数理解析研究所講究録
第 839 巻 1993 年 20-62

20

A REPORT ON OCNEANU’S LECTURE

YAMAGAMI SHIGERU

Department of Mathematics
College of General Education
Tohoku University
email: e22236@cctu.cc.tohoku.ac.jp

Dear Readers:This report supplies detailed proofs of the first chapter in “Quantum Symmetry, Differential Geometry of Finite Graphs and Classification of Subfactors”–seminary notes on
Ocneanu’s lectures prepared by Y. Kawahigashi–, particularly a proof of biunitarity ofconnections.
The essense of the proof is the Frobenius reciprocity, which turns out to be useful inthe topics of Markov traces and towers of algebras construction as well.
In this respect, the proof of Frobenius reciprocity via Lemma 24 and the explanations
after the heading “Markov Trace” might be new.
Compared to the treatment of string algebras, the notion of paragroups seems to berather difficult of access, partly because we cannot read the original proofs yet.
I hope the present report contribute to the improvement of such a situation.

Typeset by$A_{\Lambda\beta- TB^{X}}$21
YAMAGAMI SHIGERU

Notation and Convention
Let $N$ be a von Neumann algebra. By a left N-module, we understand a Hilbert space$X$ with a normal *-representation of $N$ which defines a left action of $N$ on $X$ . We often
write $NX$ to emphasize $X$ being a left N-module.
is defined to be a Hilbert space $X$ with a normal
Similarly a right N-module
representation of the opposite algebra of $N$ . Let $M$ be another von Neumann algebra. A
and right N-module at theHilbert space $X$ is called an M-N bimodule if $X$ is a left
same time and these two actions commute each other.
For a left N-module $X$ , we denote by End $(NX)$ the set of bounded linear operators in$X$ which commute with the left action of $N$ . If we denote by(or simply $N’$ if $X$ can beunderstood) the opposite algebra of End $(NX)$ , then $X$ is a right N’-module in an obvious,
way and $X$ becomes an $N- N’$ bimodule. Similarly, starting from a right N-module$X$ is an $xN’- N$ bimodule with $xN’=End(X_{N})$ .
The standard space $L^{2}(N)$ is a special but important example of an N-N bimodule.
$L^{2}(N)$ is
Note that, when $N$ is semifinite with a faithful normal semifinite weight
identified with the GNS-construction associated with with left and right actions are
induced from left and right multiplications in $N$ .
(resp.
I-matrix amplification of $N$ andGiven a set , let $Mat_{I}(N)$ be thebe the direct sum of left N-modules $NL^{2}(N)$ indexed by as row vector (resp.
(resp.
) from the rightcolumn vector). The algebra $Mat_{I}(N)$ acts on(resp.
‘).
(resp. left) and $Mat_{I}(N)$ is identified with. Since $Xe$ is N-invariant, it defines
Let $X$ be a left N-module and a projection ina left N-module $NXe$ . As a special case, consider a projection in $Mat_{I}(N)$ . We havea left N-module $NL^{2}(N)^{\oplus I}e$ and the commutant of the left action is naturally identified
with $eMat_{I}(N)e$ .
$*_{-}$$X_{N}$$\lambda f-$$N_{X}’$$X_{N}$$\tau,$$\tau$$I$$NL^{2}(N)^{\oplus I}$$I\cross$$I$$\oplus_{I}L^{2}(N)_{N})$$L^{2}(N)^{\oplus I}$$\oplus_{I}L^{2}(N)$$\oplus_{I}L^{2}(N)N$$N_{L^{2}(N)}’\oplus\tau$$N_{X}’$$e$$e$

Dimension(cf. (Theory of Operator Algebras’ by M.Takesaki)

Let $N$ be a semifinite von Neumann algebra with a faithful normal trace . Let $NX$
be a left N-module. Since to give a left N-module structure is nothing but to give a, by the structure theorem of normal hotohomomorphism fromprojection
find
a
set
and
a
in $End_{N}(L^{2}(N)^{\oplus}=L^{2}(N)\oplus$
momorphisms, we can, direct sum of countable copies) such that$\tau$$N\subset \mathcal{B}(L^{2}(N))$$\mathcal{B}(X)$

$I$

$e$

$\ldots$$NX\cong N(eL^{2}(N)^{\oplus I})$.

If we set $H=\ell^{2}(I),$ $L^{2}(N)^{\oplus I}\cong L^{2}(N)\otimes H$. With this identification, the left action of $N$. So we may assume thaton
is identffied with$L^{2}(N)^{\oplus I}$$N\otimes 1_{H}$$e\in(N\otimes 1_{H})’=N’\otimes \mathcal{B}(H)=N^{op}\otimes \mathcal{B}(H)$.

In other words, we can find a projection $p\in Mat_{I}(N)=N\otimes \mathcal{B}(H),$ $p=\{p_{ij}\},$ $p_{ij}\in N$

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A REPORT ON OCNEANU’S LECTURE

such that$e( \oplus_{i}x_{i})=\oplus_{i}(\sum_{j}x_{j}p_{ji})$$p_{12}p_{22}$$)$$=(x_{1}, x_{2}, \cdots)(p_{21}^{11}p...$$=(\oplus_{i}x_{i})p$for$\oplus_{i}x_{i}\in L^{2}(N)^{\oplus I}$,.

Now we set,$\dim_{N}X=(\tau\otimes tr)(p)$where $tr$ denotes the usual trace in $\mathcal{B}(H)=Mat_{I}(C)$ . The right hand side in this definition
is independent of the choice of . In fact, if $q\in Mat_{I}(N)$ is another projection satisfying$NX\cong N(L^{2}(N)^{\oplus I}p)$ , then we can find a partial isometry:$N$$\existsu\in
Mat_{I}(N),$$u^{*}u=q,$ $uu^{*}=p$ ,
with
left
of
.
which commutes
theaction
Thus
$p$

$v$

$v\xi=\xi u$,$\xi\in L^{2}(N)^{\oplus I}$$L^{2}(N)^{\oplus I}parrow L^{2}(N)^{\oplus I}q$.

In other words, $p\sim q$ in $Mat_{I}(N)$ which shows that$(\tau\otimes tr)(p)=(\tau\otimes tr)(q)$.

(or simply ) on
Extending this construction, we can define a semi-finite trace$N_{X}’\cong pMat_{I}(N)p$ , which is called a canonical trace and plays significant roles in thelater parts.
$\tau’$$\mathcal{T}_{N}’X$

Lemma 1.
(i) $(\tau’)’=\tau$ if th represen $tat$ ion $Narrow End(X)$ is $fai$ thfu 1.
(ii) For $T\in Hom(X,Y)$ ,$e$$\tau’(T^{*}T)=\tau’(TT^{*})$.

(iii) For a projection $e\in N_{X}’$ ,$\dim_{N}^{\tau}(Xe)=\tau’(e)$.

(iv)$\dim_{\acute{N}}^{\tau},$$X=\tau(1)$ .

(ii) follows from the definition of . To see (iii), consider the trace
and its
restrictions. (iv) is a consequence of (i) and (iii). To see (i), by the central decomposition,
we may assume that $N$ is a factor. Let $NN\cdot$
If $\tau(1)\leq\tau’(p)$ , subtracting projections which are equivalent to$)$$\tau’$$(\begin{array}{lll}1 0 \cdots 0 0 \cdots| | .\end{array})$$\tau_{X\oplus Y}’$,23
YAMAGAMI SHIGERU

we may assume that
$p=(\begin{array}{llll}1 \ddots 1 q\end{array})$,

where $q\in N$ . Then

$L^{2}(N’)=(\begin{array}{llll}L^{2}(N) L^{2}(N) L^{2}(N)q| \ddots | |qL^{2_{2}}(N)L(N) qL^{2_{2}}(N)L(N) qL^{2_{2}}(N)qL(N)q\end{array})$

and $NX$ is equivalent to$(\begin{array}{lll}L^{2}(N) L^{2}(N) L^{2}(N)q0 0 0| | |0 0 0\end{array})$.

Since the action of $N$ on $X$ is expressed by$N\ni arightarrow(\begin{array}{llll}a 0 \ddots 0\end{array})$in this realization, we have$\tau’’(a)=(\tau\otimes tr)(\begin{array}{llll}a 0 \ddots 0\end{array})=\tau(a)$.

If $\tau’(p)<\tau(1),$ $N’$ and hence $N$ are finite. So, to check that $\tau’’=\tau$ , it suffices to showthat $\tau’’(e)=\tau(e)$ for a suitable projection $e\neq 0\in N$ . This time, $NX\cong_{N}L^{2}(N)e$ with
$e\in N$ . Since $eL^{2}(N)e_{N’}$ is equivalent to $L^{2}(eNe)_{N’}$ ,$\tau’’(e)=\tau’(1_{N’})=\tau(e)$$\square$.

