# 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 $V$.

1. 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

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

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

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1. 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 $x' \in X^{\perp}$.

2. 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 $x' \in X^{\perp}$.

3. Yes, same proviso as in $2$.

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in 1) you mean the map is bijective. in 2) I understand the $G-$invariant condition but i don't see what do you mean by orthogonal action? in 3) by same proviso you mean same conditions on the action : orthogonal action and $X$ is $G-$invariant? – palio Jun 30 '11 at 11:46
@palio: in 1) I said injective (equivalent to bijective iv $V$ is finite-dimensional as the map is linear), so yes you can replace that. I assumed you're working on a finite-dimensional space $V$. If $V$ fails to be finite-dimensional, you should in addition assume that $X$ is closed and your objection is valid, then I should say bijective. 2) By an action by orthogonal maps I (and everyone else) means that $v \mapsto gv$ is an orthogonal map of $V$. 3) $X$ is automatically $G$-invariant, as it is fixed, so you need only assume that you have an orthogonal/unitary representation. – t.b. Jun 30 '11 at 11:50

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 $g(X')=X'$.

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Thomas: just a little piece of advice. I don't mind if you add an answer explaining parts of the other answers, but maybe you could make that a bit more explicit. Nevertheless, I voted your answer up since you're rather new here (but I admit that I hesitated a moment to give a downvote). Other people could react quite a bit more aggressively. – t.b. Jun 30 '11 at 12:26

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.

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If $G$ is a subgroup of $\operatorname{GL}(V)$ then it need not respect the decomposition. Let $\mathbb{Z}$ act on $\mathbb{C}^2$ by $\begin{bmatrix} 1 & n \\ 0 & 1\end{bmatrix}$ and $X = \mathbb{C} \times \{0\}$, for example. You need to assume that $G \subset \operatorname{U}(V)$. I don't understand your answer to 3) - it doesn't make sense. It can't be transitive, as it fixes $X$. – t.b. Jun 30 '11 at 11:03
@Theo Buehler: I understood the question incorrectly. About 2) I meant that the summation is preserved, but clearly not the decomposition into direct sum – Dennis Gulko Jun 30 '11 at 11:16
Well, I'm not sure I understood it correctly, but my answer corresponds to my way of trying to make sense out of it. – t.b. Jun 30 '11 at 11:18