PRODUCT INVOLUTIONS WITH 0 OR 1-DIMENSIONAL FIXED POINT SETS ON ORIENTABLE HANDLEBODIES

Let h be an involution with a 0 or 1-dimensional fixed point set on an orientable handlebody M. We show that obvious necessary conditions for fibering M as A x I so that h r x " with r an involution of A and r reflection about the midpoint of l also turn out to be sufficienl. We also show that such a "product" involution is determined by its fixed point set.

where h(a x 01)= A x OI (I [-1, 1]), then there exists an involution 9 of surface a such that h is equivalent to the involution h' of a x I given by h'(x,t) (9(z), (t)) for (z,t) E a x I and (t) or -t.That is, h is a product involution.When A is a closed surface the condition always holds; all involutions are then product involutions.When A has boundary, A x I is a handlebody.For a handlebody M, one naturally asks for which involutions a fibering exists so that the condition holds, or equivalently, which involutions are products.
The question is answered by Nelson  [3] for involutions of orientable handlebodies with 2-  dimensional fixed point sets.In that case, obvious necessary conditions are shown to be sufficient as well.In this paper, we show that such is also the case for orientable handlebodies carrying involutions with 0 or 1-dimensional fixed point sets.In this setting, we show that clear necessary conditions are also sufficient to guarantee that an involution h of an orientable handlebody M is equivalent to r x r on A x I, where r is an involution of A and r(t) -t.We also show that these involutions are completely determined by their fixed point sets.
In this paper all spaces and maps are piecewise linear (PL) in the sense of Rourke and Sanderson [5].Notation generally follows [5,3].An involution h on X is a period two homeomorphism h" X X.The orbit space of involutionh isX, X/ x wherex c yifand only ify x or h(x); the projection map is denoted by ph.The fixed point set of h is Fzx(h) {x c= Xlh(x x}.Involutions h and h' on X and X', respectively, are equivalent, denoted h h' if and only if there exists a homeomorphism f X .' such that h f-1 o h' o f.Recall that a handlebody of n is a space obtained by choosing 2n disjoint disks in the boundary of a 3-ball and identify,g in pairs by n PL homeomorphisms.Two handlebodies are homeomorphic if and only if lhey the same genus and are both orientable or both nonorientable.
If h is to be a product involution of form r x r, then certainly Fzx(h) is contained in A x {0}.
Accordingly, we say that a subset S of M is A-horizontal in M if M A x I and S can be isotope(t into A x {0}.Of course, not all systems of points or arcs and loops can be the fixed point s('! of some involution r on A. If a subset of M is isomorphic to the fixed point set of some involutio1 A, we say it is A-fixable.Note that all involutions of a campact bordered surface A (and hence all A-fixable sets) are easily listed [3].Clearly, if H is equivalent to r x r, then Fx(h) must be both A-horizontal and A-fixable.
Constructing a fibering for handlebody M as A x I, with respect to which which involution h is a product, is an inductive process.We decompose M into "smaller" h-invariant pieces which may be assumed, by induction hypothesis, to carry such a structure.The fibering of M is reconstructed from the fiberings on the pieces.The following lemma enables us to carry out the crucial step of breaking M into h-invariant pieces.
Fixed point sets of involutions on compact surfaces have the property that one can always raise the Euler characteristic of the surface by cutting along an arc which either misses the fixed point.set or crosses a fixed point loop in exactly one point.If h is an involution on handlebody M, then Fix(h) being both A-horizontal and A-fixable clearly implies that the genus of M can be lowered by cutting along a properly embedded disk missing Fix(h) or cutting a fixed point loop in exactly one point.The following lemma shows that this can be accomplished equivariantly.Following Theorem 1, it will also be clear that this cutting property implies that Fix(h) is both A-horizontal and A-fixable.
LEMMA 1.Let M be an orientable handlebody with involution h having either 0 or 1- dimensional fixed point set Fix(h).Suppose also that Fix(h) is both A-parallel and A-fixable.Then there exists a properly embedded non-boundary-parallel disk D such that either D or else D is h-invariant and cuts a fixed point loop in a single point.
PROOF: A disk D in M is in h-general position modulo Fix(h) if D and hD are both in general position with respect to Fix(h) and if D-Fix(h) and hD-Fix(h) are in general position.We always presume that a disk D is in h-general position and also that (D-Fix(h))V OM and (hD F,x(h)) V10M are in general position.We may, as noted above, assume that there exists properly embedded non-boundary-parallel disk D in h-general position.