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A REPORT ON OCNEANU’S LECTURE

Proposition 2.
(i) $\dim_{N}H_{1}=\dim_{N}H_{2}$ if(ii) $\dim_{N}(\oplus_{i}H_{i})=\sum_{i}\dim_{N}H;$ .
an(iii) If${}_{N}H_{1}\cong {}_{N}H_{2}$$(N, \tau)=\int^{\oplus}d\lambda(N_{\lambda}, \tau_{\lambda})$.

$d \int^{\oplus}d\lambda H_{\lambda}$

is the correspon $ding$ decom position of $H$ , th en$\dim_{N}^{\tau}H=\int d\lambda\dim_{N_{\lambda}^{\lambda}}^{\tau}H_{\lambda}$.

(iv) If $N$ is a (semi-fin $ite$) factor, $\dim_{N}H<+\infty\Leftrightarrow End_{N}(H)$ is a finite von Ne umann
algebra, an the $con$ verse of (i) holds: $\dim_{N}H_{1}=\dim_{N}H_{2}\Rightarrow {}_{N}H_{1}\cong {}_{N}H_{2}$ .
$d$(i), (ii) are clear from the defintion and (iv) is a refrase of a classical result. For the, see the section of coupling
proof,of (iii), including the measurability of
constant (\S V.3 in ‘Theory of Operator Algebras’).
$)$$\lambdarightarrow\dim_{N_{\lambda}^{\lambda}}^{\tau}H_{\lambda}$$\square$We present here two examples, abstract one and concrete one.
the norm alized trace. By multiplicat ion,
Example 3. Let $N$ be a finite factor with$L^{2}(N)$ is an N-bimodule an $dNL^{2}(N)$ is aand $ard$ representation. $So$ we an identify
End $(NL^{2}(N))wi$ th the right multiplication algebra of N. Th us, for a projection $e\in N$ ,$L^{2}(N)e$ is a left N-su bmodule of $L^{2}(N)$ an
we $have$$\tau$

$st$

$c$

$d$

$\dim_{N}L^{2}(N)e=\tau(e)$.

Example 4. $Consid$ erlgebra A generated by two unit aries$A$
vu. On , we an define a trace by$a^{*}- a$$uv=e^{2\pi i\theta}$

$c$

$u,$

$v$

satisfyi $ng$ the relat ion$\tau$$\tau(u^{m}v^{n})=\{\begin{array}{l}1ifm=n=00otherwise\end{array}$

There are several realiz ations of such an a $lge$ bra.
1’ Group algebra.
Consider a $cen$ tral $ext$ en sion of an additive $gro$ up$G$$1arrow Tarrow Garrow Z^{2}arrow 0$$Z^{2}$

by the torus,

where the group structure of $G=Z^{2}\cross T$ is defined by$(m, n, z)(m’, n’, z’)=(m+m’, n+n’, zz’e^{-2\pi i\theta m’n})$ .

Thro $ugll$ th iden tification$e$$u\Leftrightarrow(1,0,1)$$v\Leftrightarrow(0,1,1)$$z\Leftrightarrow(0,0, z)$

,

$T$

;

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YAMAGAMI SHIGERU

of such thatthe algebra $A$ is realized (or represen $ted$) by a unit $ary$ representation$\pi(0,0, z)=z1$ . As an example of such a , we $take$ an induced representation,$H$$\chi(0,0, z)=z$ . Its representation spaceis given by
$\pi$

$G$

$ind_{T}^{G}\chi$$\pi$$H=\{f$ : $Garrow C;f(gz)=z^{-1}f(g),$ $g\in G,$ $z\in T$ ,$\sum|f(g)|^{2}<+\infty\}$.

$g\in c/T$

Takin the restriction of
$g$

$f$

to

$Z^{2}\subset G$,$H\cong\ell^{2}(Z^{2})$We define an an ti-unit ary involution$J$.

on $H$ by$(Jf)(g)=\overline{f(g-1)},$ $g\in G$

and a unit vector$\delta$,by$\delta(m, n, z)=\{\begin{array}{l}z^{-1}ifm=n=00otherwise\end{array}$defines a andard representation of with a tracial vector. In part icuThen$lar$ , the von Neumann algebr a generated by $\pi(G)$ is finite. A bit more algebraic calcula tionshows that $\pi(G)’’$ is a factor if is irration al.
$G$

$st$

$(H, \delta, J, \pi)$$\delta$$\theta$

2’ Crossed product.
In the above represen tation $H$ , consider the $Fou$ rier transform of $f\in l^{2}(Z^{2})$ with respect
to the second variable:$\ell^{2}(Z^{2})\ni frightarrow i\in\ell^{2}(Z)\otimes L^{2}(T)$$;(m, z)= \sum_{n}z^{n}f(m, n)$ .

In th is new realization of $H$ ,$(u\hat{f})(m, z)=f(m-1, z)$$(vf)(m, z)=e^{-2\pi i\theta m}zf(m, z)$$(J\hat{f})(m, z)=f(-m, ze^{-2\pi i\theta m})$$\hat{\delta}(m, z)=\delta_{m,0}$

If we in terpret
generates$v$$L^{\infty}(\mathbb{T})$.

as a multiplication operator by the function, we define an automorphism by. On$L^{\infty}(T)$$zrightarrow z$

in$L^{\infty}(T)$, then$v$$\alpha$$\alpha(z)=e^{2\pi i\theta}z$.

Then the above realization is nothing but the regular rep resentati on of the crossed productZ.
a lgebra$L^{\infty}(\mathbb{T})\rangle\triangleleft\alpha$26
A REPORT ON OCNEANU’S LECTURE

Since$(JuJf)(m, z)=f(m+1, ze^{2\pi i\theta})$$(JvJ;)(m, z)=\overline{z}\hat{f}(m, z)$,

and(in this crossed product representation,) on sees that $\pi(G)’$ cont ains$fu$
defines a projection $e\in\pi(G)’$
Borel $su$ bset $B\subset T$ the $ch$ aract eristic nction$1_{t^{2}(Z)}\otimes L^{\infty}(T)$$e$$1_{B}\in L^{\infty}$$\forall$$by$$e=1_{l^{2}(Z)}\otimes 1_{B}$$Sin$ce the natur

$al$

trace

$\tau’$

on$1_{l^{2}(Z)}\otimes L^{\infty}(T)$.

is given by$\tau’(1_{\ell^{2}(Z)}\otimes f)=\int_{T}dzf(z)$,we fin ally ave$h$$\dim_{Me}eH=|B|$ .

Definition 5. Let $M$ be a finite factor and $N\subset M$ be a nite $su$ bfactor. The index
of $N\subset M$ is defined by $[M : N]=\dim_{N}L^{2}(M)$ where the dimension is measured withrespect to the ormalized trace of $N$ .
$ii$

$n$

Proposition 6. $[M:N]\geq 1$ and the $equ$ ality holds if and only if $M=N$ .
Since $L^{2}(N)\subset L^{2}(M)$ and $L^{2}(N)=L^{2}(M)$ if and only if $N=M$ (cf. the conditional
expectation $E:Marrow N$ ), this follows from the following calculation:$)$$\dim_{N}L^{2}(M)=\dim_{N}(L^{2}(N)\oplus(L^{2}(M)\ominus L^{2}(N)))$$=\dim_{N}L^{2}(N)+\dim_{N}(L^{2}(M)\ominus L^{2}(N))$$=1+\dim_{N}(L^{2}(M)\ominus L^{2}(N))$$\square$.

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YAMAGAMI SHIGERU

Operator valued inner productOperator valued inner product is an important notion when one considers relations between
two algebras (see [Paschke], [Rieffel]).
Let $N$ be a semifinite von Neumann algebra with a faithful trace . Given a left
N-module $X$ , we define the set of $(N, \tau)$ -bounded (or simply N-bounded if is fixed)
elements by$X_{0}\equiv\{x\in X;\exists\lambda>0,\forall a\in N, ||ax||\leq\lambda\tau(aa^{*})^{1/2}\}$.
$\tau$$\tau$

Lemma 7.
(i)
is an N-invari ant dense $linearsubsp$ ace of $X$ .
: $NXarrow NY$ is an intertwiner, $T(X_{0})\subset Y_{0}$ . In(ii) If $NY$ is another N-module andis End $(NX)$ -invari ant.
part icul $ar,$$X_{0}$$T$$X_{0}$(i) N-invariance follows from the trace property of . To see the density, we may assume
that $N\sim X$ is faithful(consider $X\oplus L^{2}(N)$ ). Then we can take a family
such
that
.
$\tau(a)=\sum_{:}(ax_{i}|x_{i}),$$)$$\tau$$\{x_{i}\}_{i\in I}\subset X$$\forall a\in N^{+}$are in
since
. Let
These
$X$
$e\in
N$
,
by the N’-invariance ofin . Since$x_{i}’ s$$||ax_{i}||^{2}\leq\tau(aa^{*})$$X_{0}$$e$be the projection onto the closure of$X_{0}$$X_{0}$$(1-e)x_{i}=0,$$\forall i\Rightarrow\tau(1-e)=0$.

is faithful, $e=1$ .
(ii) This property is an immediate consequence of the definition.

