (Take $a=z(Q)$ and note that $z^\prime = t_a\circ z$.)

share|improve this question
    
If $Q$ does not depend on $x$, they're the same. –  Ryan Budney Sep 30 '11 at 14:35
    
Your other question was about comparing $dz$ to $d(z(x-Q))$ where here you tread $z$ as the variable. $z(x-Q) \neq z(x) - z(Q)$ unless your function $z$ is linear. –  Ryan Budney Sep 30 '11 at 14:37
    
I think my question is very unclear...Q is a fixed point, whereas I use $x$ to denote an arbitrary element of $V$. Moreover, my other question is about the differential of the function $x \mapsto f(x) -a$. Here my question is about the differential of the function $x\mapsto z(x) - z(Q)$. Thus, this is just the previous question in some sense with $a= z(Q)$. –  shaye Sep 30 '11 at 15:03

1 Answer 1

As $z'(x) = z(x) - z(Q)$,

$$ (dz')(x) = d(z'(x)) = d(z(x) - z(Q)) = d(z(x)) - d(z(Q)) = d(z(x)) = (dz)(x).$$

Therefore $dz = dz'$ on $V$.

As Ryan Budney pointed out in the comments, the answer to your previous question also shows that the $dz$ and $dz'$ are the same (the $a$ in that question is $z(Q)$ here). In a previous version of that answer, Sasha was interpreting the composition notation in a different (less common) way; with that interpretation, you see that $d(z(x)) \neq d(z(x - Q))$ in general.

To be clear, using the notation from your previous question, $d(t_a\circ f) = df$ but $d(f \circ t_a) \neq df$ in general.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.