Deformation retraction of plane RP2

[semigroups]

Let RP2 denote the real projective plane (it can be obtained from glueing a Mobius band and a disk whose boundary is the same as the boundary of the Mobius band).

I know if one punches a hole off RP2 then the punched RP2 is homotopy equivalent to a Mobius band which is in turn deformable to a circle, I also know that RP2 is NOT deformable to a circle. Does this imply RP2 is not deformable to any of its proper subspaces? Otherwise by trasitivity of deformability one can imply RP2 is defromable to a circle which contradicts to the fact that it's not?

Morever, how about retractions? I know RP2 can NOT be retracted to a circle

 

[Vargo]

Any proper subspace is deformable to a subspace A of a circle. So by composition, that would imply that RP2 is deformable to A. If A is less than a full circle, then it in turn deformation retracts to a point. So can you prove that RP2 cannot deformation retract to a point?

Lets see, a retraction is a continuous map r such that r^2 = r I believe. Every topological space is retractible to a point...

 

[Jamma]

this imply RP2 is not deformable to any of its proper subspaces? Otherwise by trasitivity of deformability one can imply RP2 is defromable to a circle which contradicts to the fact that it's not?

Morever, how about retractions? I know RP2 can NOT be retracted to a circle

What do you mean by saying "not deformable to any of its proper subspaces"? Do you mean a map from RP2 to a subspace of it (f, say?) such that f followed by the inclusion is homotopic to the identity?

Couldn't you, for example, prove that if you are retracting to a subspace with no homology in H_2 [with Z/2Z coefficients] (e.g. when you remove a point and the remaining subspace is homotopic to the circle)? In that case, you'd need to factor the identity through the trivial group. I'm sure a similar thing goes for retractions.

Lets see, a retraction is a continuous map r such that r^2 = r I believe. Every topological space is retractible to a point...

Well yes, trivially for any space the map sending everything to some point is a retract, this is trivially true but doesn't say all that much. One would usually talk about deformation retractions, which are special cases of homotopy equivalences.

 

[Vargo]

Well, I think he was asking about retracts and deformation retracts independently.

So, if you rephrase the question to exclude retracts to a point, then how would you prove that there are no retracts from RP2 to a subspace A which contains more than one point? A purely algebraic answer would not suffice since a disk is homotopy equivalent to a point, and retracts to a point do exist. The image of such a retraction would have to be compact, path connected, and I think with trivial fundamental group. So does that mean it is homeomorphic to a disk?

 

[Jamma]

Well my thinking above was to use the fundamental class of the homology of the manifold.

What I said indeed proves that no deformation to a subspace will exist as long as the subspace you have has a trivial H_2 (or, indeed, an H_2 with no 2 torsion) - this includes deformation retracts as a sub-case (as I said, I wasn't sure what the OP meant by "deformable to a subspace" - my definition was a little weaker than a deformation retract since I wasn't assuming that the homotopy sending everything to the subspace kept the subspace fixed). Your objection to this answer is based on that retracts to a point exist, but I wasn't talking about retracts there.

Is it easy to prove that no proper subset of a 2-manifold (or just our RP2) has non-trivial H_2? You will probably need the manifold property - if you have a homotopy equivalent space with a "stick" pointing out of it, then this clearly won't have that property, so you need more than just the homotopy type of the space and something to do with the manifold property at a local level.

Perhaps you can get somewhere with relative homology groups - probably with the property that H_i(RP2,RP2-x)= Z[i=2],0[o'wise] (i.e. the property that makes RP2 a homology manifold).

As for the question of retracts ... clearly your subspace needs to be path-connected, and since RP2 is Hausdorff it needs to be closed (so the thing is going to be htpy equivalent to a closed disk or a point). Examining the retract of RP2 to this disc - it fixed the disc, and sends the remaining Mobius band into this disc in a way such that the boundary is mapped to the boundary, which seems fairly obviously to be possible.

 

[lavinia]

Any proper subspace is deformable to a subspace A of a circle.

What about the projective plane minus two points?

 

[Vargo]

What about the projective plane minus two points?

Good point. I was wrong about that.

 

[lavinia]

I feel - but don't see a proof - that no proper subset of the projective plane is a deformation retract of the entire plane. Similarly, this seems true of any closed surface or more generally any closed manifold (without boundary)

 

[Jamma]

I agree. A deformation retract is a htpy equivalence and it feels as though any proper subset of a n-dim manifold should have trivial homology in degree n (Z/2Z coefficients).

Could this be made explicit? You need to use closed subsets to have retracts on Hausdorff spaces (I think) and it will obviously need to be path connected. Surely such a subset has trivial homology in degree n? (given that it's a proper subset, won't this also make it homeomorphic to a closed, path connected subset of Euclidean space too?)

Of course, you need a closed manifold, since the property won't work for non-compact surfaces (e.g. the plane) or surfaces with boundary (e.g. the disc).

 

[lavinia]

I agree. A deformation retract is a htpy equivalence and it feels as though any proper subset of a n-dim manifold should have trivial homology in degree n (Z/2Z coefficients).

If x is a point of RP[itex]^{2}[/itex] not in the proper subset,A, then the inclusions

A -> RP[itex]^{2}[/itex] - {x} -> RP[itex]^{2}[/itex] show that the pull back of the fundamental Z/2Z-cohomology class of RP[itex]^{2}[/itex] to A is zero, this because RP[itex]^{2}[/itex] - x has zero second cohomology.