Since$\tau$$\square$Example 8. If $X=L^{2}(N),$ $X_{0}=N\cap L^{2}(N)$ .
For $x\in X_{0}$ , consider a positive linear functional $N\ni a\vdasharrow(ax|x)$ . Since$(ax|x)=(a^{1/2}x|a^{1/2}x)\leq||a^{1/2}x||^{2}\leq\lambda^{2}\tau(a)$,the Radon-Nikodym theorem assures that there is a positive element $h\in N$ such that$\tau(ah)=(ax|x),$$\forall a\in N^{+}$.

From this relation, is in the definition ideal of . Moreover, given $y\in X_{0}$ , the polar
identity shows that we can find a (unique) $h\in N$ which is in the definition ideal of andsatisfies$\tau(ah)=(ax|y)$ ,.
$h$$\tau$$x,$$\tau$$\forall a\in N$

In the following, this$h$is denoted by$N[x, y]$.

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A REPORT ON OCNEANU’S LECTURE

Lemma 9.
(i) $N[x, y]$ is an N-val $ued$ sesqui-linear form on(ii) $N[x, y]^{*}=N[y, x]$ .
$y\in X_{0}$ .
(iii) $N[ax, y]=a_{N}[x, y]$ , $a\in N,$$X_{0}$.

$x,$Remark. (i) The operator-valued inner product $N[, ]$ depends on the choice of a trace of$N$.

(ii) For a $N[x, y]\in N$ , let $[x, y]_{2}$ be its image in $L^{2}(N)$ . Then for $x\in X_{0},$$[x, y]_{2}$ isontinuously extended to a bounded linear map of $X$ into $L^{2}(N)$ .

$X_{0}\ni yrightarrow$$c$$N$ be a
Example 10. Let $M$ be a finite von Neuman algebra with a faith $ful$ trace
von Neum $ann$ subalge $bra$ of $M$ , an $dE$ : $Marrow N$ be a $condit$ ion expectation un iquelydetermin $ed$ by . For the left N-module $NL^{2}(M)$ , we $have$$n$$\tau,$$al$$\tau$$N[[a]_{2}, [b]_{2}]=E(ab^{*})$

where

$a,$

$b\in M$an $d[a]_{2},$$[b]_{2}$,are their $im$ ages in $L^{2}(M)$ .

Basis
Definition 11. A$family\lambda=\{\lambda_{i}\}_{i\in I}$in$X_{0}$is alled$c$$a$basis if(i)(algebr $aicsum$ ) is dense in $X$ ,(ii) $\forall x\in X,$$\sum_{i}N\lambda_{i}$$\forall i\in I$$(x| \lambda_{i})=\sum_{j}(x|_{N}[\lambda_{i}, \lambda_{j}]\lambda_{j})$Let$\Lambda=\{[\lambda_{i}, \lambda_{j}]\}_{i,j\in J}\in Mat_{I}(N)$(a bsolutely convergent).

.

Proposition 12.
is a projection in $Mat_{I}(N)$ (in particular, it is bounded).
is inand the map

(i)
(ii)

$\Lambda$$\forall i\in I,$$\oplus_{j}[\lambda;, \lambda_{j}]$$L^{2}(N)^{\oplus I}$$\sum_{i\in I}N\lambda_{i}\ni\sum_{i}a_{i}\lambda_{i}arrow\succ\oplus_{j\in I}(\sum_{i}a_{iN}[\lambda_{i}, \lambda_{j}])\in L^{2}(N)^{\oplus I}$

is extended to an isometry fro $mX$ to(iii),$\forall x\in X_{0},$$L^{2}(N)^{\oplus I}\Lambda$.

$\forall y\in X$$(x|y)= \sum_{i}([x, \lambda_{i}]\lambda_{i}|y)$(a bsolutely $con$ vergent).

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YAMAGAMI SHIGERU

We first note that$)$$\oplus_{j}[\lambda_{i}, \lambda_{j}]\in L^{2}(N)^{\oplus I}$(1):$\sum_{j}\tau([\lambda_{i}, \lambda_{j}][\lambda_{j}, \lambda_{i}])=\sum_{j}([\lambda_{i}, \lambda_{j}]\lambda_{j}|\lambda_{i})$$=(\lambda;|\lambda_{i})<+\infty$.

So, if we set $D=\oplus_{i\in I}L^{2}(N)$ (algebraic sum), the right multiplication of the operator
matrix defines a linear operator with domain $D$ . Then the reproducing kernel propertyof yields$(h\xi|h\xi)=(\xi|h\xi)$ ,,$h$$\Lambda$$\lambda$$\forall\xi\in D$

which means that is a projection (see the appendix below), proving (i).
The formula (1) also shows that the map in (ii) is an isometry. So the only thing we. Since $N\cap L^{2}(N)$ is weakly dense inneed to prove is that its image is equal to$N$ , the image is the closure of, i.e.,.
(iii) Since. This set, as, we first examine
can be easily seen, is identified with$h$$L^{2}(N)^{\oplus I}\Lambda$$(\oplus_{i}N\cap L^{2}(N))\Lambda$$\Lambda^{2}(N)^{\oplus I}\Lambda$$(L^{2}(N)^{\oplus I}\Lambda)_{0}=(L^{2}(N)^{\oplus I})_{0}\Lambda${$\oplus_{i}a;\in L^{2}(N)^{\oplus I};a;\in N\cap L^{2}(N)$$(L^{2}(N)^{\oplus I}\Lambda)_{0}$and$\sum_{i}a_{i}a_{i}^{*}$

is bounded}and the N-valued inner product is given by(2)$[ \oplus_{i}a_{i}, \oplus_{i}b;]=\sum_{i}a;b_{i}^{*}$,$\oplus;a;,$ $\oplus_{i}b;\in(L^{2}(N)^{\oplus I}\Lambda)_{0}$.

Here the convergence in the right hand side is with respect to weak operator topology. On
andare in the definition ideal of and hence in $L^{2}(N)$ . So
the other hand,
we can talk about the convergence in $L^{2}(N)$ and, in fact, the relation (2) holds as elements
in $L^{2}(N)$ . To see this, we may assume $a_{i}=b_{i}$ by the polar identity. Let
be a finite
subset and consider$[\oplus_{i}a_{j}, \oplus_{i}b_{i}]$$a_{i}b_{i}^{*}$$\tau$$F\subset I$$\tau((\sum_{i\not\in F}a_{i}a_{i}^{*})^{2})\leq\tau((\sum_{i\not\in F}a_{i}a_{i}^{*})^{1/2}||\sum_{i\not\in F}a_{i}a_{i}^{*}||(\sum_{i\not\in F}a_{i}a_{i}^{*})^{1/2})$$=|| \sum_{i\not\in F}a_{i}a_{i}^{*}||\tau(\sum_{i\not\in F}a_{i}a_{i}^{*})$$\leq||\sum_{i\not\in F}a_{i}a_{i}^{*}||\sum_{i\not\in F}\tau(a;a_{i}^{*})$

Since$\oplus_{i}a_{i}\in L^{2}(N)^{\oplus I}$, and

Now letis identified withis bounded, this converges to, within, we have$\sum_{i\not\in F}a_{i}a_{i}^{*}$$x_{0}=\oplus_{i}b_{i}=(\oplus;a_{i})\Lambda\in(L^{2}(N)^{\oplus I}\Lambda)_{0}\Lambda$$\lambda_{i}\in X$$\oplus_{j}[\lambda_{i}, \lambda_{j}]$.

as$F\nearrow I$.

$\oplus_{i}a_{i}\in(L^{2}(N)^{\oplus I}\Lambda)_{0}$$L^{2}(N)^{\oplus I}$$[x_{0}, \lambda_{i}]=\sum_{j}b_{j}[\lambda_{j}, \lambda_{i}]$$0$.

. Since30
A REPORT ON OCNEANU’S LECTURE

If we consider this relation in $L^{2}(N)$ , the right hand side is equal to the i-th component of$(\oplus;b_{i})\Lambda=(\oplus_{i}a_{i})\Lambda^{2}=(\oplus_{i}a_{i})\Lambda$.

From this, we havein $L^{2}(N)$ .

