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Inspired by this question entitled When does the shape operator commute with a derivative? we ask the following question:

Assume that $S,H$ are two surfaces whose corresponding Gauss maps are denoted by $n_S:S\to S^2, n_H:H\to S^2$ respectively.

Let $\phi:S\to H$ be a smooth map.

Under what conditions there exist a smooth map $\psi:S^2 \to S^2 $ with $n_H\circ \phi=\psi \circ n_S$ namely the following diagram commute:

$\require{AMScd}$\begin{CD} S @>\phi>> H\\ @V n_S V V @VV n_{H} V\\ S^2 @>>\psi> S^2 \end{CD}

Please compare the above diagram with the diagram in the above linked question. Are these two questions somehow equivalent?

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    $\begingroup$ Certainly they are not equivalent. In the linked question it is about pointwise behavior, and here you are asking about a global problem. $\endgroup$
    – Willie Wong
    Commented Sep 2, 2023 at 2:27
  • $\begingroup$ @WillieWong Can one find a possible reation between these two questions? $\endgroup$
    – Ali Taghavi
    Commented Sep 2, 2023 at 2:42
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    $\begingroup$ Not really. The best you can hope for is that an example satisfying your diagram will also satisfy the diagram in the linked question. But let $S$ be the unit sphere and $H$ be the sphere of radius 2, and $\phi$ be the obvious map. Then $\psi$ is the identity map. Taking the first jet of your diagram gives $$ \begin{CD} T_pS @>D\phi_p>> T_{\phi(p)}H\\ @V Dn_S V V @VV Dn_{H} V\\ T_pS^2 @>>\mathbf{Id}> T_pS^2 \end{CD}$$ But $D\phi_p = 2 \mathbf{Id}$ so the diagram in the linked question doesn't work. $\endgroup$
    – Willie Wong
    Commented Sep 2, 2023 at 3:01
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    $\begingroup$ In fact, if $S$ and $H$ are the smooth boundaries of two compact sets in $\mathbb{R}^3$, then the Gauss maps are smoothly invertible and your diagram can be formed. But for the linked question's diagram to hold, you need $S$ and $H$ to have the same second fundamental forms, and this won't hold for generic $S$ and $H$. $\endgroup$
    – Willie Wong
    Commented Sep 2, 2023 at 3:04
  • $\begingroup$ @WillieWong Yes I see thanks. $\endgroup$
    – Ali Taghavi
    Commented Sep 2, 2023 at 3:07

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