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 $f:U\longrightarrow \mathbf{C}$ 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?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

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 $\mathrm{d} (t_a \circ f) = \mathrm{d}(f-a) = \mathrm{d} f$.

share|improve this answer
    
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 math.stackexchange.com/questions/68282/… 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

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.