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Consider a manifold $N$ and a submanifold $M$. We're assuming that $N$ and $M$ can be non-compact. Is it possible to find a metric $g$ of $N$ such that $M$ is totally geodesic with respect to $(N,g)$ and such that the metric $g$ has positive injectivity radius ?

This question came to me because I am trying to parametrize a infinite dimensional banach manifold of curves, hence I need the positive injectivity radius, and I also need that these maps satisfy specific boundary conditions, hence the totally geodesic condition. I have been thinking about this but I am have no idea on how to proceed.

Edit: This problem came to mind because I wanted to see if I could find a metric in $T^*M$ such that it has positive injectivity radius and with totally geodesic fibers. An idea would be to consider a metric in the conformal class of the Sasaki metric $g_S$. We know that the Sasaki metric has totally geodesic fibers. Then if we could consider something like $e^{2f}g_{S}$ such that it would have positive injectivity radius and $\text{grad}(f)$ is tangent to the fibers I belive we could consider this metric to get the desired result. However I have not yet been able to find this smooth function $f$.

Any help is appreciated, thanks in advance.

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    $\begingroup$ Can you write down a metric on a tubular neighborhood of $N$ that works? $\endgroup$ Dec 21, 2021 at 18:19
  • $\begingroup$ Yes I can construct one which has the property that $M$ will be totally geodesic , and then one can use partitions of unity. However I am not sure how I can ensure that the injectivity radius stays positive ? @TedShifrin $\endgroup$
    – whatever
    Dec 22, 2021 at 13:05
  • $\begingroup$ @Something: I believe Ted is asking you to construct a metric on the tubular neighborhood that already has positive injectivity radius. $\endgroup$ Dec 27, 2021 at 18:28
  • $\begingroup$ Also, do you assume $M$ is a closed subset of $N$? $\endgroup$ Dec 27, 2021 at 18:40
  • $\begingroup$ I see I guess I am not seeing how I can do that @JasonDeVito. I assume $M$ is closed in $N$ yes , but it does not need to be compact. $\endgroup$
    – whatever
    Dec 27, 2021 at 19:04

2 Answers 2

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Let $E\xrightarrow{\pi} M$ be a vector bundle. We will construct a metric on $E$ for which $M$ (viewed as the zero-section) is totally geodesic and for which the injectivity radius is positive.

We'll use a construction of Matthew Kvalheim (see here). Let $E'$ be a complementary vector bundle in the sense that $E\oplus E' \cong M\times \mathbb{R}^k$ for some $k$. (Such compliments exist even if $M$ is non-compact, see here.)

Let $g$ be a complete metric on $M$ with positive injectivity radius and bounded curvature. The existence of such metrics is shown in

R. E. Greene, Complete metrics of bounded curvature on noncompact manifolds, Archiv der Mathematik 31 (1978), no. 1, 89-95".

Consider a product metric on $M\times \mathbb{R}^k$. Pull this back to $E$ via the obvious inclusion $E\rightarrow M\times \mathbb{R}^k$ to get a metric $h$ on $E$.

I claim that $M$ (viewed as the zero section) is totally geodesic in $(E,h)$ and that the injectivity radius of $E$ is positive.

The first statement is easy: it's enough to show that a curve which minimizes distance in $M$ also minimizes in $E$. But $M$ is totally geodesic in $M\times \mathbb{R}^k$, so a minimizing curve in $M$ is minimizing even in the larger manifold $M\times \mathbb{R}^k$.

What about the injectivity radius? Well, as is well known, we have $inj(E) > \min\{ \pi/\sqrt{K}, \frac{1}{2} \ell(E)\}$ where $K$ is an upper bound for the curvature of $E$ and $\ell(E)$ denotes the length of the shortest closed geodesic. So we simply need to bound both.