Although D f3 hD is formally a subgraph of the 1-skeleton of the triangulation of M, we will not need to consider D f3 hD with this degree of detail.Instead, we regard D f3 hD has consisting of A-objects where (a) a A-loop g is a cycle of edges forming a simple closed curve component or such that g (3 hg is a component consisting of two simple closed curves intersecting at one point, and (b) a A-arc is a non-self-intersecting chain of edges forming a component of the graph or a single edge of a component which is not an arc, loop or two loops joined at a point.
We partition the A-objects into (i) loops, (ii) arcs with both endpoints in Fix(h)-OM, (iii)   arcs with neither endpoint in Fiz(h)UOM, (iv) arcs with both endpoints in OM and (v) arcs with one endpoint in OM and the other in Fix(h).Note that arcs of type (ii) do not actually occur.A A-object is called extremal on D (or hD) if it either bounds or cobounds with other A-objects a disk E on D (or hD) the interior of which intersects no other A-objects.This proof requires the repeated "simplification" of a disk D by removal of A-objects from D f3 hD.Although removal of a A-object can always be accomplished without adding any points to DVlhD, the nature of the components of DnhD may be changed so that a A-loop is created by the removal of a A-arc.Accordingly, define the complexity of a disk D by c(D) (a, l), where a is the number of A-arcs and the number of A-loops in D fl hD.The ordered pairs are lexicographically ordered so that removal of a A-arc always lowers the complexity.
Let E be the collection of non-boundary-parallel disks properly embedded in M and missing fixed point arcs.Choose D E E such that DfqFiz(h) is minimal.The hypotheses regarding Fi:r(h) make this intersection contain at most one point.The replacements of D by a "simplified" D' E E will not add to this intersection.
If an extremal A-object on hD is of type (i) or (iii), the the object either bounds, or cobounds with other A-arcs, a disk E in the interior of hD.These cases are handled in Kwun and Tolefson [2].D is isotopic to disk D" E where c(D") < c(D) by the procedure of [2, Lemma 1].The only remaining A-objects, of types (iv) and (v), are treated below.
If the extremal A-object on hD is a type (iv) arc a, then a U /', fl' an arc in hD V10M,bounds a disk E' on hD.Also, aU , an arc on DVIOM, bounds a disk E on D, where E hE'.Suppose U /' bounds a disk E" on OM.Then E U E'U E" bounds a 3-cell C with face E" in OM such that N(C) f3 hN(C) , where N denote a small regular neighborhood.Let Gt be an isoptoy of M, with support in N(C), which moves E through C', /across E" and E off of E'.Let D' GI(D) and then c(D') < c(D).
If /U/' does not bound a disk on OM, then D' E U E' is a non boundary-parallel disk properly embedded in M such that D' Cl hD' t.
To this point we have demonstrated that we can find D E such that either D tq hD I or else D Cl hD consists entirely of type (v) arcs.We need only show that from the latter case we can construct an h-invariant disk in P containing exactly one fixed point.
Suppose DfqhD consists only of type (v) arcs.Since D was originally chosen so that Df3Fiz(h) was minimal, and since the above simplifications of D add no new points to D f'l hD, all of the type (v) arcs emit from a single fixed point.D is therefore partitioned into "wedge shaped" regions E, bounded by type (v) arcs a,_ and a, and arc fl, on OD.Similarly, regions E on hD are bounded by a,_, a,, and fl, where fl C OhD.
Define an equivalence relation on {E,} by E, E iff E hE,.Each equivalence class has two members.For each class, replace either E, by E or E by E. The result is an h-invariant disk D containing exactly one fixed point.We need now only note that D is not boundary-parallel.If it was boundary-paralled, it would cobound an invariant ball with an h-invariant disk on OM.This disk contains a fixed point.Bu this is clearly not possible unless D cuts a fixed point arc instead of a loop.
[] 3. MAIN THEOREM.THEOREM 1.Let M be an orientable handlebody with involution h having 0 or 1-dimensional fixed point set Fix(h).Suppose also that Fix(h) is B-horizontal and B-fixable where M % B 1.
Then there exist a compact, bordered surface A such that M A I and h is equivalent to r r.
where r is a nonidentity involution of A and r(t) -t for all E I [-1, 1].
PROOF: Let D be the disk provided by Lemma 1.Either D hD or D 3 hD .I n either case, let M' M-N(D)-N(hD) where N(D) and N(hD) are small regular neighborhoods of D and hD chosen to coincide (be h-invariant) when D hD and to be disjoint when D 3 hD .
The proof now follows by induction on the genus g(M).We first handle independently the cases where the desired product structure cannot be assumed by induction hypothesis.This is the case when M is a 3-ball B with involution h such that Fix(h) # or when g(M) > and h is free.
When M -Bz, it is well known that h is equivalent to either reflection through the center point or revolution about a central axis.In either case M B I and h 3 r where fl is reflection through the center point or a central axis on B2.
