4
$\begingroup$

I am trying to compute the de Rham cohomology group $H^p(\mathbb{RP}^{n+1}\#\mathbb{RP}^{n+1})$ and I am stuck at computing $H^1(\mathbb{RP}^2\#\mathbb{RP}^2)$. ($\#$ stand for the connected sum)

Let $U\simeq \mathbb{RP^2\setminus \{p\}}$ and $V\simeq \mathbb{RP}^2\setminus \{q\}$ such that $M=\mathbb{RP}^2\#\mathbb{RP}^2=U\cup V$ and $U\cap V\simeq S^1$. Then by using Mayer Vietoris, I get \begin{align*} 0&\xrightarrow{a} H^0(M)=\mathbb{R}\xrightarrow{b} H^0(U)\oplus H^0(V)=H^0(\mathbb{RP}^1)\oplus H^0(\mathbb{RP}^1)=\mathbb{R}\oplus \mathbb{R}\xrightarrow{c} H^0(S^1)=\mathbb{R} \\& \xrightarrow{d} H^1(M)\xrightarrow{e} H^1(U)\oplus H^1(V)=H^1(\mathbb{RP}^1)\oplus H^1(\mathbb{RP}^1)=\mathbb{R}\oplus \mathbb{R} \xrightarrow{f} H^1(S^1)=\mathbb{R}\\&\xrightarrow{g} H^2(M)\xrightarrow{h} H^2(U)\oplus H^2(V)=H^2(\mathbb{RP}^1)\oplus H^2(\mathbb{RP}^1)=0. \end{align*}

But the fact that it is exact sequence does not give me specific feature of $H^1(M)$ and $H^2(M)$. I have computed two possibilities.

Note that $d$ is zero map, so $e$ is injective. So only possible choice of $H^1(M)$ is $0$, $\mathbb{R}$ and $\mathbb{R}\oplus \mathbb{R}$. But $H^1(M)$ cannot be $0$ since otherwise $f$ is injective and it is impossible.

Then observe the following two cases.

(1) If $H^1(M)$ is $\mathbb{R}$, $g$ is surjective zero map which means $H^2(M)=0$.

(2) If $H^1(M)$ is $\mathbb{R}\oplus \mathbb{R}$, $f$ is zero map so $g$ is isomorphism. Thus, $H^2(M)=\mathbb{R}$.

As we can see, either ways does not make any contradiction.

I don't know where I am missing. I would be very appreciated for any help toward this. Thank you in advance :)

$\endgroup$
  • 2
    $\begingroup$ The top cohomology of a non-orientable manifold must be $0$. $\endgroup$ – Cheerful Parsnip Dec 24 '18 at 23:41
  • $\begingroup$ @CheerfulParsnip Thank you for the comment! That is helpful! So that means I should make a contradiction for the case (2). Could you let me know where is the theorem from? $\endgroup$ – Lev Ban Dec 24 '18 at 23:42
  • $\begingroup$ See the computation here: math.stackexchange.com/questions/187413/… $\endgroup$ – anomaly Dec 24 '18 at 23:43
  • 2
    $\begingroup$ @LeB: yes, any connect sum with at least one non-orientable component is non-orientable. A path along which orientation reverses survives to the connect sum. $\endgroup$ – Cheerful Parsnip Dec 25 '18 at 0:04
  • 2
    $\begingroup$ Also, in case you are curious, the connect sum of two projective planes is homeomorphic to the Klein bottle. $\endgroup$ – Cheerful Parsnip Dec 25 '18 at 0:08

Your Answer

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

Browse other questions tagged or ask your own question.