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Let $f:U\longrightarrow \mathbf{C}$ be a holomorphic function, where $U$ is a Riemann surface, e.g., $U=\mathbf{C}$, $U=B(0,1)$ or $U$ is the complex upper half plane, etc.

For $a$ in $\mathbf{C}$, let $t_a:\mathbf{C} \longrightarrow \mathbf{C}$ be the translation by $a$, i.e., $t_a(z) = z-a$.

What is the difference between $df$ and $d(t_a\circ f)$ as differential forms on $U$?

My feeling is that $df = d(t_a\circ f)$, but why?

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1 Answer 1

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The forms will be different if $a\not=0$, namely if $\mathrm{d} f = w(z) \mathrm{d}z$ locally, then $\mathrm{d}\left( t_a \circ f\right) = w(z-a) \mathrm{d} z$.

Added: Above, I was using the following, unconventional definition for the composition, $(t_a \circ f)(z) = f(t_a(z)) = f(z-a)$.

The conventional definition, though, is $(t_a \circ f)(z) = t_a(f(z)) = f(z)-a$. With this definition $\mathrm{d} (t_a \circ f) = \mathrm{d}(f-a) = \mathrm{d} f$.

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That's all I needed to hear. I've been confused about something for the past couple of days. I tried to ask about it before in this question… but didn't get any replies. Could you take a look at that question too, please? – shaye Sep 30 '11 at 13:30
Shouldn't that be $w(z) - a$? – shaye Sep 30 '11 at 14:37
No, $\mathrm{d} (t_a \circ f) = \mathrm{d}(f(z-a)) = w(z-a) \mathrm{d}(z-a) = w(z-a) \mathrm{d} z$. See wikipedia article on functional composition. – Sasha Sep 30 '11 at 14:53
Sorry, are you saying that $(t_a \circ f)(z)$ isn't $f(z) - a$? – Dylan Moreland Sep 30 '11 at 17:05
The definition of composition $(g \circ f)(z) = f(g(z))$. Now let $g=t_a$. – Sasha Sep 30 '11 at 18:28

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