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Let $X$ be the space given by joining two disjoint tori at a single point p.

By Van Kampen this can be split into the spaces of the two tori, with their intersection the point (n.b. You can also consider adding a "piece" of the other torus to each subspace, but it still deforms to these sets)

Van Kampen then gives us:

$$\pi_1(U) = \langle a,b ; ab=ba \rangle$$ $$\pi_1(V) = \langle c,d ; cd=dc \rangle$$ $$\pi_1(U\cap V) \cong 0$$

Now, from my notes this gives us that $\pi_1(X=U\cup V) = \langle a,b,c,d ; ab=ba, cd=dc \rangle$. However in thinking about this it occurs to me that by our group action surely $ac = ca$ (informally, attaching two loops around the seperate tori is the same regardless of which "direction" you go around), so how do I include this relation, considering it can't be inherited from the inclusion on the intersection? Additionally, how do I write it?

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    $\begingroup$ The comment by @Balaji sb is wrong. As quickquestion4 said in the answer, $a$ and $c$ do not commute in the free product $\mathbb Z^2*\mathbb Z^2$. And this free product is not isomorphic to the direct product $\mathbb Z\times\mathbb Z\times\mathbb Z\times\mathbb Z$. $\endgroup$ May 16 at 1:19
  • $\begingroup$ Sorry my bad... i see that they dont commute. $\endgroup$
    – Balaji sb
    May 16 at 1:35
  • $\begingroup$ @AndreasBlass thanks for the point out the error..i deleted my wrong comment and posted an answer with intuitive explanation. Hope this helps. $\endgroup$
    – Balaji sb
    May 16 at 4:10
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As you correctly noted, Van Kampen with this decomposition would say:

$$ \pi_1(X) \cong \pi_1(U)\ast \pi_1(V) \cong \mathbb{Z}^2 \ast \mathbb{Z}^2. $$

However, this is as far as the induced relations will take us (note that the inclusion of $\pi_1(U\cap V)$ has zero image). Indeed, you will unfortunately have $ac\neq ca$ (that is, there is no homotopy which takes one path to the other).

Consider the slightly easier problem that shows that $\pi_1(S^1\vee S^1) \cong \mathbb{Z}\ast\mathbb{Z}$. For a more in depth look, see Chapter 1 of Hatcher's Algebraic Topology or this question. The more general fact is that the wedge sum behaves like:

$$ \pi_1(\bigvee_{\alpha} X_\alpha) = \ast_{\alpha} \pi_1(X_\alpha), $$ the free product of all the fundamental groups.

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Here is an intuitive explantion as to why $ac \neq ca$: enter image description here

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