For curvature, I first claim that $\pi$ is a Riemannian submersion. This follows because the projection $M\times \mathbb{R}^k\rightarrow M$ is obviously a Riemannian submersion, and the horizontal spaces of $M\times \mathbb{R}^k$ and $E$ agree. Since Riemannian submersions can only increase curvature (from O'Neill's formula), an upper bound for curvature in $(M,g)$ provides an upper bound for curvature in $(E,h)$. But, by assumption, $(M,g)$ has curvature bounded above.

What about a bound on $\ell(E)$? Well, first note that $\ell(M)$ is bounded below by a positive number because $M$ has positive injectivity radius.

Now, let $\gamma$ be a closed geodesic in $E$. Writing $\gamma(t) = (\alpha(t), v(t))$ for a curve $\alpha$ in $M$ and $v$ in $\mathbb{R}^k$, I claim that $\alpha$ must be a geodesic.

Indeed, writing $\nabla$ to denote the Levi-Civita connection, $\gamma' = (\alpha',v')$ and $\nabla^E_{\gamma'}\gamma' $ is simply the projection of $\nabla^{M\times \mathbb{R}^k}_{\gamma'}\gamma'$ to $E$. Because we have a Riemannian product on $M\times \mathbb{R}^k$, $\nabla^{M\times \mathbb{R}^k}_{\gamma'} \gamma' = \nabla^M_{\alpha'}\alpha' + v''$. As $\nabla^M_{\alpha'}\alpha'$ is horizontal and $v''$ is vertical, the only way the projection to $E$ can be zero is if both projections are zero. But the horizontal space projects isomorphically, so $\nabla^M_{\alpha'}\alpha' = 0$.

Now, because $\gamma$ is a closed geodesic, $\alpha$ must be closed as well. Thus, so long as $\alpha$ is a non-trivial, $\ell(\gamma)\geq \ell (\alpha) \geq \ell(M)$.

All that's left is the case that $\alpha$ is trivial, that is, that $\alpha$ is constant. In this case, $\gamma$ is a geodesic in the fiber in $E$ above $\alpha(0)$. But this fiber is is isometric to Euclidean space with flat metric, so has no closed geodesics.

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  • $\begingroup$ This is not a full answer to your question - it simply establishes that the normal bundle to $M$ in $N$ (which exists since $M$ is closed) has the kind of metric you want. I'm not sure how to propagate it to the full manifold while still keeping some kind of bound in the injectivity radius, but no matter how you propagate it, at least $M$ will still be totally geodesic, and the injectivity radius near $M$ will be positive. $\endgroup$ Dec 28, 2021 at 4:41
  • $\begingroup$ Thanks for the answer ! "Extending" the injectvity radius does seem to be a problem. However I am interested in the case where we are working with the cotangent bundle and it's fibers. Maybe there is something that can be done there. $\endgroup$
    – whatever
    Dec 28, 2021 at 10:52
  • $\begingroup$ Not clear to me; There should be common tangentiality along with tangential curvature continuity ? like a contacting sphere-cylinder pair with common geodesics ? $\endgroup$
    – Narasimham
    Jan 3 at 0:22
  • $\begingroup$ I am now analysing the answer again and I came up with a question. You claim that by O' Neill's formula an upper bound for the curvature of $(M,g)$ provides an upper bound for curvature in $(E,h)$, however I don't see why this is the case. O'Neill's formula only gives us information about the sectional curvature of the horizontal lifts of vector fields, we don't know how the sectional curvature on planes generated by vertical lifts or an horizontal and vertical lift is behaving . Or is there something that I am missing ? @JasonDeVito $\endgroup$
    – whatever
    Jan 16 at 10:47
  • $\begingroup$ @Something: I agree that this is a gap in the argument. The curvature of the vertical planes is easy to control since the fibers are totally geodesic, but I don't see how to control the vertizontal curvatures. $\endgroup$ Jan 17 at 16:58
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Not clear fully to me; There should be common tangentiality along with tangential curvature continuity ? like a contacting cylinder-ellipsoid pair with common geodesics ?

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