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I'm study the papper "H. Bray, S. Brendle, M. Eichmair, and A. Neves, Area-minimizing projective planes in three-manifolds, Comm. Pure Appl. Math". (see

Let be $$\mathrm{sys}(M,g):=\inf \{L(\gamma): \gamma \mathrm{\ is\ a\ non-contractible\ loop\ in}\ M\}$$

The Pu's inequality say that

$$\mathrm{sys}(\mathbb{RP}^2,g) = \frac\pi2 \mathrm{area}(\mathbb{RP}^2,g),$$

if $g$ is the round metric on $\mathbb{RP}^2$.

Why $$\mathrm{sys}(\mathbb{RP}^3,g) = \pi,$$ if $g$ is the round metric on $\mathbb{RP}^3$?

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Pu's inequality cannot say that $\mathrm{sys}(\mathbb{RP}^2,g) = \frac\pi2 \mathrm{area}(\mathbb{RP}^2,g),$ because the latter is an equality, not inequality. Furthermore, Pu's inequality is concerned with arbitrary metrics rather than round ones. – Mikhail Katz Jan 23 '14 at 16:57
up vote 1 down vote accepted

The homotopy group of ${\Bbb R}{\rm P}^3$ is ${\Bbb Z}/2$, the generator of which can be taken as the image of an equatorial circle in ${\Bbb S}^3$. Since the length of the of that circle is $2 \pi$, and the map ${\Bbb S}^3 \to {\Bbb R}{\rm P}^3$ identifies the anipodal points, the length of the generator in the induced metric has to become equal to $\pi$.

It remains to understand that any shorter loop can be lifted to ${\Bbb S}^3$ where it is contractible.

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