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This is a follow-up question to: Degree 2 branched map from the torus to the sphere

This is a silly computation, but for whatever reason this is taking me much, much longer than it should. So hopefully more geometrically-oriented people can answer this easily.

Let's say we have a 2-cover of S2 branched at 4 points. We can visualize this, as stankewicz aptly put it, as having a torus, putting it on a skewer (that meets it in four points), and quotienting by the action of turning the torus by a 180 degrees around that skewer. If you prefer to think of the torus as the plane quotiented by a lattice, it's the same as identifying vectors with their minuses (the ramification points being the 2-torsion).

As we know $\pi_1$(Torus - 4 points,basept)$\cong$< a,b,c,d,e,f| [a,b]cdef=1>, and $\pi_1$(S2 - 4 points,basept)$\cong$< g,h,x,w| ghxw=1>. This 2-cover corresponds to an identification, therefore, of < a,b,c,d,e,f| [a,b]cdef=1> with an index 2 subgroup of < g,h,x,w| ghxw=1>. For the life of me, I can't figure out how this identification would go! I've been drawing tori to no avail for two days now.

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To be precise: the question is what would be an assignment for A, b, d, e and f that would give an index 2 subgroup? –  Randy Brown Feb 4 '10 at 20:39

4 Answers 4

In standard topological terms, the exact sequence that relates homotopy groups of the base $B$, fiber $F$ and total space $E$ of topological fibration gives

$$\pi_1(F) \to \pi_1(E) \to \pi_1(B) \to \pi_0(F),$$

that is,

$$ 0\to \pi_1(T-4) \to \pi_1(S^2-4) \to \mathbb Z_2.$$

The middle map works by taking a loop above and pushing it to the base; the right map works by taking a loop on $S^2-4$, lifting it as a path, and taking 0 or 1 depending on whether the resulting path is closed or not.

I believe the OP described the fundamental group of the torus as $\left< a, b, c, d, e, f| [a, b]cdef = 1\right>$ where $a, b$ are two circles of the torus and $c, d, e, f$ are four loops around the holes. For $S^2-4$ my suggestion would be to use $\left< C, D, E, F| CDEF = 1\right> = \left< C, D, E\right>$ where $c$ is over $C$ etc (rather then $g, h, x, w$).

Now, since the loop around $c$ on torus has to wind twice around it when projected to the sphere (think about complex $z\mapsto z^2$ map) it's easy to see that $c\mapsto C^2$, $d\mapsto D^2$, $e\mapsto E^2$, $f\mapsto F^2$.

What about $a$ and $b$? Careful observer should note that it's a bit tricky to define the loops. E.g, if you move $a$ parallel to itself, you'll get a new $a'$ which would differ by something like $cd$ depending on which points are where and depending on how you draw the basepoints on the loops.

For the exact calculations one should fix the torus to be $\mathbb R\times \mathbb R/\mathbb Z\times\mathbb Z$ so that fixed points of $z\mapsto z$ are the vertices $c = (0, 0)$, $d = (1/2, 0)$, $e = (0, 1/2)$, $f = (1/2, 1/2)$. Moreover, you should now select some basepoint and draw the cycles around $c, d, e, f$ so that $cdef = 1 $.

Unfortunately, from the picture it's not easy to say where some simple loops like horizontal or vertical go. While in homology they seem to be $C+D$ and $D+F$, one has to draw them really carefully with the basepoints. I couldn't do that, but here's something different instead.

I tried to exhibit some expressions $a, b$, which may be not exactly the cycles above, but which nevertheless satisfy $[a, b]cdef \mapsto 1$. In other words, these $a, b$ will be generators, but different ones.

I was able to make $a = CDEC^{-1}, b = CCDC^{-1}$ work:

$$CDEC^{-1}CCDC^{-1}(CDEC^{-1})^{-1}(CCDC^{-1})^{-1}CCDDEE(CDECDE)^{-1} = $$

$$ = CDEC^{-1}CCDC^{-1}CE^{-1}D^{-1}C^{-1}CD^{-1}C^{-1}C^{-1}CCDDEE(CDECDE)^{-1} = $$

$$ = CDEC^{-1}CCDE^{-1}D^{-1}D^{-1}DDEE(CDECDE)^{-1} = $$

$$ = CDECDE (CDECDE)^{-1} = 1 $$

I think this is more-or-less the explicit map you're asking for!

