Exercise 8.4(a) of Do Carmo, Riemannian Geometry

Question

Identify $\mathbb R^4$ with $\mathbb C^2$ by letting $(x_1,x_2,x_3,x_4)$ correspond to $(x_1+ix_2,x_3+ix_4)$. Let $$ S^3 = \{ (z_1,z_2) \in C^2 \mid |z_1|^2+|z_2|^2=1 \}, $$ and let $h : S^3 \to S^3$ be given by $$ h(z_1,z_2)=(e^{\frac{2\pi i}q}z_1,e^{\frac{2\pi i r}q}z_2), \quad (z_1,z_2) \in S^3, $$ where $q$ and $r$ are relatively prime integers and $q > 2$.

(a) Show that $G = \{\text{id},h,\ldots,h^{q-1}\}$ is a group of isometries of the sphere $S^3$, with the usual metric, which operates in a totally discontinuous manner. The manifold $S^3 / G$ is called a lens space.

(b) Consider $S^3 / G$ with the metric induced by the projection $p : S^3 \to S^3 / G$. Show that all the geodesics of $S^3 / G$ are closed but can have different lengths.

My question comes from only part (a). I was already able to show that $G$ is a group which operates in a totally discontinuous manner. So my only question is how to show that $h : S^3 \to S^3$ is an isometry.

Intuitively, I can see that it is an isometry because the moduli of both $(z_1,z_2)$ and $h(z_1,z_2)$ are the same.

But I have trouble trying to show this explicitly, with the definition of isometry given by p. 35 of do Carmo:

Let $M$ and $N$ be Riemannian manifolds. A diffeomorphism $f : M \to N$ is called an isometry if $$ \langle u,v\rangle_p = \langle df_p(u),df_p(v)\rangle_{f(p)} $$ for all $p \in M$ and $u,v\in T_pM$.

Answer 1

If $k \in \{0,\ldots,q-1\}$ and if $u,v \in T_pS^3 \subset \mathbb C^2$ are given by $$ u=(u_1,u_2)=(u^1_1+iu^2_1,u^1_2+iu^2_2) \quad \text{and} \quad v=(v_1,v_2)=(v^1_1+iv^2_1,v^1_2+iv^2_2) $$ and we define $\alpha=\frac{2\pi k}q$ and $\beta=\frac{2\pi k r}q$, then \begin{align} (e^{ik\alpha}u_1, e^{ik\beta}u_2) &= ((\cos(k\alpha)+i\sin(k\alpha))(u_1^1+iu_1^2),(\cos(k\beta)+i\sin(k\beta))(u_2^1+iu_2^2)) \\ &= (u_1^1 \cos (k\alpha)-u_1^2 \sin(k\alpha))+i(u_1^2 \cos (k\alpha)+u_1^1 \sin(k\alpha)), \\ &\qquad \qquad (u_2^1 \cos (k\beta)-u_2^2 \sin(k\beta))+i(u_2^2 \cos (k\beta)+u_2^1 \sin(k\beta))) \\ &= (u_1^1 \cos (k\alpha)-u_1^2 \sin(k\alpha),u_1^2 \cos (k\alpha)+u_1^1 \sin(k\alpha),\\ &\qquad \qquad u_2^1 \cos (k\beta)-u_2^2 \sin(k\beta),u_2^2 \cos (k\beta)+u_2^1 \sin(k\beta)), \end{align} where the last equality was due to associating $\mathbb R^4$ with $\mathbb C^2$.

Similarly, \begin{align} (e^{ik\alpha}v_1, e^{ik\beta}v_2) &= (v_1^1 \cos (k\alpha)-v_1^2 \sin(k\alpha),v_1^1 \cos (k\alpha)+v_1^1 \sin(k\alpha),\\ &\qquad \qquad v_2^1 \cos (k\beta)-v_2^2 \sin(k\beta),v_2^2 \cos (k\beta)+v_2^1 \sin(k\beta)). \end{align} Therefore, \begin{align} \langle dh^k(u),dh^k(v) \rangle_{h^\ell(p)} &= \langle(e^{ik\alpha}u_1, e^{ik\beta}u_2),(e^{ik\alpha}v_1, e^{ik\beta}v_2)\rangle \\ &= (u_1^1 \cos (k\alpha)-u_1^2 \sin(k\alpha))((v_1^1 \cos (k\alpha)-v_1^2 \sin(k\alpha)) \\ &\qquad + (u_1^2 \cos (k\alpha)+u_1^1 \sin(k\alpha))(v_1^2 \cos (k\alpha)+v_1^1 \sin(k\alpha)) \\ &\qquad + (u_2^1 \cos (k\beta)-u_2^2 \sin(k\beta))(v_2^1 \cos (k\beta)-v_2^2 \sin(k\alpha)) \\ &\qquad + (u_2^2 \cos (k\beta)+u_2^1 \sin(k\beta))(v_2^2 \cos (k\beta)+v_2^1 \sin(k\alpha)) \\ &= u_1^1v_1^1 + u_1^2 v_1^2 + u_2^1 v_2^1 + u_2^2 v_2^2 \\ &= \langle (u_1^1,u_1^2,u_2^1,u_2^2),(v_1^1,v_1^2,v_2^1,v_2^2) \rangle \\ &= \langle (u_1^1+iu_1^2,u_2^1+iu_2^2),(v_1^1+iv_1^2,v_2^1+iv_2^2) \rangle \\ &= \langle (u_1,u_2),(v_1,v_2) \rangle \\ &= \langle u,v \rangle \end{align} and hence $h^k : S^3 \to S^3$ is an isometry.

