Picture below is from the Tao's blogs, I understand the Killing form as $$ B(x,y)=\operatorname{trace}(\operatorname{ad}(x)\operatorname{ad}(y)) $$ $SO(4)$ is the group acting on $S^3$.

I don't understand the content above red line.

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1 Answer 1


Suppose $G$ is a compact Lie group which acts freely and isometrically on a Riemannian manifold $M$ with metric $\langle \cdot, \cdot \rangle$. Then the orbit space $M/G$ is a manifold which inherits a Riemannian metric from the projection $\pi:M\rightarrow M/G$ in the following way.

For each $m\in M$, $d_m \pi:T_m M\rightarrow T_{mG} M/G$ has a kernel, which gives an orthogonal decomposition $T_m M = \ker d_m \pi \oplus H_m$ where $H_m$ is the so-called horizontal space. Then $d_m\pi|_{H_m}:H_m\rightarrow T_{mG} M/G$ is an isomorphism of vector spaces. Then, for $v,w\in T_{mG} M/G$, we define an inner product $(v,w) = \langle (d_m \pi)^{-1} v, (d_m \pi)^{-1} w\rangle.$

One possible issue: suppose $mG = m' G$, do we get a consistent definition of $(\cdot, \cdot)$?

Well, if $m = m'$, then $m = f(m')$ for some $f\in G\subseteq Iso(M)$. Then note that $\pi \circ f = \pi$. Thus, by the chain rule, we have $d_m \pi = d_{m'}\pi \circ d_m f$. It follows that $d_m f$ maps $H_m$ isomorphically onto $H_{m'}$ and further, this map is an isometry because $f$ is. This gives $\langle (d_m \pi)^{-1} v, (d_m \pi)^{-1} w\rangle = \langle (d_{m'} \pi)^{-1} v, (d_{m'}\pi)^{-1} w\rangle$, so the metric is well defined.

In your particular instance, $H = SO(3)$ acts by right multiplication on $M = SO(4)$ which is equipped with a bi-invariant Riemannian metric, the negative of the Killing form. Then the quotient space $M/G = SO(4)/SO(3)$ inherits a metric as above.

Finally, the map $\pi:SO(4)\rightarrow S^3$ given by mapping the last column of a matrix in $SO(4)$ is $SO(3)$-invariant and induces a diffeomorphism $SO(4)/SO(3)\rightarrow S^3$. Use this diffeomorphsm to transport the Riemannian metric from $SO(4)/SO(3)$ to $S^3$.


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