# A reflection group is Euclidean if and only if it has an invariant of degree 2

Let $$G \subset GL(V)$$ be a FINITE IRREDUCIBLE complex reflection group acting on its natural representation $$V$$. Then $$G$$ acts on the dual representation $$V^*$$ contragrediently which induces an action on the polynomial ring $$k[V]$$, the subalgebra, $$G$$-invariant polynomials are finitely-generated. The claim is that the natural representation of $$G$$ can be realized over the real, that is $$G$$ is a Euclidean reflection group if and only if it has an invariant of degree 2.

Now suppose $$G$$ is a Euclidean reflection group, then it leaves the norm polynomial invariant, since it is a subgroup of the orthogonal group.

The converse have been a headache for me, what I'm thinking is that to derive a invariant quadratic form from this invariant of degree 2, and somehow to show that the representation of this group can be define over the real numbers.

Can someone help me with this problem please? Thanks

A Hint [From 'Unitary Reflection Group' by Lehrer and Taylor], use the invariant of degree $$2$$ to construct a non-degenerate $$G$$-invariant quadratic form. From the quadratic form and the hermitian inner product to construct a semi-linear map $$f:V \to V$$ and prove that there is a positive real number $$\lambda$$ such that $$f^2(v) = \lambda v$$ for all $$v\in V$$. Show that the set $$V_0$$ of fixed points of a suitable multiple of $$f$$ is a real vector space such that $$V = \mathbb{C} \otimes V_0$$.

• Are you assuming $G$ is finite? – user10354138 Aug 1 at 12:40
• @user10354138 yes I'm assuming $G$ is finite – kkkkk Aug 1 at 12:51
• There seems to be some missing irreducibility condition here. Take any complex reflection group on $V$ and embed it as a complex reflection group on $W=\mathbb{C}\oplus V$ by doing nothing to the $\mathbb{C}$, and we would have an invariant of degree 2. – user10354138 Aug 1 at 14:20
• @user10354138 Yes you are right, we require the reflection group to be irreducible. – kkkkk Aug 1 at 16:51
• So can you get the quadratic form from the invariant? That's a very good start. – David A. Craven Aug 1 at 20:03