Is the orthogonal group convex?

Question

Is the group of real orthogonal matrices convex?

I've read that this space has two connected components, and I don't think that this set is convex, since all convex sets must be path connected. However, I'm just not a specialist in Algebra. Could someone please provide an explanation?

Answer 1

To show it is not convex, consider the identity matrix $I$ and the negative identity matrix $-I$. Both are orthogonal matrices.

However $$\frac{1}{2}I + \frac{1}{2}(-I) = 0$$ which is not an orthogonal matrix.

Answer 2

By 'matrix' I will mean a square $n \times n$ matrix.

The orthogonal group has indeed two components: the component of the orthogonal matrices with determinant $1$ and the component of the orthogonal matrices of determinant $-1$.

As the determinant is a continuous function on the set of matrices, it is continuous on any segment in the space of matrices.

Now suppose $A$ and $B$ are orthogonal matrices such that the segment $[A,B]$ is contained in the orthogonal group. Since the determinant of an orthogonal matrix can only be $1$ or $-1$, the determinant is constant along the segment and necessarily $A$ and $B$ have the same determinant.

Therefore the orthogonal group is not convex.

Answer 3

From a purely algebraic perspective, $I + -I = 0$ disproves this.

Intuitively, the special orthogonal group roughly looks like a spherical shell. And the orthogonal group is two copies of it. So it is definitely not convex.