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In 1903, W. Boy showed that the real projective plane $\mathbb{R}P^2$ can be immersed in the Euclidean space $\mathbb{E}^3$ (see Werner Boy, Math. Ann. 57 (1903), no. 2, 151-184.). Suppose a Riemannian surface ($\mathbb{R}P^2$, g) has positive sectional curvature $K(x)$ everywhere. $K(x)$ needs not to be a constant. The question is: can this surface isometrically immersed in $\mathbb{E}^3$. I guess that there is no such an immersion. But I did not find a relative reference or a direct proof by myself.

As is well known that D. Hilbert proved that a hyperbolic plane can not be isometrically immersed in $\mathbb{E}^3$ (see D. Hilbert, Trans. Amer. Math. Soc. 2 (1901), no. 1, 87-99.). This is the motivation of the above problem.

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    $\begingroup$ I don’t think so: the immersion of the double cover of the sphere would be locally convex, hence globally convex, hence embedded. Or more easily, the projective plane would have a well-defined normal from mean-curvature, giving a section of the normal bundle. But since the pullback of the tangent bundle to R^3 is trivial, this would give a contradiction to the tangent bundle to RP^2 being nontrivial. $\endgroup$
    – Ian Agol
    Commented Jan 29, 2020 at 1:00
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    $\begingroup$ Ian is correct. In fact, the open problem is whether $\mathbb{RP}^2$ can be immersed in $\mathbb{R}^4$ so that the induced metric has non-negative curvature. (It can be embedded in $\mathbb{R}^4$ smoothly, and it can be embedded in $\mathbb{R}^5$ with a metric of positive curvature, in fact with constant positive curvature.) $\endgroup$
    – Robert Bryant
    Commented Jan 29, 2020 at 9:42
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    $\begingroup$ If you allow immersions which are only C^1, you can use the Nash-Kuiper theorem to deform Boy's embedding to an isometric one. $\endgroup$
    – Thomas Richard
    Commented Jan 29, 2020 at 12:00
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    $\begingroup$ @LaiyuanGao: I don't know any recent references. There is an embedding of $\mathbb{RP}^2$ into $\mathbb{R}^5$, equivariant under an irreducible representation of $\mathrm{SO}(3)$ on $\mathbb{R}^5$, that has constant (positive) Gauss curvature. Gromov, in his 1986 book, Partial Differential Relations, asks whether $\mathbb{RP}^2$ can be embedded into $\mathbb{R}^4$ with positive Gauss curvature (p. 279, Question (e'')). I saw a preprint in 2005 that claimed to construct such an immersion, but it had a serious error and was never published, as far as I know. That's all I know. $\endgroup$
    – Robert Bryant
    Commented Jan 30, 2020 at 12:51
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    $\begingroup$ @LaiyuanGao: That's true, but it can't happen for a compact surface in $3$-space if the Gauß curvature is positive everywhere. As Ian pointed out, the surface would have to be orientable, and then the Gauss map will make the surface into a covering space of the sphere, which ensures (if the surface is connected) that the surface is a $2$-sphere and that the Gauß map is 1-to-1. (This is what is different from the locally-convex-curves-in-the-plane case.) $\endgroup$
    – Robert Bryant
    Commented Jan 30, 2020 at 20:10

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Ian's argument of mean curvature is wonderfully simple. Here is another one. Rotate your surface to put it in generic position with respect to the heigth function z; then, the preimage of z is a Morse function f on RP^2, which has no critical point of index 1 (saddle point) since the surface is locally convex. Hence, every critical point of f has index 0 or 2 (local extremum). By Morse theory (or by the Poincaré-Hopf formula applied to a gradient-like vector field), the number of local extrema is the Euler characteristic of RP^2, hence 1. This is a contradiction, since there are at least one minimum and one maximum.

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