A Not-So-Nice Submanifold


As shown in evaluation fibration of mapping spaces and tubular neighbourhoods of mapping spaces, if we carve out a submanifold of a mapping space by specifying “coincidences”, we often get a tubular neighbourhood. On this page, we shall give an example of a submanifold with no tubular neighbourhood. The example is simple to describe. To make it concrete, we shall fix as our source space the circle, S 1S^1. For our target space, we shall take a finite dimensional smooth manifold, MM. The full smooth mapping space, C (S 1,M)C^\infty(S^1, M) is known as the smooth loop space. For simplicity, let us take based loops within this, which we write as ΩM\Omega M. Within that, we consider the space of based smooth maps S 1MS^1 \to M which are infinitely flat at the point 1S 11 \in S^1. Let us write this as Ω Unknown characterUnknown characterUnknown characterM\Omega_♭ M. As we are using based loops, we can identify the tangent space of MM at the basepoint with n\mathbb{R}^n and so we have a sequence, which is exact by Borel's theorem:

(1)Ω Unknown characterUnknown characterUnknown characterMΩM j=1 n \Omega_♭ M \to \Omega M \to \prod_{j = 1}^\infty \mathbb{R}^n

It is easy to show that this does not admit a tubular neighbourhood. If it did, there would be a splitting of the induced map on tangent spaces:

(2)T αΩ Unknown characterUnknown characterUnknown characterMΩ αM i=1 n T_\alpha \Omega_♭ M \to \Omega_\alpha M \to \prod_{i = 1}^\infty \mathbb{R}^n

but as the second map is surjective, this cannot split as a splitting map would induce a continuous injection from i=1 n\prod_{i=1}^\infty \mathbb{R}^n to a normed vector space and that is impossible.

Revised on June 27, 2011 11:43:04 by Andrew Stacey (