Equivariant Diffeomorphism $S^2 times S^3$ to itself with respect to the following $mathbb{Z}_4$ action

Mathematics Asked by TuoTuo on November 9, 2020

Let $$mathbb{Z}_4$$ be the cyclic group generated by $$(R,j)$$ where $$R in$$ SO$$(3)$$ is the rotation matrix $$R = begin{pmatrix} -1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 end{pmatrix}$$

Consider the $$mathbb{Z}_4$$ action on $$S^2 times S^3$$ generated by $$(x,q) mapsto (Rx,qoverline{j})$$. Here we are thinking of $$S^2$$ as the standard subset of $$mathbb{R}^3$$ with elements represented as column vectors and $$S^3$$ as unit quaternions.

I am looking for an equivariant diffeomorphism $$S^2 times S^3 rightarrow S^2 times S^3$$ that takes the $$mathbb{Z}_4$$ action above, and moves the action to the $$mathbb{Z}_4$$ action generated by $$(x,q) mapsto (x,qoverline{j})$$. My issue is that you cannot just use the diffeomorphism $$(x,q) mapsto (R^{T}x,q)$$ because $$R$$ is only being multiplied by the fist coordinate when half of the elements of $$mathbb{Z}_4$$ are acting.

Here's an example that makes use of the fact that the unit quaterions act by rotations on $$S^2$$.

Consider $$S^2times S^3$$ as a subset of $$mathbb{H}_{im}timesmathbb{H}$$, where $$mathbb{H}_{im}$$ denotes the subspace of purely imaginary quaterions. Note that the unit quaternions act on $$mathbb{H}_{im}$$ by conjugation $$alphacdotbeta=alphabetaalpha^*$$, which have the effect of Euclidean rotations. We can choose the identification $$mathbb{R}^3congmathbb{H}_{im}$$ so that the matrix $$R$$ is given by the action of $$j$$ (since unit imaginary quaterions act by $$180^circ$$ rotations). Define a map $$f:mathbb{H}_{im}timesmathbb{H}tomathbb{H}_{im}timesmathbb{H}$$ by $$f(alpha,beta)=(betaalphabeta^*,beta)$$ This map can be smoothy restricted to $$S^2times S^3subsetmathbb{H}_{im}timesmathbb{H}$$, and has an inverse given by $$f^{-1}(alpha,beta)=(beta^*alphabeta,beta)$$. Equivariance is straightforward to verify.

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