# Conditions for tubular hypersurfaces to be a Riemannian product

Let $(M,g)$ be a Riemannian manifold of dimension $n$ and $P$ a submanifold of dimension $k.$ Let us define the tube of radius $r$ about $P$ by

$$T(P,r):=\{x\in M: d(x,P)\le r\}$$ and the tubular hypersurface at distance $t$ from $P$ by

$$P_t=\{x\in T(P,r):d(x,P)=t\}.$$

Let $p\in P$ be a point of the submanifold, $u\in T^{\perp}P$ a unit vector field and $\gamma(t)=exp_p(tu)$ the geodesic defined by $\gamma(0)=p$ and $\gamma'(0)=u.$

Let $\omega$ be the volumen element of $M,$ $dP$ that of $P,$ and $du$ that of the unit sphere $S^{n-k-1}.$ The infinitesimal change of volume function in the direction $u$ is the real function $\theta_u(t)$ defined by $\omega(\gamma(t))=\theta_u(t) du\wedge dP\wedge dt.$

It is knwon that $\theta_u$ satisfies

$$\frac{\theta_u'(t)}{\theta_u(t)}=-\left(\frac{n-k-1}{t}+\mathrm{tr}(S(t))\right),$$ where $S(t)$ is the second fundamental form of the hypersurface $P_t.$

It is known $\theta_u(0)\equiv 1$ for any $u.$ Can we say something if $\theta_u(t)\equiv 1$ independently of $t$ and $u?$ That is, if $\mathrm{tr}(S(t))=\frac{n-k-1}{t}$ independently of $t$ and $u.$ This would imply $\omega(\gamma(t))= du\wedge dP\wedge dt.$

The only example I can think of, satisfying these conditons, is when the tubular hypersurfaces are Riemannian products, that is, $P_t=P\times \mathbb{S}^{n-k-1}(t).$ To me the assumptions are very strong, and I think this could be the only possibility, at least locally. But, is this the only possible situation? Or is it a naive idea to expect such a situation? Any hint to show that tubular hypersurfaces must be Riemannian products or to construct a counterexample is welcome.

## 1 Answer

Here is the counterexample. Consider the metric on the tangent bundle $TS^2$ of the 2-sphere, given by the Riemannian submersion $S^3\times (R^2,g) \to S^3\times (R^2,g)/S^1$, where $S^3$ is the 3-sphere of unit quaternions, and $S^1$ acts on it (as usual) by multiplications by unit complex numbers (same action on $R^2$ identified with $C$). Submanifold $P(0)$ is the zero-section of $TS^2$ diffeomorphic to $S^2$, while the boundary $P(t)$ of its tubular neighborhood is diffeomorphic to a 3-dimensional sphere (which is already, not a direct product). If we choose the metric $g$ on $R^2$ in polar coordinates $(t,\phi)$ given by $dt^2 + t^2(1+4t^2)d\phi^2$, then the fibers (images of $(R^2,g)$ under submersion above) of the metric projection $TS^2$ to its zero section $P(0)$ (which is another Riemannian submersion) are totally geodesic submanifolds isometric to flat plane, and spheres $S^{n-k-1}(t)=S^1(t)$ would be of length $2\pi t$, so that the condition $\theta_u(t)\equiv 1$ holds. Also $P(t)$ (which is diffeomorphic to $S^3$, homogeneous and of positive sectional curvature) is not a direct product even locally, since the holonomy of the bundle $TS^2$ is not trivial.

• Could you explain me a bit why the boundary $P(t)$ is diffeomorphic to a $3$-dimensional sphere? – mfl Mar 21 '15 at 21:40
• @mfl P(t) is known as Berger sphere (kind of), may be [Cheeger, Some examples of manifolds ... J Diff G v8] consider it. To prove that P has positive curvature, take two points p(t) and q(t) over p(0) on zero section S^2 and two vectors X_p(t) and X_q(t) (horizontal lifts of X tangent to S^2 at p(0)) Now "move" with a parameter s the triangle p(0)p(t)q(t) in directionS X, X_p, X_q along three geodesics - you have a prism = family of isometric triangles due to the construction. Since the upper side p(t,s)q(t,s) has constant length on s - dirst and second variation formulas imply that the – valeri Mar 21 '15 at 23:51
• curvature of P(t) in 2-direction (X_p, W) (W - direction of the upper side p(t)q(t)) vanishes only if the holonomy of X along the triangle p(0)p(t)q(t) vanishes - but the last is given as t->0 by the same curvature component R(X,Y)V,W as the holonomy of the tangent bundle (see Ambrose-Singer formula), which is constant non-zero. Mean, that P(t) is 3-sphere for small t. QED. Yet another way - you may employ [O'Neil ... 1966?] formulas for the submersion directly. – valeri Mar 22 '15 at 0:01
• One more way - prove that P(t) is simply connected. In order to do this, note that the parallel transport of vertical V=p(0)p(t) along big geodesic l(s) bounding half-sphere B in zero section S^2 gives again V. Now take the center O of B and transport parallel V first along l till l(s) and then along radius to O. The obtained family V*(s) gives you a circle S^1(t) over O, proving that vertical S^1 id homotopic to l in S^2 - null homotopic... – valeri Mar 22 '15 at 0:09
• @mfl I did not included the (obvious) proof that 3-dim P(t) of positive curvature, simply connected is S^3, is it ok? – valeri Mar 22 '15 at 8:37