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Consider the $n$-sphere $$ S^n = \{x\in\mathbb{R}^{n+1}: 1 - \sum_{k=1}^{n+1} x_k^2 = 0\}, $$ and let $g_1$ be the induced metric. Given $\lambda\in\mathbb{R}^{n+1}_{>0}$, we have the ellipsoid $$ E_\lambda \longrightarrow \{x\in\mathbb{R}^{n+1}: 1 - \sum_{k=1}^{n+1} \lambda_k x_k^2 = 0\}. $$ Let $g_\lambda$ be the induced metric on $E_\lambda$. A basic fact is that $S^n$ and $E_\lambda$ are diffeomorphic, with the map $\psi_\lambda : S^n \longrightarrow E_\lambda$ given by $$\psi_\lambda(x_1,\ldots, x_{n+1}) = \left(\frac{x_1}{\sqrt{\lambda_1}},\ldots, \frac{x_{n+1}}{\sqrt{\lambda_{n+1}}}\right).$$

Now let $H=\text{diff}(S^n)$ be the diffeomorphism group of $S^n$, a group whose definition is independent of the choice of metric on the sphere. $H$ acts (on the right) on the space of Riemannian metrics on $S^n$ via pullback: if $g$ is a metric on $S^n$, then $\varphi^\ast g$ is another Riemannian metric. I am interested in the orbit of $g_1$ under the action of $H$ on the space of Riemannian metrics.

Question: Does there exist $\varphi\in\text{diff}(S^n)$ such that $\varphi^\ast g_1$ is equal to $\psi_\lambda^\ast g_\lambda$?

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Answer: Not unless all of the $\lambda_i >0$ are equal to $1$ (or else, when $n=1$ and $\lambda_1$ and $\lambda_2$ are chosen so that the length of the curve $E_\lambda$ is $2\pi$; the condition for this is given by evaluating an elliptic integral).

For $n>1$, this follows from a calculation of the eigenvalues of the Riemann curvature tensor of $g_\lambda$ which is a diffeomorphism invariant.

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  • $\begingroup$ Thank you! My naive suspicion was that the answer was no, but I wasn't sure about the right way to go about justifying it. $\endgroup$
    – user41626
    Aug 17 '15 at 7:46

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