Introduction E9F5FC

Questions FFFFC0

• View
• Edit
• History
• Print

Linear Complex Structure

Unitary group and Orthogonal group

The 2-out-of-3 property

The unitary group {$U(n)$} is equivalently the intersection of the orthogonal group {$O(2n)$}, the symplectic group {$Sp(2n,\mathbb{R})$} and the complex general linear group {$GL(n,\mathbb{C})$}, all understood inside the real general linear group {$GL(2n,\mathbb{R})$}. Any intersection of two of these groups is contained in the third group and so includes it as well.

The 2-out-of-3 property decomposes a Hermitian form into its real and imaginary parts. The real part is symmetric (orthogonal), and the imaginary part is skew-symmetric (symplectic)—and these are related by the complex structure (which is the compatibility).

Thus starting with the orthogonal group {$O(2n)$} we can think of the linear complex structure as intersecting it with the complex general linear group understood within the real linear group.

Alternatively, we can think of the linear complex structure as intersecting it with the symplectic group.

Dimensional analysis

{$O(2n)$} has {$2n^2-n$} dimensions. {$U(n)$} has {$n^2$} dimensions. Skew-symmetric matrices {$A(2n)$} have dimension given by their upper triangle, which has {$1+2+\dots +2n-1 = \frac{2n(2n-1)}{2}=2n^2-n$} elements. Note that {$O(2n)/U(n)\cong O(2n)\cap A(2n)$} has dimension {$2n^2-n-n^2=n^2-n$}. Thus this intersection must reduce the {$2n^2-n$} dimensions of {$O(n)$} with the {$2n^2-n$} dimensions of {$A(n)$}. In each case, the intersection does not include {$n^2$} dimensions.

Complex structures on {$\mathbb{R}^n$}

Let {$n=2m$}. We are dividing up {$O(2m)$}, as a manifold, into a symmetric space {$S$} and the unitary group {$U(m)$}. {$S$} is not a group because it does not contain the identity {$I$}. This suggests that {$U(m)$} is not a normal subgroup of {$O(n)$}. I need to understand the cosets,

Consider even {$n=2m$}. {$S$} is the set of orthogonal complex structures in {$\mathbb{R}^n$}. The elements {$j\in S$} satisfy three relations:

Any two of these relations imply the third. Note that {$A(n)$} is the Lie algebra of of {$O(n)$}, its tangent space at {$I$}.

All {$j\in S$} satisfy the same equation {$j^2=I$}. Thus they have the same possible eigenvalues: {$+i, −i$}. The matrix {$j$} is skew-symmetric and so its diagonal elements are all {$0$} and thus trace is {$0$}, which means the sum of eigenvalues is {$0$}. Thus the eigenvalues must pair up and each has multiplicity {$m$}.

Let {$i$} be the "standard" linear complex structure of {$\mathbb{R}^{2m}= \mathbb{C}^m$}, the block diagonal {$2m\times 2m$} matrix whose {$m$} diagonal blocks are {$2\times 2$} matrices {$\begin{pmatrix}0 & -1 & | & 0 & 1 \end{pmatrix}$}. For all {$j_1, j_2\in S$} there exist {$B_1,B_2\in O(n)$} such that {$j_1=B_1iB_1^{-1}$} and {$j_2=B_2iB_2^{-1}$}. Consequently, {$j_1=B_1B_2^{-1}j_2B_2B_1^{-1}$}, so any {$j_1,j_2\in S$} are conjugate. {$S$} is a single conjugacy class in {$O(n)$}. The conjugacy action of {$O(n)$} acts by conjugation {$g\cdot j:= gjg^{-1}$}. The orbit of {$j_1$} (the elements to which it can be moved) is all of {$S$}.

But consider the conjugacy class of {$j$} with respect to {$SO(n)$}. is strictly smaller than with respect to {$O(n)$}.

