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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 $E \setminus E_0$).

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  • $\begingroup$ 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 $\endgroup$
    – hunter
    Oct 5, 2015 at 12:48
  • $\begingroup$ (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.) $\endgroup$
    – hunter
    Oct 5, 2015 at 12:49
  • $\begingroup$ @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. $\endgroup$ Oct 5, 2015 at 14:58
  • $\begingroup$ 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 $\endgroup$
    – hunter
    Oct 5, 2015 at 15:09
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    $\begingroup$ @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. $\endgroup$ Jan 4, 2021 at 1:34

1 Answer 1

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The answer is positive.

Proof:

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.

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