If A were a retract of RP[itex]^{2}[/itex] then there would be a sequence of maps

A [itex]\stackrel{i}{\rightarrow}[/itex] RP[itex]^{2}[/itex] [itex]\stackrel{r}{\rightarrow}[/itex]A

where r[itex]\circ[/itex]i is the identity. Thus if A had a non-zero mod 2 cohomology class its pull back under the retraction to H[itex]^{2}[/itex](RP[itex]^{2}[/itex]:Z/2Z) would have to be non-zero. But this would have to equal the fundamental class of RP[itex]^{2}[/itex].

 

[Jamma]

Nice proof, that wraps it up. Thanks Lavinia!

In fact, doesn't this prove the general statement that deformation retracts onto proper subspaces don't exist for all manifolds - I assume that punctured manifolds will always be homotopy equivalent to graphs? You can then use the above argument again to show that the top Z/Z2 cohomology class of a proper subspace is trivial so that a deformation retract can't exist.

 

[lavinia]

Nice proof, that wraps it up. Thanks Lavinia!

In fact, doesn't this prove the general statement that deformation retracts onto proper subspaces don't exist for all manifolds - I assume that punctured manifolds will always be homotopy equivalent to graphs? You can then use the above argument again to show that the top Z/Z2 cohomology class of a proper subspace is trivial so that a deformation retract can't exist.

Yes you are right. The same proof seems to work for any compact manifold without boundary.

I do not think though that a punctured 3 or higher dimensional manifold is necessarily homotopy equivalent to a graph. Doesn't RP[itex]^{3}[/itex] minus a point deform onto to RP[itex]^{2}[/itex]?

A punctured manifold does not have a fundamental cycle because the puncture can be widened into an open ball. So the punctured manifold is homotopy equivalent to a manifold with boundary.

 

[Jamma]

I think you are right - given the standard picture of a Klein Bottle, RP2 or torus as a square with identifications, one can "stretch the puncture" out to the boundary so that you have a graph. Thinking about a 3d analogue with a cube with sides identified, it would seem that you can make the space homotopy equivalent to the faces of the cube with some identifications.

So I imagine that a punctured closed n-dim manifold (which is an open n-dim manifold) will often be homotopy equivalent to some n-1 dim manifolds, possibly glued together in some way.

Having a play around with how to prove that the punctured manifold loses its top homology class, I think this does the trick:

Take a small open ball around the puncture in the manifold. Look at the Mayer Vietoris sequence (with Z/2Z coefficients, in the unoriented case). This decomposes our manifold into the punctured manifold and an open ball, the intersection giving an n-1-sphere.

This sequence seems to give isomorphisms of the homology of the manifold and the punctured manifold in dimensions below n-1 (which makes sense - in high enough codimensions we can move "around" the puncture). The fundamental class of the manifold seems to map onto the fundamental class of the (n-1)-sphere, so I think this shows that the homology of the punctured manifold is killed in dimension n.

I'm not certain about what can be said about the homology in dimension n-1. It seems to me that in the oriented case, the inclusion of the fundamental class of the (n-1)-sphere into the punctured manifold will be nullhomologous (pull it back around the rest of the fundamental class). This means that the homology of the punctured manifold will be unchanged except the killed homology in dimension n. In the unoriented case, it doesn't seem that you can use the same argument with Z coefficients (in fact, I think this inclusion in the example of the punctured RP2 will give the fundamental class of the circle), but I think with Z/Z2 the same will be true - so the Z/Z2 homology will be the same except with trivial top homology.

 

[lavinia]

Geometrically the n-1 sphere is a boundary.

Homologically it is the boundary of the fundamental cycle minus an open ball.
If the manifold is triangulated, the sum of the n-simplices is the fundamental cycle mod 2. Each n-1 face of an n-simplex appears twice - once on each of the two n-simplices that it is a face of. If one of the n-simplices is removed -topologically the same as removing an n-ball - then the rest of the fundamental cycle no longer has zero boundary since the n-1 faces of the removed n-simplex only appear once now.

I like the idea that the punctured manifold always deforms onto a complex that contains no n-cells. This would give another proof if true.

 

[Jamma]

This seems to be the case - if we at least assume our manifold is a CW complex with contractible n-cells, it seems to be that we can nibble away our n-cells touching the boundary one by one until none are left and at each stage we will have a space of the same homotopy type.

I know this isn't a very rigorous proof!

 

[lavinia]

This seems to be the case - if we at least assume our manifold is a CW complex with contractible n-cells, it seems to be that we can nibble away our n-cells touching the boundary one by one until none are left and at each stage we will have a space of the same homotopy type.

I know this isn't a very rigorous proof!

But what about a single puncture? Is it still true?


If you allow a puncture for each n cell then you argument works. Just retract one at a time.

 

[Jamma]

I meant with just one puncture. Working on intuition, removing one of the cells at the boundary one by one, you shouldn't change the homotopy type at any stage. The justification for this is that if a cell sits on the boundary, we should be able to deformation retract it to the boundary it shares with its neighbouring cells (this boundary will need to be connected).

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).

 

[lavinia]

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?

 

[Jamma]

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/servlets/DerivateServlet/Derivate-6161/mjm_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

 

[lavinia]

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/servlets/DerivateServlet/Derivate-6161/mjm_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.

 

[Jamma]

I'm thinking of a torus - how does the flow continue past the first critical point?

 

[lavinia]

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

 

[lavinia]

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

 

[Jamma]

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.