# Mathematical Physics – Breakdown of the Bogoliubov Transformation Group

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

defining $$G: = {A in U ( mathfrak {h} oplus mathfrak {h} ^ *) mid A mathcal {J} = mathcal {J} A }$$, actually $$Phi$$ 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: is this sequence divided (or, in other words, is $$mathrm {Bog} ( mathcal {F}) s mathbb {S} ^ 1 times G$$)?

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