Extended plane as a topological sphere

[TrickyDicky]

The extended plane (E2 U ∞) is a non-orientable surface, and yet topologically is a sphere which is orientable, can someone comment on how this is reconciled?

 

[lavinia]

A topological sphere which is a smooth manifold is always orientable.

One proof is that the 1 dimensional mod 2 cohomology is zero so the first Stiefel Whitney class is zero. This will be true of any simply connected manifold.

The plane can be extended by attaching a circle rather than a point. If attached in the proper way the result is the projective plane which is non orientable. Its first mod 2 cohomology group is Z/2 and the non zero class is the first Stiefel Whitney class.

 

[TrickyDicky]

A topological sphere which is a smooth manifold is always orientable.

One proof is that the 1 dimensional mod 2 cohomology is zero so the first Stiefel Whitney class is zero. This will be true of any simply connected manifold.

Ok.

The plane can be extended by attaching a circle rather than a point. If attached in the proper way the result is the projective plane which is non orientable. Its first mod 2 cohomology group is Z/2 and the non zero class is the first Stiefel Whitney class.

A circle? All my references say the extended euclidean plane is compactified by a point at infinity, i.e. wiki:"The one-point compactification of n-dimensional Euclidean space Rn is homeomorphic to the n-sphere Sn. As above, the map can be given explicitly as an n-dimensional inverse stereographic projection." from examples in http://en.wikipedia.org/wiki/Alexandroff_extension

Anyway you are agreeing that it is non-orientable, and it is homeomorphic to the sphere as can be consulted in m any texts. My question was whether orientability is preserved by homeomorphisms or not. Apparently it isn't? On the other hand the extended euclidean plane is not simply-connected, this is quite puzzling to me.

 

[TrickyDicky]

I'm actually in doubt about orientability being a topological invariant, but I'm quite positive simply-connectedness is.

 

[TrickyDicky]

So maybe they are not homeomorphic after all? But I've seen it stated in more than one source: for instance:"A sphere Σ of ˆEn(extended En) is defined to be either a Euclidean sphere S(a, r) or an extended plane ˆ P(a, t) = P(a, t) ∪ {∞}. It is worth noting that ˆ P(a, t) is topologically a sphere." In Rattcliffe's "Foundations of hyperbolic geometry"

 

[lavinia]

I'm actually in doubt about orientability being a topological invariant, but I'm quite positive simply-connectedness is.

It is a topological invariant which can be seen using the topological definition of orientability. Also for compact manifolds without boundary ,orientability is equivalent to the top dimensional Z homology being equal to Z.

 

[lavinia]

So maybe they are not homeomorphic after all? But I've seen it stated in more than one source: for instance:"A sphere Σ of ˆEn(extended En) is defined to be either a Euclidean sphere S(a, r) or an extended plane ˆ P(a, t) = P(a, t) ∪ {∞}. It is worth noting that ˆ P(a, t) is topologically a sphere." In Rattcliffe's "Foundations of hyperbolic geometry"

The projective plane can be thought of as the Euclidean plane together with a circle at infinity. This circle can be thought of as the set of pencils of parallel lines.

Topologically one obtains the projective plane from a closed disk by identifying antipodal points on the boundary circle. Note that this is a disk with a circle attached.

The projective plane is not a topological sphere. For instance it is not simply connected. Its fundamental group is Z/2.

 

[TrickyDicky]

It is a topological invariant which can be seen using the topological definition of orientability. Also for compact manifolds without boundary ,orientability is equivalent to the top dimensional Z homology being equal to Z.

The projective plane can be thought of as the Euclidean plane together with a circle at infinity. This circle can be thought of as the set of pencils of parallel lines.

Topologically one obtains the projective plane from a closed disk by identifying antipodal points on the boundary circle. Note that this is a disk with a circle attached.

The projective plane is not a topological sphere. For instance it is not simply connected. Its fundamental group is Z/2.

Ok, what you are saying makes sense, thanks. They cannot be homeomorphic. So I guess they are being sloppy in the quotes I referenced?

 

[lavinia]

Ok, what you are saying makes sense, thanks. They cannot be homeomorphic. So I guess they are being sloppy in the quotes I referenced?

