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What is the maximum number $c$ (cut-number) of non-intersecting (edit: two-sided) circles on a Klein bottle $N_2$ and, in general, a surface $N_h$ with $h$ Möbius strips, such that cutting by these circles leaves $N_h$ connected?

This should be (or is it?) the same as co-rank (maximum rank of a free homomorhic image) of its fundamental group $$G=\pi_1(N_h)=\langle a_1,a_2,\dots,a_h\mid a_1^2a_2^2\dots a_h^2=1\rangle.$$ Obviously $corank\ G\ge\left[\frac h2\right]$, but is it larger?

For an orientable surface of genus $M_g$, this is $g$.

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It is true that $c=\lfloor \frac{h}{2} \rfloor$, here's a proof.

Cutting along a circle does not change the Euler characteristic. So the connected surface-with-boundary that you get from $N_h$ by cutting along those circles, which I will denote $P_h$, has the same Euler characteristic as $N_h$: $$\chi(P_h) = \chi(N_h) $$ This Euler characteristic equals $2-h$ because, as you say, $N_h$ "has $h$ Mobius strips" which I interpret as saying that the maximal number of pairwise disjoint Mobius bands in $N_h$ equals $h$, which is equivalent to saying that $N_h$ is obtained from a sphere with $h$ holes by identifying the boundaries of those holes with the boundaries of $h$ Mobius strips.

Also, the surface $P_h$ has $2c$ boundary components. When you cap off each of these circles by sewing on a disc then the resulting closed surface, which I'll denote $Q_h$, has Euler characteristic $$\chi(Q_h) = \chi(P_h) + 2c = 2-h+2c $$

Now there's two cases.

Suppose $Q_h$ is orientable. It must be a sphere and so $\chi(Q_h)=2$, because otherwise $Q_h$ has a nonseparating 2-sided circle, leading to an additional nonseparating 2-sided circle in $N_h$, a contradiction. We have $2-h+2c=\chi(Q_h)=2$ and so $h$ is even and $$c = \frac{h}{2} = \biggl\lfloor \frac{h}{2} \biggr\rfloor $$

Suppose $Q_h$ is nonorientable. It must be a projective plane and so $\chi(Q_h)=1$; this is for the same reason as before, because otherwise $Q_h$ contains an additional nonseparating 2-sided circle (since $Q_h$ must contain a one-holed Klein bottle). We have $2-h+2c=\chi(Q_h)=1$ and so $h$ is odd and $$c = \frac{h}{2} - \frac{1}{2} = \biggl\lfloor \frac{h}{2} \biggr\rfloor $$

One thing I find interesting about this question is that it reveals something I did not know, namely that cutting $N_h$ along the maximal 2-sided curve system results in an orientable surface if and only if $h$ is even.

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  • $\begingroup$ Great! "Cutting $N_h$ along the maximal 2-sided curve system results in an orientable surface if and only if h is even" -- this is seen from the fact that $N_h$ is a sphere with $[\frac h2]$ inverted handles plus (for odd $h$) one Möbius strip. $\endgroup$ – Alexander Gelbukh Jan 1 '15 at 4:53
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Corank $r$ cannot be larger than $c$. Otherwise, consider a PL map $f$ from $N_h$ to the bouquet $B$ of $r$ circles which induces the epimorphism $\pi_1(N_h)\to F_r$. Next, take preimages of generic points in the edges of $B$ under $f$. Due to genericity of the points, the preimages form a 2-sided 1-dimensional submanifold $L$ in $N_h$. Some of the circles in $L$ can be homologous to each other or null-homologous. Removing the homologous duplicates and null-homologous cicrles, we obtain a system $L'$ of $\ge r$ circles in $N_h$ cutting along which results in a connected surface. To relate to $h$, use Lee Mosher's answer or the one I gave to this question.

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