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For homework, I was told to prove the following equalities. $$\begin{gather} \chi _{i=1}^n(\# T^2 _i)=2-2n \\ \chi _{i=1}^n(\# \mathbb RP^2 )=2-2n \end{gather}$$ First of all, the notation is strange, but I take them as: $$\begin{gather} \chi (\#_{i=1}^n T^2 )=2-2n \\ \chi(\#_{i=1}^n \mathbb RP^2 )=2-2n \end{gather}$$

I know $\chi(M_1 \#M_2)=\chi(M_1)+\chi(M_2)-2$ and I'm trying to use this to solve my problems, but I'm getting different answers: Since $\chi(\mathbb RP^2)=1$ and $\chi(T^2)=0$ I find $$\begin{gather} \chi (\#_{i=1}^n T^2 )=2-2n \\ \chi(\#_{i=1}^n \mathbb RP^2 )=2-n \end{gather}$$What is my mistake?

popo
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2 Answers2

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The given inequality for $\Bbb R P^2$ is incorrect: it should be $2 - n$ (as you compute). Indeed, $\chi(\Bbb R P^2) = 1$, as we can quickly compute using the polygonal presentation in this answer.

Community
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Travis Willse
  • 108,056
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Your computations are good. The mistake is in the question. And just in order to propose an alternative way of computing it, you don't need to use that $\chi(M_1 \#M_2)=\chi(M_1)+\chi(M_2)-2$. You can just compute the Euler Characteristic if you know the cell structure of that connected sums and probably you know (if not they are in Hatcher's Book which is freely available in his webpage).

D1811994
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