Download Isotopy lemma. `Manifolds have no points. You can`t distinguish their

Isotopy lemma.
‘Manifolds have no points. You can’t distinguish their points until you give
them additional structure.” - Dennis Sullivan, well-known topologist, in a lecture
at Berkeley in the late 1990s.
We formalize what Sullivan meant with:
Lemma 0.1 (Isotropy lemma) If M is a connected manifold, and x, y ∈ M ,
then there is an isotopy taking x to y.
Here we have used:
Definition 0.1 An isotopy of M is a diffeomorphism φ of M which is homotopic to the identity, through diffeomorphisms. That is, there is a smooth map
FM × I → M , with Ft : M → M a diffeomorphism for each t ∈ I, and F0 = Id,
F1 = φ.
We say two points, or two subsets of M are isotopic if there is an isotopy
taking one to the other.
Flows of (complete) vector fields on M are isotopies of M , and every isotopy
can be constructed as the flow of a time-dependent vector field.
Without extra structure, the symmetry group of a manifold is the group of
isotopies. And the lemma says that from this perspective every point is the same
as every other. The exercises you are assigned says more: any two collections of
N points on M are isotopic. Any two open arcs are isotopic. Any two embedded
discs are isotopic. Etc. A manifold with no extra structure is devoid of features.
However, as soon as we put a vector field or some other structure on M , we
give some of M ’s points or curves of M the chance of distinction: for example,
the zeros of the vector field are distinguished points. Or if we give M a metric,
then geodesic arcs become distinguished curves. The manifold has some scenery,
some feattures.
The proof of the Isotopy lemma is itself based on two lemmas. Call an
equivalence class on a topological space “open” if its equivalence classes are all
open sets.
Lemma 0.2 If a connected topological space is endowed with an open equivalence relation, then every point of the space is equivalent to every other: that is,
there is only a single equivalence class: the whole space.
The proof of this lemma is immediate.
The next lemma is used to show that the relation “x is isotopic to y ” is an
open equivalence relation.
Lemma 0.3 If B ⊂ Rn is an open ball, and x, y ∈ B, then there is an isotopy
of Rn which is the identity outside of B.
Proof Set v = x − y and view v as a constant vector field on Rn . The
time 1 flow for v takes x to y and indeed is simply translation by v. To make
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the desired isotopy choose a bump function which is 0 outside B and 1 in a
neighborhood of the segment [x, y] joining x to y. The new vector field βv is
zero outside of B and still has [x, y] as an integral curve. Its time 1 flow φ1 is
the desired isotopy. QED
Proof of Isotopy Lemma. The relation “x is isotopic to y ” is an equivalence relation since the set of isotopies form a subgroup of Dif f (M ). By
the first lemma above it suffices to show that this equivalence relation is open.
Suppose x is isotopic to y. We must show x is isotopic to all points in some
neighborhood of y0 . Use a coordinate chart to identify a small neighborhood of
y0 with an open ball U ⊂ Rn and choose a V ⊂ B ⊂ U with y0 ∈ V . By the
second lemma we can, for any z ∈ V , find an isotopy of U which maps y0 to z
and is the identity outside B. Extend φ to the whole manifold M by having it
be the identity everywhere outside of U . We have shown x is isotopic to any
point z ∈ V and hence the relation “is isotopic to” is an equivalence relation.
QED
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