given six real-analytic arcs in the unit disk $D$, each of which connects the origin to a boundary point, and no two arcs meet anywhere except at the origin, and the arcs meet at equal (60 degree) angles at the origin, is there a homeomorphism $\phi$ of $D$ that takes the six arcs onto six straight line segments connecting the origin to equally spaced points on the boundary (like a sliced pizza, hence the name "pizza lemma")? And if so can $\phi$ be made $C^2$ or even smoother?

I can get it $C^2 $if I assume the original arcs are (all the) zeroes of some harmonic function. I can get it $C^0$ with no additional assumption if I am willing to use the Riemann mapping theorem (on each sector and then patch up the sector boundaries). But I am hoping somebody knows a theorem of which this is an immediate application so I can just quote it.

Similarly, the "cake lemma", if there are three arcs that do not meet connecting six boundary points in pairs, there should be a homeomorphism of $D$ that takes the three arcs onto three parallel straight lines. As the first commenter pointed out, that's false without an additional assumption, so assume that the points are 0,1,2,3,4,5, and points 0,1,2 are each one endpoint of one of the three arcs.

More generally, given a homeomorphism between two 1-complexes that are subsets of $D$, when does there exist a homeomorphism of $D$ that restricts to the given homeomorphism of subsets?

  • 4
    $\begingroup$ Regarding the «cake lemma», I think it depends on the non-crossing permutation on 3 elements induced by the pairing between the 6 points of the boundary $A_0,\ldots,A_5$. Assume for instance that $A_j$ are ordered as successive points on the boundary, and that you connect $A_0, A_1$, then $A_2, A_3$ and finally $A_4, A_5$, I fail to see how you can send the arcs to three parallel lines. $\endgroup$ Commented Feb 1, 2016 at 15:23
  • $\begingroup$ This question prompts the question of the regularity in the Jordan-Schoenflies theorem, in term of the regularity of the curve. A good knowledge of the later should be the main part of an answer of the former. $\endgroup$ Commented Feb 1, 2016 at 15:44
  • 4
    $\begingroup$ If one of the arcs is $\{(x,y):x>0,y=x^2\}$ and the other arc is $\{(x,y):x<0,y=-x^2\}$, then you cannot achieve $C^2$-regularity of your map. That's because a $C^2$-diffeomorhism maps any $C^2$ curve (in our case the real axis) to a $C^2$ curve. $\endgroup$ Commented Feb 1, 2016 at 17:23


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.