# Can an oriented closed $n(\geq 2)$-dimensional manifold be embedded in $\mathbb{R}^{2n-1}$?

Can anyone provide me with an example of an orientable closed manifold $M$ of dimension $n\geq 2$, which cannot be embedded in $\mathbb R^{2n-1}$?

I know these cannot exist for $n=1$, i.e. $S^1$. If we ignore orientability, then if we take $n=2^r$, $\mathbb RP^n$ cannot be embedded in $\mathbb R^{2n-1}$.

While searching on the internet, I found some sharper results here.

So can we say anything when $\dim(M)= 2^r$ and $M$ is orientable?

It would be very helpful if someone could provide me with some more references. Thanks in advance.

A closed smooth $n$-manifold embeds into $\mathbb R^{2n-1}$ if and only if the normal $(n-1)$th Stiefel-Whitney class vanishes. This is due to Hirsch-Haefliger in dimensions $\neq 4$ and to Fang in dimension $4$. Massey showed that if the normal $(n-1)$th Stiefel-Whitney class is nonzero, then $M$ is non-orientable and $n$ is a power of $2$. Thus any smooth closed orientable $n$-manifold smoothly embeds into $\mathbb R^{2n-1}$. For references see here and here.
• I wish to add that Hirsch-Haefliger's result assumes $n>4$. The case $n=3$ has to be handled separately, and it was treated independently by Wall (see Rivin's answer below), or by Hirsch (quoted in the Hirsch-Haefliger's paper). – Igor Belegradek Jan 1 '16 at 23:43
Every closed orientable (smooth) $n$-manifold (smoothly) embeds in $\mathbb{R}^{2n-1}.$ This is true for $n> 4$ by Haefliger-Hirsch, $n=3$ by C.T.C. Wall (Wall's paper has the reference to Haefliger-Hirsch)
Wall, C. T. C.

For $n=4$ this is true in the topological category, and is open (as far as I know) in the smooth category [which means that no counterexample is known] - see Bruno Martelli's answer to this question.