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The Euler characteristic of a two-dimensional disk is $\chi=1$. If one blindly interprets the disk as a closed, orientable surface, then $\chi = 2 - 2g$, and the genus is $g=\frac{1}{2}$.

Is there some way to view a disk as possessing "half a hole" or "half a handle"?

My students asked me and I didn't have a good answer.

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Bring something shiny to class and distract them... That's the best approach. –  Mariano Suárez-Alvarez Nov 18 '11 at 2:34
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Things work differently for surfaces with boundary, is all I would say. –  Greg Martin Nov 18 '11 at 2:51
    
I guess if you glue two of them together along the boundary, you get a sphere (single handle) so it could kind of make sense in that way, but that isn't consistent with the application of the genus formula (the genus doesn't become 1). If you glue two together at a point, its the same thing up to homotopy, so it definitely doesn't become a handle in that situation. –  Carl Nov 18 '11 at 2:51

2 Answers 2

up vote 9 down vote accepted

The connected sum of two disks is an annulus. If you think of an annulus as being a hole, then I suppose a disk is half a hole.

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Very clever to connect to annulus! –  Joseph O'Rourke Nov 18 '11 at 14:43

Trying to use $\chi = 2 - 2g$ to describe things that aren't closed orientable surfaces is missing the point, I think. In my opinion one should think of the Euler characteristic of a compact space as a homotopy-invariant refinement of the cardinality of a finite set; see this blog post. A closed disk is contractible, so has Euler characteristic $1$, and that's the most transparent interpretation of it. You might also be interested in the argument in the blog post that derives $\chi = 2 - 2g$ from homotopy-invariance and inclusion-exclusion.

The thing that possesses "half a hole" isn't the closed disk; if anything, it's $\mathbb{R}P^2$, which also has Euler characteristic $1$. And this is totally sensible as it can be described as the quotient of $S^2$ by an action of $\mathbb{Z}/2\mathbb{Z}$.

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Isn't a disk a quotient of $S^2$ by a (different) action of ${\bf Z}/2{\bf Z}$? –  Gerry Myerson Nov 18 '11 at 5:10
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@GerryMyerson, fixed points make an enourmous difference! –  Mariano Suárez-Alvarez Nov 18 '11 at 5:28
    
Nice point about $\mathbb{R}P^2$. And wonderful insights in your blog post---Thanks! –  Joseph O'Rourke Nov 18 '11 at 14:40

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