0
$\begingroup$

Definition of cocycle: Let $(\Omega,G)$ be a dynamical System with $G=\mathbb{R},\mathbb{Z}$ or $\mathbb{N}_0$ (written as right Transformation Group). and $\Gamma$ be a semi-Group with composition $\circ$. A cocycle is a map $\gamma\colon\Omega\times G\to\Gamma$ with te property that for any $g_1,g_2\in G$ and $x\in\Omega$, the relation $$ \gamma(x,g_1g_2)=\gamma(xg_2,g_1)\circ\gamma(x,g_2) $$ is fulfilled.

Now, I would like to verify that the following is a cocycle:

Let $f\colon\Omega\to\mathbb{R}^d$ be a function. Then $$ \gamma(x,n)=S_nf(x)=f(x)+f(T(x))+\ldots +f(T^{n-1}(x)) $$ is a cocycle over $\mathbb{N}_0$ with values in $\Gamma=\mathbb{R}^d$.

So, when I got it right, here we have

  • $(\Omega,G)=(\Omega,\mathbb{N}_0)$,

  • $(x,n)\mapsto xn$,

  • $\gamma\colon \Omega\times\mathbb{N}_0\to\mathbb{R}^d$ with composition $\circ$

  • $T\colon\Omega\to\Omega$

But how to Show that $$ \gamma(x,n_1n_2)=S_{n_1n_2}f(x)=S_{n_1}f(xn_2)\circ S_{n_2}f(x)=\gamma(xn_2,n_1)\circ\gamma(x,n_2)? $$

I am confused about all the different compositions:

We have

  • $n_1n_2$ with $n_1, n_2\in\mathbb{N}_0$,

  • $xn$ with $x\in\Omega, n\in\mathbb{N}_0$ and

  • and $\circ$

1 Answers 1

1

You need to consider $\mathbb R^d$ with addition (that is, in this case the group operation $\circ$ is addition)

We have $$ \gamma(x,g_2)=f(x)+\cdots +f(T^{g_2-1}(x)) $$ and since $xg_2=T^{g_2}(x)$ (you need to choose the action like this), we also have $$ \begin{split} \gamma(xg_2,g_1) &=f(xg_2)+\cdots +f(T^{g_1-1}(xg_2))\\ &=f(T^{g_2}(x))+\cdots +f(T^{g_1+g_2-1}(x)). \end{split} $$ Hence, $$ \gamma(xg_2,g_1)+\gamma(x,g_2)=f(x)+\cdots +f(T^{g_1+g_2-1}(x))=\gamma(x,g_1+g_2). $$