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1. • CommentRowNumber1.
   • CommentAuthorMurat_Aygen
   • CommentTimeJan 11th 2017
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   Author: Murat_Aygen
   Format: TextThe identity RP^2 − {0} = μ is as real as the familiar C + {∞} = S^2 where RP^2 is the Projective Plane, 0 is its natural origin, μ is the Mobius Strip, C is the Complex Plane and S^2 is the sphere. I can give a formal proof of this if anyone is interested.

   The identity RP^2 − {0} = μ is as real as the familiar C + {∞} = S^2 where RP^2 is the Projective Plane, 0 is its natural origin, μ is the Mobius Strip, C is the Complex Plane and S^2 is the sphere. I can give a formal proof of this if anyone is interested.
2. • CommentRowNumber2.
   • CommentAuthorTobyBartels
   • CommentTimeJan 13th 2017
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   Author: TobyBartels
   Format: MarkdownItexI see, $\mathbb{RP}^2 = S^2/2 = (\mathbb{C} + \{\infty\})/2 = (\mathbb{R} \times S^1 + \{0\} + \{\infty\})/2 = (\mathbb{R} \times S^1)/2 + \{0,\infty\}/2 = \mu + \{[0]\}$. So besides $\mathbb{C} + \{\infty\} = S^2$, we also need to know that $\mathbb{R}^1 \times S^1 + \{0\} = \mathbb{C}$. (The $/2$ is modding out by a free action of the group of order $2$, so there\'s nothing fishy going on there.) So treating the Möbius strip as being open along its edge, its [[Alexandrov compactification]] is the projective plane. A neat fact.

   I see, $\mathbb{RP}^2 = S^2/2 = (\mathbb{C} + \{\infty\})/2 = (\mathbb{R} \times S^1 + \{0\} + \{\infty\})/2 = (\mathbb{R} \times S^1)/2 + \{0,\infty\}/2 = \mu + \{[0]\}$. So besides $\mathbb{C} + \{\infty\} = S^2$, we also need to know that $\mathbb{R}^1 \times S^1 + \{0\} = \mathbb{C}$. (The $/2$ is modding out by a free action of the group of order $2$, so there's nothing fishy going on there.)

   So treating the Möbius strip as being open along its edge, its Alexandrov compactification is the projective plane. A neat fact.

3. • CommentRowNumber3.
   • CommentAuthorDavidRoberts
   • CommentTimeJan 13th 2017
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   Author: DavidRoberts
   Format: MarkdownItex@Toby might the example then go at [[Alexandrov compactification]]? Nice to see something familiar (a manifold, even!) that isn't a sphere.

   @Toby might the example then go at Alexandrov compactification? Nice to see something familiar (a manifold, even!) that isn’t a sphere.

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeJan 13th 2017
   • (edited Jan 13th 2017)
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   Author: Todd_Trimble
   Format: MarkdownItexNice, Toby. I believe what is true more generally is that for any field $k$, removal of a point from the projective plane $\mathbb{P}^2(k)$ gives the tautological line bundle over $\mathbb{P}^1(k)$, and in the real case this bundle is identifiable with the open Moebius strip as a bundle over $\mathbb{P}^1(\mathbb{R}) \cong S^1$ (e.g., consult this [WP note](https://en.wikipedia.org/wiki/Tautological_bundle#cite_note-5)). In slightly greater detail: using homogeneous coordinates $[x: y : z]$ to represent points in $\mathbb{P}^2$, we can identify a copy of $\mathbb{P}^1$ with those points that are represented by $[x: y: 0]$ ("line at infinity"). The tautological line bundle is identified with the projection $\mathbb{P}^2 \setminus \{[0: 0: z]: z \neq 0\} \to \mathbb{P}^1$ sending $[x: y: z] \mapsto [x: y: 0]$; geometrically, this takes a point $[x: y: z]$ in the projective plane (not equal to the origin $[0: 0: z]$) and maps it to the point where the line through that point and the origin meets the line at infinity. If you remove the zero section (where $z = 0$), then this restricts to the familiar projection $\mathbb{A}^2 \setminus \{(0, 0)\} \to \mathbb{P}^1$ where each fiber is $k^\times$. If my memory is correct, you can see a nice picture of this somewhere in Hartshorne, where he draws a picture of the blowing up of the origin in the projective plane, as the locus of $\{((x, y), [z: w]) \in \mathbb{A}^2 \times \mathbb{P}^1: x z = y w\}$. If you remove the part where $(x, y) = (0, 0)$, then his picture looks just like an open Moebius strip. Edit: Found it [here](https://books.google.com/books?id=7z4mBQAAQBAJ&pg=PA37&lpg=PA37&dq=hartshorne+blowing+up+point+in+plane&source=bl&ots=OO82mstinO&sig=OrzDSxKwUUujAtsZqH3TrSuu5PU&hl=en&sa=X&ved=0ahUKEwi3l6nPob_RAhWm44MKHQCdCG0Q6AEILjAD#v=onepage&q=hartshorne%20blowing%20up%20point%20in%20plane&f=false), page 29 if you can see it. For this picture, you have to mentally paste together the endpiece where the horizontal coordinate $x = +\infty$ with the endpiece where $x = -\infty$, as those points are identified in $\mathbb{P}^1(\mathbb{R})$.

