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We have the volume form $\mu=i^*(\omega)$ on $S^n$, where $$\omega=\sum_{j=1}^{n}(-1)^{j-1}x_{j}dx_{1}\wedge\cdots\wedge \hat{dx_{j}}\wedge\cdots dx_{n+1}$$ is a k-form on $\mathbb R^{n+1}$ and $i:S^n\to \mathbb R^{n+1}$ is the inclusion.

Define the antipodal map $A:S^n\to S^n$ by $x \mapsto -x$. Indeed, $A^*(\mu)=(-1)^{n+1}\mu$. Does this gives us a volume form on $\mathbb R P^n$ for n odd? It seems we have the projection map $p:S^n\to \mathbb R P^n$, but it's the wrong direction to pull back the form. Any help will be appreciated.

user136592
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HINT: Pull back by local inverses of $p$, and use your computation to check that you get a well-defined form on $\Bbb RP^n$. Alternatively, define a form $\eta$ on $\Bbb RP^n$ by setting its value $\eta([x])(v_1,\dots,v_n)$ to be something in terms of $\mu$, and check that everything is well-defined.

Ted Shifrin
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  • Sorry to raise an old thread, but is the $\eta$ defined by $\eta([x])(v_1,\dots, v_n)=\mu(x)(z_1,\dots, z_n)$ where $\pi_z_i=v_i$, and $\pi(x)=[x]$? Clearly there's no choice in $z_i$, but one can still show that if there was it's indepndent of the choice. Then using that, $\omega_{-x}=(F^\omega)_x=(-1)^{n+1}\omega_x$ so it should work out right? – Chris Nov 08 '23 at 02:19
  • @Chris Of course there’s choice in the $z_i$. Compare $x$ and $-x$. – Ted Shifrin Nov 08 '23 at 02:45
  • @TedShiffrin I meant once we’ve made a choice in $x$ there’s no choice in $z_i$ because the projection is a local diffeomorphism. I was thinking fix $x$ show the tangent vectors are independent of choice. Then show the $x$ is independent of choice. – Chris Nov 08 '23 at 03:35