# Existence of geometric Tubular Neighborhoods in Finsler spaces

I have not found any reference among the well-known books about the existence of a geometric tubular neighborhood in the Finsler spaces. I am wondering if there exists such a neighborhood for any closed submanifold of a Fisnler Manifold $M$ (maybe with some extra hypothesis on $M$ ,like compactness ....).

FYI: by a geometric tubular neighborhood, of a pre-compact submanifold of a manifold $M$, I mean: $Tub(P)_r:=\{\gamma(1)|\gamma:[0,1]\longrightarrow M$ is a minimizing geodesic with $\gamma'(0)\in\mathfrak{C}_{\gamma(0)}(P)\cap B(r)\}$. where $B(r)$ is the ball of radius $r$ and by $\mathfrak{C}_x$ we mean the subset of vectors that each one is orthogonal to $T_xP$ at the direction of itself.

• What's your definition of a tubular neighborhood? – Deane Yang May 22 '17 at 16:18
• The tubular neighborhood theorem holds quite generally in the category of smooth manifolds. Unless you want special compatibility conditions with the Finsler structure, the result is the same. – Willie Wong May 22 '17 at 16:58
• In which case, Willie's answer applies. The definition does not depend on any metric or other geometric structure in addition to the manifold structure itself. – Deane Yang May 23 '17 at 1:31
• @WillieWong Notice that in Finsler spaces we always should be careful with the direction. It meansif a vector $v$ is orthogonal to a submanifold in some direction, $-v$ may not be orthogonal to that submanifold. So for a submanifold $L$ of $M$, the subset $\mathfrak{C}_x\subset T_xM$ that contains the vectors orthogonal to $L$ at point $x$, may not be a vector space (We say that a vector $v$ is orthogonal to $L$ if it is orthogonal at the direction of itelf.). As a result we may not have any decomposition as $T_xM=T_xL\bigoplus \mathfrak{C}_x$ and so any normal bundle. – Majid May 23 '17 at 1:38
• The definition in en.wikipedia.org/wiki/Tubular_neighborhood makes no reference to and does not need orthogonality. A submanifold in a Finsler manifold $M$ has a tubular neighborhood, because $M$ is a smooth manifold. The Finsler metric isn't needed at all. If a manifold has a distance function defined on it, then you can define a tubular neighborhood as the set of all points within distance $\delta$ of the submanifold. That works for a Finsler manifold, because the Finsler metric defines a distance function. – Deane Yang May 23 '17 at 3:02

Here's a possible approach that goes backwards, where we define the tubular neighborhood first and the normal bundle second.

Let $M$ be a Finsler manifold and $S \subset M$ a smooth submanifold. Given any $x \in M$, we can define the distance $d(x,S)$ from $x$ to $S$ to be the shortest length of curves from $x$ to $S$. We'll call a curve segment $S$-minimizing if one endpoint lies in $S$ and the length of the curve equals the distance from the other endpoint to $S$.

Then one can define a tubular neighborhood of $S$ of radius $r$ to be the set of all possible endpoints of $S$-minimizing geodesics. However, this is simply the set of all points within distance $r$ from $S$.

We can now define a subset $N_*S \subset T_*M$ with respect to the Finsler metric as follows: $v \in T_pM$ lies in $N_pS$ if there exists an $S$-minimizing curve starting at $p$ such that $v$ is tangent to the curve at $p$.

Using the existence and uniqueness of a geodesic with given starting point and velocity, we can define the exponential map $e: N_*S \rightarrow M$.. Moreover, this should show that $N_*S$ is in fact a vector bundle over $S$ and the exponential map defines a diffeomorphism from a neighborhood of the zero section of $N_*S$ onto a neighborhood of $S$ in $M$. In particular, given any precompact subset of $S$, there exists $r>0$ such that the exponential map is a diffeomorphism onto the tubular neighborhood of radius $r$ from its preimage.

• As I already mentioned, $N_pS$ is not necessarily a vector space and so $N_* S$ is not a vector bundle. Moreover, the exponential map in the case of Finsler is not smooth at zero. – Majid May 23 '17 at 21:55
• As far I know, the exponential map is not smooth only along $S$ itself, and this is due to the geodesics coming in on opposite directions not meeting at $S$ smoothly. It will, however, be $C^1$. That's also presumably why $N_*S$ is not a vector bundle. It's really a "ray" bundle, where there is scalar multiplication but not vector adition. – Deane Yang May 23 '17 at 23:01
• As you mentioned the exponential map is not a diffeomorphism in a neighborhood of the zero section and so we cannot accept this as an answer. – Majid Jun 22 '17 at 23:56