(i) Every endomorphism of V commuting with $\rho(G)$ is a homothety, and $\rho$ is absolutely irreducible.

(ii) Assume that V is finite dimensional. Let B be a non-zero bilinear form on V invariant under $\rho(G)$. Then B is non-degenerate, either symmetric or skew-symmetric, and every bilinear form on V invariant under $\rho(G)$ is proportional to B.

Let u be an endomorphism of V commuting with $\rho(G)$. Let g be an element of G such that $\rho(g)$ is a pseudo-reflection and let D be the image of $1 - \rho(g)$. Since D is of dimension 1 and $u(D) \subset D$, there exists $\alpha$ in K such that $u - \alpha . 1$ vanishes on D; the kernel N of $u - \alpha . 1$ is then a vector subspace of V invariant under $\rho(G)$ and is non-zero as it contains D; since $\rho$ is irreducible, $N = V$ and $u = \alpha . 1$. The second part of (i) follows from the first by Algebra, Chap. VIII, § 13, no. 4, Cor. of Prop. 5.

Let N (resp. $N'$) be the subspace of V consisting of those x such that $B(x, y) = 0$ (resp. $B(y, x) = 0$) for all y in V; since B is invariant under $\rho(G)$, the subspaces N and $N'$ of V are stable under $\rho(G)$ and distinct from V since $B \neq 0$. Since $\rho$ is irreducible, $N = N' = 0$ and B is non-degenerate.

Since V is finite dimensional, every bilinear form on V is given by the formula

for some endomorphism u of V. If $B'$ is invariant under $\rho(G)$, the endomorphism u commutes with $\rho(G)$. Indeed, let $x, y$ be in V and let g be in G; since B and $B'$ are invariant under $\rho(G)$, we have hence $u(\rho(g)(x)) = \rho(g)(u(x))$ since B is non-degenerate. By (i), there exists $\alpha$ in K with $u = \alpha.1$, so $B' = \alpha.B$.

In particular, we can apply this to the bilinear form $B'(x, y) = B(y, x)$; then $B(y, x) = \alpha.B(x, y) = \alpha^2.B(y, x)$ for all $x, y$ in V, and since B is nonzero we have $\alpha^2 = 1$, hence $\alpha = 1$ or $\alpha = -1$. Thus, B is either symmetric or skew-symmetric.

(i) If s is a reflection, V is the direct sum of the hyperplane $V_s^+$ and the line $V_s^-$.

(ii) Conversely, assume that V is the direct sum of a hyperplane H and a line D such that $s(x) = x$ and $s(y) = -y$ for $x \in H$ and $y \in D$. Then s is a reflection and $H = V_s^+$, $D = V_s^-$. Finally, D is the image of $1 - s$.

(i) If s is a reflection, $V_s^+$ is a hyperplane. If x belongs to $V_s^+ \cap V_s^-$, then $x = s(x) = -x$, so $x = 0$ since K is of characteristic $\neq 2$. Finally, for $x$ in V, the vector $x' = s(x) + x$ (resp. $x'' = s(x) - x$) belongs to $V_s^+$ (resp. $V_s^-$) since $s^2 = 1$, and $2x = x' - x''$. Thus V is the direct sum of $V_s^+$ and $V_s^-$, and $V_s^-$ is necessarily of dimension 1 since $V_s^+$ is a hyperplane.

(ii) Under the stated hypotheses, every element of V can be written uniquely in the form $v = x + y$ with $x \in H$ and $y \in D$ and we have $s(v) = x - y$; assertion (ii) follows immediately from this.

Let s be a reflection in V. Prop. 2, (i) shows that there exists a basis $(e_1, \ldots, e_n)$ of V such that $s(e_1) = e_1, \ldots, s(e_{n-1}) = e_{n-1}$ and $s(e_n) = -e_n$, hence $\det(s) = -1$.

(i) A subspace V' of V is stable under s if and only if $V_s^- \subset V'$ or $V' \subset V_s^+$.

(ii) An endomorphism u of V commutes with s if and only if $V_s^+$ and $V_s^-$ are stable under u.

(i) If $V' \subset V_s^+$, then $s(x) = x$ for all $x$ in $V'$, so $s(V') \subset V'$. Assume that $V_s^- \subset V'$; then, for any $x$ in $V'$, $s(x) - x \in V_s^- \subset V'$, hence $s(x) \in V'$; thus $s(V') \subset V'$. Conversely, assume that $s(V') \subset V'$; if $V' \not\subset V_s^+$, there exists $x$ in $V'$ with $s(x) \neq x$; the non-zero vector $a = s(x) - x$ belongs to the line $V_s^-$, and hence generates this space; since $a \in V'$, we have $V_s^- \subset V'$.

