Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

• Forgot your password?
• Apply for membership

1. • CommentRowNumber1.
   • CommentAuthorMurat_Aygen
   • CommentTimeJan 11th 2017
   • PermaLink
   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
   • PermaLink
   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, . So besides , we also need to know that . (The is modding out by a free action of the group of order , 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
   • PermaLink
   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)
   • PermaLink
   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 , removal of a point from the projective plane gives the tautological line bundle over , and in the real case this bundle is identifiable with the open Moebius strip as a bundle over (e.g., consult this WP note).

   In slightly greater detail: using homogeneous coordinates to represent points in , we can identify a copy of with those points that are represented by (“line at infinity”). The tautological line bundle is identified with the projection sending ; geometrically, this takes a point in the projective plane (not equal to the origin ) 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 ), then this restricts to the familiar projection where each fiber is .

   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 . If you remove the part where , 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 with the endpiece where , as those points are identified in .

5. • CommentRowNumber5.
   • CommentAuthorTobyBartels
   • CommentTimeJan 13th 2017
   • PermaLink
   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)
   • PermaLink
   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 is the Thom space of the tautological line bundle over ).

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