Compatible Peirce decompositions of Jordan triple systems - MSP

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COMPATIBLE PEIRCE DECOMPOSITIONS OF JORDAN TRIPLE SYSTEMS 63 COROLLARY 1.8. Tripotents e, f are compatible iff f=fi+fi+fo for elements f e Jle) Γ) J 2(f) This will be the case under any of the following conditions: ( i ) e,f are orthogonal: P(e)f = P(f)e = L(e, e)f = L(jf, f)e = 0 (ii) e ff are collinear: P(e)f=P(f)e = 0, L(e, e)f=f, L(f ff)e = e (iii) one lies in a single Peirce space of the other, feJê) for i = 2, 1, 0 (iv) / = / 2 + /I + /o for orthogonal tripotents f e Proof If e, f are compatible the Peirce i-component f = Ei(e)f of / remains in J 2(/); conversely, if /
64 KEVIN MCCRIMMON patibility reduces to 1.13. COMPATIBILITY CRITERION FOR IDEMPOTENTS. TWO idempotents e, f in a Jordan algebra are compatible iff / = / 2 ® / 0 for orthogonal idempotents f tGJi(e). Proof. The condition that / = α 2 + α L + a Q is idempotent is a 2 = at + E 2(e)a\ (1.10') a1 — (α2 + αo) o α! α 0 = al + E 0(e)al and compatibility (1.11) becomes (1.110 αo 0 ^ + αχoα2 = 0 , hence a λ — 0 and α 2 = α^ α 0 = αξ are orthogonal idempotents. Conversely, if /= / 2 + /o then e, / are compatible by (1.8iv). Π Note that the strong compatibility condition that the operators P(e), L(e, e) commute with P(f), L(f f) (not merely P(e) 2 and P(/) 2 ) is not an intrinsic condition: it depends on how e, f are imbedded in J. For example, e = 1[12] and / = 1[13] are collinear and strongly compatible in £>[12]+D[13] = MU2(D), but not in HID) since P(e)P(f) 1[33] - P(β)l[ll] = 1[22] Φ 0 = P(/)P(e)l[33]. The most important examples of compatible tripotents are either orthogonal e _L / (each lies in the 0-space of the other) or collinear eTf (each lies in the 1-space of the other). In the remainder of this section we investigate what collinearity amounts to in basic examples of triple systems. Recall that tripotents e, f are collinear if P(e)f - P(/)e - 0, L(e, e)f = /, L(f f f)e = e. Let us note that in a Jordan algebra we cannot have collinear idempotents; collinearity is strictly for tripotents. PROPOSITION 1.14. Two nonzero idempotents in a Jordan algebra can never be collinear. Proof. If eeJjif) and feJê) are idempotents then f—{eef)~ e 2 of = e of=eof= {eff} = e, so /= P(/)/= P(f)e - 0 and dually e = 0. (Alternately, if feJ,(e) then / 2 6 J 2(e) + J 0(e), so the only idempotent in Jê) is / = 0. Or yet again, the result follows directly from (1.13).) D
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Compatible Peirce decompositions of Jordan triple systems - MSP

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