$[x_{0}, \lambda_{i}]=\sum_{j}b_{j}[\lambda_{j}, \lambda_{i}]$-sum and operator sum gives the same results forand(see the above argument), this relation holds as weak operator convergence.
Now we haveSince

$L^{2}$$\oplus_{j}[\lambda_{i}, \lambda_{j}]\in(L^{2}(N)^{\oplus I})_{0}$$\oplus_{j}a_{j}$$[x_{0}, \lambda_{i}]\lambda_{i}=\sum_{j}a_{j}[\lambda_{j}, \lambda_{i}](\oplus_{k}[\lambda_{i}, \lambda_{k}])$$=\oplus_{k}(a_{j}[\wedge\lambda_{i}][\lambda_{i},\wedge\lambda_{k}])$$= \oplus_{k}(L\bigwedge_{j}^{2}\sim_{i}^{N}$$=\oplus_{k}(b_{i}[\lambda_{i}, \lambda_{k}])$From this expression, we see that$\sum_{i}[x_{0}, \lambda_{i}]\lambda_{i}$.

is weakly convergent in$L^{2}(N)^{\oplus I}$$\sum_{i}[x_{0}, \lambda_{i}]\lambda_{i}=\oplus_{k}(\sum_{i}b_{i}[\lambda_{i}, \lambda_{k}])$$=(\oplus_{i}b_{i})\Lambda$$=\oplus_{i}b_{i}$$=x_{0}$(since$\oplus_{i}b_{i}\in L^{2}(N)^{\oplus I}\Lambda$).

$\square$Corollary 13.
,(i)$\forall a’\in N_{X}’$$\tau_{N}’(a’)=\sum_{i}(\lambda_{i}a’|\lambda_{i})$In articular,$y\in X_{0}$ ,
(ii)
$p$

$\dim_{N}X=\sum_{i\in I}(\lambda_{i}|\lambda_{i})$.

.

$\forall x,$$\sum_{i}N[x, \lambda_{i}]_{N}[\lambda_{i)}y]=N[x, y]$$)$(i) Since $a’\in N_{X}’,$$a’$(weakly convergent).

has a matrix expression$A=(a_{ij})\in\Lambda Mat_{I}(N)\Lambda\subset Mat_{I}(N)$

.

and

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YAMAGAMI SHIGERU

We show the formula (i) for $A=\Lambda B\Lambda$ , where $B\in Mat_{I}(N)$ is of finite size and its
components are in the definition ideal of . Once this is proved, the formula of the generalcase follows by continuity (both sides in the above relation define normal weights). Thus
. Using $B=(b_{kl})$ , we have
we need to consider$\tau$$A=\Lambda B\Lambda$$a_{ij}= \sum_{kl}[\lambda_{i}, \lambda_{k}]b_{kl}[\lambda_{l}, \lambda_{j}]$(note that the summation
trace class of $N$ ), and then$\sum_{kl}$is taken over finite number of indices and$\tau_{N}’(a’)=\sum_{i}\tau(a_{ii})=\sum_{i}\sum_{kl}\tau([\lambda;, \lambda_{k}]b_{kl}[\lambda_{k}, \lambda_{i}])$$= \sum_{k,l}\tau(b_{kl}\sum_{i}[\lambda_{l}, \lambda_{i}][\lambda_{i}, \lambda_{k}])$$= \sum_{k,l}\tau(b_{kl}[\lambda_{l}, \lambda_{k}])$$= \sum_{k,l}(b_{kl}\lambda_{l}|\lambda_{k})$.

On the other hand, using the correspondence$\lambda;\Leftrightarrow\oplus_{j}[\lambda_{i}, \lambda_{j}]\in\oplus_{j}L^{2}(N, \tau)$,

we have$\lambda_{i}a’\Leftrightarrow(\oplus_{j}[\lambda_{i}, \lambda_{j}])\Lambda B\Lambda$$=(\oplus_{j}[\lambda_{i}, \lambda_{j}])B\Lambda$$= \oplus;\sum_{k,l}[\lambda_{i}, \lambda_{k}]b_{kl}[\lambda_{l}, \lambda_{j}]$$\lambda_{j}$$= \oplus_{j}[\sum_{k,l}[\lambda_{i}, \lambda_{k}]b_{kl}\lambda_{l},$$\Leftrightarrow\sum_{k,l}[\lambda_{i}, \lambda_{k}]b_{kl}\lambda_{l}$

]

.

Thus
$\sum_{i}(\lambda_{i}a’|\lambda_{i})=\sum_{:}\sum_{k,l}([\lambda_{i}, \lambda_{k}]b_{k1}\lambda_{l}|\lambda_{i})$$= \sum_{k,l}(b_{kl}\lambda_{l}|\lambda_{k})$$=\tau_{N}’(a’)$.

$b_{kl}’ s$are in the

32
A REPORT ON OCNEANU’S LECTURE(ii) By the polar identity, we may assume $x=y$ . For $a\in N_{+}$ ,$\tau(a^{1/2}\sum_{i}N[x, \lambda;]_{N}[\lambda_{i}, x]a^{1/2})=\sum_{i}\tau(a^{1/2}[x, \lambda_{i}][\lambda_{i}, x]a^{1/2})$$= \sum_{i}\tau(a[x, \lambda_{i}][\lambda_{i}, x])$$= \sum_{i}(a[x, \lambda_{j}]\lambda_{i}|x)$$=(ax|x)$$=\tau(a[x, x])$.

Since,this holds for any $a\in N+$ , we get the desired formula.

$\square$Remark. The proof of (i) for the special case $a’=1_{X}$ is much easier. We just calculate$\dim_{N}X=(\tau\otimes tr)(\Lambda)=\sum_{i\in I}\tau([\lambda_{i}, \lambda_{i}])=\sum_{i\in I}(\lambda_{i}|\lambda_{i})$Appendix. Let$h$be a hermitian $op$ erator $wi$ th dense domain $Dsuch$ that$(h\xi|h\xi)=(\xi|h\xi)$

Then

$h$

,

$\forall\xi\in D$.

is a projection.

We only need to check that

$)$

.

$h$

is bounded. This follows from$||h\xi||=(h\xi|h\xi)^{1/2}=(\xi|h\xi)^{1/2}$$\leq||\xi||^{1/2}||h\xi||^{1/2}$$=||\xi||^{1/2}(h\xi|h\xi)^{1/4}=||\xi||^{1/2}(\xi|h\xi)^{1/4}$$\leq||\xi||^{1/2}||\xi||^{1/4}||h\xi||^{1/4}$$\leq||\xi||^{1/2+1/4+\cdots+1/2^{n}}||h\xi||^{1/2^{n}}$$arrow||\xi||$.

$\square$

Change of Bases
be another basis. Let
Let
satisfies
From Corollary (ii),$\theta=\{\theta_{j}\}_{j\in J}$$V$

be an operator matrix given by$V$$V^{*}V=\Lambda$,$VV^{*}=\Theta$.

$\{N[\theta_{i}, \lambda;]\}_{i\in J,i\in I}$.

33
YAMAGAMI SHIGERU

Since$[x, \lambda_{i}]=\sum_{j}[x, \theta_{j}][\theta_{j}, \lambda_{i}]$both in operator sense and$L^{2}$,$\forall i,$$\forall x\in X_{0}$-sense, we have,$\iota_{\lambda}(x)=\iota_{\theta}(x)V$denotes the natural imbedding
where. Thus we have obtained thefollowing commutative diagram of N-module isomorphisms:$Xarrow L^{2}(N)^{\oplus I}\Lambda$$\iota_{\lambda}$

X$arrow^{\iota_{\lambda}}L^{2}(N)^{\oplus I}\Lambda$$\uparrow V$$\Vert$$Xarrow^{\iota_{\theta}}L^{2}(N)^{\oplus J}\Theta$

In this sense,$V$gives a matrix system for change of bases.

Existence of Basis$i\in I$ isis called orthogonal iffor $i\neq j$ . Note that
A basisa projection in this case (because
is a projection). Given an N-module $X$ ,we can construct a basis for $X$ . To this end, we first present$[\lambda_{i}, \lambda_{j}]=0$$\{\lambda_{i}\}_{i\in I}$$\Lambda=\{[\lambda_{i},$$[\lambda;, \lambda_{i}],$$\lambda_{j}]\}$

Lemma 14.
,(i)$\forall x\in X_{0}$$[x] \equiv\inf\{p\in proj(N);px=x\}$

is the support projection $of_{N}[x, x]$ .
(ii) If $\tau([x])<+\infty$ ,$\lambda=\lim_{\epsilon\backslash 0}(N[x, x]+\epsilon)^{-1/2}x$exists in $X,$$\lambda\in X_{0}$

, an

$d$

$N[\lambda, \lambda]=[x]$(iii) If$\lambda\in X_{0}$and$N[\lambda, \lambda]$

is a projection, then.

$\forall y\in X_{0}\cap\overline{N\lambda}$$N[y, \lambda]=y$,.

Conversely, if $p_{N}[x, x]p=N[x, x],$and hence $(px-x|px-x)=\tau$ ( [px–x, px–x]) $=0$ .
(ii)$)(i)$ If $px=x,$ $p_{N}[x, x]p=N[x, x]$ .

