Re: complex orthogonal group
- From: Ilya Zakharevich <nospam-abuse@xxxxxxxxx>
- Date: 18 Oct 2005 07:39:48 -0400
I wrote in article <dho39g$1gbf$1@xxxxxxxxxxxxxxxxxx>:
> Any good thorough book on linear algebra (i.e., designed not for
> teaching students) will contain the spectral theorem over an
> algebraically closed case.
As I said, this does not look to be the case. I provided this
classification in another message.
> There is also a classification in the real case.
Since this also does not look to be the case, I feel obligated to
provide the classification too. Below (x) stands for tensor product,
and you may need to have fixed-width font to read the matrices.
[The "proper" way to treat thsi problem may be Galois cohomology,
but I'm not fluent enough in this language, so provide an
"elementary" approach. First proceed as in complex case; but
Observation 2 is slightly more general now.]
Observation 1:
For any real invertible n x n matrix A, the transformation with the matrix
0 A is an automorphism of a real 0 Id(n)
1/A^t 0 quadratic form with Gram matrix Id(n) 0.
Observation 2:
The transformation with the matrix J(n) is an infinitesimal automorphism
for the bilinear form with Gram matrix antidiag(1,-1,1,...); the latter
form is symmetric for odd n, and skewsymmetric for even n. Here
antidiag() is the matrix with only non-0 entries on the antidiagonal
(connecting lower-left to upper-right corners); J(n) is the Jordan
matrix: 1's immediately over the principal diagonal, the rest is 0.
Thus exp J(n) is an automorphism of this bilinear form.
===================================================
The orthogonal transformations of these forms were almost enough to
obtain the full classification in complex case. (Recall that in the
second observation one doesn't need even n, and one needs to allow
multiplication of the block by -1).
In real case, one needs one more type of block; actually, the following
construction not only provides this missing type of block, but also
allows getting blocks of Observation 1 out of those from Observation 2.
===================================================
Construction:
If vector spaces V and W are both symplectic, or both with non-degenerated
quadratic forms, then the tensor product V (x) W carries a natural
quadratic form. The inertia index is 0 in symplectic case, and the
product of inertia indices of V and W in quadratic case.
If A and B are transformations of V, W which preserve the bilinear forms,
the A (x) B is an orthogonal transformation of V (x) W. If A is a Jordan
cell with eigenvalue 1, and B has simple eigenvalues mu1,...,muK, then
A (x) B has K Jordan cells (of the size dim A) with eigenvalues mu1,...,muK.
[One can also show that if the transformation B is indecomposable (among
invariant subspaces of B there is no non-trivial complementary orthogonal
subspaces), then A (x) B is also indecomposable.]
Observation 3:
Combining two preceeding Observations, one can construct following examples
of orthogonal transformations (d=dim W, i=inertia V (x) W):
d inertia(W) B eigenvalues(B) dim V dim V(x)W i
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
1 +1 or -1 1 or -1 1 or -1 odd odd 1,-1
2 2, or -2, or skew rotation mu, 1/mu (|mu|=1) any even 2,0,-2
2 0 or skew diag(mu,1/mu) mu, 1/mu (real) any even 0
4 0 or skew see below mu,mubar,1/mu,1/mubar any 4n 0
Here inertia(W)=skew means that the form in W is skew. If mu=a+ib,
mubar=a-ib, 1/mu = c + id, then the 4-dimensional matrix B can be written as
a b 0 0 0 0 1 0 0 0 1 0
-b a 0 0 0 0 0 1 0 0 0 1
0 0 c d -1 0 0 0 1 0 0 0
0 0 -d c 0 -1 0 0 0 1 0 0
[On the right we put Gram matrices of symplectic and quadratic forms
preserved by this matrix.]
===================================================
One can immediately recognize that the first line of the table corresponds
to "Observation 2" (possibly multiplied by -1). The last two lines
correspond (for |mu|>1) to "Observation 1": the invariant isotropic
subspaces correspond to sum of blocks with eigenvalues nu with |nu|>1.
They can't be decomposed into orthogonal direct sum of two transformations:
for third line: unless mu=1 or mu=-1, and dim V is odd. For fourth line:
unless mu is real.
The second line provides the remaining type of block needed for the
classification. One may assume that mu is not 1 or -1 (otherwise it
is the same as in the third row); in this case this orthogonal
transformation is indecomposable.
===================================================
The bases we use for the blocks above are not orthogonal; however, the
Gram matrices are so simple, that it is easy to find explicit orthogonal
bases. For the purpose of the following theorem, assume that the
transformations explicitly defined above are rewritten in such orthogonal
bases.
Theorem:
for any real orthogonal transformation there is an orthogonal basis
in which it is block-diagonal, with blocks given by 4 lines in the
table above. One may assume that pairs (mu,d) are subject to the
restrictions of indecomposability (stated above). If we require that
|mu| >= 1 and Im mu >= 0, then the list of blocks which appear in the
decomposition is uniquely defined.
