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Show that $\mathbf x:\mathbb R^2 \to \mathbb R^4$ given by $$ \mathbf x(\theta,\varphi)=\frac 1{\sqrt 2}(\cos \theta,\sin \theta,\cos \varphi,\sin\varphi), \quad (\theta,\varphi) \in \mathbb R^2 $$ is an immersion of $\mathbb R^2$ into the unit sphere $S^3(1) \subset \mathbb R^4$, whose image $\mathbf x(\mathbb R^2)$ is a torus $T^2$ with sectional curvature zero in the induced metric.

How can I complete the part of the proof that the torus $T^2$ has sectional curvature of zero?

I had the following from Gauss' Theorem, with $x=\frac{\partial}{\partial \theta}$ and $y=\frac{\partial}{\partial \varphi}$: \begin{align} &K\left(\frac{\partial}{\partial \theta},\frac{\partial}{\partial \varphi}\right)-\overline K\left(\frac{\partial}{\partial \theta},\frac{\partial}{\partial \varphi}\right) \\ &\qquad = \left\langle B\left(\frac{\partial}{\partial \theta},\frac{\partial}{\partial \theta}\right),B\left(\frac{\partial}{\partial \varphi},\frac{\partial}{\partial \varphi}\right) \right\rangle -\left|B\left(\frac{\partial}{\partial \theta},\frac{\partial}{\partial \varphi}\right) \right|^2 \\ &\qquad =\left\langle \overline \nabla_{\overline{\frac{\partial}{\partial \theta}}} \overline{\frac{\partial}{\partial \theta}}-\nabla_{\frac{\partial}{\partial \theta}} \frac{\partial}{\partial \theta}, \nabla_{\overline{\frac{\partial}{\partial \varphi}}} \overline{\frac{\partial}{\partial \varphi}}-\nabla_{\frac{\partial}{\partial \theta}} \frac{\partial}{\partial \varphi} \right\rangle - \left|\nabla_{\overline{\frac{\partial}{\partial \theta}}} \overline{\frac{\partial}{\partial \varphi}}-\nabla_{\frac{\partial}{\partial \theta}} \frac{\partial}{\partial \varphi}\right|^2 \end{align} And I think the curvatures $\overline K$ and $K$ are both identically zero since $\mathbf x$ is an immersion between Euclidean spaces.

I think this should help me prove that the sectional curvature of the torus $T^2$ is also zero, but I'm not sure how to proceed from here.

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  • $\begingroup$ For the record, I already proved that $\mathcal x : \mathbb R^2 \to \mathbb R^4$ is an immersion into the unit sphere $S^3(1)$, by showing that $d\mathcal x$ is injective and that $\|\mathbf x\|=1$. $\endgroup$ Jun 16, 2017 at 5:06
  • $\begingroup$ You can finish up the argument by a direct computation. Just consider the extensions $\overline{\partial_\theta} = \frac{1}{\sqrt2}\left(-x_2\frac{\partial}{\partial x_1} + x_1\frac{\partial}{\partial x_2}\right)$, and $\overline{\partial_\varphi}=\frac{1}{\sqrt2}\left( -x_4\frac{\partial}{\partial x_3} + x_3\frac{\partial}{\partial x_4}\right)$, and use basic facts about the connection (i.e. it is tensorial in the lower index, $\mathbf{R}$-linear and satisfies a product rule in the upper index). $\endgroup$ Jun 16, 2017 at 10:05
  • $\begingroup$ @YousufSoliman What about the $\partial_\theta = \nabla_{\frac{\partial}{\partial \theta}}$ and $\partial_\varphi = \nabla_{\frac{\partial}{\partial \varphi}}$ terms (the ones without the bar)? Are they zero because $\theta=\theta(t)$ and $\varphi=\varphi(t)$ are straight lines (geodesics) in $\mathbb R^4$? $\endgroup$ Jun 16, 2017 at 18:01

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An induced metric on $T$ is in fact an induced metric on $T$ from $\mathbb{R}^4$. That is $T=S^1(\frac{1}{\sqrt{2}} )\times S^1 (\frac{1}{\sqrt{2}})\subset \mathbb{R}^4$. Since it is a product, so it has a sectional curvature $0$.

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