   Nice, Toby. I believe what is true more generally is that for any field $k$, removal of a point from the projective plane $\mathbb{P}^2(k)$ gives the tautological line bundle over $\mathbb{P}^1(k)$, and in the real case this bundle is identifiable with the open Moebius strip as a bundle over $\mathbb{P}^1(\mathbb{R}) \cong S^1$ (e.g., consult this WP note).

   In slightly greater detail: using homogeneous coordinates $[x: y : z]$ to represent points in $\mathbb{P}^2$, we can identify a copy of $\mathbb{P}^1$ with those points that are represented by $[x: y: 0]$ (“line at infinity”). The tautological line bundle is identified with the projection $\mathbb{P}^2 \setminus \{[0: 0: z]: z \neq 0\} \to \mathbb{P}^1$ sending $[x: y: z] \mapsto [x: y: 0]$; geometrically, this takes a point $[x: y: z]$ in the projective plane (not equal to the origin $[0: 0: z]$) and maps it to the point where the line through that point and the origin meets the line at infinity. If you remove the zero section (where $z = 0$), then this restricts to the familiar projection $\mathbb{A}^2 \setminus \{(0, 0)\} \to \mathbb{P}^1$ where each fiber is $k^\times$.

   If my memory is correct, you can see a nice picture of this somewhere in Hartshorne, where he draws a picture of the blowing up of the origin in the projective plane, as the locus of $\{((x, y), [z: w]) \in \mathbb{A}^2 \times \mathbb{P}^1: x z = y w\}$. If you remove the part where $(x, y) = (0, 0)$, then his picture looks just like an open Moebius strip.

   Edit: Found it here, page 29 if you can see it. For this picture, you have to mentally paste together the endpiece where the horizontal coordinate $x = +\infty$ with the endpiece where $x = -\infty$, as those points are identified in $\mathbb{P}^1(\mathbb{R})$.

5. • CommentRowNumber5.
   • CommentAuthorTobyBartels
   • CommentTimeJan 13th 2017
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   Author: TobyBartels
   Format: MarkdownItexNice, Todd! Here is a more direct link to the picture in Hartshorne: <https://books.google.com/books?id=7z4mBQAAQBAJ&pg=PA29>.

   Nice, Todd! Here is a more direct link to the picture in Hartshorne: https://books.google.com/books?id=7z4mBQAAQBAJ&pg=PA29.

6. • CommentRowNumber6.
   • CommentAuthorTodd_Trimble
   • CommentTimeJan 14th 2017
   • (edited Jan 14th 2017)
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   Author: Todd_Trimble
   Format: MarkdownItexActually, under [[one-point compactification]], we *do* sort of have this example already; look under Thom space (projective space $\mathbb{RP}^{n+1}$ is the [[Thom space]] of the tautological line bundle over $\mathbb{RP}^n$). Edit: I went ahead and stuck that in at [[one-point compactification]].

   Actually, under one-point compactification, we do sort of have this example already; look under Thom space (projective space $\mathbb{RP}^{n+1}$ is the Thom space of the tautological line bundle over $\mathbb{RP}^n$).

   Edit: I went ahead and stuck that in at one-point compactification.