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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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    @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.2011-06-30
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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 $\forall g \in G =$, so g(X')=X'.

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    Thomas: just a little piece of advice. I don't mind if you add an answer e$x$plaining parts o$f$ 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.2011-06-30
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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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    Well, I'm not sure *I* understood it correctly, but my answer corresponds to my way of trying to make sense out of it.2011-06-30