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I have a Riemannian manifold $(X,g)$ which is compact, simply connected and with sectional curvature upper bounded by $k>0$ everywhere. Let $p\in X$ be any point and $q\in Cut(p)$ the nearest cut point (i.e. $d_g(p,q)=d_g(p,Cut(p))$), so I know that $q$ is a conjugate point or I have a closed geodesic $\gamma$ respect to which $p$ and $q$ are 'opposite' (i.e. $d_g(p,q)=\frac{1}{2}length(\gamma)$).

Now, I want to show that also in the second case $q$ is a conjugate point of $p$, or in alternative that $length(\gamma)\ge \frac{2\pi}{\sqrt{k}}$. In the second case clearly it's enough to prove that if $\gamma$ is the shortest closed geodesic of $X$ then $length(\gamma)\ge \frac{2\pi}{\sqrt{k}}$ (in this case one maybe can use Klingenberg's Lemma - see the version in 'Riemannian Geometry' of Gallot-Hulin-Lafontaine, pag.158).

Is it true what I want to show? How could I do?

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Weinstein proved that if $M$ is any closed manifold other than $S^2$ (of any dimension), then $M$ has a metric and a point $p$ for which the conjugate locus and cut locus in $T_pM$ don't intersect. See jstor.org/stable/1970592?seq=1. Unfortunately, it's still possible that when exponentiated, the images intersect. None the less, this provides a lot of evidence that the answer should be "no, it's not true" –  Jason DeVito Jan 29 '13 at 22:04
    
I find the question hard to parse. "I know that A or B. I want to show that also in the second case A, or in alternative C. In the second case it's enough to prove D (in this case one maybe can use...)." Could you state exactly what you want to prove? –  user53153 Feb 2 '13 at 4:19
    
My final objective is to prove that in the mentioned hypothesis (first sentence) $diamX\ge\frac{\pi}{\sqrt{k}}$. For this, I want to prove that every couple of cut-points are far $\frac{\pi}{\sqrt{k}}$ at least. So I take a point $p$, its nearest cut-point ($q$) which is a conjugate point (for $p$) or there is a closed geodesic $\gamma$ respect to which $p$ and $q$ are 'opposite' (i.e. $d_g(p,q)=\frac{1}{2}length(\gamma)$) - the two statements are not mutually exclusive. By the hypothesis on the sectional curvature I know that if $q$ is conjugate point then $d_g(p,q)\ge\frac{\pi}{\sqrt{k}}$... –  AX.J Feb 4 '13 at 9:56
    
...so I want to prove that even in the second case $q$ is conjugate point for $p$. ok? - I don't know if this is true, so I said that in the second case another possibility is to show that $lenght(\gamma)\ge\frac{2\pi}{\sqrt{q}}$, which is weaker then '$q$ is always a conjugate point of $p$'. –  AX.J Feb 4 '13 at 10:02
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