What is the lower $n$ such that $\mathbb{CP}^8$ can be embedded in $\mathbb{R}^n$? Using basic cohomological calculus it is possible to  obtain that $\mathbb{CP}^8$ cannot be embedded in a Euclidean space of dimension 23 or less.
From other side, using spinorial cohomological calculus (Mayer integrality theorem), I am obtaining that $\mathbb{CP}^8$ cannot be embedded in a  Euclidean space of dimension 29 or less.
Also it is well known that $\mathbb{CP}^8$ can be immersed into $\mathbb{R}^{31}$.
Then my question is:
What is the lower $n$ such that $\mathbb{CP}^8$ can be embedded in $\mathbb{R}^n$?
 A: In Embedding complex projective spaces in Euclidean space, Mukherjee proved that $\mathbb{CP}^n$ embeds in $\mathbb{R}^{4n-\alpha(n)}$, and if $n$ is odd with $n > 1$, then $\mathbb{CP}^n$ embeds in $\mathbb{R}^{4n - \alpha(n) - 1}$; here $\alpha(n)$ denotes the number of ones in the binary expansion of $n$. In particular, we see that $\mathbb{CP}^8$ embeds in $\mathbb{R}^{31}$. As anomaly points out in their answer, this is the smallest such dimension of Euclidean space into which $\mathbb{CP}^8$ embeds.
I don't know if the dimension provided by Mukherjee's result is always optimal. If $N(n)$ denotes the smallest dimension of Euclidean space into which $\mathbb{CP}^n$ embeds, then by the result of Feder and Segal mentioned in anomaly's answer (namely Theorem $2$), we have $N(n) > 4n - 2\alpha(n)$. So for $n$ even we have the double-sided inequality
$$4n - 2\alpha(n) < N(n) \leq 4n - \alpha(n).$$
Note that when $\alpha(n) = 1$ (i.e. $n = 2^k$ with $k > 0$), we see that $N(n) = 4n - 1$; that is, Mukherjee's result is optimal. On the other hand, for $n$ odd with $n > 1$ we have the double-sided inequality
$$4n - 2\alpha(n) < N(n) \leq 4n - \alpha(n) - 1.$$
Note that when $\alpha(n) = 2$ (i.e. $n = 2^k + 1$ with $k > 0$), we see that $N(n) = 4n - 3$; that is, Mukherjee's result is again optimal.
A slight improvement can be made by incorporating a theorem from Nonimmersion theorems for differential manifolds by Sanderson and Schwarzenberger which states that if $n$ is even with $\alpha(n) \equiv 2, 3 \bmod 4$, then $\mathbb{CP}^n$ does not immerse in $\mathbb{R}^{4n-2\alpha(n)+1}$. For such $n$ we have the improved double-sided inequality
$$4n-2\alpha(n) + 1 < N(n) \leq 4n - \alpha(n).$$
Note that when $\alpha(n) = 2$ (i.e. $n = 2^k + 2^l$ with $k > l > 0$), we see that $N(n) = 4n - 2$; that is, Mukherjee's result is again optimal.
In summary, for all $n$ with $\alpha(n) \in \{1, 2\}$, the quantity $N(n)$ is given by Mukherjee's result. The first case not covered by the above theorems is $n = 7$ since $\alpha(7) = 3$. All that we can say is $22 < N(7) \leq 24$, i.e. $N(n)$ is either $23$ or $24$.
A: I haven't read either paper in full, but there's a result from Feder and Segal (and I've also seen it, or at least a variant of it, attributed to Atiyah) that $\mathbb{CP}^8$ doesn't embed in $\mathbb{R}^{30}$, and another from Davis that $\mathbb{CP}^8$ embeds (not immerses, despite the title of the preprint) in $\mathbb{R}^{31}$. If they're accurate, they give $n = 31$ as optimal.
