5
$\begingroup$

In theorem 1.2 of Brian Conrad's handout Operations with Pseudo-Riemannian metrics, the author writes

Theorem 1.2. Every $C^p$ vector bundle $E\to M$ over a $C^p$ manifold with corners $0\leq p\leq \infty$ admits a Riemannian metric. In particular, every smooth manifold with corners admits a structure of a Riemannian manifold with corners.

The main tool in the proof will be $C^p$ partitions of unity, so it is amazing that Morrey and Grauert proved that every real-analytic vector bundle over a real-analytic manifold admits a real-analytic Riemannian metric tensor. The real-analytic case rests on serious input from the theory of several complex variables in order to circumvent the lack of real-analytic partitions of unity.

What is the idea of Morrey and Grauert's proof? What's the formalism that allows circumventing partitions of unity? What are the miracles of complex analysis that make things work?

Improve this question
$\endgroup$
4
  • $\begingroup$ It seems to me that a real analytic embedding of the unit disc bundle of a vector bundle into $\Bbb R^n$ gives you a real analytic metric on $E$, since $M$ the normal bundle of $M$ inside of $E$ (inside of $\Bbb R^n$) is the same as $NM \cap TE$. Because $M$ is an analytic submanifold of $\Bbb R^N$, $NM$ is an analytic subbundle of $T\Bbb R^n$, as is $TE$ for the same reason, so their intersection is also an analytic subbundle, and thus the Riemannian metric on this - that is, the Riemannian metric on $E$ - is analytic. $\endgroup$
    – mme
    Commented Mar 5, 2017 at 23:27
  • $\begingroup$ Thus it reduces to proving that every compact manifold with a real analytic structure admits an embedding into $\Bbb R^N$. Apparently this was done by showing that if $M$ is closed, $TM$ is a Stein manifold with the zero section $M$ a real analytic submanifold, so admits a complex analytic embedding of $TM$ into $\Bbb C^N$; of course restricting we get a real analytic embedding of $M$. $\endgroup$
    – mme
    Commented Mar 5, 2017 at 23:29
  • $\begingroup$ @MikeMiller thank you for the outline. I don't know anything about complex analysis, analytic manifolds, or Stein manifolds, so I will take a long time to digest your comments. If you feel like recommending a quick roadmap to understanding the idea, I will gladly accept it as an answer. $\endgroup$
    – Arrow
    Commented Mar 7, 2017 at 19:03
  • $\begingroup$ I would just advise you take the $C^\omega$ result on faith. $\endgroup$
    – mme
    Commented Mar 7, 2017 at 19:04

1 Answer 1

Reset to default
3
$\begingroup$

I'm not sure this should be an answer, but here is a proof that might be more conceptual resting on one non-trivial fact: if $E\rightarrow M$ is a real analytic fiber bundle over a real analytic manifold $M$ which admits a continuous section, then $E\rightarrow M$ admits a real analytic section.

If you take this one black box, the proof is identical to the standard "reduction of structure" interpretation of a metric on a vector bundle. Specifically, let $V\rightarrow M$ be a real analytic vector bundle. If $E\rightarrow M$ is the principal $GL(n,\mathbb{R})$-frame bundle of $V$, it has an induced real analytic structure. A $C^{r}$-metric on $V$ is equivalent to a $C^{r}$-reduction of structure of $E$ to the orthogonal group, which is equivalent to a $C^{r}$-section of the associated bundle of homogeneous spaces $E\times_{GL(n,\mathbb{R})}GL(n, \mathbb{R})/O(n).$

This bundle of homogeneous spaces has a $C^{0}$-section since the fibers are contractible, and therefore by our black box it has a real analytic section. This proves $V\rightarrow M$ admits a real analytic metric.

See https://www.math.uni-augsburg.de/prof/geo/Dokumente/stein106.pdf section 5.8 for a discussion of these matters. There they verify that one can even approximate as well as one likes a $C^{k}$ metric by an analytic one in the $C^{k}$-topology. Of course, as you see there, the proofs of the black box rely on embedding $M$ real analytically into some Euclidean space, and therefore this proof is really not "different" than the one explained by Mike Miller in the comments.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .