How to prove that $CP^4$ cannot be immersed in $R^{11}$

Question

Please let me know how to prove that $CP^4$ cannot be immersed in $R^{11}$. I know a proof using an integrality theorem for differentiable manifolds but I want to know if a more direct and simple proof exists. Many thanks.

Answer 1

A necessary condition for an $n$-dimensional manifold $M$ to be immersible into $\mathbb{R}^{n+k}$ is that there must be a $k$-dimensional vector bundle $N$ on $M$ (the normal bundle) such that $T \oplus N$ is trivial, where $T$ denotes the tangent bundle. (By the Hirsch-Smale theorem, this necessary condition is actually sufficient when $k \ge 1$.)

So we can get straightforward lower bounds on $k$ by computing stable characteristic classes of the stable normal bundle: for example, if the $d^{th}$ Stiefel-Whitney class of $N$ doesn't vanish then $k \ge d$, and this is how one derives the standard lower bounds for immersions of real projective spaces.

Edit #2: Okay, here is the computation just for the Pontryagin classes. Recall that, letting $\alpha \in H^2(\mathbb{CP}^n, \mathbb{Z})$ denote the first Chern class of $\mathcal{O}(1)$, the total Chern class of $\mathbb{CP}^n$ is $(1 + \alpha)^{n+1}$, so it follows that the total Pontryagin class of $\mathbb{CP}^n$ is

$$p(\mathbb{CP}^n) = (1 + \alpha^2)^{n+1}$$

and hence that the total Pontryagin class of the stable normal bundle of $\mathbb{CP}^n$ is

$$p(N) = (1 + \alpha^2)^{-n-1}.$$

When $n = 2m$ is even, the coefficient of $\alpha^{2m}$ (which is the largest nonzero term) is ${-2m-1 \choose m}$ and in particular it does not vanish, so $p_m(N) \neq 0$, and hence the dimension of $N$ is at least $2m$. In other words, $\mathbb{CP}^{2m}$ does not immerse into $\mathbb{R}^{6m - 1}$. In particular, $\mathbb{CP}^4$ does not immerse into $\mathbb{R}^{11}$ as desired.

When $n = 2m + 1$ is odd, the coefficient of $\alpha^{2m}$ (which is still the largest nonzero term) is ${-2m-2 \choose m}$ and in particular it does not vanish, so $p_m(N) \neq 0$, and hence the dimension of $N$ is again at least $2m$. In other words, $\mathbb{CP}^{2m+1}$ does not immerse into $\mathbb{R}^{6m+1}$.

Edit #3: If I'm not mistaken, the computation with Stiefel-Whitney classes is essentially identical to the case of real projective spaces except that the degrees of the relevant Stiefel-Whitney classes are all doubled. We get that $\mathbb{CP}^n$ does not immerse into $\mathbb{R}^{4n - 2 \alpha(n) - 1}$ where $\alpha(n)$ is the number of $1$s in the binary expansion of $n$. In particular, we get a stronger result when $n = 4$:

$\mathbb{CP}^4$ does not immerse into $\mathbb{R}^{13}$.

On the other hand, by Cohen's immersion theorem, $\mathbb{CP}^n$ immerses into $\mathbb{R}^{4n - \alpha(2n)} = \mathbb{R}^{4n - \alpha(n)}$. When $n$ is a power of $2$ this almost pins down the exact minimal dimension in which $\mathbb{CP}^n$ immerses, and the gap between the two results is worse the more $1$s are in the binary expansion of $n$ (so it's worst when $n$ is one less than a power of $2$).