$N\lceil_{I}ox-x$, px-x] $=0$$N$$||([x, x]+\epsilon)^{-1/2}x-([x, x]+\eta)^{-1/2}x\Vert^{2}=(([x, x]+\epsilon)^{-1}x|x)+(([x, x]+\eta)^{-1}x|x)$$-2(([x, x]+\epsilon)^{-1/2}x|([x, x]+\eta)^{-1/2}x)$$=\tau(([x, x]+\epsilon)^{-1}[x, x])+\tau(([x, x]+\eta)^{-1}[x, x])$$-2\tau(([x, x]+\epsilon)^{-1/2}([x, x]+\eta)^{-1/2}[x, x])$.

34
A REPORT ON OCNEANU’S LECTURE

Since$([x, x]+\epsilon)^{-1/2}([x, x]+\eta)^{-1/2}[x, x]\nearrow[x]$,.

$\lim_{\epsilon,\eta\backslash 0}||([x, x]+\epsilon)^{-1/2}x-([x, x]+\eta)^{-1/2}x||^{2}=\tau([x])+\tau([x])-2\tau([x])=0$

Thus$\lambda\in X$

exists.

$\lambda\in X_{0}$follows from$||a \lambda||=\lim_{\epsilon\backslash 0}||a([x, x]+\epsilon)^{-1/2}x||^{2}$$= \lim\tau(a([x, x]+\epsilon)^{-1}[x, x]a^{*})$$=\tau(a[x]a^{*})$$\leq\tau(aa^{*})$,

The above calculation also shows that
. Then(iii) Let$a\in N$$[\lambda, \lambda]=[x]$$y= \lim_{i}a_{i}\lambda\in X_{0}\cap\overline{N\lambda}$.

, because,$||a\lambda||^{2}=\tau(a[\lambda, \lambda]a^{*})$.

$\forall x\in X_{0}$$([y, \lambda]\lambda|x)=\tau([y, \lambda][\lambda, x])$$=\tau([\lambda, x][y, \lambda])$$=([\lambda, x]y|\lambda)$$= \lim_{i}([\lambda, x]a;\lambda|\lambda)$$= \lim_{i}\tau([\lambda, x]a_{i}[\lambda, \lambda])$$= \lim_{i}\tau(a_{i}[\lambda, \lambda][\lambda, x])$$= \lim_{i}(a_{i}[\lambda, \lambda]\lambda|x)$$= \lim_{i}(a_{i}\lambda|x)$$([\lambda, \lambda]=[\lambda])$$=(y|x)$ .
$\square$Returning to the construction of basis, we first decompose an N-module $X$ into cyclic
such thatpieces; we can find a family$\{x_{i}\}\subset X_{0}$$X=\oplus;\overline{Nx_{i}}$Cutting down

$x_{i}$(orthogonal direct sum).

by a projection in $N$ with finite trace, we may assume that $\tau([x_{i}])<+\infty$ .

Let$\lambda;=\lim_{\epsilon\backslash 0}([x_{i}, x_{i}]+\epsilon)^{-1/2}x_{i}\in\overline{Nx_{i}}$.

is decomposed according to the above decomis invariant under $N’,$
Since
position and the component-wise application of Lemma (iii) insures that
$X_{0}$$\forall x_{0}\in X_{0}$$x_{0}= \sum_{:}[x_{0}, \lambda_{i}]\lambda;$.

The other properites for basis can be also checked by Proposition 12 (ii). Thus
a desired basis.

$\{\lambda_{i}\}_{i\in I}$

is35
YAMAGAMI SHIGERU

Proposition 15. Let $N$ be a semi-fi nite factor, $X$ be a left N-module, and suppose that$\dim_{N}X<+\infty$ . Thenan orthogon basis consist $ing$ of finite elemen .
$al$$\exists$$ts$

Suppose that $N$ is properly infinite. If we write as$)$$H\cong L^{2}(N)^{\oplus I}q$,$q=\{q_{ij}\}\in Mat_{I}(N)$ ,then$\dim_{N}H=(\tau\otimes tr)(q)<+\infty$

implies that there are at most countably many $i\in I$ such that$\tau(q_{ii})\neq 0$

So we may assume that$I$.

is countable. This case, since $N$ is properly infinite,$NL^{2}(N)^{\oplus I}\cong NL^{2}(N)$and we can find a projection $q\in N$ such that${}_{N}H\cong NL^{2}(N)q$

Then $\tau(q)=\dim_{N}H<+\infty$ and, if we denote by
to , then $x\in H_{0}$ . Since$x$.

an element in $L^{2}(N)$ which corresponds$q$$\tau(a[x, x])=(ax|x)=\tau(qaq)=\tau(aq)$ ,$[x, x]=q$ and therefore$[ax, x]x=aqx=ax$ .

is a basis.
Thus
Suppose that $N$ is finite and$\{x\}$$\tau$

is normalized. Express as${}_{N}H\cong NL^{2}(N)^{\oplus I}q$

Since $\tau\otimes tr(q)<+\infty,$$\exists$

projections$q_{1},$$\ldots,$.

$q_{n+1}\in Mat_{I}(N)$such that$q=q_{1}+\cdots+q_{n+1}$ ,$\tau\otimes(q_{1})=\cdots=\tau\otimes(q_{n})=1,$

Since $q_{i}(1\leq i\leq n)$ are equivalent toof$\{\begin{array}{ll}1 00 0\end{array}\},$$\exists p\in N$$\{\begin{array}{ll}1 00 0\end{array}\}$and$\tau\otimes(q_{n+1})<1$$q_{n+1}$.

is equivalent to a sub-projection

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A REPORT ON OCNEANU’S LECTURE

Now if we set$(1 \leq i\leq n)$$x_{i}=0\oplus\cdots\oplus 1\oplus 0\oplus i\ldots\in {}_{N}H$$x_{n+1}=0\oplus\cdots\oplus 0\oplus p\in {}_{N}H$

is a basis.

$\{x_{i}\}_{1\leq i\leq n+1}$,$\square$Tensor Product
Let $B$ be a von Neumann algebra with a faithful trace . Take a right and a left Bbe the sets of B-bounded elements. As in the case of left
mudules
BY and letbymodules, we define a B-valued inner product $[, ]_{B}$ in$\tau$$X_{B},$$X_{0},$$Y_{0}$$X_{0}$$\tau([x_{1}, x_{2}]_{B}b)=(x_{2}b|x_{1})$.

This time, $[, ]_{B}$ is linear in the second variable.
The following lemma is essentially known in [Paschke] and [Rieffel].
Lemma 16. Let
is positive and$x_{1},$$\ldots,$$x_{n}\in X_{0}$. Then the operator matrix$\{[ax_{i}, ax_{j}]_{B}\}_{1\leq i,j\leq n}\leq||a||^{2}\{[x;, x_{j}]_{B}\}_{1\leq i,j\leq n}$

For a sequence$)$$b_{1},$$\ldots,$$b_{n}\in B\cap L^{2}(B)$,$\{[x_{i}, x_{j}]_{B}\}_{1\leq i,j\leq n}\in M_{n}(B)$$\forall a\in End(X_{B})$.

,$\{b_{i}\}\{[ax_{i}, ax_{j}]_{B}\}\{b_{j}\}=\sum_{i,j}\tau(b_{i}^{*}[ax;, ax_{j}]_{B}b_{j})arrowarrow$$= \sum_{i,j}(ax_{i}b;|ax_{j}b_{j})$$=(a \sum_{i}x_{i}b_{i}|a\sum_{j}x_{j}b_{j})$$\leq\Vert a||\sum_{i,j}(x;b_{i}|x_{j}b_{j})$.

$\square$Now define an inner product $(|)$ in the algebraic tensor product$X_{0}\otimes Y_{0}$by.

