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 $\mathbb{R}^4$](https://mathoverflow.net/questions/231495/examples-of-two-dimensional-riemannian-manifolds-that-cant-be-isometrically-emb) 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 $



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


