Split Orthogonal Group


Indefinite orthogonal group of signature $(n,n)$, or equivalently the isometry group of $\begin{pmatrix}0 & E_n \\ E_n & 0\end{pmatrix}$.

In components, $g=\begin{pmatrix}A & B\\ C& D\end{pmatrix}$, where $A^T C + C^T A = B^T D + D^T B =0$ and $A^T D + C^T B = 1$. The group is not compact.


As the symmetry group of $V\oplus V^\ast$

Let $V$ be an $n$-dimensional real vector space and consider $V\oplus V^\ast$. The space has a natural symmteric bilinear form given by $$ \langle X+\xi,Y+\eta \rangle = \frac{1}{2}(\iota_Y \xi + \iota_X \eta) $$ for $X,Y\in V$ and $\xi,\eta \in V^\ast$. The subspaces $V$ and $V^\ast$ are null under the pairing. The pairing $\langle\cdot,\cdot\rangle$ can be easily verified to be nondegenerate with signature $(n,n)$. The symmetry group of $(V,\langle\cdot,\cdot\rangle)$ is therefore $O(V\oplus V^\ast) \cong O(n,n)$.

The Lie algebra of $O(n,n)$ is $$ \mathfrak{so}(V\oplus V^\ast) = \{Q: \langle Q\cdot,\cdot\rangle + \langle\cdot,Q\cdot\rangle = 0\}, $$ or upon identifying $V\oplus V^\ast$ with its dual using the non-degenerate bilinear form, $$ \mathfrak{so}(V\oplus V^\ast) = \{ Q: Q+Q^\ast =0\}. $$ Upon decomposition of $Q$ in terms of the splitting $V\oplus V^\ast$, $$ Q=\begin{pmatrix}A & B\\ C & D\end{pmatrix} $$ this means $$ Q^\ast = \begin{pmatrix} D^\ast & B^\ast \\ C^\ast & A^\ast \end{pmatrix} = -Q. $$ Therefore $A,B,C,D$ defines an element of $\mathfrak{so}(V\oplus V^\ast)$ iff $$ A\in \operatorname{End}(V), B \in \wedge^2 V, C\in \wedge^2 V^\ast, D=-A^\ast $$ which identifies $\mathfrak{so}(V\oplus V^\ast)$ with $\operatorname{End}(V)\oplus \wedge^2 V\oplus \wedge^2 V^\ast$.

Remark. Alternatively this can be seen as the decomposition of $\mathfrak{so}(E) = \wedge^2 E$ where $E$ is a vector space endowed with a nondegenerate symmetric bilinear form. When $E=V\oplus V^\ast$, $$ \mathfrak{so}(V\oplus V^\ast) = \wedge^2(V\oplus V^\ast) = \wedge^2 V \oplus (V\otimes V^\ast) \oplus \wedge^2 V^\ast $$ and $V\otimes V^\ast$ is just $\operatorname{End}(V)$.

For $A\in\operatorname{End}(V)$ there is an element $$ Q_A = \begin{pmatrix} A & 0 \\ 0 & -A^\ast \end{pmatrix} \in \mathfrak{so}(V\oplus V^\ast) $$ acting on $V\oplus V^\ast$ as linear transformations $$ e^{Q_A} = \begin{pmatrix} e^A & 0 \\ 0 & [(e^A)^\ast]^{-1}\end{pmatrix} \in SO(V\oplus V^\ast) $$ determining a subgroup $GL^+(V)\subset SO(V\oplus V^\ast)$ since any transformation $T\in GL^+(V)$ of positive determinant is $e^A$ for som $A\in \operatorname{End}(V)$. In fact the map $$ A \mapsto \begin{pmatrix} A & 0\\ 0 & (A^\ast)^{-1} \end{pmatrix} $$ gives an injection of $GL(V)$ into $SO(V\oplus V^\ast)$. For a 2-form $B\in \wedge^2 V^\ast$, there is $$ Q_B = \begin{pmatrix} 0 & B\\ 0& 0\end{pmatrix} \in\mathfrak{so}(V\oplus V^\ast) $$ acting on $V\oplus V^\ast$ via the linear transformation $$ e^B = e^{Q_B} = \exp\begin{pmatrix} 0 & B \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 1 & B \\ 0 & 1\end{pmatrix} $$ which maps $\begin{pmatrix} X \\ \xi \end{pmatrix}$ to $\begin{pmatrix} X+\iota_\xi B \\ \xi\end{pmatrix}$.

