Prerequisites:
A little Scheme,
My Scheme module system
Sylvester’s notions of signature.
The non ‘positive definite’ metric, ala Minkowski.
The metric η (eta) for flat Minkowski space.

This dense code needs some careful explanation. ((fileVal "IndCliff") sig) produces a list of tools for the Clifford algebra generated by the vector space with a quadratic form whose signature is sig. (lambda (C+ C- C* … supplies local names for each of those tools explained here.

The function Om takes a Clifford group element and returns a possibly indefinite orthogonal matrix. The matrix A= (OM g) performs the transformation on V associated with g. Matrix A can be taken as a transformation on V using ordinary n-tuples of reals to identify members of V.

The expression (m= (mm A (mm eta (tr A))) eta) in conventional notation is:
A η A† = η
The expression evaluates to #t which is printed in the results. The metric in the transformed space is unmodified. This is the simplest definition I know that A is an isometry.

For curved spaces the metric tensor depends on coördinates but for flat spaces we can usually find a constant metric tensor, as here. A metric tensor is classically positive definite and then we would write
A g A† = g
but here we allow Minkowski’s indefinite generalization.

To turn a vector x in V by a member t of the Clifford group we compute:
tx(α(t−1))
The Clifford group is not connected. There are the even and odd components to the Clifford group. 1 is in the even component and anything in V is in the odd component. The Clifford group is not closed under addition. Neither component is simply connected. The product of elements from the same component is in the even component and the product from different components is in the odd component. The determinant of (Om g) is 1 if g is in the even component or –1 if in the odd component.

Line of code with text “Simple” shows that the reals within the Cliffords, the space V and products of those all pass the Clifford group test. For spaces of degree 4 line with “even vals” shows that some even Clifford numbers fail test.