$( \sum_{i}x_{i}\otimes y;|\sum_{j}x_{j}’\otimes y_{j}’)=\sum_{i,j}\tau([x_{j}’, x_{i}]_{BB}[y_{i}, y_{j}’])$For $x_{j}’=x_{j},$matr ces$y_{j}’=y_{j}$, the right hand side is the trace of the product of two positive operator$\{[x;, x_{j}]_{B}\}$,$\{B[y_{\{}, y_{j}]\}$

37
YAMAGAMI SHIGERU

becomes a (degenerate) pre-Hilbert space,and hence non-negative. In this way,its completion. Note that, from the property of operator valued
and we denote byinner product, the above inner product is well-defined on the algebraic tensor product as
B-modules. Also, the above lemma shows that for $a\in EndX_{B}$ ,$X_{0}\otimes Y_{0}$$X\otimes_{B}Y$$x\otimes_{B}yrightarrow ax\otimes_{B}y$defines a bounded linear operator in $X\otimes_{B}Y$ and this correspondence gives a normalin
Y. A similar result holds for $End_{B}Y$ . In particular, if
representation of End$X$ is an A-B module andis a B-C module, then $X\otimes_{B}Y$ is an A-C module in a naturalway.
$X_{B}$$X\otimes_{B}$$Y$, thenandRemark. (i) If one takes a basis
gives an orthogonal decomposition and, from this, one sees that the inner.
isproduct non-degenerate on
depends on the choice of a trace .
(ii) The inner product in$X_{0}\otimes_{B}Y_{0}=$$\{\mu i\}_{j\in J}\subset Y_{0}$$\{\lambda_{i}\}_{i\in I}\subset X_{0}$$\sum_{i,j}\lambda;\otimes_{B}\mu_{j}$$X_{0}\otimes_{B}Y_{0}$$X\otimes_{B}Y$$\tau$

Proposition 17.
(i) $(X_{1}\oplus X_{2})\otimes_{B}Y=(X_{1}\otimes_{B}Y)\oplus(X_{2}\otimes_{B}Y)$ .
(ii) $X\otimes_{B}(Y\otimes_{C}Z)=(X\otimes_{B}Y)\otimes_{C}Z$ .
$)$(i) is clear.
(ii) Since$(x\otimes_{B}y|x’\otimes_{B}y’)=\tau([x’, x]_{BB}[y, y’])$$=([x’, x]_{B}y|y’)$ ,$x,$$x’\in X_{0},$$y,$ $y’\in Y_{0}$,. The above
this inner product is continuously extended to an inner product inis again dense
formula also shows that if $Y’$ is a dense linear subspace in , thenin $X\otimes_{B}Y$ . With these observations, to see (ii), it suffices to show that $X_{0}\otimes_{B}(Y\otimes_{C}Z_{0})=$. Of course, the identification is given by$X_{0}\otimes_{B}Y$$Y$$X_{0}\otimes_{B}Y’$$(X_{0}\otimes_{B}Y)\otimes_{C}Z_{0}$$x_{0}\otimes_{B}(y\otimes_{C}z_{0})=(x_{0}\otimes_{B}y)\otimes_{C}z_{0}$.

To complete the proof, we nned to check the equality of the norms:$(x_{0}\otimes_{B}(y\otimes_{C}z_{0})|(x_{0}\otimes_{B}y)\otimes_{C}z_{0})=([x_{0}, x_{0}]_{B}y\otimes_{C}z_{0}|y\otimes_{C}z_{0})$$=(y|[x_{0}, x_{0}]_{B}y_{C}[z_{0}, z_{0}])$$=([x_{0}, x_{0}]_{B}y|y_{C}[z_{0}, z_{0}])$$=(x_{0}\otimes_{B}y|x_{0}\otimes_{B}y_{C}[z_{0}, z_{0}])$$=((x_{0}\otimes_{B}y)\otimes_{C}z_{0}|(x_{0}\otimes_{B}y)\otimes_{C}z_{0})$$\square$.

38
A REPORT ON OCNEANU’S LECTURE

Remark. By the proof of (ii), the inner product in $X\otimes_{B}Y$ is given by$(\xi\otimes_{B}\eta|\xi’\otimes_{B}\eta’)=(\xi_{B}[\eta, \eta’]|\xi’)=(\eta|[\xi, \xi’]_{B}\eta’)$and this formula reveals that the linear maps$\xi\vdasharrow\xi\otimes_{B}\eta_{0}$defined for$\xi_{0}\in X_{0},$$\eta_{0}\in Y_{0}$,$\etarightarrow\xi_{0}\otimes_{B}\eta$,

are bounded. This fact will be used later.

Example 18. Let $N$ be a semi-finite von Neuman algebra. Then$n$

$L^{2}(N)\otimes_{N}L^{2}(N)\cong L^{2}(N)$.

The isomorphism is given by$N\cap L^{2}(N)\otimes_{N}N\cap L^{2}(N)\ni a\otimes_{N}b-\rangle$

ab E $L^{2}(N)$ .

and BY be bimodu les. As described aft er Lemma 16, we an imbed
Lemma 19. Let$(X_{B})$$(BY)$. Then, in, we a $ve$
in to
and EndEnd$X_{B}$$c$$h$$\mathcal{B}(X\otimes_{B}Y)$$\mathcal{B}(X\otimes_{B}Y)$

End $(X_{B})’=End(BY)$ .
$)$Express$X_{B}$

and $BY$ as$X_{B}=pL^{2}(B)_{B}^{\oplus I}$

with

$p,$

$q$

,

$BY=BL^{2}(B)^{\oplus I}q$,projections in $Mat_{I}(B)$ . Then

End $(X_{B})=pMat_{I}(B)p$ ,End $(BY)^{op}=qMat_{I}(B)q$ ,on$X\otimes_{B}Y=pL^{2}(Mat_{I}(B))q$ ,and the assertion is reduced to the usual relations of commutants in the induction andreduction operations.
$\square$Remark. The results in this part including Lemma 19 can be formulated and proved without the assumption of the existence of traces but with much effort on modular operators.
See [Sauvageot] for the details.
Bimodule
be finite sums of finite factors. According to Popa (Connes ?), we call a
Let $A,$bimodule $X=AXB$ being of finite type if $\dim_{A}X<+\infty$ and $\dim X_{B}<+\infty$ (note that
this definition is independent of the choice of faithful traces on and ).
andare finite factors and we consider
In the rest of this part, we assume thatandon and .
respect
thenormalized
totracesdimensions with
$B$

$A$

$A$

$B$

$A$

$B$

$B$

$\tau_{A}$$\tau_{B}$39
YAMAGAMI SHIGERU

Proposition 20. Let
be a bimodule of finite type and BY be a finit $ely$ gener a $ted$
module $(i.e., \dim_{B}Y<+\infty)$ . Then
$AX_{B}$.

$\dim_{A}X\otimes_{B}Y=\dim_{A}X\dim_{B}Y$By a decomposition of $Y$ into cyclic submodules, the problem is reduced to the case
$BY=BL^{2}(B)p$ with $p\in B$ a projection. This case, $AX\otimes_{B}Y=AXp$ and the left handside is$)$$\dim_{A}Xp=\tau_{A}’(p)=\frac{\tau_{A}’(p)}{\tau_{A}(1)}\tau_{A}’(1)=\frac{\tau_{A}’(p)}{\tau_{A}(1)}\dim_{A}X$

while $\dim_{B}Y=\tau_{B}(p)=\tau_{A}’(p)/\tau_{A}’(1)$ becauseB.

,is the normalized trace of$\tau_{A}’(1)^{-1}\tau_{A}’|_{B}$$\square$For a bimodule $X=AX_{B}$ (A and$B$

are assumed to be factors), we define its index by$[X]=[X]=\dim_{A}X\dim X_{B}$ .

Corollary 21. Let $C$ be another finite factor.
(i) $If_{B}Y_{C}$ is a bimodule of finite type, then $AX\otimes_{B}Y_{C}$ is of fi nite type and$[AX\otimes_{B}Y_{C}]=[AX_{B}][Y]$.

(ii) $If_{A}Y_{B}$ is a bimodule of finite type,$[X\oplus_{A}Y_{B}]\geq[AX_{B}]+[Y]$ .

(iii) Any bimodule $AX_{B}$ of fin $ite$ type has index $[X]\geq 1$ , with $[X]=1$ if and on $ly$ if $A’=B$($i.e.,$ $B$ is realized as the commutant of ).
(iv) If $AX_{B}$ and $AYB$ are bimodules of finite type, then $Hom(XY_{B})$ is finite dimension al.
$A$(ii) is immediate from the definition. (i) follows from Proposition 20.
(iii): Since is a finite factor and $\dim_{A}X<+\infty,$
is a finite factor and$A’$,with $\tau_{A}’(1)=\dim_{A}X$ . Since
is a trace of$\dim X_{A’}=(\tau_{A}’(1)^{-1}\tau_{A}’)’(1)=\tau_{A}’(1)^{-1}\tau_{A}(1)=1/\dim_{A}X$.
Using this formula and Proposition 20,
$)$

$A$

$A’\subset \mathcal{B}(X)$$\tau_{A}’$$\tau_{A}’’=\tau_{A}$$\dim_{A}X\dim X_{B}=\dim_{A}X\dim(X\otimes_{A’}L^{2}(A’)_{B})$$=\dim_{A}X\dim X_{A’}\dim L^{2}(A’)_{B}$$=\dim L^{2}(A’)_{B}$.

Now the as sertion follows from $[A’ : B]\geq 1$ and $[A’ : B]=1\Leftrightarrow A’=B$ .
(iv): Since $Hom(XY_{B})$ is a subspace of End $(X\oplus AY_{B})$ , it suffices to considerby (iii), the inequality in (ii)the case $X=Y$ . Since any submodule of $AX_{B}$ has index
shows that the number of components does not exceed [X].
$\geq 1$$\square$Frobenius Reciprocity
is naturally a B-A module.
Let $H$ be an A-B module. The conjugate Hilbert space$x\inH$
the element in.
corresponding to
We denote by$H^{*}$$x^{*}$$H^{*}$40
A REPORT ON OCNEANU’S LECTURE

Example 22.