The isotropy group is the subgroup of {$O(2m)$} that fixes {$i$}, namely {$G_i=\{g\in O(2m):g\cdot i=i\}$}, which is to say, {$G_i=\{g\in O(2m):gig^{-1}=i\}$}, which means {$G_i=\{g\in O(2m):gi=ig\}$}, which is {$U(m)\subset O(m)$}.

{$O(2m)$} has two connected components, one of which is {$SO(2m)$} and the other is its complement. Whereas {$U(m)$} is connected. Neither of the halves of {$O(2m)$} is contained by {$U(m)$}. Consequently, {$S\cong O(2m)/U(m)$} has two connected components.

Moreover, it is again a fixed point set component of an isometry, the map x 7→ −x T on Rn×n which preserves the submanifold O(n). Though S is again not a regular preimage (the differential of the defining map F : j 7→ j 2 + I is certainly not onto, by reasons of dimension), we have TjS = ker dFj = {v ∈ A(n); vj+jv = 0}; in fact, ker dFj is the space of elements of A(n) which anticommute with j; this is complementary to the subspace TIU(m) ⊂ A(n) containing the elements commuting with j, so it has the right dimension. Now the symmetry sj is just the conjugation with j: it fixes j and maps v ∈ TjS to −v since jvj−1 = −vjj−1 = −v. The space of quaternionic structures on C 2m is treated similarly; it is embedded into the space of antihermitean complex n × n-matrices and is an orbit of the group U(2m) acting by conjugation; the isotropy group is Sp(m) ⊂ U(2m).

Math Stack Exchange. Fundamental group of the special orthogonal group SO(n).

Cosets of {$U(n)$}

The cosets of {$U(n)$} in {$O(2n)$} are the points of {$O(2n)/U(n)$}.

Perturb {$U(n)$}

Consider ways that a unitary matrix {$U$} can be multiplied by an orthogonal nonunitary matrix {$O$} to yield an orthogonal nonunitary matrix {$O'$}.

First, suppose that {$U$} is a {$2\times 2$} block diagonal matrix and that {$O$} is likewise a {$2\times 2$} block diagonal matrix.

If {$O$} has a single nontrivial {$2\times 2$} block diagonal matrix, a reflection {$(1\; 0 \;|\; 0\; -1)$}, then {$OU$} includes a rotoreflection, and is not unitary.

{$O(2n)/U(n)\cong O(2n)\cap A(2n)$}

Consider the skew-symmetric matrices that are diagonal {$2\times 2$} blocks. Then they can't be eigenvalue pairs {$1,-1$} as with reflections because the diagonal elements must be all {$0$}. The only possible blocks can be rotations of the forms:

{$$i\equiv\begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix}, \bar{i}\equiv\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$$}

Let {$A$} be a skew-symmetric real invertible matrix.

• The trace (the sum of the eigenvalues) of {$A$} is {$0$}.
• The eigenvalues of {$A$} are not real.
• {$A^2$} is symmetric negative semi-definite. For all column vectors {$x$}, {$x^TMx\leq 0$}.

Skew-symmetric matrices do not form a group. The identity matrix is not skew-symmetric. And the square of a skew-symmetric matrix is a symmetric matrix.

Skew-symmetric matrices {$A(n)$} do form the Lie algebra of the orthogonal group.

Suppose {$A$} is also orthogonal.

• {$A$} must be diagonalizable in terms of rotations whose eigenvalues are not real.

Decomposition of unitary matrices

How to decompose a unitary transform into two level unitary matrices It is possible to conduct Gaussian elimination, relating two elements and then setting the other two elements in a {$2\times 2$} matrix so that it is unitary. Applying such actions to {$U$} yields the identity. This leads to {$U$} equaling a product of such unitary matrices.

References

• nLab: G-structure
• Physics Stack Exchange. Coset space of Lie groups.
  • R. Glimore, Lie Groups, Lie Algebras, and some of Their Applications
  • A.O. Barut R. Raczka, Theory of group representations and applications.