Whitney's embedding theorem states that any smooth $n$-manifold $M$ can be smoothly embedded into $\Bbb R^{2n}$. And we know $\mathbb{R} {\text P}^{2n+1} $ is a $2n+1$-manifold, it can be embedded into $\mathbb R^{4n+2}$ according to the theorem. But I want to know if it can be embedded into $\mathbb R^{4n+1}$, is there any method to construct the embedding?
Construct a embedding from $\mathbb{R}{\text P}^{2n+1}$ to $\mathbb{R}^{4n+1}$
characteristic-classesdifferential-geometrysmooth-manifolds
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Just to add something to Jesse's answer, the idea behind the proof of the Easy Whitney Embedding Theorem is to place different pieces of the given $ n $-dimensional smooth manifold in 'general position' in $ \mathbb{R}^{2n + 1} $. The proof is not very hard to follow; I think that Munkres does a pretty good job in his book Topology. The Hard Whitney Embedding Theorem, which tries to embed a smooth $ n $-dimensional manifold in $ \mathbb{R}^{2n} $, requires a more technical proof. A clever idea, called 'Whitney's trick' nowadays, is the main idea behind the proof. Notice that we have no notion of distance on a general smooth manifold $ M $ unless some metric on $ M $ is specified. Hence, both versions of the Whitney Embedding Theorem do not talk about preserving distances between points when constructing the required smooth embedding.
The Nash Embedding Theorem, however, is much harder. Not only must you embed the given Riemannian manifold in Euclidean space, you must do so isometrically, i.e., in a way that preserves distances between points. This requires the solution of a formidable system of partial differential equations that yields the required isometric embedding. Nash solved this PDE system using a special version of Newton's iteration method, called Newton's method with post-conditioning. When unmodified, Newton's iteration method, in general, fails to converge to a solution because each step of the iteration might result in the loss of derivatives, i.e., the order of differentiability is reduced. Nash recovered the lost derivatives by applying smoothing operators (defined via convolution) at each step of the iteration. This ensures that Newton's iteration method does actually converge to a solution. The application of a smoothing operator at each step is called post-conditioning. As you can see, Nash's result is definitely much harder and requires more technology to prove than Whitney's results.
These two results also have different natures. The Whitney Embedding Theorem is more topological in character, while the Nash Embedding Theorem is a geometrical result (as it deals with metrics). However, the structure of smooth manifolds is sufficiently rigid to ensure that they are also geometrical objects (cf. my comment below Jesse's answer), to which the Nash Embedding Theorem can be applied.
It's a nontrivial result, and it's probably not something amenable to a proof going back to first principles. In very broad terms, the idea of the proof is:
Show that the manifold $M^n$ embeds into some $\mathbb{R}^N$. This part is straightforward in the compact case: Choose a finite good cover, construct an embedding on each element of the cover (which is trivial), and patch them together using a partition of unity to get a smooth embedding into $\mathbb{R}^N$ for large $N$.
Show that we can take $N = 2n + 1$. The idea here is to use Sard's theorem to construct a nice map onto lower dimension, but there are some tricks involved in controlling the behavior of the resulting map.
Show that we can take $N = 2n$. This is nontrivial, and it uses what's now known as the Whitney trick. Basically, the idea is that you reduce to an embedding with nice singularities, then use the fact that the dimension is high to show that the singularities can be pulled together in pairs through an embedded disk and removed. This trick fails (spectacularly; it's the origin of a lot of the weirdness of $4$-manifold topology) below dimension $5$, but the case $n = 2$ is easy anyway.
You might also be interested in looking up the $h$-cobordism theorem, which contains a lot of the same ideas (including the Whitney trick) and is a more modern result.
Best Answer
For each $n > 0$, $\mathbb{R}P^{2n+1}$ embeds into $\mathbb{R}^{4n+1}$. For $n=1$, this is a result of Wall that all $3$-manifolds embed into $\mathbb{R}^5$:
Further, in
Thomas states that by combining results of Haefliger, Hirsch, Massey, and Petersen, it follows that all orientable $m$-manifolds embed into $\mathbb{R}^{2m-1}$, provided $m> 4$. Of course, $\dim \mathbb{R}P^{2n+1} = 2n+1 > 4$ when $n\geq 2$, and $\mathbb{R}P^{2n+1}$ is orientable. So these results cover the case of emedding $\mathbb{R}P^{2n+1}$ into $\mathbb{R}^{4n+1}$.
I do not know of any explicit embeddings.