group action in orthogonal decomposition let $V$ be an inner product space. Let $X$ a subspace of $V$ and $X'$ its orthogonal complement i.e, $V=X\oplus X'$. Let $G$ be a group $G$ acting on $V$. 


*

*an element in $X\oplus X'$ is it a couple $(x,x')$ or a sum $x+x'$, i'm asking because the map $V\times V \rightarrow V; \,(x,x')\mapsto x+x'$ is not injective

*if each $v\in V$ is written $v=x+x'$ so how is the action corresponding to this decomposition? can we write $g(x+x')=gx+gx'$

*supppose $G$ fixes all elements of $X$, can we write $V/G$ is homeomorphic to $X\times (X'/G)$?
 A: *

*Both. You should check that $X \times X^{\perp} \to V$ given by $(x,x') \mapsto x+x'$ is injective. This is equivalent to saying that every $v \in V$ can be written uniquely as $v = x + x'$ with $x \in X$ and $x' \in X^{\perp}$.

*Yes, provided that you assume that the action is by orthogonal (or unitary) maps $V \to V$ (equivalently, you deal with an orthogonal/unitary representation of $G$ on $V$). Moreover, you need to assume that $X$ is $G$-invariant, that is $gx \in X$ for all $x \in X$, then $gx' \in X^{\perp}$ for all $x' \in X^{\perp}$.

*Yes, same proviso as in $2$.
A: 1) The map $X\oplus X' \rightarrow V; (x,x') \mapsto x+x'$ is injective, because $\forall v \in V,  \exists ! x \in X, \exists ! x' \in X'$ such that $v=x+x'$, but the map $V \times V \rightarrow V; (v,w) \mapsto v+w$ is not, because $\forall v \in V, (v,0)\mapsto v$ and $(0,v) \mapsto v$. 
2) As Dennis said, it depends on the action. If it is a representation, the answer is yes.
3) If the map is unitary, I'd say yes, because $\forall g \in G <g(x),g(y)>=<x,y>$, so $g(X')=X'$. 
A: 1) An element of $X\oplus X^\perp$ is $x+x'$ (and this decomposition is clearly unique).
2) This depends on the action. If $G$ is a subgroup of ${\rm GL}(V)$ then clearly it holds that $g(x+x')=gx+gx'$, but even then it isn't necessary that $gx\in X$ or $gx'\in X^\perp$. In the more general case you can't say. The action of $G$ on $V$ doesn't have to preserve the vector-space structure.
