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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 $\#_k(\mathbb{CP}^2 \# -\mathbb{CP}^2)$.

Martin Tancer
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$\def\RR{\mathbb{R}}\def\CC{\mathbb{C}}\def\PP{\mathbb{P}}$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.

David E Speyer
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    It's worth noting that $\mathcal{L} = \mathcal{O}(1)$ and $U$ is its total space. In your last line, did you mean to say $KO(S^2)$? – Michael Albanese May 25 '23 at 18:18
  • @MichaelAlbanese Fixed, thanks. – David E Speyer May 25 '23 at 18:49
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    Phrased a little differently, the 2nd Stiefel-Whitney class of $\mathbb{C}P^2$ is nontrivial and since the inclusion of $\mathbb{C}P^1$ induces an isomorphism on second cohomology, we can deduce that the tangent bundle of the punctured $\mathbb{C}P^2$ is stably nontrivial, so does not admit a codimension 1 embedding into Euclidean space. – Connor Malin May 26 '23 at 02:57
  • Thank you for the answer and the comments. Now, I am trying to digest it (whether I am able to fully verify). I am not too familiar with K-theory, thus I am perhaps closer to verifying the comment on Stiefel-Whitney classes. Some double checks: The answer above provides no in "smooth" case but with Stiefel-Whitney classes, this seems to give "no" also in piecewise linear case? The isomorphism on second cohomology is meant of $T\mathbb{C}P^2$ and $TU$? If so, shouldn't there be still said something about the relation to $T\mathbb{C}P^1$? Is the second cohomology sufficient? – Martin Tancer May 26 '23 at 13:14
  • Continued: This I do not know, but could it be the case that the Stiefel-Whitney classes depend on whole cohomology ring? – Martin Tancer May 26 '23 at 13:19
  • The K theory here is pretty trivial: Rank $r$ vector bundles on $S^2$ are classified by homotopy classes of maps $S^{1} \to GL(r) \simeq O(r)$, in other words, by $\pi_1(SO(r))$, which is trivial for $r=1$, $\mathbb{Z}$ for $r = 2$ and $\mathbb{Z}/2 \mathbb{Z}$ for $r>2$. Taking direct sum of vector bundles adds the classes of the summands, reducing modulo $2$ if necessary. So the claim is that $TS^2$ is trivial, $\underline{\mathbb{R}}$ is trivial but that $\mathcal{L} \cong \mathcal{O}(1)$, which is nontrivial mod $2$. – David E Speyer May 26 '23 at 13:27
  • See https://math.stackexchange.com/questions/1923402/understanding-vector-bundles-over-spheres and https://ncatlab.org/nlab/show/basic+complex+line+bundle+on+the+2-sphere for some possibly useful background. – David E Speyer May 26 '23 at 13:29
  • I don't know about the PL/topological questions. – David E Speyer May 26 '23 at 13:29
  • Thank you again. Hopefully, now I understand reasonably well thus I have accepted the answer. – Martin Tancer May 29 '23 at 14:17