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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 $z^\prime = t_a\circ z$.)

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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

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.

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