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I know that $\beta_{1}(\sharp_{h}\mathbb{RP}^{2})=h-1$. Also it is clear: $$\beta_{i}(\sharp_{h}\mathbb{RP}^{2n})=0 \mbox{ for }0<i<2n-1\mbox{ and }\beta_{2n}(\sharp_{h}\mathbb{RP}^{2n})=0.$$ is it true $\beta_{2n-1}(\sharp_{h}\mathbb{RP}^{2n})=h-1$?

Michael jordan
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1 Answers1

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The $\mathbb{RP}^{2n}$ is closed manifold, so we have: $$\chi(\mathbb{RP}^{2n}\sharp \mathbb{RP}^{2n})=\chi(\mathbb{RP}^{2n})+\chi(\mathbb{RP}^{2n})-\chi(S^{2n}).$$ This implies $\chi(\mathbb{RP}^{2n}\sharp \mathbb{RP}^{2n})=0.$ So $\beta_{2n-1}(\mathbb{RP}^{2n}\sharp \mathbb{RP}^{2n})=1.$ Continue this process to get $\beta_{2n-1}(\mathbb{RP}^{2n}\sharp_{h} \mathbb{RP}^{2n})=h-1.$

King Khan
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