I need a little help with understanding homology groups.

In particular, consider a simplicial complex with three 1-cycles and one 1-boundary (so we have two 1-holes?). Then the first homology group is the second order cyclic group, i.e. $H_1\tilde{=}\frac{\mathbb{Z}^3}{\mathbb{Z}}\tilde{=}\mathbb{Z}^2$. Is this correct?

enter image description here

If so, the problem is I'm getting $\frac{\mathbb{Z}^3}{\mathbb{Z}}=\mathbb{Z}$ here.

  • 1
    $\begingroup$ I don't quite understand your description of the simplicial complex. Normally you have a list of simplices, and what their boundaries are. In your case, are you saying you have one $2$-simplex whose boundary is the sum of the three $1$-simplices, who in turn join up to make a triangle? (i.e. you've taken your complex to be the $2$-simplex and all its faces.) If so, then all the homology should vanish except for $H_0$, because it's contractible. In any case, you need to be more specific about your complex! $\endgroup$
    – Josh Hunt
    Commented Sep 25, 2016 at 11:46
  • $\begingroup$ Sorry about that, I have limited knowledge in the area. What I'm trying to describe is the 3 triangles with one of them sharing 2 edges with two others. Also two triangles are hollow. $\endgroup$
    – Kosm
    Commented Sep 25, 2016 at 11:56
  • $\begingroup$ @Kosm: Ok, that's homotopically equivalent to 3 copies of a circle with a point i common $\endgroup$
    – AdLibitum
    Commented Sep 25, 2016 at 12:11
  • $\begingroup$ @Kosm could you draw a picture or describe more precisely? As in, triangle A has two edges in common with triangle B, and also has two edges in common with triangle C? Or triangle A has one edge in common with triangle B and also one edge in common with triangle C? Which of the triangles is hollow? $\endgroup$
    – Josh Hunt
    Commented Sep 25, 2016 at 12:15
  • $\begingroup$ @JoshHunt Added a picture. $\endgroup$
    – Kosm
    Commented Sep 25, 2016 at 12:26

1 Answer 1


You should indeed be getting $\mathbb Z^2$ here. One way to see this is that we can shrink the "filled" triangle down to a point, leaving you with two circles meeting at a single point, whose homology is $\mathbb Z^2$.

The cycles are the free abelian group generated by $[ab]+[bc]+[ca]$, $[bc]+[cd]+[db]$ and $[db]+[be]+[ed]$. The boundaries are the subgroup generated by $[bc]+[cd]+[db]$, so you get $\mathbb Z^2$ as expected.

I don't know exactly where you went wrong in this calculation because you didn't provide much of your working, but if you're more explicit I can try to answer your question more thoroughly.

  • $\begingroup$ I figured it now, thank you. $\endgroup$
    – Kosm
    Commented Sep 25, 2016 at 14:54

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