Compatible Peirce decompositions of Jordan triple systems - MSP

74 KEVIN MCCRIMMON
component in J tucJ 2.)
For (2.6), if {&, 1)0 {i, j} = 0 then in particular 0 cannot appear
in both pairs. If i, j Φ 0 then as above J i$ is spanned by J 2's with
J klc.J 0, so P{J iβ)J kι = {JijJkiJ} = 0 follows from P(J 2)J 0 = {JWl^O.
Similarly if k, I Φ 0 it follows from PWΛ = {JJJΓ} = 0. If {fe, Z}£
ίi i> n s}, say kΦi f j, r, s, then the product falls in P(J 0(ί? k)){J i(&k) +
j-^g 7 ^)} = 0 as long as k Φ 0, while if & = 0 by the above we must
have 1 6 {i, j} Π {r, s) so by symmetry we may take 1 = j = r, therefore
the only possible unlinked products are (replacing s by fc) of
the form
(U) PGWo/ or {JijJojJn} (ί, j , k Φ θ , k Φ i ) .
Starting from (A) in all instances, we analyze these unlinked
products. In the first, either ί — j, in which case (Ul) results from
(B) (r = i), or else i Φ j, in which case (U2) results from (E) (r=£).
In the second product either i = j Φ k, whence (U3) results from
(C) (r = k), (E) (r = i) and dually if i Φ j — k, or else
whence (U4) results from (C) (r = ΐ and r = k).
i,j,kΦ,
To analyze the linked products (Pl-11) we again start from (A)
in all instances. (PI) results from P(J0)JocJ0 and P(J2)J2ciJ2 + Jt; (P2) results from (D) (r = j) when i ^ 0; (P3) results from (B)
when j Φ 0; (P4) results from (B) (r = s = i; r = s = i); (P5) results
from (B) (r — s = ΐ) plus, when i ^ 0, (C) (r = j); (P6) results from
(B) (r = β = £); (P7) results from (B) (r = s = i; r = i, s = fc) plus,
when & ^ 0, (C) (r = fc); (P8) results from (C) (r = £; r = fc) when
i,kΦθ, and (C) (r = ί) when & = 0; (P9) results (see below) from
(C) (r = k) plus (B) (r = ί, s = j) when k Φ 0, and from (B) (r = i,
s = i; r = i, 8 = 0) when λ; = 0; (P10) results (see below) from (D)
(r = i) when i =£ 0, and from (B) (r = 0, 8 = j; r — k, s = 0) when
i — 0; (Pll) results from (C) (r = i; r = Z) when i ^ 0, and from
(C) (r = i) plus (B) (r = k, s = 0) when Z = 0.
To see there are no components of {JijJijJik } in Jjk in (P9)
(if j Φ 0 but fc = 0 allowed) we may assume xih ytj , zik lie in Peirce
spaces Juv ... tin ) of (2.1). Then ytj lies in some Jx (eά ) for ^ e ^ ,
whence α^ does too (otherwise it lies in J0 (e/) with ziki and {xyz} e
{JQJJO} = 0), in Jwhich case {CCT/^} e {J^Jo} c J (βi). 0 This cannot be
true for all β 6 g y 7 ^ if there is to be a component in J , ίk so some
βy6^ has VijβJoie'j), whence x jGJι(e'j) t (otherwise x eJ (ej) iS 0 and
again {xyz} e {Jo«Vo} c J (e'/)). 0 At the same time x lies in some Jfa)
for e