Projective special orthogonal group for bilinear form of positive Witt index is simple

Statement
Suppose $$k$$ is a field and $$V$$ is a finite-dimensional vector space over $$k$$, of dimension $$n$$. Suppose $$b$$ is a nondegenerate symmetric bilinear form on $$V$$. Further, suppose that the Witt index $$\nu$$ of $$b$$ is positive. In other words, the hyperbolic subspace in the Witt decomposition of $$V$$ has dimension $$\nu > 0$$.

Then, the projective special orthogonal group for $$b$$ is a simple group, except in the following cases: $$n = 4, \nu = 2$$ and $$n = 3, |k| = 3$$.

Related facts

 * Projective special orthogonal group over reals is simple
 * Projective special orthogonal group over non-Archimedean ordered field is simple