Don't know

BTW: The projective plane is not only non orientable but also is not the boundary of any solid(three dimensional manifold). In this it differs from the Klein bottle which is not orientable but is the boundary of a 3 manifold. A manifold which is a boundary must have even Euler characteristic. The Klein bottle has Euler characteristic zero,the projective plane has Euler characteristic 1. The sphere has Euler characteristic 2.

 

[WWGD]

It is a topological invariant which can be seen using the topological definition of orientability. Also for compact manifolds without boundary ,orientability is equivalent to the top dimensional Z homology being equal to Z.

Besides, homeomorphisms ( actually, even homotopies ) preserve (co)homology, homotopy, and orientation is defined in terms of (co) homology classes.

 

[WWGD]

I'm actually in doubt about orientability being a topological invariant, but I'm quite positive simply-connectedness is.

For simple-connectedness, I think the same argument as the previous works: the fundamental group is preserved by homotopy equivalences, so in particular by homeomorphisms. Same for connectedness. I think path-connectedness is also preserved. Then a homeomorphism h: X-->Y will preserve π₁(X) and, if X is connected, then so is Y. So if X is simply-connected, i.e., path-connected and with trivial fundamental group, Y will also have a trivial fundamental group. Now it just remains to see if the homeo h preserves path components.

 

[homeomorphic]

If the extended plane is what I think it is (namely, the one point compactification of R^2), then, yes, it is homeomorphic to a sphere. The mistake is to assert that it's not orientable.

 

[WWGD]

There are different types of compactifications; there is the 1-point (a.k.a, Alexandrov ), 2-point., etc. This works, i.e., it is definable only on locally-compact Hausdorff spaces. Then there are 2-point compactifications, like that of the Real line tgiven by the extended Real line. Then there is the Stone-Cech compactification, which works for a broader collection of spaces; the Tychonoff spaces, and it is categorical, i.e., it is described through a universal property. In all cases, a compactification of a space X is defined as a space Y so that X is dense in Y and Y is compact, usually you want Hausdorff together with compact. I never heard of adding a sphere in order to compactify; do you know the name of that type of compactification?

http://en.wikipedia.org/wiki/Compactification_(mathematics )

 

[WWGD]

To show simple-connectedness is a topological invariant, we can show both the fundamental group and path-connectedness are preserved by homeomorphisms.

First is straightforward: homotopy equivalence preserves fundamental group ( by definition).

Well, continuity is not enough to preserve path-connectedness, but a homeomorphism is enough ; the topologists' sine curve is a(n) (counter) example for continuity alone. Naively, it would seem that if X is path connected and I have a path p between x_1, x_2 in X, and
$f: X \rightarrow Y$ is continuous, them f(p) is a path between$f(x_1) , f(x_2)$ . But of course, we would need a bijection to guarantee that every y in Y is the image of some x in X. And a homeomorphism gives us that.
So, to show homeos. preserve path connectedness; basically by pulling back paths bijectively:
Let h: X-->Y be a homeomorphism , with X path connected. Consider $y_1, y_2 \in Y$ . Then there are $x_1, x_2 \in X$, with $f(x_1)=y_1 , f(x_2)=y_2$ . By path-connectedness of X, there is a path $p : I \rightarrow X$ with $p(0)=x_1, p(1)=x_2$ ( let's assume I = [0,1] to simplify ). The image $h(p)$ is a path joining $y_1, y_2$ in $Y$ . Of course we need a bijection to use t5his result, which we get since $h$ is a homeomorphism.

 

[homeomorphic]

I never heard of adding a sphere in order to compactify; do you know the name of that type of compactification?

I was talking about the 1-point compactification. If you do that to a 2-d Euclidean plane, you get a sphere. The sphere is not added. The plane becomes a sphere when you add the point at infinity.

 

[TrickyDicky]

I was talking about the 1-point compactification. If you do that to a 2-d Euclidean plane, you get a sphere. The sphere is not added. The plane becomes a sphere when you add the point at infinity.

