Homology group of Real projective plane

Question

I know the homology group of Real Projective plane $\mathbb{RP}^2$

$H_i(\mathbb{RP}^2) = 0$ for $i>2$, $\mathbb{Z}$ for $i=0$ , $\mathbb{Z}/2\mathbb{Z}$ for $i=1$ (non-reduced case).

Proving when $i \neq 2$ is easy but $i=2$ case is slightly hard for me. $\mathbb{RP}^2$ has CW complex structure with one of each $0,1,2$ cells so this takes care of $i>2$ case and $\mathbb{RP}^2$ is connected so it takes care of $i=0$ case and finally I know the fundamental group of real projective plane and I know the relation between first homology group and fundamental group so that part is done too.

I also understand that we can use simplicial homology tool to calculate it as well as using the degree formula to find out the boundary map for CW complex. But is there any other way (for instance using Mayer-Vietoris sequence or directly working out the boundary map $\delta_2$ explicitly in CW complex case) to show $H_2(\mathbb{RP}^2)=0$?

Answer 1

I might have come up with something so it would be grateful if you tell me this is correct. So I am assuming every result I know about $H_i(\mathbb{RP}^2)$ for $i \neq 2$.

We have sequence $0\rightarrow H_2(X_2,X_1)=\mathbb{Z}\rightarrow H_1(X_1,X_0)=\mathbb{Z} \rightarrow H_0(X_0)=\mathbb{Z}$.

But we know the map from $H_1(X_1,X_0)$ to $H_0(X_0)$ has to be trivial map and this means we can deduce that the map from $H_2(X_2,X_1)$ to $H_1(X_1,X_0)$ has to be $\times 2$ map due to the fact that we know $H_1(X) =\mathbb{Z}/2\mathbb{Z}$. (the part I'm not 100 percent sure about.) and hence we get $H_2$.

I guess this gives absolutely no topological intuition but I think this is correct at least?

Answer 2

The real projective plane is the space of all lines in $\mathbb R^3$. One can also treat it as $\mathbb S^2/\sim$, where $x\sim -x$. So, it can also be treated as the upper hemisphere together with the equator with the identification $z\sim z^2$ on the equator. So the gluing map $\delta_2$ is of degree two.

Answer 3

You'll want to use the fact that $\mathbb{R}P^n$ can be written as $\mathbb{R}P^{n-1}\cup_f D^n$ where $D^n$ is the $n$-dimensional ball, and $f\colon S^{n-1}\to\mathbb{R}P^{n-1}$ is a 2-fold covering map, so we are gluing the $n$-ball along its boundary to $\mathbb{R}P^{n-1}$ via this map. You can then use Mayer-Vietoris and induced maps to explicitly work out the connecting map. In your case, you have $\mathbb{R}P^2=M\cup_f D^2$, where $M$ is the Mobius strip and $f\colon S^1\to M$ is the doubling map up to homotopy, or just the inclusion of the boundary into the Mobius strip.

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To be more explicit, via Mayer-Vietoris, we get a long exact sequence $$\cdots\to H_2(M)\oplus H_2(D^2)\to H_2(\mathbb{R}P^2)\to H_1(S^1)\to H_1(M)\oplus H_1(D^2)\to\cdots$$ which, using the fact that $H_2(M)=H_2(D^2)=H_1(D^2)=0$ and $H_1(S^1) \cong H_1(M) \cong\mathbb{Z}$, reduces to the exact sequence. $$\cdots\to 0\to H_2(\mathbb{R}P^2) \stackrel{g}{\to} \mathbb{Z} \stackrel{\times 2}{\to} \mathbb{Z} \to\cdots$$

where we get that $\times 2$ map in the above sequence from the fact that the inclusion of the intersection of the two spaces (homotopy equivalent to a circle) into the Mobius strip is (up to homotopy) the degree-$2$ covering map, which induces multiplcation by $2$ in first homology.

By exactness, the image of $g$ must be $0\subset\mathbb{Z}$ as the doubling map in injective, but $g$ must itself be injective by exactness because the map $0\to H_2(\mathbb{R}P^2)$ has trivial image. The only way both of these conditions on $g$ can be satisfied is if $H_2(\mathbb{R}P^2)$ is trivial.

Answer 4

In the sequal we in the algebraic topology course computing homology of projective plane in as following way by covering property: The paths in $P^2$ have lifts to paths in $D^2$, and the loops in $P^2$ are quotients of paths in $D^2$ for which the endpoints are either equal or diametrically opposite. The path $\alpha$ = $q\circ \beta$ in $P^2$, where is the path in $D^2$ that takes the interval to the upper semicircle on the boundary $S^1$ of $D^2$, is not homologous to zero, however $2\alpha$ is. Let $h$ be a continuous map takes $\Delta_2$ to $D^2$ such that the side opposite to the vertex $0$ goes to the above upper semicircle, the side opposite to the vertex $1$ goes to one point $a \in S^1 \subseteq D^2$, and the side opposite to the vertex 2 goes to the lower semicircle. And let $\hat{a}$ be the constant map that takes $\Delta_2$ to the point $a$. Then $\partial_2( \hat{a} + q\circ h) = 2\alpha$. It follows that $H_1 (P^2) = \mathbb Z_2$.

Answer 5

There's a way to do this only involving diagram chasing, using the pushout square

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} S^1 & \ra{\sigma_2(z)=z^2} & S^1 \\ \da{inc} & & \da{J}\\ D^2 & \ras{\hspace 0.4cm F \hspace 0.4cm} & \mathbb{R}P^2\\ \end{array} $$ and that $H_k(D^2,\,S^1) \cong H_k(\mathbb{R}P^2,\,J(S1))$, because $(D^2, S^1)$ is a good pair: Take the LESes for both pairs $(\mathbb{R}P^2,\,J(S1)),\, (D^2,\,S^1)$ and 'connect' them using the characteristic map $F$. Diagram chasing and some basic properties of singular homology should then do the trick.