As a ‘symmetry group’ of quantum torus

An analogy

For an antisymmetric matrix $\theta$ there is a corresponding quantum torus $A_\theta$. Under a ‘fractional linear transformation’ $$ \theta \mapsto g\theta = (A\theta + B)(C\theta + D)^{-1} $$ where $g\theta$ is defined when $C\theta + D$ is invertible, $g\theta$ is again an antisymmetric matrix, and defines a quantum torus $A_{g\theta}$.

The tori $A_\theta$ and $A_{g\theta}$ are strongly Morita equivalent. When the smoothness of the tori are taken into account, two quantum torus are in a same $O(n,n;\mathbb{Z})$-orbit iff they are completely Morita equivalent.

There is a subgroup $\mathfrak{so}(n;\mathbb{Z}) \subset O(n,n;\mathbb{Z})$ and for $g\in \mathfrak{so}(n;\mathbb{Z})$, $A_\theta \cong A_{g\theta}$. Here the group $\mathfrak{so}(n;\mathbb{Z})$ is the additive group of $\mathbb{Z}^{n\times n}$, embedded into $O(n,n;\mathbb{Z})$ via $B\mapsto \begin{pmatrix} 1 & B \\ 0 & 1 \end{pmatrix}$.

Here is a strange analogy:

It is highly possible that $A$ should be realized not as a functor but an anafunctor, and the antisymmetric matrices $\theta$ should be realized as 2-cocycles whose classes are in $H^2(\mathbb{Z}^n,U(1))$, pulled back from $H^3(X,\mathbb{Z})$ where $X$ is a principal torus bundle. The ‘Fourier development’ of the realization of $A$ should be given by an $\mathfrak{so}(n;\mathbb{Z})$-principal bundle for each $\theta$; these should come from a foliation of $X$ by some Dirac-like structure.

Realizing the analogy via the orbit method:

The ‘higher’ phenomena come from the fact that quantum tori as twisted group $C^\ast$-algebras are algebras encoding the information about projective representations of $\mathbb{Z}^n$.

Heisenberg group, metaplectic representation, theta correspondence

[Under Construction]

A possible upper half plane $\mathcal{H}_n$

Definition(Narain upper half plane) . Let the Narain upper half plane be the set $$ \mathcal{H}_n = \{\tau\in\operatorname{Mat}(n\times n,\mathbb{C})|\tau^\mathsf{t} = -\tau\}. $$
  1. $\gamma\tau$ is agiain an antisymmetric matrix, when defined. $$ (C\tau + D)^\mathsf{t}(\gamma\tau + (\gamma\tau)^\mathsf{t})(C\tau + D) = \tau + \tau^\mathsf{t} = 0. $$

Strange connection with antisymmetric matrices

It has been pointed out that the “fractional linear transformation” takes an antisymmetric matrix to another antisymmetric matrix. Now let’s look at a larger group $GO(n,n;\mathbb{Z})$, it is given by a group extension $$ 1 \to \mathbb{Z}^{2n}_{2} \to GO(n,n;\mathbb{Z}) \to O(n,n;\mathbb{Z}) \to 1 $$ More precisely it is the group that preserves the quadratic form $$ \begin{pmatrix}0 & E_n \\ E_n & 0\end{pmatrix} $$ only up to a sign. It can be shown that this group also takes an antisymmetric matrix to another antisymmetric matrix. Now

  1. Does it takes a quantum torus to another Morita equivalent quantum torus?
  2. Are all the Morita equivalent quantum tori in the same orbit of the group $GO(n,n;\mathbb{Z})$?
  3. Maybe in the definition of the upper half plane we can force the upper-trangular entries of the antisymmetric matrices to be non-negative, since with $GO(n,n;\mathbb{Z})$ signs can always be reversed. The thing to check is whether signs are invariant under a fractional linear transformation.
  4. Identifying an antisymmetric matrix $\Gamma$ with the map $\mathbb{R}^{n\ast}\to \mathbb{R}^n$, what happens to the graph $\Gamma_{\Pi} = \{(\Pi\xi,\xi)|\xi\in\mathbb{R}^{n\ast}\}$ under a $GO(n,n;\mathbb{Z})$ action?