Finally, note that the map $ \left< C, D, E, F| CDEF = 1\right> \to \mathbb Z_2$ is given by counting all the letters modulo 2 (consistent because $F = 1 = 3 = (CDE)^{-1}$), so the image of the map discussed above should contain exactly expressions with even number of letters.

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The sphere minus 4 points is doubly covered by the torus minus 4 points. This double cover gives a representation of $\pi_1(S^2\setminus\mbox{4 points})$ in the symmetric group on two letters. Namely, each of your generators $$g,h,x,w$$ gives a transposition. The kernel of this representation is the image of the fundamental group pf the torus minus 4 points. (Usually there is an ambiguity coming from the choice of base point in the covering space, but not in this example, since the covering is regular.) The kernel is formed by all words containing an even number of letters.

[had to make one of the formulas display style, since it wouldn't show otherwise]

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So what are you saying a, b, c, d, e and f go to? –  Randy Brown Feb 4 '10 at 20:12
    
Randy, that would depend on how exactly you define these (which you don't explain), but I would imagine that $c,d,e,f$ are loops around the singular points of the branched covering. Note that they can be chosen in many ways, but we can choose them so that each is taken to the square of a generator of $\pi_1$ of the sphere minus 4 pts. The choice of $a$ and $b$ is also ambiguous, but one can choose them so that (in the picture you mention) the image of $a$ would separate two top points from two bottom ones and the image of $b$ would separate the top one and the bottom one from the remaining two –  algori Feb 4 '10 at 20:32
    
Can you give an explicit example of such an assignment? I can't make it work. –  Randy Brown Feb 4 '10 at 20:40

Draw a square, and view the torus using the standard glueing of opposite edges. Say the hyperelliptic involution is given by a half rotation about the center of the square, and remove the 4 fixed points. Choose a basepoint somewhere in the interior of the resulting object (say, halfway between a corner and the center), and consider the set of paths from the basepoint to its image under the half rotation. These are the loops in the quotient space that map to the nontrivial element in the quotient group of order two. There are 8 classes of such paths of roughly minimal length (corresponding to generators of $\pi_1$ of the punctured sphere), but the half rotation puts pairs of them into correspondence. If you choose the correct elements g,h,x,w in each pair, they will compose to identity. You can then draw your preferred generators of the fundamental group of the torus as composites of these paths.

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Perhaps I'm misunderstanding, but $G=\pi_1(S^2-4\text{ pts})$ is free on 3 generators, and $H=\pi_1(T^2-4\text{ pts})$ is free on 5 generators. By Nielsen-Schreier, every subgroup of index 2 in $G$ is free on 5 generators, and by considering set maps $\{g,h,x\}\rightarrow \{0,1\}$ there are seven such subgroups. The one Ilya considers above is the subgroup generated by (in his notation) $\langle CD,CE,DC,DE,EC\rangle$.

The point of saying these groups are free is that there's nothing to check once you've defined the maps on the generating set.

Steve

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"The point of saying these groups are free is that there's nothing to check once you've defined the maps on the generating set." There are plenty of maps from F_5 to F_3 that don't realise a subgroup of index two. So there is something to check. –  HJRW Feb 4 '10 at 23:23
    
The point I was making is that you can simply map {a,b,c,d,e} to {CD,CE,DC,DE,EC} and rest assured it's a group isomorphism. –  Steve D Feb 4 '10 at 23:26
    
It's true, but the question was stated in geometric terms, so the answer should also include identification of letters with cycles. You can't have this for free: if your map is that simple, the expressions for cycles must be a bit more involved then mine :) –  Ilya Nikokoshev Feb 4 '10 at 23:34
    
So I was misunderstanding :) I thought the question was about an explicit embedding of (my) H into (my) G, based on the covering given in the question. Both your and algori's answer describe this subgroup as a kernel of a particular map, and I gave a presentation for the kernel. So together that yields an explicit embedding. –  Steve D Feb 4 '10 at 23:40

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