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This is an example of triangulation of simplex from the book by M. Nakahara. This example is of an unoriented simplex. It says that $\langle p_0\rangle\cup\langle p_2\rangle$ is not a simplex, why not, they both are points and points are valid simplex. I am assuming that points taken together do not constitute a simplex.

To further argue, following is a triangulation of Mobius strip. enter image description here

$(p_0p_1p_2) \bigcap (p_1p_4p_2)=p_1\cup p_2\cup(p_1p_2)$

Here also I am getting two points ($0$-simplex) in union form but it is valid, but in the first diagram it is invalid, why? The only way I can understand it as $p_1\cup p_2\cup(p_1p_2)=(p_1p_2)$ - can we do things like this. Because if this is so, we are "unionizing" simplexes of different dimensions and putting it as a single simplex. Is it even valid?

What am I getting wrong here? Your support is greatly appreciated.

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  • $\begingroup$ In the second case the intersection of the two $2$-simplices is $(p_1,p_2)$, which is a $1$-simplex. $\endgroup$
    – s.harp
    Commented Nov 20, 2017 at 13:03
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    $\begingroup$ In $(b)$ shouldn't the intersection of the two simplices $\sigma_2$ and $\sigma_{2'}$ be $(p_0,p_2)$? $\endgroup$ Commented Sep 8, 2020 at 7:11
  • $\begingroup$ @SHASHANKPATHAK did you figure out why the intersection is not $(p_0,p_2)$? $\endgroup$
    – Ivan
    Commented Oct 26, 2022 at 15:09
  • $\begingroup$ @Ivan if you imagine joining the identified sides to form the actual cylinder, then we have the 1-simplex $(p_0, p_2)$ corresponding to half of the "upper" boundary circle of the cylinder, and the 1-simplex $(p_2, p_0)$ corresponding to the other half of the "upper" boundary circle. The intersection of the two 2-simplices is just two vertices, rather than $(p_0, p_2)$, because $(p_0, p_2)$ and $(p_2, p_0)$ are different 1-simplices. Hopefully this helps! $\endgroup$ Commented Feb 6 at 11:42

1 Answer 1

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For this ($K$) to be a triangulation note that

If $\sigma$ and $\sigma'$ are two simplexes of $K$, the intersection $\sigma \cap \sigma'$ is either empty or a common face of $\sigma$ and $\sigma'$, that is, if $\sigma,\sigma'\in K$ then either $\sigma\cap\sigma' = \emptyset$ or $\sigma\cap\sigma' \leq \sigma$ and $\sigma\cap\sigma' \leq \sigma'$

In this example $\sigma\cap\sigma = \langle p_0\rangle \cup \langle p_2\rangle$. You are right in the sense that both $\langle p_0\rangle$ and $\langle p_2\rangle$ are simplexes on their own. But $\langle p_0\rangle \cup \langle p_2\rangle$ is not a face of neither $\langle p_0p_1p_2\rangle$ or $\langle p_2p_3p_0\rangle$

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  • $\begingroup$ Could you explain why the intersection is $\langle p_0\rangle \cup \langle p_2\rangle $ instead of $\langle p_0,p_2\rangle$? $\endgroup$
    – Ivan
    Commented Oct 26, 2022 at 15:11

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