Matrices mod p, O4 Kernel and Quotient
Quaternion Subgroup of O4
Review the embedding of the unit quaternions
into the 4 by 4 orthonormal matrices.
This defines a subgroup of O that is isomorphic to the unit quaternions,
over any ring or field.
Inside O′
Let q be a quaternion that gives rise to the matrix Q.
Premultiply by a row vector v and the result is vq, under quaternion multiplication.
Unless q is 1, vq is not equal to v.
In fact Q fixes nothing, nor Q2, nor Q3 etc, until you get back around to 1.
If Q has order d then the cycle decomposition consists entirely of disjoint d cycles.
Let d be even, so that each cycle is an odd permutation.
We then need an odd number of cycles.
Thus d consumes all the powers of 2 in the group.
Either Q or some odd power of Q has order 2l,
where l is the largest power of 2 dividing w-1 or w+1.
Pull this back to the quaternions, and then the group G2.
Mod out by the center C and find a cycle of length l.
Yet the 2 sylow subgroup of G/C is dihedral D½l,
and cannot support a cycle of length l.
The permutation is even,
and the entire group of quaternion matrices lives in O′.
Normal Subgroup
The quaternion group is normal in O.
Let q be a quaternion that leads to the matrix Q,
and consider XQ/X, where X is a matrix in O4.
Let X have a top row of r.
This is a quaternion, and it leads to the matrix R.
Let R be the "standard" matrix with top row r.
Recall that every matrix in O4
with top row r
is some block matrix B times R,
where B has 1 in the upper left,
and a matrix of O3 in the lower right.
Write the conjugate as BRQ/R/B.
Note that RQ/R is a quaternion matrix based on rq/r.
So the problem has been reduced to conjugating a quaternion matrix by a block matrix B.
At this point I'm going to hit you over the head with algebra.
Let an arbitrary quaternion be s+ti+uj+vk, and build a generic matrix in O3 using the
variables a through i.
Thus B has 1 in the upper left and these 9 variables in a block in the lower right.
Remember that the inverse of B is its transpose.
Evaluate BQ/B,
cancel some terms along the way,
and get the following matrix.
| s |
at+bu+cv |
dt+eu+fv |
gt+hu+iv |
| -at-bu-cv |
(a2+b2+c2)s |
(ad+be+cf)s + (ce-bf)t + (af-cd)u + (bd-ae)v |
(ag+bh+ci)s + (ch-bi)t + (ai-cg)u + (bg-ah)v |
| -dt-eu-fv |
(ad+be+cf)s + (bf-ce)t + (cd-af)u + (ae-bd)v |
(d2+e2+f2)s |
(dg+eh+fi)s + (fh-ei)t + (di-fg)u + (eg-dh)v |
| -gt-hu-iv |
(ag+bh+ci)s + (bi-ch)t + (cg-ai)u + (ah-bg)v |
(dg+eh+fi)s + (ei-fh)t + (fg-di)u + (dh-eg)v |
(g2+h2+i2)s |
This looks complicated, but I have grouped terms together to make it all work out.
Since a2+b2+c2 = 1,
the entry in position 2,2 is just s.
The same holds for positions 3,3 and 4,4.
Thus the main diagonal is s, as you would expect from a quaternion matrix.
Move to position 2,3 and look at the first term.
Rows are orthogonal, hence ad+be+cf = 0.
The first term goes away.
The next three terms can be simplified by realizing that each is a cofactor times t u or v.
In an orthogonal matrix, the cofactor
equals the entry itself.
Thus (ce-bf)t becomes -gt, and so on.
Make these substitutions throughout, and the matrix is indeed quaternion, as shown below.
Conjugation maps our subgroup into itself,
and the quaternion matrices are normal in O.
| s |
at+bu+cv |
dt+eu+fv |
gt+hu+iv |
| -at-bu-cv |
s |
-gt - hu - iv |
dt + eu + fv |
| -dt-eu-fv |
gt + hu + iv |
s |
-at - bu - cv |
| -gt-hu-iv |
-dt - eu - fv |
at + bu + cv |
s |
Quotient Group
Given a matrix X in O, read the top row as a quaternion q,
and rewrite X as BQ,
where B is a block matrix with 1 in the upper left and a matrix from O3 in the lower right.
The lower right block is the cosrep of X.
This will become the image in the quotient group.
Consider the product of two matrices X*Y.
Write this as BQCR, where Q and R are quaternion matrices, and B and C are block matrices
that define the images of X and Y in the quotient group.
Since the quaternion group is normal,
replace QC with CZ, where Z is some other quaternion matrix.
Now XY = (BC)(ZR), and BC becomes the image of XY.
The map from O4 into O3 is a group homomorphism.
A block matrix, such as B, proves the map is onto.
Therefore O3 is the quotient of O4 mod the quaternions.
Note that the sizes work out.
The quaternion group, and O3, both have order w3-w,
while |O4| = (w3-w)2.
This is more than a homomorphism.
A copy of the quotient group lives in O4,
hence we have a semidirect product.
O4′ and O3′
We showed above that the quaternions live in O4′.
Mod out by the quaternions and find a subgroup of O3 having index 2.
We already proved
that such a group has to be O′.
Therefore O′ upstairs maps onto O′ downstairs.
Quadratic Homomorphism
Given a matrix X in O4, write BQ = X, and multiply by Q inverse on the right.
Each entry of B is now a quadratic function in the 16 entries of X.
This is the explicit group homomorphism from O4 onto O3.
When the field is the reals, this map becomes a continuous projection.
O4/O3 becomes a
fiber bundle, with the unit quaternions,
essentially S3, acting as the fiber.
Automorphisms
Given a matrix X in O3,
place it in the lower right of a block matrix B, with 1 in the upper left, and use B to conjugate the quaternion matrices.
This in turn conjugates the quaternions themselves.
As you can seee from the formulas, we are really running a column vector,
consisting of the coefficients on i j and k, through X,
thus giving the new coefficients on i j and k.
Turn the page for more on quaternion automorphisms.