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How to determine the triangulation of these two objects? can we use the above to compute Fundamental Group of Klein bottle and Real Projective plane?

I can use the van kamen theorem to prove one is $Z/2Z$ and the other is $Z^2/<x^2-y^2>$, but it is hard for me to use the triangulation method. Thank you!

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  • $\begingroup$ Many thanks for your help! $\endgroup$ – Kevin May 4 '15 at 7:56
  • $\begingroup$ Since you're new here, don't forget to accept an answer (click on the check mark by it) if it works for you. Also, it's nice to upvote any answers that are beneficial. Regards, $\endgroup$ – user12802 May 8 '15 at 13:43
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$\Bbb RP^2$ and $K$ has quite easy $\Delta$-complex structure obtained from cutting the fundamental square through the diagonal.

Claim : Barycentrically subdividing any $\Delta$-complex $X$ twice gives a simplicial complex $X'$ homeomorphic to $X$.

This is Hatcher's exercise $2.1.23$. Sketch of a proof is to note that a $\Delta$-complex might have two $2$-simplices pasted together along their sides, so that neither of them are uniquely determined by their boundaries. After the first triangulation, since the barycenter lies in interior of each of the simplices and the interiors are left unidentified while constructing a $\Delta$-complex, each of the $2$-simplicies in the subdivided $X$ are uniquely determined by their sides. Similarly, to resolve this problem for $1$-simplices with two of it's boundary points identified, one needs a second barycentric subdivision. It is not hard to see that all the higher dimensional simplices in the resulting $\Delta$-complex are uniquely determined by their faces. Thus, one obtains a simplicial complex $X' \cong X$.

Using this statement, one can obtain triangulations of $\Bbb RP^2$ and $K$.


One doesn't need a simplicial structure to compute $\pi_1$. $\Bbb RP^2$ and $K$ has the CW structure (obtained from the fundamental square) $e^2 \cup e^1 \cup e^0$ and $e^2 \cup e^1 \cup e^1 \cup e^0$, where in the case of $\Bbb RP^2$, $e^2$ is pasted to the circle $e^1 \cup_{\partial} e^0$ via the map $x \mapsto x^2$ and in the case of $K$, $e^2$ is attatched to the figure eight $e^1 \cup_\partial e^0 \cup_\partial e^1$ by attatching map given by the word $aba^{-1}b$.

Applying van Kampen theorem, one obtains $\pi_1(\Bbb RP^2) \cong \Bbb Z/2$ and $\pi_1(K) \cong \langle a, b | ab = b^{-1} a\rangle$

Alternatively, an easier way to do this is to note that there are regular covering maps $S^2 \to \Bbb RP^2$ and $S^1 \times S^1 \to K$ with deck transformation $\Bbb Z/2$ each, and use the following lemma

Claim : $X$ be a path connected, locally path connected and semi-locally simply connected space with a group $G$ acting freely and properly discontinuously on $X$. Then there is a short exact sequence $$1 \to \pi_1(X) \to \pi_1(X/G) \to G \to 1$$

Since deck transformation group acts properly discontinuously and freely on covering spaces, we can use this to obtain $\pi_1(\Bbb RP^2) \cong \Bbb Z/2$ (since $S^2$ is simply connected) and that $\pi_1(K)$ is an extension of $\Bbb Z \times \Bbb Z$ by $\Bbb Z/2$, which reduces to the presentation above after knowing the explicit action of deck transformation.

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