I saw from wikipedia (https://en.wikipedia.org/wiki/Real_projective_space) that $P(\mathbb{R}^n)$ admits a positive scalar curvature metric. Then it also describes that "For the standard round metric, this has sectional curvature identically $1$."
My question is to show that "$P(\mathbb{R}^n)$ admits a positive sectional curvature which are constant for two planes $\sigma$ at any $Tp(P(\mathbb{R}^n))$. I am not sure what metric I should consider. Is it the "standard round metric" ? (I am not sure either if this is the metric induced from $S^{n+1}$ ?) Also, sectional curvature is quite hard to calculate for any possible $2-$plane. How can one show the statement like "all sectional curvature is positive, and moreover, they are all equal !"
The standard projection $\pi : S^n \to P(\mathbb{R}^{n+1})$ is the universal covering. Given the round metric on the sphere, there exists a unique metric on the right that makes this projection a riemannian covering, that is, a local isometry (this is because the round metric is unvariant under $x\mapsto -x$.) This is what we actually mean by "the round metric of the projective space". It has constant sectional curvature $1$, because so has the sphere and because the projection is a local isometry, hence it has constant (positive) scalar curvature.