# Group fixes $U^\perp$ implies only $e$ fixes $U$

I'm having difficulty with this result (given as 2 lines in my book):

Let $\Phi$ be a root system as defined http://en.wikipedia.org/wiki/Root_system and let $W$ be a group generated by the reflections $s_\alpha$ for $\alpha \in \Phi$.

Suppose that we know each element of $W$ fixes pointwise the orthogonal complement, $U^\perp$ of the subspace spanned by $\Phi$, $U$.

Somehow, this implies that only $e$ can fix $\Phi$.

I can kinda see how we would want this to be true, but I'm unsure how to prove it.

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How is the action of $U$ being extended to all of $\mathbb{R}^n$? If $W$ acts on $\Phi$, then the action naturally extends to $U$, but how do you extend it to $U^{\perp}$? Or is $W$ already acting on all of $\mathbb{R}^n$, and it just turns out that it acts on the set $\Phi$? – Arturo Magidin Oct 17 '11 at 18:04
What is $e$? The identity? And what do you mean by "fix"-ing a set or a subspace? Do you mean $U^\perp$ is invariant under $W$, or something else? – user1551 Oct 17 '11 at 18:10
@ArturoMagidin $W$ acts on $R^n$ – Zeophlite Oct 17 '11 at 18:13
@user1551 Sorry, pointwise, $\forall w \in W \hspace{3mm} \forall \lambda \in U^\perp \hspace{3mm} w(\lambda) = \lambda$ – Zeophlite Oct 17 '11 at 18:15
You are missing some important hypothesis: consider any non-trivial group acting trivially on $\mathbb R^{27}$ and let $\Phi$ be any finite subset of $\mathbb R^{27}$. – Mariano Suárez-Alvarez Oct 17 '11 at 18:25

Clearly if a linear morphism fixes pointwise $\Phi$ and the orthogonal complement of $U$ then it fixes the whole space, being generated by $\Phi$ and the complement of $U$. So if one element in $W$ does that it is $e$. Perhaps I missed something but it seems quite easy ...
The fact after updating your question that $W$ appears to be a Weil group does not change my answer in my mind : it only ensures the fact that it actually fixes the orthogonal complement of the root space ... – brunoh Oct 17 '11 at 19:14
Suppose $G$ is a group of linear transformations of a vector space $V$ and suppose that $v_1, \dots, v_n$ is a basis of $V$ such that $g \cdot v_i = v_i$ for some $g \in G$. Then $g$ is the identity transformation.