Cohomogeneity 3 Actions on Spheres J. McGowan C. Searle

History A.Classification of actions of low cohomogeneity Hsiang & Lawson: minimal submanifolds; classification of maximal linear cohomogeneity 1 & 2 actions – Uchida (some) examples omitted from H&L’s work Grove & Halperin: Dupin hypersufaces; descriptions of cohom. 1 actions and diameters & 1996 –Straume: Complete classification of cohom. 0,1 & 2 actions on true spheres and homology spheres

History B.Work on diameters of manifolds and orbit spaces – M. lower bound on diameters of manifolds of constant curvature – Flach showed that there is a lower bound on the diameters of positively curved manifolds whose sec. curvature K>δ> – Greenwald found a lower bound on diameters of curvature 1 orbifolds – Dunbar, Greenwald, M, Searle found lower bounds on diameters of quotients of S 3 by finite groups

Step one: identifying cohomogeneity 3 actions on spheres From work by Dadok,  G x M  M, M/G polar cohom n-1  irreducible symmetric space of rank n corresponding to the action Tables of cohomogeneity 0, 1, and 2  Hsiang & Lawson, addendum by Uchida  Straume List of symmetric spaces Helgason

Some irreducible polar cohomogeneity 3 S n actions GroupRepresentationDimension SO(5)9 SU(5)Ad24 Sp(4)Ad36 SO(8)Ad28 SO(9)Ad36 Sp(4)27 4k U(4)×SU(k)8k Sp(4)×Sp(k)12k U(8)30 U(9)42 U(4)12 SO(4)×SO(k)

SO(4)  SO(4) Let A and B  SO(4). Represent x  R 16 as a 4  4 matrix. Then x  S 15 if ||x|| = 1. Let SO(4)  SO(4) act on S 15 by This action can diagonalize any 4  4 matrix. Since it is an isometric action, this diagonalized matrix has at most 4 nonzero entries and norm 1: the orbit space is contained in a 3- dimensional subset of S 15.

M, a 4x4 R-matrix, ═> is diagonal. We choose such an M to represent each orbit.

Isotropy Subgroup:

That is, where det(A) = 1. (It is this subset that we mean by S(O(1) 4 ).)

So, for example, if an orbit has a diagonal matrix with the first two entries equal, then the isotropy subgroup for that orbit contains matrices of the type A  S(O(2)O(1) 2 )

and If where Then

Similarly, If isotropy subgroup: S(O(1)O(2)O(1)) If S(O(1) 2 O(2)). If S(O(1) 2 O(2)), where the action is now

We get even larger isotropy groups if more entries are equal: S(O(2)O(2)), S(O(3)O(1)) or SO(4) as our isotropy group. We summarize these results in a diagram:

S 15 /(SO(4)  SO(4)) S(O(2)O(1) 2 ) SO(4) S(O(2)O(2)) S(O(1)O(3))  SO(3) S(O(1)O(3)) S(O(1)O(2)O(1)) S(O(1) 2 O(2)) S(O(1) 2 O(2))xSO(2) S(O(3)O(1)) S(O(2)O(2))  SO(2)

We can find the diameter of the orbit space by looking at the distances between the vertices; for this orbit space, the diameter is  /3.  /3  /4

Nonpolar and Exceptional Actions For nonpolar actions and polar actions by exceptional Lie groups, these methods do not suffice. For most cases, we must look at the roots and weights of the representation to decipher the orbit space. Hsiang’s Cohomology Theory of Topological Transformation Groups outlines a method of finding the connected component of the principal isotropy subgroup, given the group and the representation.

Hsiang’s Method Let G be a group acting on X Let T be the maximal torus in G and Δ(G) its system of roots Let S be the maximal torus in the principal isotropy group H. (We may assume S is a subset of T.) Let Ω be the weight system of a complexification of X, and Ω’ the non-zero weights

Hsiang’s method, continued 1.Is Ω’ contained in Δ(G)? If yes, then S=T, Δ(H)= Δ(G)\ Ω’. Done. 2.If not, choose w 1 in Ω’ \Δ(G), and let S 1 = w 1 ┴ =exp{X in Lie(T)| w 1 X=0} 3.Is [Ω| S 1 ]’=[Δ(G)| S 1 ]’? 4.If yes, S=S 1, Δ(H)=[Δ(G)| S 1 ]’\ [Ω| S 1 ]’. Done 5.If not, pick w 2 in [Δ(G)| S 1 ]’\ [Ω| S 1 ]’ and let S 2 = w 2 ┴ ∩S 1 6.Eventually, the process terminates, and you have the roots of the principal isotropy group.

Another case… Once we know all our angles, how do we find the diameter of the orbit space? For an exceptional Lie group with the adjoint action, we can read off information about the orbit space from the Dynkin diagram Consider the following orbit space under a the adjoint F 4 action:  Four surface boundaries, making the following angles with one another: Hyperplane 1: π/2, π/2, π/3 Hyperplane 2: π/2, π/2, π/3 Hyperplane 3: π/2, π/3, π/4 Hyperplane 4: π/2, π/3, π/4

We know we have the following type of tetrahedron:

Finding the orbit space We may position the first hyperplane in the (x,y,z,0) subspace, having unit normal (0,0,0,1). The second hyperplane makes an angle of π/2 to the first, so we position it in the y=0 subspace, with unit normal (0,1,0,0) We may position the first vertex at (1,0,0,0) Plane 3 meets the others at angles π/2, π/3, π/4, has an equation ax+by+cz+dw=0, with normal vector (a,b,c,d). Plane 4 has the same constraints

Putting it all together Put the bottom of the tetrahedron in the w=0 plane with a vertex at (1,0,0,0) Put the back of the tetrahedron in the y=0 plane Now determine the last 2 planes

Normal to the “front” plane Suppose the front plane has unit normal (a,b,c,d). Then its equation is ax+by+cz+dw = 0 Since it contains the point (1,0,0,0), a=0 Since it is orthogonal to the y=0 plane, b=0. Since its angle with the w=0 plane is π/3, d=1/2. Since we want a unit vector, we solve for c, and find

Finishing After we find all the unit normals in this fashion, we solve the planar equations simultaneously to find the 3 remaining vertices of the tetrahedron. Now we can find the distances between vertices, and hence the diameter of the tetrahedron.

We find the following distances And the diameter is π/4.

Cohomogeneity 3 classification Using these methods and the classification of lower dimensional actions, we can classify all the actions and their orbit spaces. From there, we can also find diameters.