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deformation retraction of plane RP2

 
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Jun9-12, 05:54 AM   #18
 
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deformation retraction of plane RP2


Quote by Jamma View Post

e.g. tile your square with identifications for RP2 with 9-cells (3x3 grid). For the punctured version, lose the middle one (you are left with 8 squares). Removing a square (but not the boundary of the remaining squares!), you should get a homotopy equivalent space. Do the same again with each of them and you are left with the circle.

As I said, this is just intuition, and may be wrong for the general picture. Another problem is that not all manifolds have CW-structures, let alone CW structures with contractible cells (although all manifolds are homotopy equivalent some CW-complex).
Right.

Since any surface is the quotient of a polygon with identifications on the boundary, a single puncture in the middle of the polygon allows a retraction onto the boundary that preserves the identifications.

For a compact smooth manifold, I would guess that the exponential map at an point - with respect to any Riemannian metric - maps a connected open set in the tangent space onto the manifold minus a set of measure zero - the cut locus perhaps. The exponential flow defines a deformation retraction of the manifold minus this point onto the set of measure zero. I will research this.

An example would be the flat torus. Here the exponential map is just radial flow along straight lines. It retracts the punctured torus onto the cut locus which is a figure 8.

What is an example of a manifold that does not have a CW structure?
Jun11-12, 06:06 AM   #19
 
Hmm, I may have been wrong. Some browsing reveals this:

All manifolds are homotopy equivalent to a CW complex
[--interestingly, there is kind of an inverse too: all countable CW complexes of dim n are homotopy equivalent to a differentiable manifold of dimension 2n+1
http://miami.uni-muenster.de/servlet...m_vol_3_01.pdf ]

All smooth manifolds have a CW structure
Not all manifolds are triangulable (e.g. the E8 manifold)
Whether or not all manifolds have CW structure, I think is an open problem, although I'm having difficulties looking this one up.

It seems we were basically correct about a punctured n-manifold being htpy equivalent to an n-1 complex, this gives an affirmative answer in the smooth case:

EDIT BY ANDY PUTMAN: Mohan isn't registered and thus isn't able to comment, but he sent me an email with more details. The result is true in all dimensions : any noncompact smooth n-manifold is homotopy equivalent to an n-1 complex. The key is to construct a strictly subharmonic morse exhaustion function. The subharmonicity prevents the function from having local maxima. Details of this can be found in his paper "Elementary Construction of Exhausting Subsolutions of Ellitpic Operators", which was joint with Napier and was published in L’Enseignement Math´ematique, t. 50 (2004), p. 1–24.
From this page:
http://mathoverflow.net/questions/18454
Jun12-12, 07:15 AM   #20
 
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Quote by Jamma View Post
Hmm, I may have been wrong. Some browsing reveals this:

All manifolds are homotopy equivalent to a CW complex
[--interestingly, there is kind of an inverse too: all countable CW complexes of dim n are homotopy equivalent to a differentiable manifold of dimension 2n+1
http://miami.uni-muenster.de/servlet...m_vol_3_01.pdf ]

All smooth manifolds have a CW structure
Not all manifolds are triangulable (e.g. the E8 manifold)
Whether or not all manifolds have CW structure, I think is an open problem, although I'm having difficulties looking this one up.

It seems we were basically correct about a punctured n-manifold being htpy equivalent to an n-1 complex, this gives an affirmative answer in the smooth case:



From this page:
http://mathoverflow.net/questions/18454
Interesting. I tried to find info on the cut locus but the references were all specific cases. In general, I guess it can be pretty wild but I find it hard to believe that it can ever contain an open set. Will work more on it.

BTW: In the smooth case, what happens with the gradient flow of an arbitrary Morse function? Puncture the manifold at a relative minimum and follow the flow from there.
Let's make up some examples.
Jun12-12, 10:44 AM   #21
 
I'm thinking of a torus - how does the flow continue past the first critical point?
Jun12-12, 10:41 PM   #22
 
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Quote by Jamma View Post
I'm thinking of a torus - how does the flow continue past the first critical point?
Not sure maybe it doesn't work. Have to think more.

BTW: The cut locus can not contain an open set since an interior point would be reached by a geodesic that minimizes length until it reaches the point
Jun15-12, 10:53 AM   #23
 
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Quote by Jamma View Post
I'm thinking of a torus - how does the flow continue past the first critical point?
With the height function on the torus, I think the deformation works past the first critical point if the circles connecting the remaining 3 critical points, all those except the minimum, are left stationary. On the rest of the torus just follow the gradient flow. In the end, you are left with a figure 8
Jun17-12, 02:02 PM   #24
 
So it's the gradient flow just "going upwards" but multiplied by a scalar?

Just drew a picture, and I think I see how it goes, you at least get to a very small retract of the figure 8.
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