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. 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 welcomed.

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.