(ii) Assume first that $u$ commutes with $s$. If $x$ is a vector such that $s(x) = \varepsilon.x$ (where $\varepsilon = \pm 1$), then $s(u(x)) = u(s(x)) = \varepsilon.u(x)$, so $V_s^+$ and $V_s^-$ are stable under $u$. Conversely, assume that $V_s^+$ and $V_s^-$ are stable under $u$; it is clear that $us - su$ vanishes on $V_s^+$ and on $V_s^-$, and since V is the direct sum of $V_s^+$ and $V_s^-$, we have $us - su = 0$.

If $V_s^- \subset V_u^+$ and $V_u^- \subset V_s^+$, Prop. 3, (i) shows that $V_u^+$ and $V_u^-$ are stable under $s$, hence $su = us$ by Prop. 3, (ii).

Conversely, if $su = us$, the subspace $V_s^-$ is stable under $u$ by (ii); by (i), there are two possible cases:

a) We have $V_u^- \subset V_s^-$: since they are both of dimension 1, these spaces are therefore equal, hence $V_s^- \not\subset V_u^+$; since $V_u^+$ is stable under $s$, we have $V_u^+ \subset V_s^+$ and so these two hyperplanes are equal. But then $s = u$, contrary to our assumptions.

b) We have $V_s^- \subset V_u^+$: the image of $1 - s$ is thus contained in the kernel of $1 - u$, so $(1 - u).(1 - s) = 0$. Since $s$ and $u$ commute, we have $(1 - s).(1 - u) = 0$, in other words $V_u^- \subset V_s^+$.

(i) X is stable under s. (ii) $X^0$ is stable under s. (iii) H contains X or $X^0$.

We have $V_s^+ = H$, and by what we have said, $V_s^-$ is the orthogonal complement $H^0$ of H with respect to B. By Prop. 3, X is stable under s if and only if $X \subset H$ or $H^0 \subset X$; but the relation $H^0 \subset X$ is equivalent to $X^0 \subset H$ by Algebra, Chap. IX, § 1, no. 6, Cor. 1 to Prop. 4. This proves the equivalence of (i) and (iii); that of (ii) and (iii) follows by interchanging the roles of X and $X^0$, since $(X^0)^0 = X$.

The existence and uniqueness of $v$ are immediate. The automorphism of $V$ associated to $s_{H'} s_H$ is the identity; thus $s_{H'} s_H$ is a translation. On the other hand, let $a \in H'$; then $a - v \in H$ and

showing that $s_{H'} s_H$ is the translation by the vector $2v$.

Indeed, by Prop. 2 of Chap. IV, § 1, no. 2, it suffices to show that $s_{H'} s_H$ is of infinite order (resp. order $2p$), which is clear.

Let $u$ be a rotation transforming $\Delta$ into $\Delta'$. Since $su$ is an orthogonal transformation of V of determinant $-1$, it is a reflection and thus $(su)^2 = 1$. Consequently, $u^{-1} = sus^{-1}$ transforms $s(\Delta)$ into $s(\Delta')$, hence the proposition.

We know that $s_{D'} s_D$ is a rotation since it is of determinant 1. Let $\Delta$ and $\Delta'$ be two half-lines starting at the origin carried by D and D'. We have

hence the corollary.

Now let $\Delta$ and $\Delta'$ be two half-lines of V such that

and let s and $s'$ be the orthogonal reflections with respect to the lines D and D' containing $\Delta$ and $\Delta'$. Let $\theta$ be the principal measure of the angle $(\widehat{D}, \widehat{D'})$. If $\theta \in \pi \mathbf{Q}$, denote by m the smallest integer $\geq 1$ such that $m\theta \in \pi \mathbf{Z}$. If $\theta \notin \pi \mathbf{Q}$, put $m = \infty$. Let W be the group generated by s and $s'$.

The Corollary to Prop. 6 shows that $s's$ is of order m, which gives the first assertion. The elements of W are thus of the form $(s's)^n = \rho_{2n\theta}$ or $(s's)^n s = \rho_{2n\theta} s$; the last assertion follows from this, since $D' = \rho_\theta(D)$.

If $m = \infty$, the image of the set $n \theta$ ($n \in \mathbf{Z}$) is dense in $\mathbf{R}/2\pi \mathbf{Z}$ (General Topology, Chap. VII, § 1, Prop. 11); the union of the transforms of D by the elements of W is thus dense in V and meets C. If m is finite and if $\theta = k \pi / m$ with $1 < k < m$, the integers k and m being relatively prime, there exists an integer h such that $hk \equiv 1$ mod. m; then $(D, \rho_{h \theta}(D)) \equiv \pi / m$ (mod. $\pi$), and $\rho_{h \theta}(D)$ meets C. This shows that the condition is necessary. The converse is immediate.