Describing $\mathbb{RP}^n$ using the orthogonal and special orthogonal groups.

Question

There is a nice geometric way of interpreting the rotation group $SO(3)$ as the real projective space $\mathbb{RP}^3$. We do this by noting that every rotation is characterized by its axis of rotation and its angle of rotation. Thus we can represent a rotation by a vector with in $\mathbb{R}^3$ with its direction representing the axis of rotation, and its length representing the angle. This gives multiple representations for the same rotation, so we can restrict us to the ball with radius $\pi$ with antipodal points on the boundary identified, which is homeomorphic with $D^3$ with antipodal points on the boundary identified, which gives us the identification with $\mathbb{RP}^3$.

It is clear that this does not hold for arbitrary $n$, that is, we do not generally have $SO(n)\cong \mathbb{RP}^n$. I was wondering if there is some way to describe $\mathbb{RP}^n$ in terms of the special orthogonal groups or the orthogonal groups modulo some equivalence relation. In the book Lie Groups and Invariant Theory, by E. vinberg, he mentions that for $n\geq 3$, we can identify $\mathbb{RP}^{n-1}$ with $SO(n)/S(O(1)\times O(n-1))$, but I did not find any clear explanation or reference for that fact. So I was wondering if someone here can help me with a explanation, or a reference to some literature considering these identifications.

Answer 1

If you interpret $SO(n)$ as the set of all positively oriented orthonormal bases, then the quotient by $S(O(1)\times O(n-1))$ means that you identify two o.n. bases $u_1,\ldots, u_n$ and $w_1,\ldots, w_n$ if

• $w_1=\pm u_1$
• $(w_2,\ldots, w_n)^t=Q(u_2,\ldots, u_n)^t$ with $Q\in O(n-1)$
• the two bases have the same orientation.

So, the elements of your quotient can be identified with straight lines through the origin: given a straight line $\ell$, we write $\ell=\mathrm{Span}{v_1}$ with $\|v_1\|=1$, we consider the hyperplane $V=\ell^\perp$ and we take an o.n. basis of $V$ $v_2,\ldots, v_n$ such that $v_1,\ldots, v_n$ is positively oriented; once we change this with another o.n. basis as described above, we still get the same line $\ell$ as the span of the first vector.

Now, by its very definition, $\mathbb{RP}^{n-1}$ is the set of the lines through the origin of $\mathbb{R}^n$.

In a way, you are considering all the pairs $(\ell, \ell^\perp)$ as orthogonal decompositions of $\mathbb{R}^n=\ell+\ell^\perp$ and you are identifying every pair with the set of o.b. bases (positively oriented) which are "adapted" to such a decomposition.

A final note: taking the special orthogonal groups instead of the orthogonal groups is just a matter of taste... you would end up with the same space considering $O(n)/(O(1)\times O(n-1))$. Moreover, as William points out in the comments, this is actually a way of describing the various Grassmannian manifolds $\mathrm{Gr}(n,d)$ as quotients $O(n)/(O(d)\times O(n-d))$.