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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.

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Isn't it just the 2-fold cyclic branch cover of $S^4$ branched over the embedded $\mathbb RP^2$ ? That should be the construction, because $\mathbb CP^2$ is the branched cover over the unknotted embedding. Regarding Q2, yes the orientation issue is precisely given by the normal euler class, at least in the unknotted case. –  Ryan Budney Sep 10 '13 at 21:45
    
Ryan, if you can reiterate your comment as an answer and flesh it out a bit more, then I'll indicate that the answer is correct. Thanks! –  Scott Carter Sep 11 '13 at 14:08

2 Answers 2

up vote 5 down vote accepted

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:

  • 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
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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 $CP^2$.

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Presumably Melvin knows something about that involution that I don't know. :) –  Ryan Budney Sep 12 '13 at 2:13
    
A fundamental group calculation shows that the fundamental group of $\tau_{2n+1}(K) \sharp CC{\mathbb R}P^2}$ (where $\tau$ denotes the twist-spin and $CC$ means the cross-cap is ${\mathbb Z}_2$. So I guess that if it is not standard, then the involution is crazy. –  Scott Carter Sep 12 '13 at 14:05

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