Proposition 3

Suppose that A is a commutative field and that E is of finite dimension $> 0$. Let $\Phi$ be a non-degenerate hermitian form on E. The mapping $u \to \det u$ is a homomorphism of the unitary group $\mathbf{U}(\Phi)$ associated with $\Phi$ onto the multiplicative subgroup H of A formed by the elements $\rho$ such that $\rho \overline{\rho} = 1$ (subgroup reduced to $\{1, -1\}$ when J is the identity).

So let $u$ be an element of $\mathbf{U}(\Phi)$, $U$ its matrix with respect to a basis of E, and $R$ the matrix of $\Phi$ with respect to this basis. The relation $R = {}^t U \cdot R \cdot \overline{U}$ ($§ 1$, n° 10, formula (48)) shows that $(\det U)(\det \overline{U}) = 1$ since $R$ is invertible; whence $(\det u)(\det u) = 1$. The homomorphism $u \to \det u$ maps $\mathbf{U}(\Phi)$ onto H. Indeed, when A has characteristic 2 and J is the identity, H is reduced to the element 1. Otherwise there exists an orthogonal basis $(e_i)$ ($i = 1, \ldots, n$) of E (th. 1); for every $\rho \in A$ such that $\rho \overline{\rho} = 1$, let $u$ be the automorphism of E defined by $u(e_1) = \rho e_1$ and $u(e_i) = e_i$ for $i = 2, \ldots, n$; then $u$ is unitary and $\det u = \rho$, whence the proposition.

Under the conditions of prop. 3, the kernel of the homomorphism $u \to \det u$ is a normal subgroup of $\mathbf{U}(\Phi)$, called the special unitary group associated with $\Phi$; it is sometimes denoted by $\mathbf{SU}(\Phi)$.

When $J$ is the identity and $A$ is not of characteristic 2, this group is also called the special orthogonal group associated with $\Phi$ (or with the quadratic form $Q(x) = \Phi(x, x)$) and is sometimes denoted by $\mathbf{SO}(Q)$.

If $E = A^n$ and $\Phi$ is the form whose matrix with respect to the canonical basis of $E$ is the unit matrix, the notations $\mathbf{SU}(n, A)$ or $\mathbf{SU}_n(A)$ and $\mathbf{SO}(n, A)$ or $\mathbf{SO}_n(A)$ are used.