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Consider the group $\mathbb{C}[[z]]_1$ of the power series of the form $a_1 z + a_2 z^2 + \cdots$, with $a_1\neq 0$, under the operation of composition, and $\mathbb{C}[[z]]$ as a $\mathbb{C}[[z]]_1$-module.

I want to know what are the crossed homomorphisms $\beta: \mathbb{C}[[z]]_1 \rightarrow \mathbb{C}[[z]]$, that is, the functions that satisfy $\beta(h_2\circ h_1)=\beta(h_1)+\beta(h_2)\circ h_1$

This question is related with group cohomology, since the group $H^1(\mathbb{C}[[z]]_1, \mathbb{C}[[z]])$ is the quotient of the crossed homomorphisms over principal crossed homomorphisms.

That principal crossed homomorphisms are of the form $\beta_g (h) = g\circ h - g$. I found that there is a crossed homomorphism $\beta (h) = \ln(h/z)$ that is not principal, so, my conjecture is that the group $H^1(\mathbb{C}[[z]]_1, \mathbb{C}[[z]])$ is generated by that element (as a $\mathbb{C}$-module). Can anyone help me to prove this (probably using group cohomology techniques)? what moreover results can be extracted using these tools?

Thank you!

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    $\begingroup$ Crossposted: math.stackexchange.com/questions/1403457/… $\endgroup$
    – Qiaochu Yuan
    Aug 20, 2015 at 3:21
  • $\begingroup$ This looks strange... How do you prove that $h\mapsto\ln(h/z)$ is a crossed homomorphism? $\endgroup$
    – Mikhail Borovoi
    Aug 24, 2015 at 11:07
  • $\begingroup$ Also $\ln(h/z)$ is not defined uniquely when $h=a_1 z+a_2 z^2+\dots$ and $a_1\neq 1$. $\endgroup$
    – Mikhail Borovoi
    Aug 24, 2015 at 18:45
  • $\begingroup$ Mikhail. It's obvious that $ln(h_1/z)+ln(h_2/z)\circ h_1 =ln(h_1/z) + ln(h_2\circ h_1 / h_1) = ln((h_2\circ h_1 / h_1)\cdot(h_1/z)) = ln(h_2\circ h_1 / z)$ $\endgroup$
    – jpceia
    Sep 4, 2015 at 19:03
  • $\begingroup$ $ln(1+z)$ is well defined as power series, then $ln(h/z) = ln(a_1 + a_2 z + \cdots) = ln(a_1) + ln(1 + a_2/a_1 z + \cdots) = ln(a_1) + ln(1 + z)\circ (a_2/a_1 z + \cdots) $ is well defined also $\endgroup$
    – jpceia
    Sep 4, 2015 at 19:05

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