# Extended plane as a topological sphere

## Main Question or Discussion Point

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

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lavinia
Gold Member
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.

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.

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

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
Gold Member
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
Gold Member
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.

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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
Gold Member
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.

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WWGD
Gold Member
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
Gold Member
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.

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
Gold Member
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 [Broken])

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WWGD
Gold Member
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.

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.

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.

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
Gold Member
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
Gold Member
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
Gold Member
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
Gold Member
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.

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WWGD
Gold Member
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
Gold Member
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

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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
Gold Member
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