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Let $(M,g)$ be a Riemannian manifold. Let $S_g$ be the corresponding Sasaki metric on $TM$. For every $p\in M$, $V_p\in T_pM$, is it true and obvious that $0_p$ is the closest point of the zero section to $V_p$?

With some abuse of terminology a rephrase of the question would be: Is the height of a right triangle shorter than its hypotenuse?

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    $\begingroup$ By Sasaki metric you mean the one using the Levi-Civita connection, correct? $\endgroup$
    – Gabe K
    Dec 10, 2020 at 1:08
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    $\begingroup$ Yes. Consider the function $f\colon V\mapsto |V|$. Note that $\nabla_Vf$ has vertical part $V/|V|$ if $V\ne0$ and its horizontal part vanishes. Make a conclusion. $\endgroup$
    – Anton Petrunin
    Dec 10, 2020 at 5:44
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    $\begingroup$ @LSpice thank you for your edit! $\endgroup$
    – Ali Taghavi
    Dec 10, 2020 at 11:38
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    $\begingroup$ @AliTaghavi: In response to your question about the more general case of a Riemannian submersion (and where I assume that you meant $p=v$ and $q=w$), the answer is a definite 'no', even in the case that the source and target of the Riemannian submersion are complete. Just take the standard flat metric on $M = \mathbb{R}^2/\mathbb{Z}$ where the action is $n\cdot(x,y) = (\,x+2\pi\,n,\, (-1)^ny\,)$ and the submersion is $\sigma([x,y]) = \mathrm{e}^{ix}$ and look at the shortest path from $[0,a]$ to $[0,-a] = [2\pi, a]$ for $a>\pi$. $\endgroup$
    – Robert Bryant
    Dec 15, 2020 at 14:02
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    $\begingroup$ @AliTaghavi: Yes, there is an orientable counterexample. Just let $M=\mathbb{R}^3/\mathbb{Z}$ with action $n\cdot(x,y,z) = (x{+}2\pi\,,\,(-1)^ny\,,\,(-1)^nz\,)$ and let the submersion $\sigma:M\to S^1$ be $\sigma([x,y,z])=\mathrm{e}^{ix}$. The same construction works with $p=[0,a,0]$ and $q = [0,-a,0]=[2\pi,a,0]$, but now $M$ is orientable. $\endgroup$
    – Robert Bryant
    Dec 18, 2020 at 1:15

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Let me expand my comment to remove the question from unanswered.

Any point in $w\in\mathrm{T}M$ is a pair $w=(V,p)$ where $p\in M$ and $V\in\mathrm{T}_pM$.

Let $t\mapsto w(t)=(V(t),\gamma(t))$ be a curve in $\mathrm{T}M$. Let $V=V(0)$ and $p=\gamma(0)$. Note that $$w'(0)=\nabla_{\gamma'(0)}V\oplus \gamma'(0)\in \mathrm{T}_{p}M\oplus\mathrm{T}_{p}M=\mathrm{T}_{(V,p)}\mathrm{T}M.$$

Therefore $$ \begin{aligned} \langle V\oplus 0,w'(0)\rangle &=\langle V,\nabla_{\gamma'(0)}V\rangle= \\ &=\tfrac12\cdot\langle V,V\rangle'(0). \end{aligned} $$ It follows that $(V,p)\mapsto V\oplus 0$ is the gradient of the function $f\colon(V,p)\mapsto \tfrac12\cdot\langle V,V\rangle$. Whence the statement follows.

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  • $\begingroup$ Thank you very much Anton for your very interestinganswer. Sorry my late attention $\endgroup$
    – Ali Taghavi
    Jan 24, 2023 at 18:37

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