This was my initial understanding but after listening to the replies and looking into it there are some nuances that I think might be worth taking into account: if one just means by the Euclidean plane a surface with Euclidean geometry on it, like for intance the complex plane CP1, it is a correct statement, however if one is referring to the projective plane R2, as it was pointed out there is not a one point compactification, one needs a circle to compactify it, and it is not homeomorphic to S2 but to its quotient by +/- the identity if I understood it right.

 

[homeomorphic]

This was my initial understanding but after listening to the replies and looking into it there are some nuances that I think might be worth taking into account: if one just means by the Euclidean plane a surface with Euclidean geometry on it, like for intance the complex plane CP1, it is a correct statement, however if one is referring to the projective plane R2, as it was pointed out there is not a one point compactification, one needs a circle to compactify it, and it is not homeomorphic to S2 but to its quotient by +/- the identity if I understood it right.

I'm not sure I would call those "nuances". It's more like ambiguities and people being unclear about which thing they are talking about. In your initial post, it appeared that you were just adding one point at infinity, which would just give you a 2-sphere.

But yes, if they are adding a whole circle at infinity, that's the non-orientable projective plane, which is not homeomorphic to the sphere.

 

[lavinia]

Well, continuity is not enough to preserve path-connectedness, but a homeomorphism is enough ; the topologists' sine curve is a(n) (counter) example for continuity alone. Naively, it would seem that if X is path connected and I have a path p between x_1, x_2 in X, and
$f: X \rightarrow Y$ is continuous, them f(p) is a path between$f(x_1) , f(x_2)$ . But of course, we would need a bijection to guarantee that every y in Y is the image of some x in X. And a homeomorphism gives us that.
So, to show homeos. preserve path connectedness; basically by pulling back paths bijectively:
Let h: X-->Y be a homeomorphism , with X path connected. Consider $y_1, y_2 \in Y$ . Then there are $x_1, x_2 \in X$, with $f(x_1)=y_1 , f(x_2)=y_2$ . By path-connectedness of X, there is a path $p : I \rightarrow X$ with $p(0)=x_1, p(1)=x_2$ ( let's assume I = [0,1] to simplify ). The image $h(p)$ is a path joining $y_1, y_2$ in $Y$ . Of course we need a bijection to use t5his result, which we get since $h$ is a homeomorphism.

A surjective continuous map suffices to preserve path connectedness.You do not need a bijection no less a homeomorphism.

 

[lavinia]

This was my initial understanding but after listening to the replies and looking into it there are some nuances that I think might be worth taking into account: if one just means by the Euclidean plane a surface with Euclidean geometry on it, like for intance the complex plane CP1, it is a correct statement, however if one is referring to the projective plane R2, as it was pointed out there is not a one point compactification, one needs a circle to compactify it, and it is not homeomorphic to S2 but to its quotient by +/- the identity if I understood it right.

Absolutely correct.

 

[WWGD]

I was talking about the 1-point compactification. If you do that to a 2-d Euclidean plane, you get a sphere. The sphere is not added. The plane becomes a sphere when you add the point at infinity.

I never said the sphere was added, I said that 1-pt compactifications can be done only on locally-compact Hausdorff spaces by adding a point, and that the result would be a topological sphere by , e.g., the inverse stereographic projection. I guess some types of surfaces can be embedded in complex projective n-space; I don't know the necessary conditions for this to happen. If these surfaces can be embedded in R^n , then I guess we can compose the embedded image can be composed with the inverse stereographic projection for an embedding into S^n. Maybe we can use Whitney embedding theorem : embed a surface in R^4 ( or lower ) , then do the 1-pt. compactification, then the composition of these two would give us an embedding into S^n. Anyone know if there are results for embedding surfaces in CP^n for n>1 ?

Absolutely correct.

Not quite; the geometry of CP^1 is locally-Euclidean, CP^1 being a manifold, but is not Euclidean; for one, CP^1 is compact ( being homeo. to S^2 ) , but Eucliodean space is not.

And the topologists sine curve is an example of a continuous surjection ( onto its image ) from a path-connected space ( the unit interval) , into the non-path-connected topologists sine curve. The OP asked whether path-connected was a topological property and that is what I answered.

 

[WWGD]

How about the surjection of the unit interval into the topologists sine curve?

This is in reply to Lavinia's claim that a continuous surjection is enough to preserve path-connectedness. Still trying to figure out how to use the quote and multi-quote.