$L^{2}(M)^{*}=L^{2}(M)$as M-M$m$odule.

In the rest of this paragraph, von Neumann algebras are assumed to be finite sums offinite factors.

Lemma 23. Let $AX_{B}$ be a bimodu le of fin $ite$ type.
(i) $(AX)0=(X_{B})_{0}$ ($i.e.$ , an elemen $t\xi\in X$ is A-bounded if and on $ly$ ifB-bounded). In the.
following, this set is denoted by$ted$
module.
For finite ases
for $AX$ andfor BY,(ii) Let BY be a finitely genera
forms a basis for $AX\otimes_{B}Y$ .
$X_{0}$$b$$\{\lambda_{i}\}$$\{l^{\iota_{j}}\}$$\{\lambda_{i}\otimes_{B}\mu_{j}\}$andare factors.
(i) Decomposing $AX_{B}$ into a direct sum, we may assume that$AL^{2}(A)\oplus\cdots\oplus_{A}L^{2}(A)\oplus_{A}L^{2}(A)p$$AX$$TheI\iota A‘=End_{A}X$ is a factor and, if we realizeaswith $p\in A$ a projection (see), then $A’$ is realized by
$A$

$)$

$B$

$(\begin{array}{llll}A A Ap| \ddots | |A A AppA pA pAp\end{array})$

From this form, it is easy to see that elements in $A\oplus\cdots\oplus A\oplus Ap\subset AL^{2}(A)\oplus\cdots\oplus$$AL^{2}(A)\oplus AL^{2}(A)p$ are bounded with respect to the right action of $A’$ and hence they arebounded with respect to $B\subset A’$ as well. In particular, the basis$1\oplus 0\oplus\cdots\oplus 0\oplus 0$$0\oplus 1\oplus\cdots\oplus 0\oplus 0$.
$0\oplus 0\oplus\cdots\oplus 1\oplus 0$$1\oplus 0\oplus\cdots\oplus 0\oplus p$

is an A-linear
for $AX$ consists of B-bounded elements. Since any element in
is A-invariant, is B-bounded. Thus
combination of this basis and since$(X_{B})0$ . By the symmetry of the argument, we get the reverse inclusion as well.
, and. Since $(\otimes_{B}\mu)^{*}\in Hom(AX\otimes_{B}Y_{A}X)$ ,(ii) Let $\varphi\in(X\otimes_{B}Y)_{0},$and we infer$\xi$$(X_{B})_{0}$$\lambda\in X_{0}$$\mu\in Y_{0}$$\varphi(\otimes_{B}\mu)^{*}\in X_{0}$$(a\varphi|\lambda\otimes_{B}\mu)=(\varphi(\otimes_{B}\mu)^{*}|a^{*}\lambda)$$=\tau_{A}(a_{A}[\varphi(\otimes_{B\}}\iota)^{*}, \lambda])$which shows that$\lambda\otimes_{B}\mu\in(X\otimes_{B}Y)_{0}$and$A[\varphi, \lambda\otimes_{B}\mu]=A[\varphi(\otimes_{B})^{*}, \mu]$.

$(_{A}X)_{0}$$(_{A}X)_{0}\subset$

$\xi$

,

41
YAMAGAMI SHIGERU

Now the following calculation gives a proof.
$\sum_{ij}(A[\varphi, \lambda;\otimes_{B}\mu J]\lambda_{\hat{l}}\otimes_{B}\mu;|\xi\otimes_{B}\eta)=\sum_{ij}(A[\varphi(\mu_{j})^{*}, \lambda_{i}]\lambda;|\xi_{B}[\eta, \mu_{j}])$$= \sum_{j}(\varphi(\mu_{j})^{*}|\xi_{B}[\eta, \mu_{j}])$$= \sum_{j}(\varphi|\xi_{B}[\eta,\mu_{j}]\otimes_{B}\mu_{j})$$= \sum_{j}(\varphi|\xi\otimes_{B}[\eta, \mu_{j}]\mu_{j})$$=(\varphi|\xi\otimes_{B}\eta)$.

$\square$Lemma 24. Let $AX_{B}$ and $BYA$ be bimod les of finite $type$ . Then we have the following-preserving (i.e., the adjoin of intertwin $ers$ correspond to conjuga $te$ vectors in Hilbert$sp$ aces) linear isomorphisms:
$u$

$*$

$ts$

$(X\otimes_{B}Y)^{A}$$\varphi$$\cong$$Hom(Y_{B}^{*},X_{B})$$rightarrow$$\Phi$

where, for an A-A $bim$ odule $AZA,$$Z^{A}$$(Y\otimes_{A}X)^{B}$$\cong$is defin $ed$ by$Z^{A}=\{\zeta\in Z;a\zeta=\zeta a \forall aEA\}$

and$\varphi,$$\psi,$$\Phi$,

are related by$(\xi\otimes_{B}\eta|\varphi)=(\xi|\Phi(\eta^{*}))=(\eta\otimes_{A}\xi|\psi)$

$)$

For

$\varphi\in X\otimes_{B}Y$, define a linear map$\Phi$:$Y_{0^{*}}arrow X$$\Phi(\eta^{*})=\varphi(\otimes_{B}\eta)^{*}$

Note that,$\psi$$\otimes_{B}\eta$denotes the bounded linear map

.

by

.

$\xi->\xi\otimes_{B}\eta$$(\xi|\Phi(\eta^{*}))=(\xi\otimes_{B}\eta|\varphi)=(\eta|(\xi\otimes_{B})^{*}\varphi)$. Sincefor$\xi\in X_{0}$,the adjoint of is densely defined and hence is a closable operator. On the other hand,
the A-invariance of shows that is A-B equivariant. Using the irreducible decomposi(recall that $Hom(X,Y_{B}^{*})$ is finite-dimensional) and the polar
tions of $AX_{B}$ anddecomposition of the closure of , one sees thatis a constant multiple of an isometry on each irreducible components. In particular, it is bounded and we conclude that$\Phi$$\Phi$$\Phi$$\varphi$$AY_{B}^{*}$$\Phi$$\Phi\in Hom(XY_{B^{*}})$ .

$\Phi$42
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Conversely, suppose that $\Phi EHom(XY_{B}^{*})$ . Take a finite basis$\{\mu_{j}\}$for $BY$ andset.

$\varphi=\sum_{j}\Phi(\mu_{j}^{*})\otimes_{B}\mu_{j}\in X\otimes_{B}Y$

Note that$\Phi(\mu_{j}^{*})\in X_{0}$since$\mu_{j}^{*}EY_{0}^{*}$. For$\xi\in X_{0},$$\eta\in Y_{0}$,$( \xi\otimes_{B}\eta|\varphi)=\sum_{j}(\xi|\Phi(\mu_{j}^{*})_{B}[l^{\iota}i, \eta])$$= \sum_{j}(\xi|\Phi(\mu_{j}^{*})[\mu_{j}^{*}, \eta^{*}]_{B})$$=( \xi|\Phi(\sum_{j}\mu_{j}^{*}[\mu_{j}^{*}, \eta^{*}]_{B}))$$=(\xi|\Phi(\eta^{*}))$

from which we see that the correspondence,$\varphi-\succ\Phi$

is bijective.

$\square$Corollary 25. Let $AX_{B},$ $BYC,$ $AZC$ be bimodules of fin $ite$ type. Then
$Hom(AX\otimes_{B}Y_{C,A}Z_{C})\cong Hom(BY_{C,B}X^{*}\otimes_{A}Z_{C})$.

More precisely, an intert $wi$ ner $T\in Hom(AX\otimes_{B}Y_{C,A}Z_{C})$ and its Fro ben $ius$ transform
S E $Hom(YX^{*}\otimes_{A}Z_{C})$ is relat $ed$ by$(\xi^{*}\otimes_{A}(|S(\eta))=(\zeta|T(\xi\otimes_{B}\eta))$.

$)$

1. $h.s$ . $\cong((X\otimes_{B}Y)\otimes cZ^{*})^{A}$$=(X\otimes_{B}(Y\otimes_{C}Z^{*}))^{A}$$\cong((Y\otimes_{C}Z^{*})\otimes_{A}X)^{B}$$=(Y\otimes_{C}(Z^{*}\otimes_{A}X))^{B}$$\cong r.h.s$.

$\square$Lemma 26. Let $AX_{B},$ $BYC$ $Az_{c}$ be imodules of finite type and suppose that $BYC$
and $Az_{c}$ are irreduci $ble$. For $T\in Hom(AX\otimes_{B}Y_{C,A}Z_{C})$ , its Frobenius transform$S\in Hom(YX^{*}\otimes_{A}Z_{C})$ satisfies$b$$\dim_{B}Y||S||^{2}=\dim_{A}Z||T||^{2}$.

