# Under what conditions are two orientation-reversing involutions of a compact surface equivalent?

Let $$M$$ be a compact, connected, orientable surface and $$\varphi_1,\varphi_2$$ be two orientation-reversing involutions (i.e., diffeomorphisms for which $$\varphi^2=Id$$) such that the fixed-point set of both is non-empty. I am trying to understand what conditions guarantee the existence of an equivariant self-diffeomorphism $$f$$. (I.e., such that $$f \circ \varphi_1 = \varphi_2 \circ f$$, which is what I call "equivalen[ce]" in the title.)

The fixed-point set is a union of circles. Perhaps conditions can be derived from how those circles from $$\varphi_1$$ and $$\varphi_2$$ interact homologically. (More precisely, how the image of their fundamental classes through the inclusion are related to one another. Of course, necessary conditions can arise from this.) Since such a diffeomorphism restricts to a diffeomorphism between the fixed point sets, it is clear that this self-diffeomorphism does not always exist. (E.g., take a $$2$$-genus surface on $$\mathbb{R}^3$$ to be lied down, and consider reflection through the middle vertical circle as $$\varphi_1$$ and through the $$xy$$ plane as $$\varphi_2$$. The fixed point sets have a different number of connected components.)

I have two main questions:

1. Under what conditions on $$M$$, $$\varphi_1$$ and $$\varphi_2$$ (or perhaps their fixed point sets) does this hold?
2. If the above is a little too broad, then is it known if this always holds for $$S^2$$? What about $$S^1 \times S^1$$?
• Let me just remark that such a diffeomorphism would especially imply that the orbit spaces are homeomorphic. Now for similar questions, say we look at free actions of $\mathbb{Z}/p$ on S^3, one gets quite a few such actions with non-homeomorphic orbit spaces (these orbit spaces are called lense spaces). Thus this question asks about things which only hold in dimension 2 and only when the order of the diffeomorphism is 2. – HenrikRüping Jun 30 '20 at 9:15

Two orientation-reversing involutions of a given closed orientable surface are equivalent if and only they have the same number of fixed point circles and have the same orientation character, in the sense that the quotient surfaces (when there is non-empty fixed point set) are both orientable or both non-orientable.

Here's a sketch.

If there are no fixed points then the orbit map is just the orientation double covering of a non-orientable surface whose Euler characteristic is half that of the given surface, and hence determined by the classification of surfaces. A homeomorphism of quotient surfaces lifts to an equivariant homeomorphism.

More generally, when there are fixed point circles, the orbit map is a surjection onto a connected surface with boundary, with one boundary component for each fixed point circle. Removing annular neighborhoods of the boundary curves we see a double covering of a surface with boundary. The 2-fold covering is connected if and only if the quotient is non-orientable. In each case we know the Euler characteristic of the quotient again is half the original Euler characteristic. Therefore, knowing the number of boundary components we again have determined either the orientable double covering or a trivial double covering, of a specific surface with boundary, which is then completed by equivariantly attaching a set number of invariant annuli joining pairs of boundary components. To do this note that each boundary curve in the target is trivially covered by a pair of boundary curves in the domain. For each such pair of boundary curves one attaches an annulus $$S^1\times [-1,1]$$ with an involution that interchanges the ends and fixes $$S^1\times \{0\}$$. The effect in the quotient to to attach $$S^1\times [0,1]$$ along $$S^1\times \{1\}$$. Again a homeomorphism of orbit spaces lifts to the required equivariant homeomorphism.

Note that the orientation character condition can be reformulated as saying for each of the two involutions the fixed point set represents a nonzero element of mod 2 homology (in the non-orientable quotient space case) or each represents zero in mod 2 homology (in the orientable quotient space case).

Aside: Involutions of non-orientable surfaces are perhaps more interesting. One has to distinguish fixed point components that are themselves orientation reversing loops. Moreover, a given involution can also have both fixed circles and isolated fixed points, e.g. a standard involution on the projective plane.

• Thanks for the answer! The point which is still unclear to me is how to attach these annuli in the last part. Could you please expand a bit? – Aloizio Macedo Jul 4 '20 at 16:24
• I added some discussion about attaching the annuli. Hope that's a little clearer. I also added a note about the meaning of the mod 2 homology class represented by the fixed point set. – Allan Edmonds Jul 6 '20 at 15:28
• Thanks for the clarification! The note was also pretty useful. I've accepted the answer, but I wonder if this would also apply for $M$ a compact manifold with boundary. I don't see a point where the argument would fail, but perhaps I'm missing something. – Aloizio Macedo Jul 6 '20 at 19:08
• One issue that bothers me is that if $M$ has boundary, then removing the annuli neighbourhoods can leave corners. – Aloizio Macedo Jul 6 '20 at 22:48
• If $M$ has boundary, then the fixed point set might contain arcs. One would have to account for the number of such arcs, and whether a fixed point arc connected two different boundary components or had both end points in the same boundary component. I don't think the corners are a problem. They would actually be useful for keeping track of where to re-attach the neighborhoods of the fixed point arcs. – Allan Edmonds Jul 7 '20 at 21:35

This always holds for anti-conformal involutions of $$S^2$$. Represent $$S^2$$ as the Riemann sphere, then the fixed point set of an anti-conformal involution is a circle on the sphere. And this circle determines the involution uniquely. And the group of automorphisms of the Riemann sphere acts transitively on circles (because it acts transitively on triples of distinct points, and each such triple lies on a unique circle).

But this does not hold for tori. For example, a square torus has two essentially distinct anti-conformal involutions: one fixes two circles and another fixes one circle.

Of course, the square torus is an exception but a generic torus with an involution which fixes two circles, there is another involution which also fixes two circles and they are not conjugate: represent such a torus by a rectangle in the usual way; then one involution fixes vertical sides (and a vertical middle line), while another fixes horizontal sides (and the horizontal middle line). Unless it is a square, there is no conformal automorphism which interchanges vertical and horizontal sides.

Same applies to a generic torus with an involution fixing one circle (rhombic torus). There is always another involution, and these two involutions are not conjugate to each other.

Remark. Consider the Weierstrass form of the torus: $$y^2=4(x-e_1)(x-e_2)(x-e_3).$$ When all $$e_j$$ are real, the unvolution $$(x,y)\mapsto (\overline{x},\overline{y})$$ has two fixed circles. When $$e_1$$ is real but $$e_2=\overline{e_3}$$ are not real, the same involution has one fixed circle. Just sketch the set of real points of this curve in $$x,y$$ plane.

• Maybe I am blind, but I cant see the involution withone one fixed circle. – HenrikRüping Jun 30 '20 at 4:54
• Hello, thanks for the answer. Why can we assume that the involutions are anti-conformal? And I also can't see the involution with just one fixed circle on the torus. – Aloizio Macedo Jun 30 '20 at 10:49
• Consider a rhombus. Identifying the opposite sides by translations gives you a torus. The involution of this torus with one fixed circle comes from reflection of this rhombus with respect to a diagonal. Since a rhombus has two diagonals, there always two non-conjugate such involutions, unless the rhombus is a square. – Alexandre Eremenko Jun 30 '20 at 11:29
• @Aloizio Macedo: a conformal involution can only have isolated points fixed. Thus when the set of fixed points is infinite, the involution must be anti-conformal. – Alexandre Eremenko Jun 30 '20 at 11:32
• @AlexandreEremenko It seems that you are saying that an involution must be either conformal or anti-conformal. If so, I don't understand why. If this and other facts of the first paragraph can be found somewhere, would you have a reference? Thanks again! – Aloizio Macedo Jun 30 '20 at 17:28