Riemannian metric making a submanifold totally geodesic and with positive injectivity radius 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.
 A: 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.
A: Not clear fully to me; There should be common tangentiality along with tangential curvature continuity ? like a contacting cylinder-ellipsoid pair with common geodesics ?

