My lectures notes say that the second diagram isn't a well-defined $\Delta$-complex of the torus because the $2$-simplices aren't totally ordered.

enter image description here

I don't really understand what that means.

Let's call the upper $2$-simplex $U$, the lower one $L$, the vertical $1$-simplex $a$, the horizontal one $b$ and the diagonal one $c$.

If we orient the top $2$-simplex clockwise and the bottom one counter-clockwise. Then the boundary of the top one is $a+b+c$ and the boundary of the second one is $-a-b-c$.

So where's the problem? Why is the first diagram well-defined, and the second one not?

  • 1
    $\begingroup$ this is actually a question that I was struggling with only a day or two ago, so I'm glad I could help! feel free to let me know if there's anything else you're not sure about :) $\endgroup$
    – Tim
    Oct 21, 2015 at 14:39
  • $\begingroup$ thanks! I was confused because I was thinking that all the vertices would get identified as the same one $\endgroup$ Oct 21, 2015 at 17:32
  • $\begingroup$ They do when we actually glue the edges together, but the idea is (sort of, and somebody please do correct me if I'm wrong) that we construct the torus as a $\Delta$-complex by first constructing a square as a $\Delta$-complex, then identifying the sides. So it is the structure before the gluing that we care about, but we also want the gluing to preserve this structure, i.e. the ordering of the vertices. $\endgroup$
    – Tim
    Oct 21, 2015 at 17:37
  • 1
    $\begingroup$ Thinking about it that way means that the second diagram is 'ok' for the torus, but becomes 'not-ok' when we unglue it and look at the square we defined it on. The first diagram is 'ok' as a torus, and is still ok as a square after ungluing. $\endgroup$
    – Tim
    Oct 21, 2015 at 17:38
  • $\begingroup$ Thanks for your clear explanation! $\endgroup$ Oct 21, 2015 at 17:51

2 Answers 2


Let's ignore the side identification for a bit, and just take a look at the following picture, where the orange arrows represent the orientation of a line (a $1$-simplex): square

On the left-hand side we see that we can label the vertices $v_i$ in such a way that the numbering agrees with the arrows. That is, if we go along a line in the direction of an arrow, then we go from a smaller number to a bigger one, and vice versa. This corresponds to the fact that we require that the restriction of $n$-simplices to their faces ($n-1$-simplices) must respect the ordering of the vertices ($0$-simplices).

Try labelling the diagram on the right in the same way -- see if it's possible.


For a $\Delta$-complex the orientation of each simplex is given by a linear order on its vertices, the faces then have to be oriented accordingly. Hence, the $1$-faces of a simplex can never be oriented in a cyclic manner.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.