 

[WWGD]

I am a bit confused about your notation: the real projective 2-space is usually noted RP^2 , or PR^2. It compact to start with, because it is the continuous image of S^2 under a continuous map of identifying a point with its antipode (it is a quotient map and quotient maps are by definition continuous.). If by R2 you refer to the Euclidean 2-plane, then it does have a 1-pt compactification ( into the 2-sphere ) because it is locally-compact and Hausdorff.

Sorry, this is in reply to post #16 by Tricky Dicky, I am still learning how to use the quote function here.

 

[WWGD]

I think an interesting question that follows from Tricky Dicky's OP is that of embeddability ( Tricky, let me know if you prefer that I post this on a separate thread ) about general results on compactifications and on embeddability. We have:

1) 1-point compactification: every locally-compact +Hausdorff space admits a 1-pt compactification.

2) According to Wiki, all affine spaces and affine varieties can be embedded in some projective space

--------------------------------------------------------------------------------------------------------------------
These two embeddings are compactifications; is the list in #2 exhaustive?

For general embeddability of manifolds, we have

3) For complex manifolds, AFAIK we have only Stein manifolds are embeddable in C^n , for some n.

4)Whitney's theorem for Real Manifolds.

Are there any other major results on compactifications and embeddings?

 

[lavinia]

Absolutely correct.

I

I never said the sphere was added, I said that 1-pt compactifications can be done only on locally-compact Hausdorff spaces by adding a point, and that the result would be a topological sphere by , e.g., the inverse stereographic projection. I guess some types of surfaces can be embedded in complex projective n-space; I don't know the necessary conditions for this to happen. If these surfaces can be embedded in R^n , then I guess we can compose the embedded image can be composed with the inverse stereographic projection for an embedding into S^n. Maybe we can use Whitney embedding theorem : embed a surface in R^4 ( or lower ) , then do the 1-pt. compactification, then the composition of these two would give us an embedding into S^n. Anyone know if there are results for embedding surfaces in CP^n for n>1 ?

Not quite; the geometry of CP^1 is locally-Euclidean, CP^1 being a manifold, but is not Euclidean; for one, CP^1 is compact ( being homeo. to S^2 ) , but Eucliodean space is not.

And the topologists sine curve is an example of a continuous surjection ( onto its image ) from a path-connected space ( the unit interval) , into the non-path-connected topologists sine curve. The OP asked whether path-connected was a topological property and that is what I answered.

I think trickydicky meant the complex plane since that is what he said.

You are mistaken about the topologists sine curve. If you examine your own proof that path connectedness is a topological invariant you see that you only used that the map was surjective.

 

[WWGD]

I

I think trickydicky meant the complex plane since that is what he said.

You are mistaken about the topologists sine curve. If you examine your own proof that path connectedness is a topological invariant you see that you only used that the map was surjective.

But how can the topologist's sine curve not be a counterexample? It is a continuous surjection (clearly not an injection, since sine is periodic) between the path-connected unit interval and the non-path connected topologist's sine curve? I will review my proof see if I made an assumption I am not aware of about injectivity.

 

[Terandol]

however if one is referring to the projective plane R2, as it was pointed out there is not a one point compactification, one needs a circle to compactify it, and it is not homeomorphic to S2 but to its quotient by +/- the identity if I understood it right.

This is not quite what was said. $RP^2$ (or $CP^2$) is a quotient of a sphere so it is already compact. So if you take the one point compactification (which definitely does exist as has already been mentioned since the space is locally compact Hausdorf) you just get $RP^2$ or $CP^2$ with an isolated point added so this is not interesting at all.

What the others were talking about was compactifying the usual nonprojective plane $\mathbb{R}^2$ in different ways to get either a sphere or a projective space as a result, not compactifying projective space itself. The example of attaching a circle to the Euclidean plane in a specific way gives you the projective plane which is not homeomorphic to the one point compactification $\mathbb{S}^2$ but it is still a compactification of the plane. How you attach the circle is extremely important though since there are many ways to attach a circle to an open disc (which is the same as attaching it to $\mathbb{R}^2$ topologically) and the result is usually not the same space. Take a look at the dunce cap for another way of attaching a circle to get something topologically different.

