Galois group acts by the $i$-th power of the Teichmuller character on $H$ In the second paragraph of Soulé - Perfect forms and the Vandiver conjecture, it is written that:
For any natural integer $i \le p − 2$, let $C^{(i)}$ be the subgroup of $C$, where the Galois group of $\Bbb Q(\zeta_p)$ over $\Bbb Q$ acts by the $i$-th power of the Teichmuller character.
I know the definition of the Teichmuller character, $\omega(x)=\lim_{n \to \infty} x^{p^n}$, and I can realize it, but I can not understand the quoted paragraph: Galois group acts by the $i$-th power of the Teichmuller character on $H$, what does it mean? I want to realize this.
It is probable that I did not realize the action of the Galois group on the ideal class group, but I think I understand it well enough. So any general explanation on the action of the Galois group on the ideal class group would be welcome too.
 A: $\DeclareMathOperator\Gal{Gal}$So as you defined it, the Teichmüller character operates on $\mathbb{Z}_p^\times$, and is basically a homomorphism
$$
\mathbb{Z}_p^\times\rightarrow \mu_{p-1}.
$$
Under the isomorphism $\mathbb{Z}_p^\times \cong \Gal(\mathbb{Q}(\zeta_{p^\infty})/\mathbb{Q})$, it can be thought of as a map $\Gal(\mathbb{Q}(\zeta_{p^\infty})/\mathbb{Q})\rightarrow \mu_{p-1}$. This map factors through the quotient
$$
\Gal(\mathbb{Q}(\zeta_{p^\infty})/\mathbb{Q})\rightarrow \Gal(\mathbb{Q}(\zeta_{p})/\mathbb{Q})\cong \mu_{p-1}.
$$
This final isomorphism is also sometimes called the Teichmüller character $\omega$. Furthermore, for any $i$, we can talk about the $i$-th power of $\omega$ via $\omega^i(x) \mathrel{:=} \omega(x)^i$, and this is still a well-defined homomorphism.
Now for the paper.
$C$ can be thought of as the Galois group of an abelian extension $K/\mathbb{Q}(\zeta_p)$. So you have an exact sequence of groups
$$
0 \rightarrow C \rightarrow \Gal(K/\mathbb{Q}) \rightarrow \Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q}) \rightarrow 0.
$$
It turns out that in any such exact sequence where the first and third groups are abelian (but not necessarily the second), the third group acts on the first via lifting + conjugation. The action is independent of the lift, proof left to the reader.
So now $C$ is a finite $p$-torsion group, which means it comes with a natural action by $\mu_{p-1}$. Given this action, we can ask for the subgroup of $C$ on which $\Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q})$ acts by $\omega^i$. You can precisely define it as e.g.
$$
\{c \in C \mathrel: \omega^i(x)c = xc \quad \forall x \in \Gal(\mathbb{Q}(\zeta_p)/\mathbb{Q})\}.
$$
