- Cuntz-Pimsner algebras 151
Let Ji= A0e L^2 (B, EE) be the Hilbert D-module obtained from A0B
equipped with the D-valued semi-inner product
(a10 b1,a2 0 b2) = EEW;e(aia2)b2).
We will view Ji as a C* -correspondence over D with left action determined by
1r1-(.(d)a ----- 0 b = (a.(d)a) -- 0 b. Let F(Ji)^0 = EBn>l 1-l®n and consider the shift
isometry TA E IBS(F(H)^0 ) given by TA((10··~0 (n) = f©l 0 (1 0 · · · 0 (n.
Since A naturally acts on 1-l, we have A C IBS(F(H)^0 ). One checks that
TA aTA = ( 1r7-(. o EE o e) (a) = cp( a) for every a E A. As in the first paragraph
we have T(H~) ~ C(A,TA) c IBS(F(H)^0 ). The map B 3bf--r10b E A0B
gives rise to an isometry V E JBS(JC, Ji) with V(a ----- 0 b) = ----B(a)b. (Use
the Schwarz-type inequality [B(ai)B(aj)] = B([ai · · · a~V)e([a1 ···an]) ~
e([ai ... a~V[a1 ... an])= [B(aiaj)] to check that V is bounded.) Now one
verifies that 1r7-(. ( d) V = V 1rK ( d) for every d E D. Hence,
VJ:·: :F(JC)^0 3 (1 0 · · · 0 (n f--r (V (1) 0 · · · 0 (V (n) E :F(Ji)^0
is a well-defined adjointable isometry (like every other sentence in this proof,
we leave it as an exercise to check this). We claim that TA_aV.r = V.rTBe(a)
for every a E A. Indeed,
TA_aV_r(b;_ 0 · · · 0 b:n) =TA_((~) 0 (f®b;) 0 · · · 0 (~))
= ( 1rH(ED(B(a)b1))(l B --) 0 b2) 0 · · · 0 (10 --bn)
= (l 0 nK(EE(e(a)b1))b2)1' 0 · · · 0 (~)
= V.rTBe(a)(b;_ 0 · · · 0 b:n).It follows that V}aTA = B(a)TB V} as well. Similarly, one has V}aV.r = B(a)
for a E A. Therefore
V}(aoTAa1 · · ·TAakTA · · · TA_an)V.r
= e(ao)TBe(a1) ... TBe(ak)TB ... TBe(an)
and 8(x) = v;xv_r is the desired u.c.p. map. D
Let Ji be a C -correspondence over A and B be a C -algebra. The
algebraic tensor product Ji 0 B is naturally equipped with an A 0 B-valued
semi-inner product:
(~ 0 a, 'fl 0 b) = (~, 'fl)H 0 a*b EA 0 B.
By separation and completion, we obtain a Hilbert A 0 B-module Ji 0 B.
We observe that IBS(Ji) 0 B c IBS(H 0 B) naturally, where IBS(Ji) 0 B acts on
Ji 0 B by