Does $\overline{\mathbb{C}P^2 \setminus B^4}$ embed (smoothly/topologically) in $\mathbb R^5$?

Question

Question: Does $\overline{\mathbb{CP}^2 \setminus B^4}$ (that is the closure of complex projective plane minus a 4-ball) embed (smoothly/topologically/piecewise linearly) in $\mathbb R^5$?

Background: The question I am really interested in is whether the connected sum $\#_k(\mathbb{CP}^2 \# -\mathbb{CP}^2)$ embeds into $\mathbb R^5$. I expect that it might be already problematic to embed one summand thus I ask the question in the form above. I expect that the answer is "no" and I would be happy to rule out smooth embeddings. However, I also find it meaningful to ask more generally about topological/piecewise linear embeddings.

Some related results: $\mathbb{CP}^2$ embeds in $\mathbb{R}^7$ (by Penrose-Whitehead-Zeeman theorem) and I suspect that it does not embed in $\mathbb{R}^6$ but I did not find a reference. (I would appreciate a pointer to any reference). It follows from Rokhlin theorem that a single $\mathbb{CP}^2$ does not embed into $\mathbb{R}^5$ because $\mathbb{CP}^2$ has non-zero signature and thus it does not bound a 5-manifold. But this argument does not apply to $\#_k(\mathbb{CP}^2 \# -\mathbb{CP}^2)$.

Answer 1

$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\PP{\mathbb{P}}$I think the answer is no. Topology isn't my strength, so check this argument.

Let $\infty$ be the point $[0:0:1]$ in $\CC \PP^2$ and put $U = \CC\PP^2 \setminus \{ \infty \}$. Note that $\{ [z_1:z_2:0] \} \cong \CC\PP^1 \cong S^2$, I'll always identify $S^2$ with this particular subset of $U$.

At a point $z=[z_1:z_2:0]$ in $S^2$, let $L(z)$ be the line of $\CC\PP^2$ through $z$ and $\infty$. We have the decomposition of tangent spaces $$T_z U = T_z S^2 \oplus T_z L(z).$$ This gives an equality of vector bundles: $$TU|_{S^2} \cong TS^2 \oplus \mathcal{L}$$

Now suppose that $U$ embeds into $\RR^5$. Then we can choose a normal direction to $U$, and get $$T\RR^5|_U \cong TU \oplus \underline{\RR}$$ where $\underline{\RR}$ denotes the trivial bundle. And $T\RR^5$ is trivial, so we can rewrite this as $$\underline{\RR}^5 \cong TU \oplus \underline{\RR}.$$ Restricting to $S^2$, $$\underline{\RR}^5 \cong TS^2 \oplus \mathcal{L} \oplus \underline{\RR}.$$ But $TS^2$ is stably trivial and $\mathcal{L}$ represents the nontrivial class in $\text{KO}(S^2)$, a contradiction.