9

Not necessarily. Take small $\delta>0$. Spend time $\frac 23(1-\delta)$ at $(0,1)$. Then, in time $\delta/4$ travel the route $(0,1)\to(0,2)\to(-1,2)\to(-1,-2)$ so that the integral of the vertical coordinate is $0$. Then stay at $(-1,-2)$ for the time $\frac 16(1-\delta)$ and return to $(0,1)$ running the same movement backwards. Make a symmetric (with ...


5

If I understand your question correctly, the answer is affirmative for Ricci-flat (i.e. flat) surfaces (not necessarily embedded in a Euclidean space) according to equation (14) of Reilly, Robert C., Applications of the Hessian operator in a Riemannian manifold, Indiana Univ. Math. J. 26, 459-472 (1977). ZBL0391.53019. which is available at http://www....


1

The desired relation can be written in components as $$f_n(\mathbf{z}^{(1)},\mathbf{\bar z}^{(1)})-f_n(\mathbf{z}^{(2)},\mathbf{\bar z}^{(2)})=\int_0^1 d\tau\,\sum_{m}\left(\frac{\partial f_n}{\partial z_m}(z_m^{(1)}-z_m^{(2)})+\frac{\partial f_n}{\partial \bar{z}_m}(\bar{z}_m^{(1)}-\bar{z}_m^{(2)})\right),\;\;[1]$$ with the prescribed $\tau$ dependence: $$\...


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