The real projective plane has the property that if you divide into two different 2-manifolds, they will not be homeomorphic (i.e. one will be orientable, and the other non-orientable).

The sphere does not have this property. It can be divided into two homeomorphic disks.

My question is, what 2-manifolds can't be divided into equal (homeomorphic) parts (where the parts are 2-manifolds)?

Note: We say that $M$ can be divided into manifolds $A$ and $B$ if $A \cup B = M$ and $\operatorname{int}(A \cap B) = \emptyset$.

(The interesting thing about these spaces is you can play a game like Projex on them. Simply choose some anti-reflexive total relation on manifolds, and a tiling on the manifold. Plays take turn coloring tiles, and when the board is filled, the relation determines the winner.)

  • $\begingroup$ closed surfaces with odd genus $\endgroup$ – Max Mar 1 '18 at 21:42
  • $\begingroup$ @Max Uhm, the torus has odd genus, and it can be decomposed into two cylinders. $\endgroup$ – PyRulez Mar 1 '18 at 21:43
  • $\begingroup$ Ah, didn't realize you wanted to use multiple curves. $\endgroup$ – Max Mar 1 '18 at 21:44
  • $\begingroup$ @Max The pieces don't necessarily need to be connected, even. $\endgroup$ – PyRulez Mar 1 '18 at 21:45
  • $\begingroup$ Every closed orientable surface admits a closed curve (which intersects itself many times and) whose complement is a union of disks. If you get an odd number of disks, you can just draw an extra line segment cutting a disk into two. So it's always possible, assuming you are ok with the parts being non-connected. $\endgroup$ – Max Mar 1 '18 at 21:47

You don't want $A \cap B = \varnothing$; you want $\partial A = \partial B = A \cap B$.

Recall that the Euler characteristic is well-behaved under decomposition: if $X = A \cup B$, then $\chi(X) = \chi(A) + \chi(B) - \chi(A \cap B)$. If a (compact without boundary) surface $\Sigma$ can be written as $\Sigma = A \cup B$ where $A \cong B$ is a (compact) surface. The above formula gives $\chi(\Sigma) = 2\chi(A) - \chi(A \cap B)$. But $A \cap B$ is a disjoint union of circles, and the Euler characteristic of a circle is zero. So $\chi(\Sigma) = 2\chi(A)$, and in particular is even.

If $\Sigma_g$ is the surface of genus $g$ (that is, the connected sum of $g$ copies of $T^2$), and $N_g$ the connected sum of $g$ copies of $\Bbb{RP}^2$, then $\chi(\Sigma_g) = 2 - 2g$. Indeed, you can decompose this into two pieces (perhaps the easiest way is to slice it in half along the $xy$-plane, in the usual presentation of these surfaces: image). $\chi(N_g) = 2-g$, however, and so only $N_{2g}$ can be a 'double'. If $N_{g,1}$ is what you get when you delete a small open disc from $N_g$, so that it has one boundary circle, $N_{g,1} \cup_{\partial} N_{g,1} = N_{2g}$ (we're just doing the connected sum operation!)

Punchline: Only the connected sum of an odd number of copies of $\Bbb{RP}^2$ have this property.

  • $\begingroup$ Quick question: If you take a n-torus, and identify anti-podal points, do you get something equivalent to an odd number of real projective planes connected together? $\endgroup$ – PyRulez Mar 3 '18 at 7:17
  • $\begingroup$ I'm confused. The real projective plane can be divided into three disks, which have euler characteristic 1. Why wouldn't this make $\mathbb RP^2$ have a euler characteristic of 3? $\endgroup$ – PyRulez Mar 5 '18 at 6:50
  • $\begingroup$ Also, what if you glue $\mathbb{RP}^2$ and $T^2$ together, or other combinations like that? $\endgroup$ – PyRulez Mar 5 '18 at 6:58
  • $\begingroup$ $\Bbb{RP}^2 \# T^2 = \Bbb{RP}^2 \# K = N_3$. In fact, the "adding a tube" that happens when you increase the number of boundary components you're gluing together, is just the operation of connected sum with $T^2$, and this formula explains why the answer never changes parity. $\endgroup$ – user98602 Mar 5 '18 at 14:57
  • $\begingroup$ I'm not sure precisely what 3 discs you're thinking of, but you'd need to apply the formula above twice to find the Euler characteristic of the whole space, or an inclusion exclusion formula. When you include all of the intersections, many of which are probably arcs and points instead of circles, you will get the correct Euler characteristic. $\endgroup$ – user98602 Mar 5 '18 at 15:00

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