Actions by affine algebraic groups?

Question

I am learning geometric invariant theory from Hoskins' notes. I am really confused about notions of group actions. One particular problem is the definition of a (rational) action on an algebra.

In Definition 3.7 in the notes

Definition 3.7. An action of an affine algebraic group $G$ on a $k$-vector space $V$ (resp. $k$-algebra $A$) is given by, for each $k$-algebra $R$, an action of $G(R)$ on $V\otimes_kR$ (resp. on $A\otimes_kR$) \begin{equation}\begin{split}\sigma_R:G(R)\times (V\otimes_kR)\to V\otimes_kR\\\sigma_R:G(R)\times(A\otimes_k R)\to A\otimes_kR\end{split}\end{equation}such that $\sigma_R(g,-)$ is a morphism of $R$-modules (resp. $R$-algebras) and these actions are functorial in $R$. We say that an action of $G$ on a $k$-algebra $A$ is rational if every element of $A$ is contained in a finite dimensional $G$-invariant linear subspace of $A$.

The following notion of invariant vector is purely from my mind. Hoskins defined only for $R=k$ (Section 3.5 in the notes)

For a $G$-action on $V$, a vector $v\in V$ is $G$-invariant iff $\sigma_R(g,v\otimes 1)=v\otimes 1$ for all $R\in\mathrm{Alg}_k$ and all $g\in G(R)$. Similar for $G$-actions on $A$.

There is another notion called dual action, which does not appear in the notes.

Definition. A dual action of an affine algebraic group $G$ on a $k$-vector space (resp. $k$-algebra $A$) is given by a morphism of $k$-vector spaces (resp. $k$-algebras)\begin{equation}\begin{split}\sigma^*:V\to \mathcal O(G)\otimes_kV\\\sigma^*:A\to \mathcal O(G)\otimes_kA\end{split}\end{equation}satisfying:

• $(\mu^*\otimes\mathrm{id})\circ\sigma^*=(\mathrm{id}\otimes\sigma^*)\circ\sigma^*$, where $\mu^*:\mathcal O(G)\to\mathcal O(G)\otimes_k\mathcal O(G)$ is the co-multiplication;
• $(e^*\otimes\mathrm{id})\circ\sigma^*=\mathrm{id}$, where $e^*:\mathcal O(G)\to k$ is the co-unit.

For a dual action of $G$ on $V$ (resp. $A$), an element $v\in V$ is invariant iff $\sigma^*(v)=1\otimes v$ (resp. $\sigma^*(a)=1\otimes a$).

My main question is the following: Let $G$ be reductive. Are the following two statements true and equivalent?

1. If $G$ acts on $A$ rationally with $A$ finitely generated, then $A^G$ is also finitely generated.
2. If $G$ acts on $A$ dually with $A$ finitely generated, then $A^G$ is finitely generated.

I notice even the definitions of $A^G$ are different.

The motivation is that for affine GIT, $G\curvearrowright\mathrm{Spec}(A)$, the data is equivalent to a dual action of $G$ on $A$. It is also possible to show a $G$-action on $A$. Then there are two ways to construct $\mathrm{Spec}(A^G)$. However equivalence is not obvious.

Thanks.