Does there exist a compact smooth closed $6$-dimensional manifold $M,$ so that it admits an embedded submanifold $N$ diffeomorphic to $\mathbb{CP}^2$, and $N$ is a boundary in the homology sense? In other words, $N$ is in the trivial homology class. Many thanks!
The lowest dimension of Euclidean space that $\mathbb{CP}^2$ embeds is $7$. (Thanks to Michael Albanese for pointing out the false claim of $8$.) However, I wonder if one can lower the dimension of the target manifold if one considers compact closed manifolds. The homological triviality is to exclude trivial examples like product of $\mathbb{CP}^2$ with some surface.
A slight modification of the argument that $\mathbb{CP}^2$ doesn't immerse in $\mathbb{R}^6$, given by András Szűcs here, shows that there is no such $M$.
Let $i : \mathbb{CP}^2 \to M$ be an immersion with $i_*[\mathbb{CP}^2] = 0$. Then there is a short exact sequence of real vector bundles on $\mathbb{CP}^2$
$$0 \to T\mathbb{CP}^2 \to i^*TM \to \nu \to 0$$
where $\nu$ denotes the normal bundle. Note that $\nu$ has rank $2$, and is orientable since $\mathbb{CP}^2$ is simply connected. Therefore, we can view $\nu$ as a complex line bundle.
Note that
$$\langle p_1(i^*TM), [\mathbb{CP}^2]\rangle = \langle i^*p_1(TM), [\mathbb{CP}^2]\rangle = \langle p_1(TM), i_*[\mathbb{CP}^2]\rangle = 0.$$
Since $H^4(\mathbb{CP}^2; \mathbb{Z}) \cong \mathbb{Z}$, we see that $p_1(i^*TM) = 0$. On the other hand, since $H^4(\mathbb{CP}^2; \mathbb{Z})$ has no $2$-torsion, we have
$$p_1(i^*TM) = p_1(T\mathbb{CP}^2\oplus\nu) = p_1(T\mathbb{CP}^2) + p_1(\nu) = 3x^2 + c_1(\nu)^2$$
where $x$ denotes a generator of $H^2(\mathbb{CP}^2; \mathbb{Z}) \cong \mathbb{Z}$, and hence $x^2$ is a generator of $H^4(\mathbb{CP}^2;\mathbb{Z}) \cong \mathbb{Z}$. Writing $c_1(\nu) = ax$, we have
$$a^2x^2 = c_1(\nu)^2 = p_1(i^*TM) - 3x^2 = -3x^2$$
which is a contradiction as $a^2 = -3$ has no integral solutions.