Intuition for no isometric embedding of round projective plane into $\mathbb{R}^4 $

Question

This has been on MSE for over a month with four upvotes but no answers or even comments so I'm cross-posting:

According to Examples of two-dimensional Riemannian manifolds that can't be isometrically embedded into $\mathbb{R}^4$ there is no smooth isometric embedding of the round (=constant positive curvature) projective plane into $ \mathbb{R}^4 $.

Is there some intuition for why there is a smooth isometric embedding into $ \mathbb{R}^5 $ and even into round $ S^4 $ but not into $ \mathbb{R}^4 $?

Some things I already know:

• There is an isometric embedding of round $ \mathbb{R}P^n $ into $ \mathbb{R}^{N} $ where $ N=\frac{n(n+3)}{2} $

• This embedding is equivariant with respect to the isometry group $ O_{n+1} $ at least for $ n=1,2 $. In fact the image of the embedding arises as the orbit of a vector with respect to an $ N $ real dimensional irreducible orthogonal representation of $ O_{n+1} $.

• It is an open question if there exists a $ C^r $ embedding of $ \mathbb{R}P^2 $ into $ \mathbb{R}^4 $ for $ 1 < r <2 $

Answer 1

I'm not sure what you would accept as 'intuition' for this result. It's actually a simple consequence of two facts, which actually prove something much stronger:

The first fact is that, for any smooth surface $S\subset\mathbb{R}^n$ with positive Gauss curvature $K$, its mean curvature vector $H$ cannot vanish. This is because the Gauss equation says that for any smooth surface in flat space, we have $K = |H|^2 - |I\!I_0|^2$, where $I\!I_0$ is the trace-free part of the second fundamental form. In particular, if $K>0$ everywhere, then $H$ is a nowhere vanishing normal vector field on the surface.

The second fact is an old differential topology result of Whitney that says that, when $\mathbb{RP}^2$ is embedded smoothly in $\mathbb{R}^4$, its normal bundle has no nonvanishing (continuous) section. This is not obvious; Whitney proved it using his obstruction theory for sphere bundles. See his On the topology of differentiable manifolds, Lectures in Topology, University of Michigan Press, 1941.

Granted these facts, it is immediate that, for any smooth embdding of $\mathbb{RP}^2$ into $\mathbb{R}^4$, there must be a point of $\mathbb{RP}^2$ where the induced metric has non-positive Gauss curvature. A fortiori, a metric of constant positive Gauss curvature on $\mathbb{RP}^2$ cannot be isometrically embedded into $\mathbb{R}^4$. (This is essentially the argument that Gromov and Rohklin give in Appendix 4 of their 1970 article Embeddings and immersions in Riemannian geometry , though their argument for the first fact (from differential geometry) seems less immediate to me than just using the Gauss equation, as I did above. But, eye of the beholder, etc.)

As far as I know, it is still an open problem whether $\mathbb{RP}^2$ can be smoothly immersed into $\mathbb{R}^4$ so that the induced metric has positive Gauss curvature (let alone constant positive Gauss curvature).