# Is the unit bundle of a Finsler vector bundle a sphere bundle?

Note: By now, I have asked this question also at mathoverflow.

Let $$E$$ be a Finsler vector bundle* of rank $$k$$ over a manifold $$M$$. Does the unit "bundle" $$UE$$ admit a structure of a sphere bundle? (If it matters, I am more interested in the case where $$M$$ is compact).

*I assume the Finsler metric is reversible (i.e, $$F(v)=F(-v)$$ or $$F$$ is a norm when restricted to each fiber).

What I know:

1. It is always an embedded submanifold of $$E$$ of codimension 1. (As inverse image of a regular value).

2. It is an $$\mathbb{S}^{k-1}$$-bundle when $$E$$ is Riemannian. (i.e the Finsler function comes from a metric on $$E$$). However, the way to construct the trivializations in the Riemannian case is via orthonormal frames, which doesn't have meaning for a general Finsler norm.

3. There is no "pointwise" obstruction for $$UE$$ to be a sphere bundle. For each $$p \in M$$, the 'fiber' $$(UE)_p = \{v_p \in E_p | F(v_p) = 1 \}$$ is diffeomorphic to the standard Euclidean sphere $$\mathbb{S}_{Euc}^{n-1}$$, since it's the unit sphere of a smooth* norm. (And every such sphere is diffeomorphic to the standard one, See proof below**).

Motivation:

I am trying to find out if compactness of $$M$$ implies compactness of $$UE$$. (Is it true?)

In the Riemannian case, I can use the fact that $$UE$$ is a sphere bundle over a compact base ($$M$$), and a fiber bundle with compact base and a compact model fiber ($$\mathbb{S}^{k-1}$$) is compact.

**Here is the following proof of the above claim:

*Definition: A norm on $$\mathbb{R}^n$$ is called smooth if its restriction to $$:\mathbb{R}^n \setminus \{0\} \to \mathbb{R}$$ is smooth.

(Note that no norm can be smooth on all $$\mathbb{R}^n$$, since this will imply smoothness of its induced metric which is impossible. Hence, this is the maximal degree of smoothness one can expect).

Lemma 1: All unit spheres of norms on a finite dimensional vector space are homeomorphic.

Proof:

Let $$\| \cdot \|_1 \, , \, \| \cdot \|_2$$ be two norms on $$\mathbb{R}^n$$. look at the maps $$\alpha: S_1 \to S_2 \, , \, \alpha(x) = \frac{x}{\|x\|_2} \, , \, \beta : S_2 \to S_1 \, , \, \beta(x) = \frac{x}{\|x\|_1}$$.

Since all the norms are equivalent $$\alpha,\beta$$ are continuous (w.r.t the subspace topologies on $$S_1,S_2$$ induced by the standard topology on $$\mathbb{R}^n$$).

A trivial verification shows $$\alpha,\beta$$ are inverses, thus homeomorphisms.

Corollary 1: All Unit spheres of smooth norms on a finite dimensional vector space are diffeomorphic.

Proof: Assume $$\|\cdot \|_1,\|\cdot \|_2$$ are smooth norms. $$S_1,S_2$$ are embedded submanifolds of $$\mathbb{R}^n \setminus \{0\}$$ as inverse images of a regular value of a smooth submersion. (The norms). Let $$\alpha,\beta$$ be the homeomorphisms constructed in the proof of Lemma1 above.

Considering $$\alpha$$ as a map $$:\mathbb{R}^n \setminus \{0\} \to \mathbb{R}^n \, , \, \alpha$$ is smooth. Hence it stays smooth after restricting the domain to $$S_1$$, so $$\alpha:S_1 \to \mathbb{R}^n$$ is smooth. But $$\alpha(S_1) =S_2$$ , so restricting the codomain also preserve smoothness. This shows $$S_1,S_2$$ are diffeomorphic, as required.

Corollary 2: The unit sphere of every smooth norm on a finite dimensional space is diffemorphic to the standard unit sphere $$\mathbb{S}_{Euc}^{n-1}$$.

Corollary 3: $$\forall p \in M \, , \, (UE)_p \cong \mathbb{S}_{Euc}^{n-1}$$.

Proof:

It only remains to show the restriction $$F|_{E_p}$$ is a smooth norm. But this is trivial since $$F_{E \setminus E_0}$$ is smooth by defintion of a Finsler function, and $$E_p \setminus \{0_p\} = (E \setminus E_0)_p=\pi|_{E\setminus E_0}^{-1}(\{p\})$$ is the inverse image of the submersion $$\pi|_{E\setminus E_0}:E\setminus E_0 \to M$$. (This is a submersion as a restriction of the natural projection $$\pi:E \to M$$ to the open submanifold $$E \setminus E_0$$).

• nice question -- as for the motivation, surely by local triviality of $E$ and compactness of $M$ you're reduced to showing that the unit sphere in one space is compact, which follows from the discussion here if I am not mistaken sma.epfl.ch/~troyanov/Papers/WeakMinkowski.pdf Oct 5 '15 at 12:48
• (I have not thought that carefully, and it's an answer to the motivation, and not the question, so take it with a grain of salt. also, by "one space" I mean "an asymmetric normed space," i.e. the thing that a fiber of a Finsler vector bundle is.) Oct 5 '15 at 12:49
• @hunter: Actually, I do not see how the reduction works, if I do not know $UE$ is locally a product of the form $U \times \mathbb{S}^{n-1}$ where $U\subseteq M$ is open..., that is if $UE$ is not a sphere bundle. Oct 5 '15 at 14:58
• you don't know that $UE$ is locally a sphere bundle, but you do know that $E$ is a vector bundle (hence locally trivial) which is what I'm using Oct 5 '15 at 15:09
• @cngzz1 Please stop editing all these years-old posts. Unless you find a substantive mathematical error, you're bringing these up to the top of the heap for no good reason. It's very distracting and serves no purpose. Jan 4 at 1:34

The unit bundle of every Riemannian vecor bundle is a sphere bundle. Now take any Riemannian metric $g$ on $E$. (There always exists one via partitions of unity argument, if the base manifold $M$ is paracompact).
Now just use radial projections: using the fact that any Riemannian bundle is inparticular Finsler, the two maps $x \to \frac{x}{\|x\|_g},x \to \frac{x}{\|x\|}$ are smooth maps between $UE_{\|\cdot \|},UE_{\|\cdot \|_g}$ which are inverses of each other. Hence they ae diffeomorphisms, so $UE_{\|\cdot \|} \cong UE_{\|\cdot \|_g}$.
Now just use the fact that being a $B$- bundle (a fiber bundle over a base $B$) is a property which is diffeomorphism-invariant.