Prev: what did Abel and Galois proof, i supose only God knows, and doyou knows really what they proofed?
Next: what exactly happens when sugar is added to gasoline? When salt ?is added?
From: Ki Song on 5 Aug 2010 12:37
So... I guess the following map is an example of a degree 1 map from a
compact closed n-manifold M to S^n:

Pick a point p on M, pick a small ball around it. Map the complement
of the ball to a point.

Is there a reason why M needs to be compact for this to be degree 1?
Or am I missing something?
From: W. Dale Hall on 7 Aug 2010 18:09
Ki Song wrote:
> So... I guess the following map is an example of a degree 1 map from a
> compact closed n-manifold M to S^n:
>
> Pick a point p on M, pick a small ball around it. Map the complement
> of the ball to a point.
>
> Is there a reason why M needs to be compact for this to be degree 1?
> Or am I missing something?

One problem is that the map you've described is null-homotopic, and
it's generally a Good Thing for degree to be homotopy invariant.

The standard (well, it's standard for me, maybe for more folks than
just me) definition of degree for a mapping between oriented
manifolds of dimension n:

The degree of the map

f: M ---> N

is given by the multiple deg(f) by which the induced map
on top-dimensional homology maps the fundamental class [M]
to the fundamental class [N]:

f* : H_n(M) ---> H_n(N)
[M] |-------> deg(f) [N]

The problem with non-compact manifolds is that top
homology vanishes.

This has implications when one considers the obstruction
to a null-homotopy of a map from an open (i.e., non-
compact, without boundary) n-manifold M^n to S^n: suppose
you have the map f:M ---> S^n, and look at the problem
of extending it to tM, the cone on M. The obstructions
for constructing an extension lie in the cohomology
groups

H^j (tM, M; pi_(j-1)(S^n)).

that is, jth cohomology of (tM, M) with
coefficients in the j-1st homotopy of S^n.

Since S^n has trivial homotopy groups for j < n, the first
occurrence of an obstruction occurs for j = (n+1).

Also, H^*(tM,M) is isomorphic (by excision) to H^*(SM)
(where SM is the suspension of M ) which, in turn, is
isomorphic to H^(*-1)(M), so this cohomology vanishes
in dimensions greater than n+1.

The only possibility exists in dimension n+1.

However, H^n(M) = 0, so the only possible obstruction
to extending f:M^n ---> S^n to tM^n ---> S^n vanishes,
and thus f is null-homotopic.

Every map from an open n-manifold to S^n is then homotopic
to a constant map. So, if you want degree to be homotopy
invariant, it'll have to be 0 for non-compact connected
manifolds M.

Dale
From: Daniel Giaimo on 8 Aug 2010 17:50
On 8/7/2010 6:09 PM, W. Dale Hall wrote:
> Ki Song wrote:
>> So... I guess the following map is an example of a degree 1 map from a
>> compact closed n-manifold M to S^n:
>>
>> Pick a point p on M, pick a small ball around it. Map the complement
>> of the ball to a point.
>>
>> Is there a reason why M needs to be compact for this to be degree 1?
>> Or am I missing something?
>
> One problem is that the map you've described is null-homotopic, and
> it's generally a Good Thing for degree to be homotopy invariant.

Are you sure the map he described is null-homotopic? Just considering
the case where M is S^n seems to contradict that. In that case the map
is simply the identity map on the sphere.

--
Dan G

From: W. Dale Hall on 9 Aug 2010 02:25
Daniel Giaimo wrote:
> On 8/7/2010 6:09 PM, W. Dale Hall wrote:
>> Ki Song wrote:
>>> So... I guess the following map is an example of a degree 1 map
>>> from a compact closed n-manifold M to S^n:
>>>
>>> Pick a point p on M, pick a small ball around it. Map the
>>> complement of the ball to a point.
>>>
>>> Is there a reason why M needs to be compact for this to be
>>> degree 1? Or am I missing something?
>>
>> One problem is that the map you've described is null-homotopic, and
>> it's generally a Good Thing for degree to be homotopy invariant.
>
> Are you sure the map he described is null-homotopic? Just
> considering the case where M is S^n seems to contradict that. In that
> case the map is simply the identity map on the sphere.
>
> -- Dan G
>
Yes, for a compact n-manifold, the procedure described will yield a map
of degree 1, as can be verified by checking the induced homomorphism in
top-dimensional homology. I was referring to the case where M is non-
compact, and it was incorrect for me not to have made that explicit.

I'll think a bit more on a non-obstruction-theory proof of a null-
homotopy; I've got to review the details of ends of manifolds, so
it may take a bit of time.

Dale
 | 
Pages: 1
Prev: what did Abel and Galois proof, i supose only God knows, and doyou knows really what they proofed?
Next: what exactly happens when sugar is added to gasoline? When salt ?is added?