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This isn't a homework problem, just working through Hempel's 3-Manifolds for my own benefit. One exercise is to show that the lens space $L(2k,q)$ contains a surface of Euler characteristic $2-k$.

I've been attempting to come up with an explicit construction. I'm reasonably sure I have a construction that works for $L(2k,1)$: I think that (please, correct me if this is wrong?) this lens space can be constructed by a quotient on a closed 3-ball as follows. Draw parallel arcs on the boundary from the north to the south pole which wrap around the ball $k$ times, and identify the north and south pole, so each $S^1$ wraps $2k$ times. Now identify "$k$-antipodal" points, i.e. those lying on one curve by a homeomorphism that identifies antipodal points on the $S^1$ that is that curve.

(EDIT: I should clarify that "wraps $k$ times" means that if you project it to the equator, it represents the $k\in \mathbb{Z}\cong\pi_1(S^1)$ of the equatorial $S^1$.)

When $k=0$, this clearly just gives $S^3$, and when $k=1$, it's easy to see that this gives $\mathbb{RP}^3$. If we carve a vertical core out of this space, it seems to give a Heegaard splitting for the Lens space $L(2k,1)$, if I've drawn my pictures right (this was the motivation for the above construction).

Now, the surface is just the equatorial disk inside of the 3-ball. Along the boundary of this disk, the $k$-antipodal identification restricts to identifying $2k$ regions in such a way that makes the disk a connect sum of $k$ copies of $\mathbb{RP}^2$, so the Euler characteristic is $1+k-2k+1=2-k$.

My questions are: is the above construction of a lens space correct? Does it generalize to $L(2k,q)$ in a way that allows one to still easily see the surface in question? Is there a non-constructive proof?

Thank you for your help,

-GM

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You can also do the following. Mark $2k$ points on the equator of the 3-ball, and then connect them all to both the north and south poles. Identify the poles, and then identify these equatorial points by "adding $q$ "; that is, point $1$ is identified with point $q+1$ (reading clockwise), etc., modulo $2k$. This is the lens space $L(2k,q)$. The equatorial disk is, once again, your surface. –  user641 Dec 31 '12 at 20:20
    
Oh, I see -- that's like gluing the north hemisphere to the south by a rotation of $\pi q/k$, yes? And then the Heegaard surface is also very easy to see. Thank you! –  gmoss Dec 31 '12 at 21:47

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