Generalized Rotation

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A lengthy verification in characteristic classes.

In the construction of cell structure of Grassmannian manifold, in the Page 77 of Characteristic Classes written by John Milnor, we need the following property:

Prop. Let u,v∈Rku,v\in\mathbb{R}^k be unit vectors, such that u≠−vu\neq -v​. Then

T(u,v)x:=x−(u+v)⋅x1+u⋅v(u+v)+2(u⋅x)yT(u,v)x:=x-\frac{(u+v)\cdot x}{1+u\cdot v}(u+v)+2(u\cdot x)y

is the unique rotation in Rk\mathbb{R}^k taking uu to vv​, such that

  1. T(u,u)=idT(u,u)=\mathrm{id}, and
  2. T(u,v)∘T(v,u)=idT(u,v)\circ T(v,u)=\mathrm{id}​.

The verification of the proposition is a little bit lengthy, but I am going to do it here in the article.

Proof. (1) Keep in mind that u⋅u=v⋅v=1u\cdot u=v\cdot v=1.

T(u,u)x=x−2u⋅x1+u⋅u(2u)+2(u⋅x)u=x−2(u⋅x)u+2(u⋅x)u=x.\begin{aligned} T(u,u)x &=x-\frac{2u\cdot x}{1+u\cdot u}(2u)+2(u\cdot x)u\\ &=x-2(u\cdot x)u+2(u\cdot x)u\\ &=x. \end{aligned}

(2) Let y=T(u,v)x=x−(u+v)⋅x1+u⋅v(u+v)+2(u⋅x)vy=T(u,v)x=x-\frac{(u+v)\cdot x}{1+u\cdot v}(u+v)+2(u\cdot x)v. Then

T(v,u)∘T(u,v)x=T(v,u)y=y−(u+v)⋅y1+u⋅v(u+v)+2(v⋅y)u\begin{aligned} &T(v,u)\circ T(u,v)x\\ =&T(v,u)y\\ =&y-\frac{(u+v)\cdot y}{1+u\cdot v}(u+v)+2(v\cdot y)u \end{aligned}

Consider one by one the coefficient of uu and vv. Observe that

(u+v)2=u2+v2+2u⋅v=2(1+u⋅v).(u+v)^2=u^2+v^2+2u\cdot v=2(1+u\cdot v).

The coefficient in uu is

−(u+v)⋅x1+u⋅v−(u+v)⋅y1+u⋅v+2(v⋅y)=2(v⋅y)−(u+v)⋅x1+u⋅v−u+v1+u⋅v⋅(x−(u+v)⋅x1+u⋅v(u+v)+2(u⋅x)v)=2(v⋅y)−(u+v)⋅x1+u⋅v−(u+v)⋅x−2(u+v)⋅x+2(u⋅x)(u⋅v+1)1+u⋅v=2(v⋅y)−2(u⋅x)=2v⋅(x−(u+v)⋅x1+u⋅v(u+v)+2(u⋅x)v)−2u⋅x=2(v⋅x)−2(u+v)⋅x1+u⋅v(u⋅v+1)+4(u⋅x)−2(u⋅x)=−2(u⋅x)+4(u⋅x)−2(u⋅x)=0.\begin{aligned} &-\frac{(u+v)\cdot x}{1+u\cdot v}-\frac{(u+v)\cdot y}{1+u\cdot v}+2(v\cdot y)\\ =&2(v\cdot y)-\frac{(u+v)\cdot x}{1+u\cdot v}-\frac{u+v}{1+u\cdot v}\cdot \left(x-\frac{(u+v)\cdot x}{1+u\cdot v}(u+v)+2(u\cdot x)v\right)\\ =&2(v\cdot y)-\frac{(u+v)\cdot x}{1+u\cdot v}-\frac{(u+v)\cdot x-2(u+v)\cdot x+2(u\cdot x)\color{blue}(u\cdot v+1)}{\color{blue}1+u\cdot v}\\ =&2(v\cdot y)-2(u\cdot x)\\ =&2v\cdot \left(x-\frac{(u+v)\cdot x}{1+u\cdot v}(u+v)+2(u\cdot x)v\right)-2u\cdot x\\ =&2(v\cdot x)-2\frac{(u+v)\cdot x}{1+u\cdot v}(u\cdot v+1)+4(u\cdot x)-2(u\cdot x)\\ =&-2(u\cdot x)+4(u\cdot x)-2(u\cdot x)\\ =&0. \end{aligned}

By the above procedure, we can know that

(u+v)⋅x1+u⋅v+(u+v)⋅y1+u⋅v=2(u⋅x)=2(v⋅y).\frac{(u+v)\cdot x}{1+u\cdot v}+\frac{(u+v)\cdot y}{1+u\cdot v}=2(u\cdot x)=2(v\cdot y).

Hence the coefficient in vv​ is

−(u+v)⋅x1+u⋅v−(u+v)⋅y1+u⋅v+2(u⋅x)=−2(u⋅x)+2(u⋅x)=0.\begin{aligned} &-\frac{(u+v)\cdot x}{1+u\cdot v}-\frac{(u+v)\cdot y}{1+u\cdot v}+2(u\cdot x)\\ =&-2(u\cdot x)+2(u\cdot x)\\ =& 0. \end{aligned}

This completes the proof. □\square