I'm reading Hatcher's book of Algebraic Topology where he define the notion of $\Delta$-complex. Then he puts the following diagram


And said that this shows a $\Delta$-complex structure of the torus, the real projective plane and the Klein bottle. I'm trying to figure out how this structure is. This is my interpretation.

If I have a $\Delta$-complex structure $\mathcal{A}=\{\sigma_\alpha:\Delta^{n_\alpha}\rightarrow X\}_\alpha$ on a topological space $X$ and I have a quotient space $$\pi:X\rightarrow X'=X/\sim$$ Then we have a canonical family of maps $\mathcal{A}'=\{\pi \circ\sigma_\alpha:\Delta^{n_\alpha}\rightarrow X'\}_\alpha$ and hopefully this is going to be a $\Delta$-complex structure for $X'$. So perhaps this is the $\Delta$-complex structure that Hatcher gives.

But now if we take $X$ to be a square and we give it the $\Delta$-complex structure of the figure

enter image description here

Then this structure is explicitly given by the maps $$\mathcal{A}=\{\sigma_{ABD},\ \sigma_{BDC}, \ \sigma_{AB}, \sigma_{BD}, \ \sigma_{DA}, \ \sigma_{CB}, \ \sigma_{DC}, \ \sigma_{A}, \ \sigma_{B}, \ \sigma_{C}, \ \sigma_{D} \}$$

And then if we take the family of maps $\mathcal{A}'$ as above, this gives a $\Delta$-complex structure in the case of the torus. But it's not a $\Delta$-complex structure in the case of $\mathbb{R}P^2$ because in this case $\pi\circ \sigma_{AD} \neq \pi \circ \sigma_{BC}$ (because if $\Delta^1=[v_0,v_1]$ we have $\bar{A}=\pi\circ \sigma_{AB}(v_0)\neq \pi \circ \sigma_{DC}(v_0) = \bar{D}$) but $\text{im}(\pi\circ \sigma_{AD})=\text{im}(\pi\circ \sigma_{DC})$ so this mess up the partition part of the definition. I think there is a similar issue with the Klein bottle.

So my questions are

How you use the diagram above to find a $\Delta$-complex structure for the real projective plane and the Klein bottle and how do you do this for other quotients of $\Delta$-complexes (such as seeing the "dunce hat" as a quotient of a triangle).

Walter Simon
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2 Answers2


The identification diagrams are not quotients of a delta complex, but rather delta complex structures on the quotient space for the square itself. Delta complexes don't behave particularly well under taking quotients, which is what I believe you are observing.

For example, with $T^2$ the identification diagram is representing the torus as $X=[0,1]\times[0,1]/{\sim}$ where $(0,t)\sim(1,t)$ and $(t,0)\sim(t,1)$ for all $t\in[0,1]$. The delta complex structure is

  • $v:\Delta^0\to X$ by $v(1)=(0,0)$.
  • $a:\Delta^1\to X$ by $a(1-t,t)=(0,t)$.
  • $b:\Delta^1\to X$ by $b(1-t,t)=(t,0)$.
  • $c:\Delta^1\to X$ by $c(1-t,t)=(t,t)$.
  • $U:\Delta^2\to X$ by $U(t,1-t-s,s)=(s,1-t)$.
  • $L:\Delta^2\to X$ by $V(t,1-t-s,s)=(1-t,s)$.

The images here should be understood as being in the quotient space. If I got the coordinates and orientations correct, then the boundaries of each of these simplices should be what the diagrams would lead you to expect.

From another point of view, the identification diagrams are describing abstract delta complexes whose topological realizations are homeomorphic to the spaces Hatcher is claiming they are.

Kyle Miller
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You say that

Then this structure is explicitly given by the maps $$\mathcal{A}=\{\sigma_{ABD},\ \sigma_{BDC}, \ \sigma_{AB}, \sigma_{BD}, \ \sigma_{DA}, \ \sigma_{CB}, \ \sigma_{DC}, \ \sigma_{A}, \ \sigma_{B}, \ \sigma_{C}, \ \sigma_{D} \}$$

but $\mathcal{A}$ is not a $\Delta$-complex structure on the square. For example, by definition of $\Delta$-complex, the restriction $\sigma_{ABD}\vert\ _{[e_0,e_2]}=\sigma_{AD}$ has to be in $\mathcal{A}$, but your collection does not contain it.

The main confusion arises from the fact that the $\Delta$-complex structure on a triangle is defined in such a way that the "orientations" of the edges are not in the cyclic order, but in the order which preserves the ordering of the vertices.

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  • If we reverse the direction of $c$ in the case of $\mathbb{R}P^2$, will we still get a $\Delta$-complex structure? Why or why not? It still looks okay as far as the triangles are concerned, no? Can you please explain? – WhySee Apr 13 '21 at 19:12
  • @WhySee Yes, you can reverse the direction of $c$ for $\mathbb{R}P^2$. You can directly check this from the definition. Heuristically, reversing $c$ should not do any harm because it amounts to simply turning over the square about the axis orthogonal to $c$. – Ken Apr 14 '21 at 00:21
  • Thank you very much! I can see it now. – WhySee May 08 '21 at 21:04