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) = $ 0which would answer my question positively.