Knotted projective planes and fake complex projective space Paul Melvin gave a talk at Knots in Washington last year in which he asked whether the connected sum of an odd twist-spin of a classical knot and a standard cross-cap embedding of ${\mathbb R}P^2$ is a standard cross-cap. I believe this question to be a standard one, and I have two questions about it.
First, I think that Paul stated that if the result is knotted, then there is a non-standard ${\mathbb C}P^2$. Please may I have an outline of the argument that gives the construction of the non-standard ${\mathbb C}P^2?$
Second, for aesthetic reasons, I imagine that there is a difference in using a cross-cap with normal Euler class $2$ and using one with normal Euler class $-2$. Won't the construction give a $\pm {\mathbb C}P^2$ (or better ${\mathbb C}P^2$ or $\overline{{\mathbb C}P}^2$) depending on the normal Euler class of the cross-cap?
Any other folk-lore would be appreciated.
 A: I haven't been to Melvin's talks, but I suspect he's using the cyclic 2-sheeted branch cover construction.  Specifically, the cyclic 2-sheeted branched cover of $S^4$ branched over the unknotted embedded $\mathbb RP^2$ is either $\mathbb CP^2$ or its mirror reflection depending on the normal Euler class of the $\mathbb RP^2$. 
A sort of algebraic-geometric way to see this would be to observe that $\mathbb CP^2$ modulo complex conjugation is $S^4$.  You can check the $\mathbb CP^1$ in $\mathbb CP^2$ maps down to the standard embedding of $\mathbb RP^2$.  Checking that the orientation of $\mathbb CP^2$ corresponds to the normal Euler class is perhaps simplest in this model. 
Another way to see this is to think of $\mathbb CP^2$ remove a ball as a disc bundle over $S^2$ with Euler class $1$, and similarly as $S^4$ as the union of two mapping cylinders from $S^3 / \{ \pm 1, \pm i, \pm j, \pm k\}$ to $\mathbb RP^2$. I am apparently not the first to think of this relation, there's a paper of Terry Lawson on this:


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*T.Lawson, Splitting $S^4$ on $RP^2$ via the Branched Cover of $CP^2$ over $S^4$. Proceedings of the American Mathematical Society. Vol. 86, No. 2 (Oct., 1982), pp. 328-330

A: I can't think of another way to get $CP^2$ from an embedded $RP^2$ in the 4-sphere, so I would guess you are right.  But why would the knottedness of the $RP^2$ imply that the $CP^2$ is exotic?  After all, there are plenty of knotted 2-spheres in $S^4$ whose double branched covers are $S^4$.  In other words, a knotted $RP^2$ might produce an exotic involution on the genuine (I don't want to say real, for obvious reasons!) $CP^2$ rather than an exotic $CP^2$.
