# Fundamental group of two tori with a circle ($S^1✕${$x_0$}) identified

Compute the fundamental group of the space obtained from two tori $$S^1✕S^1$$ by identifying a circle $$S^1✕$${$$x_0$$} in one torus with the corresponding circle $$S^1✕$${$$x_0$$} in the other.

Using van Kampen's theorem, I can fairly quickly show that if $$T_1$$ is one torus and $$T_2$$ is the other, then the group has to be isomorphic to $$((π_1(T_1) ∗ π_1(T_2)))/N ≅ (\mathbb{Z} ✕ \mathbb{Z})∗ (\mathbb{Z} ✕ \mathbb{Z})/N$$, where N is generated by elements of the form $$i_{12}(w)i_{21}(w)^{-1}$$, w is in $$\pi_1(S^1✕$${$$x_0$$}$$) ≅ \mathbb{Z}$$ and $$i_{12}$$, $$i_{21}$$ are the maps from $$\pi_1(S^1✕$${$$x_0$$}$$)$$ to $$π_1(T_1)$$, $$π_1(T_2)$$ respectively determined by inclusion. My problem is, as it seems to always be with van Kampen-related problems, figuring out what the elements of N look like.

My attempt to do this in a pure-algebra way, as far as I can tell, failed me: I tried examining the case where w represents a loop that goes n (some integer) times around the circle, but then it seems like $$i_{12}(w)$$ "$$=$$" $$(n, 0)$$ in $$T_1$$ and $$i_{21}(w)$$ "$$=$$" $$(n, 0)$$ in $$T_2$$, which says to me that $$i_{12}(w)i_{21}(w)^{-1}$$"$$=$$"$$(n, 0) + (-n, 0) = 0$$. The notation here isn't right, since $$(n, 0)$$ in $$T_1$$ is not the same as $$(n, 0)$$ in $$T_2$$ and their addition (composition?) shouldn't be written quite that way, hence my quotes around the equals signs, but certainly no loop in $$S^1✕$${$$x_0$$} is going to magically include anything but the 0 loop in either torus's second component, right? But this doesn't seem right to me, since if a, b are the generators of the circles in $$T_1$$ and d, e are the generators of the circles in $$T_2$$, and c is the constant path then the identification of the circle should make, say, $$a = d$$, so for example $$(a, b)*(d, e)$$ $$= (a^2,b)*(c, e)$$ $$= (c, b)*(d^2,e)$$. This doesn't match with what I just worked out N might be, but I also can't figure out what N should look like to make this equivalence make sense.

My attempt to understand this geometrically has gone even worse. What I know for sure is that the resulting figure isn't the "2-torus", 2 tori glued together at a disc. What I don't know for sure is whether this identification should actually create the hypertorus $$S^1✕S^1✕S^1$$ or something completely different.

• It certainly can't be the hypertorus - the dimensions are different; your space is (essentially) a surface, whereas the hypertorus is a volume. Commented Sep 13, 2014 at 18:12

Here are two tori with a copy of $S^1$ identified. The picture is clearer when we move $x_0$ to the inner rim of one torus, and to the outer rim on the other torus.

The two images below show a path on one of the tori that belongs to the homotopy class of the latter generator.

Basically Van Kampen says that these two become homotopic in $\pi_1$ of the combination. Geometrically this is obvious, as you can slide that path from one torus to the other via the shared circle of contact.

More precisely, if $b$ denotes the homotopy class above and $a_1$ (resp. $a_2$) is a loop around the tube of the bigger (resp. smaller) torus, then elements of the amalgamated product are objects like $$(a_1^{k_1},b^{\ell_1})*(a_2^{m_1},b^{n_1})*\cdots*(a_1^{k_t},b^{\ell_t})*(a_2^{m_t},b^{n_t}),$$ where the exponents $k_i,\ell_i,m_i,n_i$ are all arbitrary integers, $t$ is a natural number and factors with $a_1$ (resp. $a_2$) alternate.

Because $b$ commutes with both $a_1$ and $a_2$, the above product simplifies to $$a_1^{k_1}a_2^{m_1}a_1^{k_2}a_2^{m_2}\cdots a_1^{k_t}a_2^{m_t}b^r$$ with $$r=\sum_{i=1}^t(\ell_i+n_i).$$ This is path that first does $k_1$ laps around the outer tube, then $m_1$ laps around the inner loop et cetera, and at some point also goes $r$ laps along the intersection.

Looks like $(\Bbb{Z}*\Bbb{Z})\oplus\Bbb{Z}$ as a group. A direct product of two groups where the first factor is a free product of two infinite cyclic groups, and the latter is just $\Bbb{Z}$.

In retrospect this is clear. Undoubtedly you know that the fundamental group of figure 8 is that free product of two infinite cyclic groups. Your space is the direct product $8\times S^1$.

• What is $\ast$ in the line with the $a_1$, $a_2$ and $b$? Commented Dec 18, 2020 at 1:52
• @Ramanujan It's just the group operation of the amalgamated product. I don't remember why I chose to denote it that way. Probably so that it wouldn't be confused with the group operation of the fundamental group of either torus. Commented Dec 18, 2020 at 4:49

I'll add that this means the group is isomorphic to $\mathbb{Z} \times F$, where $F$ is a free (nonabelian) group on two generators.

In particular, before modding out, there are two pairs of generators, say $\{x,y\}$ for the first $\mathbb{Z}\times \mathbb{Z}$ and $\{a,b\}$ for the other. Each pair commutes with itself, but the two pairs do not commute. Then, modding out by $N$ just says $y = b$.

So now we have three generators, $y(=b), x, a$. The first one commutes with both the other two, and $x$ and $a$ don't commute. That gives $\mathbb{Z} \times F$.

• Looks like @Jyrik Lahtonen and I found the same final answer for the description of the group! Commented Sep 13, 2014 at 19:21
• Yeah (+1). I'm a bit ashamed actually. I had been staring at that picture for quite some time. And only then it dawned on me that this is the direct product of figure eight and $S^1$. Commented Sep 13, 2014 at 19:29
• Arguably the algebraic approach is simpler. Though on the other hand, I didn't realise that the space itself is literally $8 \times S^1$. Commented Sep 13, 2014 at 19:43
• I figured out what was going on from mostly just the first picture of one torus inside another, before the other answer was expanded to a more detailed explanation. But this one is also really helpful, because my phrasing and notation in the answer as it is now is probably still kind of confusing, and will benefit from rewriting things to look more like this. Thank you! Commented Sep 14, 2014 at 1:03