Recently, I read the book Kazhdan's Property (T). There is a lemma on the page 75 (Lemma 2.2.1) as following:
Lemma. Let $\pi$ be an orthogonal representation of $G$ on $H^0$. For a mapping $\alpha: G\to Isom(H)$ the following conditions are equivalent:
(i) $\alpha$ is an affine isometric action of $G$ with linear part $\pi$;
(ii) there exists a continuous mapping $b : G → H^0$ satisfying the 1-cocycle relation $b(gh) =b(g)+π(g)b(h)$, and such that $α(g)x = \pi(g)x+b(g)$ for all $g,h\in G$ and $x \in H$.
Since $\pi$ is an orthogonal representation of $G$ on $H^0$, $\pi(g)$ is a map on $H^0$. But in (ii), there is a $\pi(g)x$ for $x\in H$. My question is what is $\pi(g)x$?
I think both $H$ and $H^0$ are the same set, the difference in notation only emphasizes the structure under consideration. Namely, $H$ is an Adobe space, while $H^0$ is a linear space, that is an affine space with a chosen zero element.
The automorphism groups are different because a diffeomorphism of a linear space is required to fix the origin. But since the elements are the same, any map on affine space is also a map on the linear space.