Answer 2

Let $\alpha=\frac{2\pi}{q}$ and $\beta=\frac{2\pi r}{q}$, and so $h^k(z_1,z_2)=\left(e^{ik\alpha}z_1,~e^{ik\beta}z_2\right),$ and in turn $$dh^k_{(z_1,z_2)}=\left(e^{ik\alpha}dz_1,e^{ik\beta}dz_2\right),$$ where $dz_i=dx_i + idy_i$.

One of your mistakes is in the expression of the differential. In your attempt you didn't apply the same linear operator (differential) to both $u$ and $v$.

Now for any $p\in\mathbf{S}^3$ and $u=(u_1,u_2),v=(v_1,v_2)\in T_p\mathbf{S}^3$, where $u_j=u_{j}^1+iu_{j}^2$ and $v_j=v_{j}^1 + i v_j^2$ for $j\in\{1,2\}$. Now recall that since we identified $\mathbf{R}^4$ with $\mathbf{C}^2$ that the inner product is the inner product on $\mathbf{R}^4$, not the natural Hermitian inner product on $\mathbf{C}^2$ (so you can't pull out the complex constants $e^{ik\alpha}$; on that note, even if the metric was Hermitian you would have to pull out the conjugate of $e^{ik\beta}$ from the right hand side of the inner product). Keeping this in mind we compute \begin{align*} \langle dh^k(u),dh^k(v)\rangle_{h^k(p)} & = \left\langle\left(e^{ik\alpha}u_1,~ e^{ik\beta}u_2\right),\left(e^{ik\alpha}v_1,~ e^{ik\beta}v_2\right)\right\rangle\\ & = \left\langle\left((u_1^1+u_1^2)(\cos(k\alpha) + i\sin(k\alpha)),~(u_2^1+ u_2^2)(\cos(k\beta)+i\sin(k\beta)\right),\right.\\ &\qquad\left.\left((v_1^1+v_1^2)(\cos(k\alpha) + i\sin(k\alpha)),~(v_2^1+ v_2^2)(\cos(k\beta)+i\sin(k\beta)\right)\right\rangle \\ & = (u_1^1+u_1^2)(v_1^1+v_1^2)\cos^2(k\alpha)+(u_2^1+u_2^2)(v_2^1+v_2^2)\cos^2(k\beta) \\ &\qquad + (u_1^1+u_1^2)(v_1^1+v_1^2)\sin^2(k\alpha)+(u_2^1+u_2^2)(v_2^1+v_2^2)\sin^2(k\beta) \\ & = (u_1^1+u_1^2)(v_1^1+v_1^2)+(u_2^1+u_2^2)(v_2^1+v_2^2) \\ & = \langle u,v\rangle. \end{align*} So we see that $h^k$ is indeed an isometry. Let me know if anything isn't clear.

Answer 3

Another way : We can extend $h$-action on $\mathbb{R}^4$. That is, $h$ can be viewed as an element in $SO(4)$. Note that $S^3$ is invariant under $h$. Since $S^3$ has a metric induced from $\mathbb{R}^4$, for $x,\ y\in S^3$, there are curves $c_i$ in $S^3$ from $x$ to $y$ s.t. $$ d_{S^3}(x,y) =\lim_i\ {\rm length}\ c_i $$

And since $h$ is an isometry, $$ d_{S^3} (h(x),h(y)) \leq \lim_i\ {\rm length}\ h\circ c_i =d_{S^3}(x,y) \ \ast$$

If $c$ is a curve in $S^3$ from $h(x)$ to $h(y)$ s.t. ${\rm length}\ c< d_{S^3}(x,y) $, then $h^{-1}\circ c$ is a curve from $x$ to $y$.

Hence there is no such curve $c$ so that an inequality in $\ast$ is an equality.