# Topological Intuition behind the 2-Homology of $RP^2$

I believe that, intuitively, one can think of the 2-homology of a space as how many ways one can wrap a surface around it and close it up such that it's injective. For example, one can take the entire torus or sphere as the surface that wraps around them to give a generator of the 2-homology.

On the other hand, one cannot do this for the Klein bottle, a cylinder/Möbius strip, a circle, and these can therefore be seen to have a trivial 2-homology.

However, this intuition is struggling when considering the real projective plane. The 2-homology of the real projective plane is trivial, however I'm struggling to see why wrapping the entire surface around it would not give such a generator. Perhaps such a wrapping does not separate two spaces, as with the Klein bottle? Is this because the real projective plane cannot be embedded into $\mathbb{R}^3$?

• Is it considered "too abstract" the claim that the top homology group of a closed mfld is non-zero iff the manifold is orientable? Anyway I find your definition of "wrapping" confusing, what does it mean to close it up such that it's injective? – Riccardo May 9 '17 at 15:47
• I don't consider that too abstract, it makes sense to me, as then there is no concept of outside and inside, and therefore one cannot wrap such a surface around to enclose an area. I wasn't trying to define it rigorously, to be honest, as it is, as I say, an intuition rather than strict definition. Consider an open 2-disc on the boundary of the sphere, a Klein bottle, etc. Then, if we can 'close this up' by expanding it outwards (as opposed to going back on itself, which we could always do, but isn't injective), then I feel we have surrounded a hole. Perhaps we only have if it's orientable. – Alex W May 9 '17 at 16:01
• I tried writing something about the interpretation of your second homology group, it was too long for a comment, hope it helps a little bit! – Riccardo May 9 '17 at 16:20

Honestly, your definition of "wrapping around a surface" is not clear to me. I will list here what geometric meaning you can give to the homology groups of a manifold.

The main "geometric" interpretation is the following theorem you can find on any book of Algebraic Top:

Thm: Let $M$ be a connected $n$-manifold without boundary. Then $H_n(M;\Bbb Z)$ is $\Bbb Z$ if and only if $M$ is closed and orientable, $0$ otherwise.

A geometric intuition (which holds in the smooth case) of this theorem is that in the orientable case, starting from an oriented triangulation of the manifold, you can build a non-exact closed $n$-simplex (could this count as a kind of wrapping?) which generates the top homology group.

you can weak the orientability requirement if you work in $\Bbb Z_2$ coefficient, and you obtain that, with the same hypothesis:

Thm: Let $M$ be a connected $n$-manifold without boundary. Then $H_n(M;\Bbb Z_2)$ is $\Bbb Z_2$ if and only if $M$ is closed. $0$ otherwise.

So you see that in your case $H_2(\Bbb RP^2; \Bbb Z)=0$ detects the non-orientability of the projective space.

$$\times \times \times$$

Cellular homology should prove another visual interpretation close to your concept of "wrapping", because it's the homology of the chain complex generated by the cells, whose differentials are the degree of certain maps, so in fact in you case you are studying the $2$-cells (the usual disks) and how they attach to the $1$-skeleton of $\Bbb RP^2$. In our case, you see that the unique $2$-cell in the standard CW-structure of $\Bbb RP^2$ is attached to the unique $1$-cell via a degree $2$ map, therefore you get the $0$ in the second homology group. Cellular homology makes cristal clear why the circle has trivial second (and higher) homology groups: it admits a CW-structure with only 0 and 1 cells

$$\times \times \times$$

In the simply connected case, $H_2(M;\Bbb Z)$ is isomorphic to the second homotopy group $\pi_2(M)$, which is the group of pointed homotopy classes of maps $S^2\to M$, via the Hurewicz map

• Could you jog my memory on what the $\mathbb{Z}$ and $\mathbb{Z}_2$ are for in the homology? – Alex W May 9 '17 at 16:27
• well, I assume you have seen the simplicity homology, as the homology of the simplicity complex. The simplicial complex in degree $n$ is the group whose elements are the linear combination of $n$ simplex with coefficients in the integer $\Bbb Z$. But nothing stops you to repeat the same construction but with coefficients in $\Bbb Z_2$, ($\Bbb Z/(2)$) and obtain the homology with coefficients $\Bbb Z_2$. Basically you do that to get rid of the orientability issues, because in $\Bbb Z_2$ $1=-1$ – Riccardo May 9 '17 at 16:50
• You can take any abelian group $A$ as coefficient. In general the convention is $H_k(M)=H_k(M;\Bbb Z)$, but it's better to specify each time, since many books have different conventions – Riccardo May 9 '17 at 16:50
• This makes sense. I've little understand when it comes to using $\mathbb{Z}_2$ in this context, but I understand the gist of it, and overall this answer satisfies my question in that it implies my intuition (which you still may not understand) simply didn't take into account orientation properly. Thank you! – Alex W May 9 '17 at 17:21
• @AlexW I'm glad that helped. Morally (i.e. via Universal coefficient theorem), if you know $H_k(M;\Bbb Z)$ for all $k$, then you know $H_k(M;A)$ for all $k$ and abelian group $A$. But computing homology with $\Bbb Z_2$ is much easier (with coefficient a field in general) and sometimes it can help answer many questions (it has less information than the usual homology with integer coefficient still). As a last remark, you can use the same techniques like Mayer-Vietoris and excision for the homology with coefficients! – Riccardo May 9 '17 at 19:52