Przytycki [4] classifies all free Z,-actions on handlebodies.By Theorem of that paper there is at most one orientation preserving and one orientation reversing free involution up to equivalence on any handlebody M with g g(M) > 1.Furthermore, these occur on orientable handlebodies only when the genus is odd.Hence, we need only present an example of each in factored form to complete this first step.
In both cases we view the handlebody of genus g as A I where A is the 2-sphere S less 29 disjoint disks and h 7" r.In the orientation preserving case, r is induced by the antipodal map on S when g invariant pairs of disks are removed to form A. In the orientation reversing case,is induced by the involution on S with two fixed points (i.e.revolution about a central axis piercing the fixed points) when an invariant disk about each fixed point and g-invariant pairs of disk are removed to form A. The induction step consists in cutting along invariant D or along D t3 hD, that is, deleting N(D) U N(hD) from M, assuming a product structure on each piece by induction hypothesis, modifying the product structures to be consistent with each other and then reassembling the pieces.We consider two cases.
Case 1: The disk D provided by Lemma is h-invariant.By induction hypothesis, N(D) and M' or M_I and M1 are each endowed with a product structure such that hIN(D and h' or h_l and hi are each equal to some r r.The difficulty is that the "scars", i.e. the images of the D, on these pieces, are not necessarily consistent with the fiberings.It is not immediately the case that D, a I where a is some arc on OA and the piece of interest fibers as A I. We show that the fibering may be so chosen by deforming the original fibering to meet this condition. Since D is h-invariant, we may assume that Fix(h) is 1-dimensional.Then each D, is also an h-invariant disk with a unique fixed point x,.Consider the disk pn,(D,) on the orbit space pt,,M' M".Note that ph,(D,) is a disk on OM" containing point p,,(z,) in its interior.Now M' factors as A' I with h' having a fixed point arc terminating in z,.Let a be an invariant arc on OA' containing x,.There is an isotopy on O(M'*) with support in a small regular neighborhood of disk p,,(a I) U p,(D,) moving p,,(a I) onto p,(D,) and constant on pt,,(x,).This extends to an isotopy K' of M'* with support in a collar (Rourke and Sanderson [5, Theorem 2.26]) of the boundary.This isotopy may be assumed constant on pt,,(Fix(h')).Hence, it lifts to an isotopy Kt of M' moving a I onto D, while fixing all points of Fix(h').We deform the fibering of M' by K -1 so that we may assume that D, a I.The same deformation can also be carried out on M,.This same argument is now applied in a neighborhood of D, in N(D) with the additional requirement that if N(D) is fibered as as B I and 13 is an invariant arc on OB containing x,, then (plN(D))( 3) is initially isotoped into a position corresponding with the adjusted position of p,(a) on p,,(M').Reassembly now gives M A I where A is constructed from the compact bordered surfaces B and A' or A,, -1, 1, by identification along arcs on the boundary.Hence A is also a compact bordered surface.Clearly, h r v where r is constructed from a' or a,, -1, 1, and the induced involution of B.
Case : DC3hD where D is the disk provided by Lemma 1.In this case we note that N(D) and N(hD) are disjoint form Fix(h).The argement is basically the same as for case except that one need not take special measures on Fix(h).Since p,(U(D)) ph(U(hD)) in the orbit space M*, the isotopy K' with support in N(D,) lifts to isotopy Kt with support in N(D,)tO N(hD,), -1, and commutes with h.
Any product involution with 0-dimensional or 1-dimensional fixed point set on an orientable handlebody may be routinely generated from an involution with identical fixed point set on an orientable, compact bordered surface.Since classifying the involutions on compact surfaces is an easy matter, all product involutions with 0-dimensional or 1-dimensioanl fixed point sets on orientable handlebodies can now be easily generated.
Our problem is that the generated list might contain many redundancies.For instance, if M is the orientable handlebody with genus three, an involution with two fixed point arcs may be generated in at least the following three ways: (i) T1 r on Ax I where A1 is the 2-sphere S less four disks.Involution 'x is constructed from the involution on S with a single simple closed curve as fixed point set by removing two invariant disks intersecting this fixed point loop and an invariant pair of disks disjoint from the fixed point set.
(ii) r r on A I where A is the torus T less two disks.Involution r2 is constructed from s x idsa on T $1 x Sx, where s is complex conjugation, by the removal of invariant disks intersecting each of the two fixed point loops.
(iii) rax r on Aa x I where Aa is again the torus T less two disks.Involution ra is constructed form the involution v on T $1 Sx, where v((x,y)) (y,x), by removal of two invariant disks from the single fixed point loop of v.
Might rl r and r r or T X r and ra r be equivalent even though rx and r are involutions of different surfaces and Fiz(T) separates while Fix(r3) does not?Similar questions are appropriate in

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