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We consider the map $\varphi_0:\mathbb{S}^2\rightarrow\mathbb{R}^3$ such that $\varphi_0(\theta_1,\theta_2)=R(\cos\theta_1\cos\theta_2,\cos\theta_1\sin\theta_2,\sin\theta_1)$. I know that the induced riemmannian metic is

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From this, how can I obtain that the second fundamental form is

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and mean curvature H=-2/R?

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    $\begingroup$ The second fundamental form cannot be obtained from the metric. $\endgroup$ Jun 25, 2023 at 10:38
  • $\begingroup$ But I know that the second fundamental form is $h_{ij}X^iY^i$ where $ h_{ij}=g\bigg(\nu,\frac{\partial^2\varphi}{\partial x^i\partial x^j}\bigg)$ . ($\nu$ is the normal vector). Even applying this formula, however, I am unable to obtain what is written above. $\endgroup$
    – Seurat
    Jun 25, 2023 at 10:49
  • $\begingroup$ That g is not the Riemannian metric of the surface, since nu is not a tangent vector. $\endgroup$ Jun 25, 2023 at 11:12

1 Answer 1

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You need to be able to compute the normal vector $n$ first, where $n$ is the unit vector along $\frac{\partial\varphi}{\partial x^1}$ x $\frac{\partial\varphi}{\partial x^2}$. Then by definition, $h_{ij}$ would be the inner product of $\frac{\partial^2\varphi}{\partial x^i\partial x^j}$ and $n$. The mean curvature is just half the trace of the Weingarten Map matrix, which can be obtained by multiplying the matrix of the second fundamental from and the inverse of the metric matrix.

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