Compatible Peirce decompositions of Jordan triple systems - MSP

COMPATIBLE PEIRCE DECOMPOSITIONS OF JORDAN TRIPLE SYSTEMS 75
and again {xyz} e {J2J2J0} = 0 leads to a contradiction. Thus there
never is a component of {xi3yi3zik} in Jjk. To get rid of the components JijΊ Jik in (P10) we may assume
by symmetry that j Φ 0. If fc = 0 then Jik reduces to the term
Jm and if k Φ 0 our argument for Ji3 will apply to Jik. Thus we
may assume ί, j Φ 0 and show there is no component in Ji3 . Ji3- is
J
spanned by the various J2{et + e3) where in (2.9ii) we saw P{ei re3f—
P(βi, βj) 2
> so Ji3- is spanned by elements {eiXi3e3 }, where {{eiXi3e3 }y3Ίczki } =
{eiXitesVsiPhi}} - {{eiXiίzkt }yjkeί } + {zkt {xijeiyik }ej } (by {$•%)) ^{eiXi3 {e3y3kzki })
(by (U), (2.6) for i, jΦθ) = {ejyjk {eixίjzki }}-{{ejy3keί }xίjzki }
(by (0.5)) - {eAyύke5xi5 }zk% ) (by
involves no index j.
(U), (2.6) again) e {Jo ( &,)JJQ (&,)} c
•
Note that when ^ = g 3
^ consists of a single family, the decomposition
J" = J« © Jio Θ Joo and Peirce rules (2.7), (2.8) reduce to
(2.2). It is easy to give examples where the unexpected Peirce
terms in (2.8) are nonzero if one of the spaces is of the form Jm since this may include elements which "ought" to belong to Ju. For
example in (Ul), in a matrix algebra Mn(Φ) if we take g^ = {eιu e22} then eneJii, e12eJί0 (not Jtil), e22eJu so {ene12e22} = e12eJίof) P(Ju)Jί0. Similar arguments apply to all components except
(U2)' P{Ji5 )JQ5 in Jί3 (P2)' P(JiS)Jtt in Ji5 (i = 0 allowed)
(P3)' P^)/^ in /„, Ji0 (P6)' {Ji^i/**} in e/- fi
(P9)' {J.^^o} in Ju. It is not clear whether such terms can actually exist.
Whatever their defects and uncertainties, such decompositions
are intrinsic.
PROPOSITION 2.10. If & = g\ U U
Any ideal K