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Let $\mathcal{V}$ be the space $C^r$ vector fields on a non-compact (smooth) manifold $M$. Being a subspace of $C^r(M, T M)$, it inherits the natural $C^r$ topology (i.e. the strong topology) of that space. Furthermore, since $C^r(M, T M)$ is Baire and $\mathcal{V}$ is closed in $C^r(M, T M)$, $\mathcal{V}$ is Baire too [See Hirsch, Differential Topology, Theorem 4.4].

If we further restrict $\mathcal{V}$ to only include the vector fields that induce locally uniformly bounded trajectories, is the restricted space still Baire?

Note: Let $f \in \mathcal{V}$ be a particular vector field and $\phi(x, t)$ be the flow obtained from the ODE $\dot{x} = f(x)$. Vector field $f$ induces locally uniformly bounded trajectories if $\displaystyle \sup_{x \in C \; , \; t \geq 0} \| \phi(x, t) \| < \infty$ for all compact $C \subseteq M$.

Some comments

The cited theorem implies the following: The space $\mathcal{V}$ has a complete metric, if we consider the weak topology instead of the strong topology. Furthermore, to prove that the restricted space is Baire in the strong topology, it is sufficient to show that the restricted space is closed in this weak topology. Therefore, if one can show that the restricted space is still complete (with the mentioned metric), it is necessarily (weakly) closed, and hence Baire in the strong topology.

An informal argument: If $M \subseteq \mathbb{R}^n$ for some $n > 0$, or (more generally) if $M$ is a uniform space, then the weak topology on $\mathcal{V}$ (i.e. the compact-open topology) coincides with the topology of compact convergence. Since the restriction criterion is formulated in terms of compact subsets of $M$, I suspect that the restricted space is closed under the weak topology. Thus, I suspect it is Baire. However, I wasn't able to turn this into a rigorous argument, and it is possible that this line of reasoning is incorrect.

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migrated from math.stackexchange.com Apr 8 '14 at 7:58

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  • $\begingroup$ Migrating by OP request. $\endgroup$ – Willie Wong Apr 8 '14 at 7:58

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