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Suppose $M$ is a surface and $g_0$, $g_1$ are two Riemannian metrics having the same constant Gaussian curvature $K$. Is there anything that can be said about the Gaussian curvature of $g_t := (1-t)g_0 + tg_1$ where $0 \leq t \leq 1$?

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  • $\begingroup$ I think it was asked before and the answer was "no." $\endgroup$ Aug 6, 2023 at 22:23
  • $\begingroup$ Any chance you can say a little about how this question arose and whether there is a conclusion you are hoping for? Locally, $g_1 = u^*g_0$ where $u$ is a diffeomorphism. Here's a suggestion: Let $h = g_1-g_0$. The difference between the connections is a tensor involving covariant derivatives of $h$, which in turn can be written in terms of second order covariant derivatives of $u$. The Gauss curvature will have a formula in terms of third order covariant derivatives of $h$. Maybe if you calculate the formulas, you might see something interesting there. Maybe what you're hoping for. $\endgroup$
    – Deane
    Aug 6, 2023 at 23:17

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Comment only.

Any chance you can say a little about how this question arose and whether there is a conclusion you are hoping for? Locally, $g_1 = u^*g_0$ where $u$ is a diffeomorphism. Here's a suggestion: Let $h = g_t-g_0$. The difference between the connections is a tensor involving covariant derivatives of $h$ with respect to the Levi-Civita connection of $g_0$, which in turn can be written in terms of second order covariant derivatives of $u$. The Gauss curvature of $g_t$ will have a formula in terms of the Gauss curvature of $g_0$ and third order covariant derivatives of $u$. Maybe if you calculate the formulas, you might see something interesting there. Maybe what you're hoping for.

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