Zagier, in his paper 'Some Surprising Consequences of the Cohomology of SL$_2(\bf{ Z})$' (link, p. 6), studies the action of $\Gamma=PSL_2(\bf Z)$ on a vector space $V$, denoting the action by $v\ |\ \gamma$. Recall the following presentation of $\Gamma=\langle S,U\ | \ S^2,U^3=1\rangle$, where: $$S=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}, U=\begin{pmatrix}1 & -1\\ 1 & 0\end{pmatrix}, T=\begin{pmatrix}1&1\\0&1\end{pmatrix}, $$ satisfying $S=UT$ and $\begin{pmatrix}1&0\\1&1\end{pmatrix}=U^2S$. Also, call $f$ a cocyle if $f(\gamma_1\gamma_2)=f(\gamma_1)|\gamma_2+f(\gamma_2).$

Now, Zagier makes the following claim:

The function $f:\Gamma\to V$ such that $f(T)=0$ and $f(S)=Q$ can be extended to a cocycle if and only if $Q$ satisfies the conditions $Q|{(1+S)=0}$ and $Q|{(1+U+U^2)}=0$.

The $\Rightarrow$ direction is clear, but the $\Leftarrow$ direction does not seem obvious by direct computation, since in that case one does not know much about $f$. Is there some general fact needed to see this?

  • $\begingroup$ What exactly seems to be the problem? Knowing $f(T)$ and $f(S)$ is enough to reconstruct a cocycle, and you just need to check that it agrees with the defining relations of the group, no? $\endgroup$ Commented Jun 26, 2015 at 23:27
  • $\begingroup$ @VladimirDotsenko That's right, but using only the relations given I (and others I've talked to) have not been able to prove the cocyle property directly. $\endgroup$
    – Tian An
    Commented Jun 27, 2015 at 1:31
  • $\begingroup$ This may well turn out to be a simple computation, but my sense from reading Zagier's papers is that often unproved assertions are in fact nontrivial statements. $\endgroup$
    – Tian An
    Commented Jun 27, 2015 at 1:50

1 Answer 1


Define a function $\widehat f$ from $\Gamma$ to the semidirect product $\Gamma \ltimes V$ by $\widehat f(\gamma) = \bigl( \gamma, f(\gamma) \bigr)$. Saying that $f$ is a cocycle is exactly the same as saying that $\widehat f$ is a homomorphism. Therefore, $f$ extends to a cocycle if and only if $\widehat f(S)^2 = 1$ and $\widehat f(U)^3 = 1$.

Under the given conditions, we have $$ \widehat f(S)^2 = (S, Q)^2 = (S^2 , Q|S + Q) = (1,0) .$$ Also, $$ \widehat f(U) = \widehat f(S) \cdot \widehat f(T)^{-1} = (S,Q) \cdot (T,0)^{-1} = (ST^{-1}, Q|T^{-1}) = (U, -Q|U) ,$$ so $$ \widehat f(U)^3 = (U, -Q|U)^3 = (U^3, -Q(U^3 + U^2 + U) \bigr) = (1,0) .$$

  • $\begingroup$ But doesn't $\hat{f}(U)=\hat{f}(ST^{-1})=\hat{f}(S)\cdot\hat{f}(T)^{-1}$ assume that $\hat{f}$ is a homomorphism? Our trouble was finding the value of $f(U)$ knowing only that $f$ is a function. $\endgroup$
    – Tian An
    Commented Jun 27, 2015 at 4:56
  • $\begingroup$ Zagier's claim is that if $Q$ satisfies the given conditions, then there is a homomorphism $\varphi \colon \Gamma \to \Gamma \ltimes V$ with $\varphi(S) = (S,Q)$ and $\varphi(T) = (T,0)$. (This implies there is a cococyle with the specified values on $S$ and $T$.) To verify the claim, check that $(S,Q)^2 = (U, -Q|U)^3 = (1,0)$, so there is a homomorphism $\varphi$ with $\varphi(S) = (S,Q)$ and $\varphi(U) = (U, -Q|U)$. Then it is easy to calculate that $\varphi(T) = (T,0)$ as desired. $\endgroup$ Commented Jun 27, 2015 at 5:11

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