# In the real analytic category, are the fibers of a proper submersion isomorphic?

Ehresmann's theorem says that a proper smooth submersion is a fiber bundle. The proofs I know rely on the existence of connections locally on the base, and this is furnished by partitions of unity.

This question gives a counterexample in the holomorphic category which is probably classic (elliptic curves), but I don't know any of the story.

I am wondering about the real analytic category. Are the fibers of a proper real analytic submersion isomorphic? If not, will it be locally trivial (in the real analytic category) when they are isomorphic, as in the linked question?

• The question in the title is just the opposite than the question in the body of the question. I suggest an edit for the coherence's sake. – Francesco Polizzi Jun 14 '19 at 10:35
• @FrancescoPolizzi done. – Arrow Jun 14 '19 at 10:47

There exists an analytic Riemann metric on a real-analytic manifold, which follows from embeddability of real analytic manifolds (see The Analytic Embedding of Abstract Real-Analytic Manifolds Charles B. Morrey, Jr.) - maybe can also be proved easier.

So, you can probably just take orthogonal connection.

• So, the answer to the question in the title is yes, right? – Francesco Polizzi Jun 14 '19 at 10:35
• Dear @Lev, I just found this comment which remarks the some references assert Ehresmann is false in the real analytic category. Just out of curiosity, have you ever stumbled upon such a claim? – Arrow Jun 14 '19 at 10:50
• Dear @Lev, your answer gives a connection on the tangent bundle of an analytic manifold, but how can it be used to obtain a connection on an analytic submersion? – Arrow Jun 17 '19 at 13:08
• Arrow, considering this comment, I wonder. I hadn't seen such claims, and I have a feeling that Ehresmann should be true in analytic setting from reasons mentioned above. The construction of the connection is similar to the smooth case - we need to chose analytic distribution which is transverse to fibers. Lets just choose it to be orthogonal (so it is even easier than what you've wrote in your answer, I suppose I should've made the proof more clear). – Lev Soukhanov Jun 23 '19 at 21:26
• Dear @LevSoukhanov, I would really appreciate some additional details since I don't see how to construct such an analytic connection on the submersion. (I would like to circumvent the tubular neighborhood proof of Ehresmann sketched in my answer, which feels like transparent to me.) – Arrow Jun 23 '19 at 21:42

I think Ehresmann's theorem does hold in the real-analytic category. A proof via tubular neighborhoods can be repeated in the real-analytic context since it makes no use of partitions of unity. Here's a sketch.

By Lev's answer, a real-analytic manifold admits a real-analytic embedding into Euclidean space with its standard real-analytic structure. I think this implies a real-analytic manifold admits a real-analytic Riemannian metric.

The exponential map of this metric is also real-analytic (as a solution to analytic ODE), so an analytically embedded submanifold admits a real-analytic tubular neighborhood. We may then repeat this proof of Ehresmann's theorem using tubular neighborhoods.

Remark. This comment along with several other places in the literature assert Ehresmann's theorem fails in the real-analytic category but provide no examples.