In fact, now that I think about it, it seems every compact connected surface (orientable or not) is the compactification of $\mathbb{R}^2$ since it can be obtained by gluing a circle (usually represented as the boundary of a polygon) to the disc in a particular way and clearly the original disc is dense in the resulting surface.

But how can the topologist's sine curve not be a counterexample? It is a continuous surjection

You need to include the point $0$ for the topologists sine curve to be non-path connected. You can't extend the map from $(0,1]$ to the topologists sine curve including $0$ in a continuous way.

2) According to Wiki, all affine spaces and affine varieties can be embedded in some projective space

Since you included affine varieties in your previous discussion of embeddings, there are certainly many more embedding type theorems of varieties into projective space. For example, every smooth projective variety of dimension $n$ can be embedded in $\mathbb{P} ^{2n+1}$ (if I recall correctly this requires you to work over an algebraically closed field but I'm not sure if it extends to more general fields or not.)

In fact the idea of very ample line bundles is precisely one way of giving a variety (the base) enough information (the sections of the sheaf) to embed it into projective space. Of course there is some nonuniqueness to take care of here if you want to set up a correspondence with very ample line bundles and embeddings of varieties into projective space but I seem to remember this being straightforward to deal with. The Kodaira embedding theorem is another such result you may want to look into if interested. I'm sure there are many more but none are coming to mind at the moment.

 

[TrickyDicky]

¿

I never said the sphere was added, I said that 1-pt compactifications can be done only on locally-compact Hausdorff spaces by adding a point, and that the result would be a topological sphere by , e.g., the inverse stereographic projection. I guess some types of surfaces can be embedded in complex projective n-space; I don't know the necessary conditions for this to happen. If these surfaces can be embedded in R^n , then I guess we can compose the embedded image can be composed with the inverse stereographic projection for an embedding into S^n. Maybe we can use Whitney embedding theorem : embed a surface in R^4 ( or lower ) , then do the 1-pt. compactification, then the composition of these two would give us an embedding into S^n. Anyone know if there are results for embedding surfaces in CP^n for n>1 ?

Not quite; the geometry of CP^1 is locally-Euclidean, CP^1 being a manifold, but is not Euclidean; for one, CP^1 is compact ( being homeo. to S^2 ) , but Eucliodean space is not.

And the topologists sine curve is an example of a continuous surjection ( onto its image ) from a path-connected space ( the unit interval) , into the non-path-connected topologists sine curve. The OP asked whether path-connected was a topological property and that is what I answered.

That was just obviously a mistake, I was referring to the complex line that is compactified to CP1.
Nowhere in the OP I asked about path-connectedness, my doubt was clarified and it is best for the rest of discussions to start their own thread to avoid further confusion like Terandol interpreting I was trying to compactify the projective plane.
Thanks everyone for your inputs.

 

[Terandol]

I was referring to the complex line that is compactified to CP1.

Sorry to drag this thread out even further than it already has been but this quote gave me the idea that the confusion may have come because of the difference between complex projective space and real projective space and everybody has been misinterpreting everybody else in this thread. I may just be misinterpreting you again and if so just ignore this post but I don't think anybody has explicitly stated that the complex projective line is exactly the same thing as the 2 dimensional sphere in this thread yet which may be what your sources were saying to begin with.

If your book/quotes stated that the one point compactification of $\mathbb{C}^1$ (or equivalently the real plane) is the complex projective line $\mathbb{C}P^1$ then this is completely accurate since $\mathbb{C}P^1 \cong S^2$ so it is orientable. Lavinia didn't say that this wasn't true, but rather he showed you how to compactify $\mathbb{C}^1$ in a different way by adding a circle to get the real projective plane $\mathbb{R}P^2$ which is a completely different object than $\mathbb{C}P^1$ even though they are both 2-dimensional. This is not the one point compactification though and this space is not orientable.

 

[WWGD]

O.K, good, thanks, Terandol, I was just having trouble reconciling two apparently-contradictory statements of the Topologists sine curve and the fact that surjective continuous maps preserve path-connectedness. I guess I could have been more careful reading on the Topologists' sine curve.