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I guess this question is more related to my understanding of the amalgamated product of free groups than to the underlying topology.

The goal is to compute the fundamental group of the 2-holed torus (i.e. the connected sum of 2 tori, $T^2 \# T^2$). I want to apply Van Kampen's theorem, and my decomposition is the following : take $U_1$ to be the first torus plus some overlap on the second one, $U_2$ to be the second torus plus some overlap on the first one, and $U_0 = U_1 \cap U_2$ to be the "tube" joining the two.

Unless I am mistaken, we have that :

  • $U_1$ and $U_2$ are homotopy equivalent to $T^2$ minus a point, and thus $\pi_1(U_1)=\pi_1(U_2)= \mathbb{Z} * \mathbb{Z}=F(\left\{a,b\right\})$, the free group on 2 generators.

  • $U_0$ retracts to a circle, and $\pi_1(U_0)=\mathbb{Z}$.

Thus by Van Kampen, $$ \pi_1(T^2 \vee T^2)= F(\left\{a,b\right\}) *_{\mathbb{Z}}F(\left\{a,b\right\}) $$ where $*_\mathbb{Z}$ denotes the free product with amalgation of $\pi_1(U_0)=\mathbb{Z}$. But this is as far as I can go in my description of this fundamental group. In a more general setting the free product with amalgation of $\pi_1(U_0)$ makes sense to me : it is simply the free product of the groups $\pi_1(U_1)$ and $\pi_1(U_2)$ with the added property that we can concatenate elements if they belong in the intersection. However in this very practical case I am unable to give any information on this product.

This question, for example, gives the Cayley graph of the fundamental group. How can such generators/relations be recovered from the expression given above ?

Any help would be appreciated.

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  • $\begingroup$ Are you familiar with the "cut and paste" diagram of a torus? Here's a link math.cornell.edu/~mec/Winter2009/Victor/part4(7).png concretely in this case you may think of it in the following way: the group for both $U_i$ is (respectively) generated by the borders of the squares, which are loops around the internal hole of the torus and "around the handle"; the generator of the group of $U_1\cap U_2$ is the blue loop. You only need to write it first using the generators of the first group, then the generators of the second group, and this gives you the final relation $\endgroup$ – TheMadcapLaughs Oct 15 '17 at 16:40
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You are correct that that $U_i=T^2-D^2\simeq S^1\vee S^1$, so $\pi_1(U_i)\cong \mathbb{Z}\ast\mathbb{Z}$. Writing $\pi_1(U_1)=F(a,b)$ and $\pi_1(U_2)=F(c,d)$ for the free groups we have that the inclusion map $\omega_i:S^1\simeq U_0\hookrightarrow U_i\simeq S^1\vee S^1$ has opposite orientations for $i=1,2$, but this is irrelevant here since we can just re-choose our generators. Now $\omega_i$ is given by the Whitehead product since this is the attaching map for the top cell of $T^2$. It is represented by $\omega_1=aba^{-1}b^{-1}$ and $\omega_2=cdc^{-1}d^{-1}$. Then

$\pi_1(T^2\sharp T^2)\cong F(a,b)\ast_{\mathbb{Z}}F(c,d)\cong F(a,b,c,d)/[\omega_1\cdot\omega_2]\cong\left<a,b,c,d\mid aba^{-1}b^{-1}cdc^{-1}d^{-1}=1\right>$

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The map $\pi_1(U_0)\rightarrow \pi_1(U_1)$ identifies the generator of $\pi_1(U_0)$ with $x_1*x_2$ if $x_1,x_2$ generate $\pi_1(U_1)$ (take the picture of the square with the hole in the middle) same with $\pi_1(U_0)\rightarrow \pi_1(U_2)$, thus the amalgamated product $\pi_1(U_1)*_{\pi_1(U_0)}\pi_1(U_2)$ is the quotient of the free group generated by $x_1,x_2,y_1,y_2$ by the normal subgroup generated by $x_1*x_2{y_1}^{-1}{y_2}^{-1}$.

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  • $\begingroup$ I don't believe that this is correct. The groups $\pi_1U_1\cong \pi_1U_2\cong \mathbb{Z}\ast\mathbb{Z}$ each have two generators. $\endgroup$ – Tyrone Oct 15 '17 at 16:26

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