Metrics on $SO(n+1)/SO(n)$

Question

I am having a question about finding the metric on $SO(n+1)/SO(n)$ such that $\pi:SO(n+1)\rightarrow SO(n+1)/SO(n)$ is a Riemannian submersion, which $SO(n)$ are equipped with the standard bi-invariant metric. I try to use Schur's Lemma to show that it is isometric to the canonical metric on $S^n$ multiplied by a constant. I am confusing about how to find this constant directly.

Answer 1

First, you are right that there is a unique (up to scaling) metric on $S^{n-1}$ that makes $\pi$ into a Riemannian submersion, as I explain in my answer here. Note that the isotropy action in this case is transitive on the unit sphere, so is definitely irreducible.

Now that we know there is a constant we can scale by, let's figure it out. I'm not exactly sure what you mean by the "standard bi-invariant metric" on $SO(n)$, but the bi-invariant metric I like to use is defined on $T_I SO(n)$ by $\langle X,Y\rangle =-Tr(XY)$.

The function $\pi:SO(n)\rightarrow S^{n-1}$ I'm going to use is $\pi(A) = A_n$ where $A_n$ denotes the last column of $A$. This means that the preimage of the point $p=(0,...,0,1)\in S^n$ corresponds to matrices of the block form $diag(B,1)$ with $B\in SO(n-1)$.

Consider the tangent vector $\alpha'(0)\in T_p S^{n-1}$ with $\alpha(t) = (0,....,\sin(t),\cos(t))$. Note that $\|\alpha'(0)\| = 1$ in the usual metric on $S^{n-1}$.

Now, the identity matrix $I\in SO(n)$ is an element of $\pi^{-1}(p)$, so let's find a tangent vector in$ (\ker \pi_\ast)^\bot\subseteq T_I SO(n) = \mathfrak{so}(n)$ which projects to $\alpha'(0)$. (The notation $\pi_\ast$ refers to the differential $\pi_\ast: T_I SO(n)\rightarrow T_p S^{n-1}$.) Then we can compute the length of this tangent vector to find out the scaling we need to have a Riemannian submersion.

To that end, first note that because $\pi$ is constant on the orbit $I \,\cdot SO(n-1)$, it follows that $\ker \pi_\ast$ contains $\mathfrak{so}(n-1)$, embedded in $\mathfrak{so(n)}$ as matrices with the block form $diag(B,0)$ with $B\in \mathfrak{so}(n-1)$. Since $\pi$ is a subermsion, the kernel of $\pi_\ast$ cannot be any larger, so $\ker \pi_\ast = \mathfrak{so}(n-1)$. A reasonable straightforward calculation now shows that $(\ker \pi_\ast)^\bot = \{M = (M)_{ij}\in \mathfrak{so}(n): M_{ij} = 0$ if both $i,j < n\}.$ In other words, $\ker \pi_\ast^\bot$ consists of matrices of the form $$M = \begin{bmatrix} 0 & \cdots & 0 & m_{1,n}\\ 0 & \cdots & 0 & m_{2,n}\\ \vdots & & \ddots & \vdots\\ -m_{1,n} & -m_{2,n} & \cdots & 0\end{bmatrix}.$$

Now, consider $\gamma:\mathbb{R}\rightarrow SO(n)$ with $\gamma(t) = diag\left(1,...,1, \begin{bmatrix} \cos t & \sin t\\ -\sin t & \cos t\end{bmatrix}\right)$. Then $\gamma(0) = I$ and $\gamma'(0)$ is a matrix whose only non-zero entries are $\gamma'(0)_{n-1,n} = -\gamma'(0)_{n,n-1} = 1$. It follows that $\gamma'(0)\in (\ker\pi_\ast)^\bot.$

Finally, note that $\pi \circ \gamma = \alpha$, so $\pi_\ast(\gamma'(0)) = \alpha'(0)$.

Now, an easy calculation shows that $\langle \gamma'(0),\gamma'(0)\rangle = 2$. Since $\langle \alpha'(0), \alpha'(0)\rangle = 1$, we see that the submersion metric on $S^{n-1}$ is the usual metric scaled by a factor of $2$.