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Assume that $M$ is a surface in $\mathbb{R}^{3}$. We denote its shape operator by $S$. A vector field $X$ is shape related to $Y$ if $S(X)=Y$. (of course it is not an equivalent relation).

Assume that $X, Y$ are two vector fields on $S^{2}$ with the same singular points. Is there a surface $M$, diffeomorphic to $S^{2}$ via a diffeomorphism $\phi: M \to S^{2}$, such that $\phi^{*} (X)$ is shape related to $\phi^{*}(Y)$.

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  • $\begingroup$ What is a shape operator? $\endgroup$
    – Fan Zheng
    Commented Nov 30, 2015 at 20:03
  • $\begingroup$ @FanZheng If $M$ is a surface in $\mathbb{R}^{3}$ with Gauss normal vector $N$, the shape operator at each point $p\in M$ is a linear map $S$ on the tangent space $T_{p} M$ with $S(X)=\nabla_{X}^{N}$. Example: this operator is the identity operator for the round sphere. $\endgroup$
    – Ali Taghavi
    Commented Dec 1, 2015 at 7:07

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