About a property of Reynolds operator (Invariant theory) I am learning about classical invariant theory, and I have a question about Reynolds operator. The book I am reading is $\ulcorner$Classical Invariant Theory$\lrcorner$ written by Hansepter Kraft, Claudio Procesi. If you want, you can download it here.
Consider an infinite field $K$ and a group $G$ and a finite dimensional G-module W and its coordinate ring $K[W]$, and assume that the representation of $G$ on $K[W]$ is completely reducible. In the textbook, they define a canonical G-equivariant linear projection $R:K[W]\rightarrow K[W]^G$ and call it a Reynolds operator. ($K[W]^G$ is a ring of invariants)
They asserts that $R(hf)=hR(f)$ holds for $h\in K[W]^G$, $f\in K[W]$, and this property is crucial in proof of Hilbert's finiteness result for invariants. (p.13 Theorem 1 in the textbook) But I cannot figure out why this relation holds. Actually, it is one of its exercise to prove the equation in a little more general setting. (p.13 Exercise 30)



*Let $A$ be a (commutative) algebra and let $G$ be a group of algebra automorphisms
of $A$. Assume that the representation of $G$ on $A$ is completely
reducible. Then the subalgebra $A^G$ of invariants has a canonical $G$-stable
complement and the corresponding $G$-equivariant projection $p:A\rightarrow A^G$ satisfies the relation $p(hf)=hp(f)$ for $h\in A^G$, $f\in A$.


I have two questions. First, what does a canonical $G$-stable complement mean? Indeed complete reducibility ensures the existence of a $G$-stable complement, but how can we 'construct' such complement of $A^G$?
Second, how can we show that $p(hf)=hp(f)$ for $h\in A^G$, $f\in A$ ? I tried to prove that the complement is stable under multiplication by $h\in A^G$, but couldn't make any progress. 
Any helps, comments, suggestions will be appreciated. Thank you for reading.
 A: First, 'constructing' this complement in such generality certainly cannot be done. The point is just that A can be broken up into a direct sum $A^G \oplus \bigoplus_{i} V_i$, where each $V_i$ is an isotypic $G$-submodule of $A$, i.e., a submodule such that any two simple submodules of $V_i$ are $G$-isomorphic. Moreover, this decomposition is unique up to relabeling, so in particular there is a unique $G$-module complement to $A^G$ in $A$, and it makes sense to define this projection without reference to an auxiliary decomposition. So in this case 'canonical' means something more like 'uniquely determined by the data' than 'directly describable from the data'.
For your second question, the main point is that it suffices to show that if $W$ is a simple $G$-submodule of $A$ not lying in $A^G$ and $a \in A^G$, then $aW$ is also a $G$-submodule intersecting $A^G$ trivially. So let $w \in W$, $a \in A^G$, and $g \in G$. We have
\begin{align}
g(aw) = g(a)g(w) = ag(w),
\end{align}
so $aW$ is a $G$-submodule and the map $W \to aW$, $w \mapsto aw$, is a surjective $G$-module homomorphism. Thus either $aW$ is trivial or it is isomorphic to $W$, and in either case it intersects $A^G$ trivially. So multiplication by elements of $A^G$ preserves the $G$-stable complement to $A^G$ in $A$ and this gives the desired property for the projection.
A: SeanC has already addressed your second question so I'll just address your first one. I think Sean is being a bit too pessimistic. In general if $V$ is any $G$-representation then we have a canonical inclusion $i : V^G \to V$ of invariants and also a canonical quotient $q : V \to V_G$ to coinvariants (the quotient of $V$ by the equivalence relation $gv \sim v$). These compose to give a canonical map
$$q \circ i : V^G \to V_G.$$
If $V$ is completely reducible then this map is an isomorphism (exercise), so we can consider its inverse $(q \circ i)^{-1} : V_G \to V^G$. This lets us define a map
$$(q \circ i)^{-1} \circ q : V \to V^G$$
which, by construction, splits the inclusion $i : V^G \to V$. This means we have a canonical idempotent
$$R = i \circ (q \circ i)^{-1} \circ q : V \to V$$
whose image is $V^G$; this is the Reynolds operator. The kernel of $R$ is then the desired canonical $G$-equivariant complement of $V^G$. ("Canonical" here can be strengthened to "unique.") To connect this up with Sean's answer, $\text{ker}(R)$ can equivalently be defined as the sum of all nontrivial simple subrepresentations of $V$, or as the maximal subrepresentation of $V$ containing no nonzero invariant vector, and this (together with the fact that $R$ is an idempotent with image $V^G$) uniquely determines $R$.
This is quite abstract, so it's worth knowing for concreteness that if $G$ is a finite group and $K$ is a field of characteristic not dividing the order of $G$ (which implies that every $G$-representation is completely reducible by Maschke's theorem) then $R$ is given explicitly by the averaging operator
$$R(v) = \frac{1}{|G|} \sum_{g \in G} gv.$$
You can also gain some intuition for the relationship between invariants and coinvariants, and for the map $(q \circ i)^{-1}$, by considering the following special case: suppose $G$ is a finite group and $K$ a field of characteristic not dividing $|G|$ as above, and let $V$ be a permutation representation given by taking the free $K$-vector space on a set $X$ on which $G$ acts. Then $V^G$ has a basis given by sums over the orbits of $G$ acting on $X$, while $V_G$ has a basis given by the orbits themselves, each member of which has been identified. The map $(q \circ i)^{-1} : V_G \to V^G$ then sends an orbit $O$ to $\frac{1}{|O|} \sum_{o \in O} o$, which is equal to $\frac{1}{|G|} \sum_{g \in G} go$ for any $o \in O$. (This special case is also a nice example of what happens if we don't assume complete reducibility: if $G$ and $X$ are infinite in this construction then $V^G$ has a basis given by sums over finite orbits while $V_G$ still has a basis given by all orbits, so $q \circ i : V^G \to V_G$ is not an isomorphism if there are infinite orbits.)
When $K = \mathbb{R}$ or $\mathbb{C}$ and $G$ is a compact Lie group acting on a finite-dimensional $V$ then there is a generalization of the above formula involving averaging over $G$ with respect to Haar measure.
