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 =
-1 & 0 & 0 \
0 & 0 & 1 \
0 & 1 & 0

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.

One Answer

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.

Correct answer by Kajelad on November 9, 2020

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