4
$\begingroup$

Suppose we have a Fredholm section S, (the differential is Fredholm at the 0-set) of some Banach vector bundle over X, transverse to the 0-section, with Fredholm index 1 and such that the 0-set of this section is a circle which is contractible in X, can S be pushed of the 0-section through Fredholm sections?

edit: as stated this is not right. Mike Usher pointed out that in finite dimensional case (say a rank k real vector bundle over X^{k+1}) there are secondary obstructions to having a non-vanishing section. Since these obstructions have to do with $\pi_k (S^{k-1})$ it is not even obvious how to extend this to Fredholm setting. I guess the right question is then can one formulate obstruction theory in the Fredholm setting?

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ It would be nice if the manifolds and bundles in question where Hilbert rather than Banach. The group of linear automorphisms of an arbitrary Banach space could be weird. Assumming this is the case, then your manifold $X$ is an open subset of a Hilbert space $H$, and The bundle is a trivializable Hilbert bundle. The section $S$ is then a map $S: X\to H$ and the section whose differential at every point is Fredholm of index $1$. $\endgroup$
    – Liviu Nicolaescu
    Commented May 9, 2012 at 17:01

1 Answer 1

Reset to default
1
$\begingroup$

Late to the party.

I would like to shamelessly advertise the work of Alberto Abbondandolo and me on non-linear Fredholm mappings [1,2]. This builds on work of Elworthy and Tromba [3].

Allow me to work in the category of Hilbert manifolds. As Liviu Nicolaescu remarks: Hilbert bundles are trivial and a consequence is that the differential of a mapping $f:M\rightarrow N$ between (infinite dimensional ) Hilbert manifolds can be viewed as a map $df:M\rightarrow L(\mathbb H)$. The map[ing $f$ is then Fredholm if the differential takes values in the space of linear Fredholm operators $\Phi(\mathbb H)$. The homotopy type of this space is well understood. This space has $\mathbb Z$ many components (indexed by the Fredholm index), and each component is homotopy equivalent to the infinite Grassmannian $BO$.

Two Fredholm mappings between Hilbert manifolds $f,g:M\rightarrow N$ are then homotopic if

$f,g:M\rightarrow N$ are homotopic as continuous maps

$df,dg:M\rightarrow N$ are homotopic as continuous maps.

This means the (non-linear) homotopy classes of maps are in bijection with $[M,N]\times [M, \Phi(\mathbb H)]$.

See Theorem 1 in [1], which is an alternative proof of Proposition 2.24 in [3]

Thus the answer becomes yes: As Liviu remarks the section can be seen as a map $f:M\rightarrow \mathbb H$. Given such a map, there exits a Fredholm map $g$, whose image avoids zero and whose differential is homotopic to $f$. Explitly take g=h\circ f, where the map $h:\mathbb H\rightarrow \mathbb H$ is given by $h(x_1,x_2,\ldots)=(x_1^2+1,x_2,\ldots)$). By the theorem $f$ and $g$ are Fredholm homotopic.

The situation becomes much more interesting if the Fredholm maps are required to be proper. There is then an invariant (framed cobordisms) is full in the case of proper Fredholm mappings $M\rightarrow \mathbb H$.

The framed cobordism sets (hence the proper homotopy classes) have been computed for index $k\leq 0$ in [1] and for index $1$ in [2] in the case that $M$ is simply connected.

[1] Abbondandolo, Alberto ; Rot, Thomas O. On the homotopy classification of proper Fredholm maps into a Hilbert space. J. Reine Angew. Math. 759 (2020), 161–200.

[2] Abbondandolo, Alberto ; Rot, Thomas O. Rot The homotopy classification of proper Fredholm maps of index one arxiv:2005.03936

[3] Elworthy, K. D.; Tromba, A. J. Differential structures and Fredholm maps on Banach manifolds. 1970 Global Analysis (Proc. Sympos. Pure Math., Vol. XV, Berkeley, Calif., 1968) pp. 45–94 Amer. Math. Soc., Providence, R.I.

Improve this answer
$\endgroup$

You must log in to answer this question.

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