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.
$\mathbb{C}P^2$ 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.