===================================================
Remark A.
Considering blocks based on exp J is an overkill in the case of the
3rd and 4th lines of the table; the approach of Observation 1 gives
much more sparse representatives. In the case of 3rd row take A to
be a real Jordan block with eigenvalue mu. In the case of 4th row,
mu = a + ib, take a block matrix with 2x2 blocks: diagonal blocks are
(a,b;-b,a); ones immediately above diagonal are Id(2); the rest are 0.
Remark B.
Stating the obvious, there are 2 ways to write a matrix of a tensor
product of matrices. In the case of 2nd line of the table, the
resulting orthogonal matrix can be written either as a block matrix
with 2x2 blocks, non-0 ones (above diagonal) being R/k!; here R is a
2x2 matrix of rotation, and k is the distance to the diagonal.
In the other form, it is a 2x2 block matrix with blocks being
c exp J(n); here c is cos phi, sin phi or -sin phi correspondingly.
Remark C.
I do not know a way to write representatives for the 1st and 2nd
lines of the table in such a way that both the transformation and
the Gram matrix are sparse.
Remark D.
If we know that the initial quadratic form has only a few minuses in
the signature, this severely restricts the configuration and the total
number of "non-definite" blocks which appear in the decomposition.
E.g., if there is only 1 minus, then there is at most one non-definite
block; it can't come from the last line of the table. It may come
either from the first line with dim V=3 [parabolic elements of
SL(2)=SO(2,1)], or from the third line with dim V=1 [elements
of SO(1,1)].
Remark E.
Warning: rotations by angles phi and -phi ARE conjugate in O(2), but
not in in the symplectic group Sp(2). Thus in the |mu|=1 case with
symplectic W (so dim V is even, and resulting inertia index is 0) one
get two non-isomorphic representatives with the same eigenvalues of
Jordan blocks and the same signature. (In all other cases the
signature of the form and GL-conjugacy class [information on sizes and
eigenvalues of Jordan blocks] uniquely determines the one of the
listed blocks.) This is similar to situation in SO(2) - but not O(2).
===================================================
It turns out that using elementary means it is easy to reduce the proof
to the "linearized" problem: classification of pairs of a quadratic and
a skewsymmetric bilinear forms (see Thompson reference in my initial
posting). Indeed, recall that two root subspaces of an orthogonal
transformation with eigenvalues mu, mu' are orthogonal unless mu' = 1/mu.
Thus for indecomposable transformations if eigenvalue mu appears, than
the only other eigenvalues which may appear are one of 1/mu, muBar, 1/muBar.
If |mu| is not 1, then combining eigenvalues outside and inside the unit
circle, one obtains 2 real complementary isotropic subspaces; thus one is
in the situation of Observation 1. Since A (of Observation 1) must be
indecomposable, it is either a Jordan block with real eigenvalue, or the
matrix described near the end of Remark A.
[The theory of blocks for real matrices is reduced to one for complex
matrices via the following observation: if real matrices A, A' are
conjugated by a complex transformation, then they are conjugated by
a real transformation. Indeed, write the complex transformation as
B + iC; then (B + t C) A = A' (B + t C) for t=i, thus for t=-i, thus
for any t. Moreover, B + t C is invertible for generic real t.
Given any real matrix, consider the complex Jordan blocks of its
complexification. Since the matrix of end of Remark A has two
Jordan blocks with complex conjugate eigenvalues, one can find
a block-diagonal real matrix with blocks of the above forms which has
the same complex Jordan blocks; thus it is conjugate to the initial
matrix.]
This finishes the "trivial" part of the classification. What remains is
the case of |mu|=1. First, consider the case mu=1. We claim that in this
case the logarithm of the orthogonal transformation is an infinitesimal
automorphism of the form. Indeed, if an element of a Lie algebra acts
non-trivially and nilpotently, then its exp acts non-trivially. Thus
the case of mu=1 is reduced to the linearized case indeed. Multiplying
by -1, one covers the case mu=-1 as well.
If |mu|=1 and mu is non-real, then eigenspaces for mu and for 1/mu are
complementary isotropic, thus dual. They are also complex conjugated.
In particular, for any nu with |nu|=1, consider the operator which
multiplies vectors in one of these subspaces by nu, in another by
1/nu. It is both real and orthogonal. It also commutes with the
initial operator. Choosing nu = 1/mu, the composition has all
eigenvalues 1; thus its log is an infinitesimal automorphism; thus
logarithm of the initial transformation is an infinitesimal
automorphism as well. Thus the classification of pairs is applicable.
Hope this helps,
Ilya
.
- References:
- complex orthogonal group
- From: Tobias Fritz
- Re: complex orthogonal group
- From: Ilya Zakharevich
- complex orthogonal group
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