# Which closed orientable $4$-dimensional manifolds cannot be embedded in $6$-space?

This question is a follow-up to my previous question . The statement of the question is the title.

Note that the $4$-dimensional real projective space is non-orientable and a characteristic class argument gives that it does not embed in $7$-space. Right now, I am more interested in orientable $4$-manifolds.

This is true if and only if $X^4$ is spin and its signature vanishes. This is on p. 345 in Gompf/Stipsicz (4-manifolds and Kirby calculus), who cite Ruberman: Imbedding four-manifold and slicing links, 1982.

EDIT Of course I mean that $X^4$ CAN be embedded in 6-dimensional space iff the conditions are met.

$$\mathbb{CP}^2$$ does not even immerse in $$\mathbb{R}^6$$. Proof: If such an immersion exists, then the normal Euler class has the property, that its square is the normal Pontrjagin class, and that is $$-3$$ times the signature (when evaluated on the fundamental homology class). But in $$H^2(\mathbb{CP}^2)$$ there is no such a class $$x$$, for which $$x^2$$ evaluated on the fundamental class is $$-3$$. QED.

Moreover the following theorem is true (It is essentially due to Hughs) Theorem: In the 4-dimensional oriented cobordism group $$\Omega_4 \cong \mathbb{Z}$$ precisely the even elements contain a manifold that admits an immersion into $$\mathbb{R}^6$$.

About embeddings: The conditions (the manifold must be spin and have zero signature) are clearly necessary: An embedded manifold in a Euclidean space has zero normal Euler class. Hence in the present case both $$p_1$$ and $$w_2$$ are zero. The opposite is non-trivial, it is the content of Ruberman's paper mentioned above.

• Thanks Andras (forgive the lack of accent). That fact is quite helpful. I have been trying for some time to use chart movies to construct an immersed braiding of CP with branch set a standardly embedded ${\mathbb R}P^2$. Your remark explains why this is doomed. Now I think I can desingularize the construction in 7-space. Jul 17 '13 at 1:35

$\mathbb{C}P^2$ does not embed in $\mathbb{R}^6$. See

Feder, S.; Segal, D. M. Immersions and embeddings of projective spaces, Proc. Amer. Math. Soc. 35 (1972), 590–592.