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Consider map from $S^{n}$ to $\mathbb{RP}^{n}$ $$\varphi:S^{n}\to\mathbb{RP}^{n}$$ which maps point $x\in S^{n}$ to corresponding direction in $\mathbb{R}^{n+1}$. This map induces map $${\varphi}^{*}:\Omega^{\bullet}(\mathbb{RP}^{n})\to\Omega^{\bullet}(S^{n})$$ and map of de Rham cogomology groups $${\psi}^{*}:H^{\bullet}(\mathbb{RP}^{n})\to H^{\bullet}(S^{n}).$$ as I see for n-th cohomology $$\dim H^{n}(\mathbb{RP}^n) = \dim\operatorname{Im} {\psi}^{*} + \dim\operatorname{Ker} {\psi}^{*}$$ Am I right that $$\dim\operatorname{Ker} {\psi}^{*} = 0$$ (why?) and $$\dim\operatorname{Ker} {\psi}^{*} = 0,1~~~\text{depending on the parity of}~n?$$

Thanks a lot!

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Дорогой Aspirin, это интересный вопрос: +1 – Georges Elencwajg Mar 10 '12 at 23:40
Спасибо! (Thanks a lot!) =) – Aspirin Mar 11 '12 at 1:50
up vote 3 down vote accepted

The involution $i:S^n\to S^n:x\mapsto -x$ induces a decomposition $\Omega^n(S^{n})=\Omega^n_+(S^{n}) \oplus \Omega^n_-(S^{n})$ where $\Omega^n_\pm(S^{n}) $ consists of the differential forms satisfying $i^*\omega=\pm \omega$.
This yields a decomposition $H^n(S^{n})=H^n_+(S^{n}) \oplus H^n_-(S^{n})\cong \mathbb R$

On the other hand the quotient map $\phi:S^{n}\to\mathbb{RP}^{n}$ induces an isomorphism $\phi^*:\Omega^n(\mathbb P^{n})\stackrel {\cong}{\to }\Omega^n_+(S^{n})$ and then an isomorphism $\phi^*:H^n(\mathbb P^{n})\stackrel {\cong}{\to }H^n_+(S^{n})$.

Finally there remains to notice that the the canonical generator $[\omega_0]\in H^n(S^{n})=\mathbb R [\omega_0]$ is in $H^n_+(S^{n})$ or $H^n_-(S^{n})$ according as $n$ is odd or even, since $i^*(\omega_0)=(-1)^{n+1}\omega_0$.
We can thus conclude that $H^n(\mathbb P^{n})=\mathbb R$ for $n$ odd and $H^n(\mathbb P^{n})=0$ for $n$ even.

The form $\omega_0\in \Omega^n(S^n)$ is defined by $$\omega_0(s)(v_1,\cdots,v_n)=det(s,v_1,\cdots,v_n)$$

Explanation of this formula: The point $s$ lies on the sphere which is itself embedded in $\mathbb R^{n+1}$, i.e. $s\in S^n\subset \mathbb R^{n+1}$.
The value $\omega_0(s) $ at $s$ of $\omega_0$ is an $n$-multilinear form on $T_s(S^n)\subset T_s(\mathbb R^{n+1})=\mathbb R^{n+1}$ and $T_s(S^n)$ consists of those $v\in \mathbb R^{n+1}$ orthogonal to the vector $s$, i.e. $T_s(S^n)=s^\perp$.
The formula says that the value of that multilinear function $\omega_0(s) $ on a $n$-tuple of vectors $v_1,\cdots,v_n\in T_s(S^n)$is the determinant of the matrix of the $n+1$ vectors $s,v_1,\cdots,v_n$ seen as column vectors in $\mathbb R^{n+1}$.

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perhaps you wanted to say that ${\omega}_{0}(s, v_1,\ldots ,v_n) = det (v_{1}(s),\ldots ,v_{n}(s))$ – Aspirin Mar 11 '12 at 1:40
Dear @Aspirin, no, no, I wrote what I meant. I have added a few words of explanation of the formula to clarify. – Georges Elencwajg Mar 11 '12 at 10:26

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