Map \begin{align*} \phi:\qquad S^2\quad &\longrightarrow\quad\mathbb{R}^4\\ (x, y, z)\quad &\longmapsto\quad (\frac{x^2-y^2}{2}, xy, yz, zx) \end{align*} is an immersion. Take the standard metric on $ \mathbb{R}^4 $: $ \operatorname{d}s^2=\operatorname{d}x_1^2+\operatorname{d}x_2^2+\operatorname{d}x_3^2+\operatorname{d}x_4^2 $ then $ \phi $ induces a Riemannian metric on $ S^2 $ \begin{align*} \phi^*\operatorname{d}s^2&=\operatorname{d}\phi_1^2+\operatorname{d}\phi_2^2+\operatorname{d}\phi_3^2+\operatorname{d}\phi_4^2\\ &=(x\operatorname{d}x-y\operatorname{d}y)^2+(y\operatorname{d}x+x\operatorname{d}y)^2\\ &\quad +(z\operatorname{d}y+y\operatorname{d}z)^2+(z\operatorname{d}x+x\operatorname{d}z)^2\\ &=\operatorname{d}x^2+\operatorname{d}y^2+(x^2+y^2-2z^2)\operatorname{d}z^2. \end{align*}

How to compute the Gauss curvature of $ S^2 $ regarding the metric induced by $ \phi $?

I have tried to use the stereographic projection: \begin{align*} X_1:\qquad\mathbb R^2\quad &\longrightarrow\quad S^2-\{ (1, 0, 0) \} \\ (u, v)\quad&\longmapsto\quad\frac{1}{1+u^2+v^2}(u^2+v^2-1, 2u, 2v) \end{align*} and compute the first and the second fundamental forms to gain the Gauss curvature.

But, the huge amount of computations made me quit doing this. I want to know whether there is another way to do this?


The trick is indeed to use a parametrisation of $S^2$. Instead of using the stereographic projection, I think the usual $$ \begin{align*} x &= \cos u \sin v \\ y &= \sin u \sin v \\ z &= \cos v, \end{align*} \qquad \qquad \text{$u \in (0,2\pi)$, $v \in (0,\pi)$,} $$ will make things a bit easier. The map $\phi$ becomes $$ \phi(u,v) = \frac{1}{2}\left(\cos (2 u) \sin ^2v,\sin (2 u) \sin ^2v, \sin u \sin (2 v),\cos u \sin (2 v)\right). $$ Then the metric becomes $\phi_u \cdot \phi_u = \sin^2 v$, $\phi_u \cdot \phi_v=0$ and $\phi_v\cdot\phi_v = \tfrac{1}{8}(5+3\cos (4v))$. The calculations will not go superswift, but will this be more manageable for you?

  • $\begingroup$ @Philip Did you manage to calculate the Gaussian curvature from the metric? $\endgroup$ – Ernie060 Oct 28 '18 at 17:43

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