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By a theorem of Livesay, the 3-sphere has a unique (up to equivariant diffeomorphism) smooth involution with exactly two fixed points. Thinking of $S^3$ as the unit sphere in $\mathbb{R}^4$, this involution is given by $(x,y,z,w) \to (-x,-y,-z,w)$. This same formula gives a smooth involution on the 4-ball with a pointwise fixed arc. Are there any other (orientation reversing) smooth involutions on $D^4$ with a pointwise fixed arc (up to equivariant diffeomorphism)?

I know there are infinitely many distinct involutions with a pointwise fixed $D^2$, but I cannot find any literature on the case where the involution is orientation reversing with a fixed arc.

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    $\begingroup$ The group action can be eliminated from the discussion by excising a tubular neighborhood of the fixed point set. Start from a given an involution of $D^4$, and attach along $\partial D^4$ the standard involution (=product of the identity map of $D^1$ and the antipodal map of $D^3$ with corners smoothed). You get a involution of $S^4$ that fixes an unknotted circle. Excising an equivariant tubular neighborhood of the circle, and passing to the quotient gives a 4-manifold with boundary $S^1\times RP^2$ whose 2-fold cover is $D^2\times S^2$. You need to smoothly classify such manifolds. $\endgroup$
    – Igor Belegradek
    Commented Mar 20, 2021 at 1:09
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    $\begingroup$ For example, is there an exotic $D^2\times RP^2$? If yes, you can reverse the above construction, and get a non-standard involution on $D^4$ with fixed point set an arc. $\endgroup$
    – Igor Belegradek
    Commented Mar 20, 2021 at 1:16
  • $\begingroup$ Thanks for the comment! I like the idea of excising the fixed set, but I'm not sure I follow the last part. What if the double cover of the exotic $D^2 \times \mathbb{R}P^2$ is an exotic $D^2 \times S^2$? Then I don't immediately see how to get a smooth involution on the standard $D^2 \times S^2$ (or on the standard $D^4$). $\endgroup$
    – Strongly Negative Amphicheiral
    Commented Mar 20, 2021 at 2:56
  • $\begingroup$ It has to be an exotic $D^2\times RP^2$ whose 2-fold cover is the standard $D^2\times S^2$. Then you can glue it back. But I suspect such exotic $D^2\times RP^2$'s are unknown. (I am not a 4d topologist of course). $\endgroup$
    – Igor Belegradek
    Commented Mar 20, 2021 at 3:06
  • $\begingroup$ Ah, thanks! Yes, that's my suspicion as well. $\endgroup$
    – Strongly Negative Amphicheiral
    Commented Mar 20, 2021 at 3:13

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