enter image description here

enter image description here

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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


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 $\langle p_2p_3p_0\rangle$

  • $\begingroup$ 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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .