As a Lie algebra, the Standard Model Gauge Boson Generators, whichin the D4-D5-E6-E7-E8 VoDou Physicsmodel act on the Internal Symmetry Space CP2 =SU(3) / U(2), can be written in several equivalent ways,including:

**SU(3)xU(2) **

**U(3)xSU(2) **

**S(U(3)xU(2)) **

However, their Global Group Structures are not equivalent, so itis reasonable to ask:

In hep-th/9802112,The Central Correlations of Hypercharge, Isospin, Colour andChirality in the Standard Model, Heinrich Saller says[here I use the notation S(U(2)xU(3)) instead ofhis U(2x3) and sometimes use = for "isomorphic to"]:

"... The correlation of the fractionally represented hypercharge group with the isospin and colour group in the standard model determines as faithfully represented internal group the quotient groupU(1) x SU(2) x SU(3) / Z2 x Z3 ... the 'synchronization' of the nonabelian isospin-colour center I(2) x I(3) with the hypercharge property shows that the group, faithfully represented in the standard model, is the quotient group

S(U(2)xU(3)) = U(1) x SU(2) x SU(3) / I(2) x I(3) ... The universal covering group for the Lie Algebra of S(U(2)xU(3)) is R x SU(2) x SU(3) with the maximal compact subgroup U(1) x SU(2) x SU(3). ...".

John Baez said in asci.physics.research post of 2003-01-17 on the subject The StandardModel Symmetry Group Is S(U(2)xU(3):

"... U(1) x SU(2) x SU(3) is a sextuple cover of the "true" gauge group of the Standard Model - i.e. the group all of whose non-identity elements act nontrivially on some particle or other. This "true" gauge group is usually called S(U(2) x U(3)), since you can think of it as 5x5 unitary matrices that have determinant one and are block diagonal, built from a 2x2 unitary block and a 3x3 unitary block.One reason why S(U(2) x U(3)) is nice is that it's a subgroup of SU(5). This was important for people who believe in the SU(5) grand unified theory. U(1) x SU(2) x SU(3) *cannot* be embedded as a subgroup of SU(5). So, only by virtue of puzzling things like the fact that quarks have charges coming in thirds can the SU(5) theory extend the Standard Model. So if you're an optimist, you might say SU(5) "explains" the fractional charges of quarks! ...".

In that sci.physics.research post JohnBaez gave details by quoting from his web page week133 ... April 23, 1999 ...,saying:

"... What's the gauge group of the Standard Model? ... U(1) x SU(2) x SU(3) ... is perhaps not the most accurate answer. ... Every particle in the Standard Model transforms according to some representation of U(1) x SU(2) x SU(3), but some elements of this group act trivially on all these representations. Thus we can find a smaller group which can equally well be used as the gauge group of the Standard Model:the quotient of U(1) x SU(2) x SU(3) by the subgroup of elements that act trivially. ... To ... figure out this subgroup ... we need to go through all the particles and figure out which elements of U(1) x SU(2) x SU(3) act trivially on all of them.

Start with the gauge bosons. In any gauge theory, the gauge bosons transform in the adjoint representation, so the elements of the gauge group that act trivially are precisely those in the *center* of the group. U(1) is abelian so its center is all of U(1). Elements of SU(n) that lie in the center must be diagonal. The n x n diagonal unitary matrices with determinant 1 are all of the form exp(2 pi i / n), and these form a subgroup isomorphic to Z/n. It follows that the center of U(1) x SU(2) x SU(3) is U(1) x Z/2 x Z/3.

.... let's see which elements of U(1) x Z/2 x Z/3 act trivially on all ... representations! Note first that the generator of Z/2 acts as multiplication by 1 on the isospin singlets and -1 on the isospin doublets. Similarly, the generator of Z/3 acts as multiplication by 1 on the leptons and exp(2 pi i / 3) on the quarks. Thus everything in Z/2 x Z/3 acts as multiplication by some sixth root of unity. So to find elements of U(1) x Z/2 x Z/3 that act trivially, we only need to consider guys in U(1) that are sixth roots of unity....

ACTION OF ACTION OF ACTION OF exp(pi i / 3) -1 exp(2 pi i / 3) IN U(1) IN SU(2) IN SU(3) e_L -1 -1 1nu_L -1 -1 1u_L exp(pi i / 3) -1 exp(2 pi i / 3)d_L exp(pi i / 3) -1 exp(2 pi i / 3) e_R 1 1 1u_R exp(4 pi i / 3) 1 exp(2 pi i / 3)d_R exp(4 pi i / 3) 1 exp(2 pi i / 3) H -1 -1 1... the element (exp(pi i / 3), -1, exp(2 pi i / 3)) in U(1) x SU(2) x SU(3) acts trivially on all particles. This element generates a subgroup isomorphic to Z/6. ...

... the "true" gauge group of the Standard Model - i.e., the smallest possible one - is not U(1) x SU(2) x SU(3), but the quotient of this by the particular Z/6 subgroup we've just found. ... it's the subgroup of U(2) x U(3) consisting of elements (g,h) with (det g)(det h) = 1. If we embed U(2) x U(3) in U(5) in the obvious way, then this subgroup ...[ S(U(2)xU(3) ]... actually lies in SU(5) ... this is what people do in the SU(5) grand unified theory. ...".

Heinrich Saller, in hep-th/9802112,continued beyond discussion of the Standard Model Internal SymmetryGroup, saying:

"... There exists a further discrete Z2-correlation between chirality and Lorentz properties and also a continuous U(1)-external-internal one between hyperisospin and chirality. ...... the Lorentz transformations, defined by the covering group SL(C2) with centr SL(C2) = I(2) of the orthochronous group SO+(1,3), and being a subgroup of the unimodular group ... UL(2) ...[subgroup of]... GL(C2) with | det | = 1, have an I(2)-correlation with the external phase group U(1_2) ...[subgroup of]... GL(C2) ...

... The internal hypercharge U(1) ...[subgroup of]... S(U(2)xU(3)) and the external chirality U(1) ...[subgroup of]... UL(2), have to use the same phase - both U(1) coincide in the Higgs field ...

... Taking into account the identification of the external and internal phase group, the symmetry group, faithfully represented in the standard model, shows three central correlations - the discrete internal one by I(2) x I(3), the discrete external one by I(2) and, finally, the continuous external-internal one by U(1)

U(1)_ext x U(1)_int x SL(C2) x SU(2) x SU(3) / U(1) x I(2) x I(2) x I(3) = = UL(2) x S(U(2)xU(3)) / U(1)

... From an esthetical standpoint, debatable of course, the external-internal group U(1) x SL(C2) x U(1) x SU(2) x SU(3) with five direct factors (chirality, Lorentz symmetry, hypercharge, isospin, colour) may look rather unnatural and complicated, even more its quotient group above with the four central correlations.

Grand unified theories hope for a correlation of the direct factors via 'nondiagonal' supplements in larger simple groups, like SU(5) or SO(10) for the internal symmetry ...".

......