Title:Computing Orthogonal Group Signs
Author:Jon Thackray
Id:$Id: nikolaus.2002.html,v 1.1 2002/12/01 13:01:04 thackray Exp $


  1. Purpose
  2. Background
  3. Caveats
  4. The algorithm
  5. Some results
  6. References
  7. Acknowledgements

Purpose of the talk

To describe an algorithm suitable for computer implementation for the computation of the isometry types of quadratic forms, and hence the group sign for groups preserving such forms. This is of interest for understanding the embedding structure of arbitrary finite groups inside classical finite orthogonal groups.

Theoretical background

The finite analogues of the classical orthogonal groups have long been known and studied (Dickson). It is well known that in odd dimensions there is only one isomorphism class of simple orthogonal group, but for even dimensions there are two classes, distinguished by the size of a maximal totally singular subspace with respect to the form (Aschbacher). These classes are labelled plus and minus, with the plus corresponding to the case of dimension n, and the minus corresponding to dimension n-1, where the overall dimension is 2n.

In the case of dimension 2, there are precisely two isometry classes of spaces with non-degenerate forms, labelled D+ (hyperbolic) and D- (definite). In a definite space, there are no non-trivial singular (of norm zero) vectors. In a hyperbolic space, a basis can be chosen of two norm zero vectors whose inner product with respect to the associated bilinear form is non-zero

We will assume throughout this talk that all forms are non-degenerate, ie that the subspace perpendicular to the whole space is {0}. Hence, given a space V with form f, and a subspace W, we have dim(W) + dim(W perp) = dim(V). A subspace W will be called totally singular if W consists entirely of singular vectors and W <= W perp. Clearly then, a totally singular subspace W of a subspace V of dimension 2n has dimension at most n. Suppose V has dimension 2n, and W is a totally singular subspace of dimension less than n-1. Then W perp = U direct sum W, where dim(U) > 2. Hence U contains a non-zero singular vector u. Further since U <= W perp, if w is an element of W, then (u+w)f = uf + wf + (u,w)b, where b is the associated bilinear form, so (u+w)f = 0. Thus W direct sum <u> is a totally singular subspace. Hence we can grow a totally singular subspace up to its maximum possible size by adding isotropic elements from its complement in its perp. Clearly this process stops when either W = W perp, or the complement U is definite. It is also the case that the termination condition (either maximum size or definite complement) is independent of the choices of vectors u that we have made (Witt).


The group sign is a property of the actual representation, including the field over which it exists. Finding a non-zero singular vector is essentially a matter of solving a quadratic equation over the representation field, so extending the field quadratically will always allow the sign to become plus. One can interpret this in terms of Chevally groups as the embedding of 2Dn(q) in Dn(q**2). Since the field of the representation is the field of the reduction of the character, it is easy to deduce which field one should look at, but nonetheless important to be sure one is looking at the right field.

The algorithm

The algorithm is a loop. We essentially consider three spaces, V, V perp and W, where V perp = V direct sum W. The algorithm terminates when dim(W) = 2, at which point the isometry type of W is the answer. Initially, V = {0}, and V perp and W are therefore the whole space. We actually hold only a basis for W, which we enumerate w1, w2, ...wm. Each iteration consists of the following steps, assuming m >= 3 (note that m is always even).

  1. Find a vector w != 0 within <w1, w2, w3> such that wf = 0. Thus w extends V by one dimension to V'.
  2. Discard one of the vectors w1, w2, w3 from the basis of W, the choice having non-zero coefficient in the expression of w. Thus we have W' such that V' direct sum W' = V perp. Let w1', ... wk' be the basis of W' so derived, where k = m-1.
  3. Compute the bra (w, )b (a vector by matrix multiplication).
  4. Now compute ui = (w,wi')b for each wi'.
  5. Choose i such that ui != 0 (such i must exist because dim(V' perp) = dim(V perp) - 1).
  6. For each j != i such that uj != 0, subtract uj * (ui inverse) * wi' from wj'.
  7. Discard wi' from the basis of W', giving a basis of W'' such that V' perp = V' direct sum W''.
  8. dim(W'') = dim(W) - 2, and V', V' perp and W'' satisfy the entry conditions for the loop. If dim(W'') > 2, start again at step 1.

Finally, given W such that dim(W) = 2, we attempt to find a non-zero isotropic vecvtor. If we succeed, the answer is plus (W'' is hyperbolic). If not, the answer is minus (W'' is definite).

Some results

I have computed the orthogonal group signs for characteristics 2, 3, 5 for HS and McL (groups for which I have some sentimental attachment). I am not prejudiced against larger primes, however, my meataxe currently does not support any field sizes larger than 5.

For HS the results are:-

p = 2

p = 3

p = 5


p = 2

p = 3

p = 5



I wish to thank Juergen Mueller for suggesting the problem, Rob Wilson for numerous useful technical conversations, and Richard Parker for a number of useful conversations in the polishing stages.