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  Decomposition of the group of Bogoliubov transformations

+ 5 like - 0 dislike

Consider the fermion Fock space $\mathcal{F}=\bigoplus_{k\ge 0}\bigwedge^k\mathfrak{h}$ of some finite-dimensional 1-particle Hilbert space $\mathfrak{h}$. The group $\mathrm{Bog}(\mathcal{F})$ of Bogoliubov transformations can be defined as the set of unitary maps $U$ on $\mathcal{F}$ for which there are linear maps $u:\mathfrak{h}\to\mathfrak{h}$ and $v:\mathfrak{h}\to\mathfrak{h}^*$ such that $$Ua^*(f)U^*=a^*(uf)+a(J^*(v(f)))\quad\forall f\in\mathfrak{h},$$ where $a^*,a:\mathfrak{h}\to\mathcal{B}(\mathcal{F})$ denote the usual fermion creation- and annihilation operators and $J:\mathfrak{h}\to\mathfrak{h}^*$ denotes the Riesz isomorphism. It is not hard to see that these $u$ and $v$ define a unitary map $\Phi(U)\in U(\mathfrak{h}\oplus\mathfrak{h}^*)$ commuting with $\mathcal{J}$, where $$\Phi(U):=\begin{pmatrix}u&J^*vJ^*\\ v&JuJ^*\end{pmatrix},\quad\mathcal{J}:=\begin{pmatrix}0&J^*\\J&0\end{pmatrix}.$$

Defining $G:=\{A\in U(\mathfrak{h}\oplus\mathfrak{h}^*)\mid A\mathcal{J}=\mathcal{J}A\}$, it turns out that $\Phi$ defines a short exact sequence of Lie groups $$1\to\mathbb{S}^1\to\mathrm{Bog}(\mathcal{F})\to G\to 1,$$ Now my question is: does this sequence split (or, put differently, is $\mathrm{Bog}(\mathcal{F})\cong\mathbb{S}^1\times G$)?

Note that, if we are working in the category of groups (as opposed to Lie groups), central extensions of the group $G$ by $\mathbb{S}^1$ are classified (upto isomorphism) by the cohomology group $H^2(G,\mathbb{S}^1)$. If this classification is also valid in the Lie group setting, there might be some general result showing that $H^2(G,\mathbb{S}^1)=0$, which would answer my question positively.

This post imported from StackExchange MathOverflow at 2019-03-12 18:54 (UTC), posted by SE-user Robert Rauch
asked Feb 26, 2019 in Mathematics by Robert Rauch (25 points) [ no revision ]
retagged Mar 12, 2019
If i give you a $v : h^* \to h$ could you give me a preimage for the matrice with coeeficients $u=0$ and $v$ ?

This post imported from StackExchange MathOverflow at 2019-03-12 18:54 (UTC), posted by SE-user Bleuderk
Well, yes: given $v:\mathfrak{h}\to\mathfrak{h}^*$ (note the order of $\mathfrak{h}$ and $\mathfrak{h}^*$), then there is a Bogoliubov transformation $U$ on $\mathcal{F}$ with $u=0$ and the given $v$. The main difficulty here is to construct the image $U\Omega$ of the vacuum $\Omega\in\mathcal{F}$. A proof can be found e.g. in Solovejs lecture Notes on Many Particle Quantum Mechanics (Theorem 9.5). Note that in the infinite-dimensional case $V^*V$ is required to be trace class ("Shale Stinespring condition").

This post imported from StackExchange MathOverflow at 2019-03-12 18:54 (UTC), posted by SE-user Robert Rauch

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