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YAMAGAMI SHIGERU
$)$

Let

$\{\lambda_{i}\}\subset AX,$$\{\mu j\}\subset BY\{\zeta_{k}\}\subset AZ$

be finite bases. We first claim$S( \mu_{j})=\sum_{i}\lambda_{i}^{*}\otimes_{A}T(\lambda_{i}\otimes_{B}\mu_{j})$.

In fact,$(\xi^{*}\otimes_{A}\zeta|S(\eta))=(\zeta|T(\xi\otimes_{B}\eta))$$=( \zeta|T(\sum_{i}A[\xi, \lambda_{i}]\lambda_{i}\otimes_{B}\eta))$$= \sum_{i}(\zeta|[\xi^{*}, \lambda_{i}^{*}]_{A}T(\lambda_{i}\otimes_{B}\eta))$$= \sum_{i}(\xi^{*}\otimes_{A}\zeta|\lambda_{i}^{*}\otimes_{A}T(\lambda_{i}\otimes_{B}\eta))$

Since

$Y$

and

$Z$

are irreducible,$TT^{*}$

and$S^{*}S$.

are scalar multiples of identity and hence$\dim_{B}Y||S||^{2}=\sum_{j}(S(\mu_{j})|S(\mu j))$$= \sum_{ijk}(\lambda_{i}^{*}\otimes_{A}T(\lambda_{i}\otimes_{B}\mu_{j})|\xi_{k}^{*}\otimes_{A}T(\xi_{k}\otimes_{B}\mu_{j}))$$= \sum_{ijk}(T(\lambda_{i}\otimes_{B}\mu_{j})|_{A}[\lambda_{i}, \lambda_{k}]T(\lambda_{k}\otimes_{B\int}\iota_{j}))$$= \sum_{1j}(T(\lambda_{i}\otimes_{B}\mu j)|T(\lambda_{i}\otimes_{B}\mu_{i}))$$= \sum_{1jk}\otimes_{B}\mu J$$= \sum_{ijk}A\otimes_{B}\mu j$$= \sum_{ijk}\tau_{A}(A[T^{*}(\zeta_{k}), \lambda_{i^{\otimes_{B}\mu i}}]_{A}[\lambda_{i}\otimes_{B}\mu i, T^{*}(\zeta_{k})])$$= \sum_{ijk}\tau_{A(A}[T^{*}(\zeta_{k}), \lambda_{i}\otimes_{B}\mu i]\lambda_{i}\otimes_{B}\mu_{j}|T^{*}(\zeta_{k})])$$= \sum_{k}(T^{*}(\zeta_{k})|T^{*}(\zeta_{k}))$$=\dim_{A}Z||T||^{2}$(since$\{\lambda;\otimes\mu j\}$is a basis).

$\square$depend on the choice of traces
andRemark. Boththe dependence of $\dim_{A}Z,$ $\dim_{B}Y$ on traces.
$||S||$$||T||$$\tau_{A},$$\tau_{B}$, which is canceled by

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A REPORT ON OCNEANU’S LECTURE

Notation. In the following, for an intertwiner , its (linear) Frobenius transform changing
the left actions of bimodules is denoted by . Similarly the Frobenius transform changing
the right actions is denoted by ‘. Furthermore, in the situation of Lemma (i.e., withthe assumption on irreducibility), the normalization of the adjoint of the left Frobenius) is denoted by(resp.
(resp. the right Frobenius transform). Fortransform$T^{l}=S$
andexample, with the notation of the above lemma,$T$$T^{l}$$T$$T^{l}$$\tau\#=\sqrt{\frac{\dim_{B}Y}{\dim_{A}Z}}s*$$T^{b}$$\tau\#$$T^{f}$.

Lemma 27. Let $T\in Hom(AX\otimes_{B}Y_{C,A}Z_{C})$ and S E $Hom(ZW_{C})$ . Then$(ST)^{l}=(1\otimes_{A}S)T^{l}$

Let

$)$

$\{\zeta_{i}\}$.

be an orthonormal basis in $X$ . For $\xi\in X,$ $\eta\in Y$ , and $\omega\in W$$((ST)^{l}(\eta)|\xi^{*}\otimes_{A}\omega)=(ST(\xi\otimes_{B}\eta)|\omega)$$=(T(\xi\otimes_{B}\eta)|S^{*}\omega)$$= \sum_{i}(T(\xi\otimes_{B}\eta)|\zeta_{i})(\zeta_{i}|S^{*}\omega)$$= \sum_{i}(T^{l}(\eta)|\xi^{*}\otimes_{A}\zeta_{i})(\zeta_{i}|S^{*}\omega)$$=(T^{l}( \eta)|\xi^{*}\otimes_{A}\sum_{i}\zeta_{i}(\zeta_{i}|S^{*}\omega))$$=(T^{l}(\eta)|\xi^{*}\otimes_{A}S^{*}\omega)$$=(T^{l}(\eta)|(1\otimes_{A}S^{*})(\xi^{*}\otimes_{A}\omega))$$=((1\otimes_{A}S)T^{l}(\eta)|\xi^{*}\otimes_{A}\omega)$.

$\square$Paragroup
Now we introduce a group-like structure (called paragroup) for a pair $N\subset ilI$ of
factor and subfactor with finite index. For $A,$ $B=M$ or $N$ , let
be the set of
equivalence classes of A-B bimodules which appear in(notethat End $(AL^{2}(M)\otimes_{N}\cdots\otimes_{N}L^{2}(M)_{B})$ is finite-dimensional due to Corollary 21). Let
be a bipartite graph having even vertices
and odd vertices
with edges between
if, a suitable M-action
and
$E$
$N$
being restricted to , one of or appears in the other with multiplicity . For example,
if $E=E$ and $O=NO_{M)}$ then $n=\dim Hom(NO_{MN}E_{M})$ . Note that, by Frobenius$\mathcal{G}_{A,B}$$AL^{2}(M)\otimes_{N}\cdots\otimes_{N}L^{2}(\Lambda I)_{B}$$\mathcal{G}_{even}^{(0)}=C_{J_{M}}^{(0)_{M}}\cup \mathcal{G}_{N}^{(0)_{N}}$$\mathcal{G}$$\mathcal{G}_{odd}^{(0)}=\mathcal{G}_{M}^{(0)_{N}}\cup \mathcal{G}_{N,}^{(0)_{M}}$$E\in \mathcal{G}_{even}^{(0)}$

$n$

$O$

$O\in \mathcal{G}_{odd}^{(0)}$

$n$

45
YAMAGAMI SHIGERU

reciprocity, this number can be also interpreted as $\dim Hom(ML^{2}(\Lambda f)\otimes_{N}O_{M,M}E_{M})$ . The
class of the identity bimodule $ML^{(}M)_{M}$ (resp. $NL^{2}(N)_{N}$ ) is distinguished and denoted).
also $admits*$ -operation induced by taking adjoint bimodules.
(resp.
by(resp.
) be two of them, which is obtained
has 4 distinguished subgraphs. Let
to(resp. to). The other two
by reducing the vertex set.
andare given by$\star_{M}$$\star_{N}$$\mathcal{G}$$\mathcal{G}_{M}$$\mathcal{G}$$\mathcal{G}^{(0)}$$\mathcal{G}_{M}^{*}$

Lemma 28.

$Gr$$\mathcal{G}_{N}$$\mathcal{G}_{N}^{(O)_{M}}\cup \mathcal{G}_{N}^{(0)_{N}}$$\mathcal{G}_{M}^{(0)_{M}}\cup \mathcal{G}_{MN}^{(0)_{1}}$$\mathcal{G}_{N}^{*}$aph$s\mathcal{G}_{M},$$\mathcal{G}_{N}$are on ected an locally finite.
$c$

$d$

$n$

Local finiteness follows from the finite dimensionality of the space of intertwiners. By) is connected with the irreducible bimodule(resp.
the definition, any vertex in$ML^{2}(M)_{N}$ (resp. $NL^{2}(M)_{M}$ ).
$)$$\mathcal{G}_{N}^{(0)}$$\mathcal{G}_{M}^{(0)}$$\square$At this stage, edges with same end points are not distinguished; we have only fixed thenumber of edges between two vertices and not defined edges themselves. Now we give a
concrete meaning to edges: Take an even vertex $X$ and a odd vertex . To be specific,
suppose that $X=MX_{M},$ $Y=NYM$ for example, and set $n=\dim Hom(NY_{M,N}X_{M})$ .
Since is the multiplicity of the irreducible bimodule $NYM$ in the bimodule $Nx_{nr}$ , wefrom $NYM$ into $NX_{M}$ .
can find pair-wise orthogonal isometric intertwiners
We regard such a choice of intertwiners represent n-edges between $X$ and , and call it arepresentation of .
It is easy to see that we can find a representation of so that -operation keeps it
invariant. Such representations are called*-representations.
$Y$

$n$

$T_{1},$$n$$T_{n}$$\ldots,$$Y$$\mathcal{G}$$*$$\mathcal{G}$Haar Measure

Definition 29. Let be a connected unoriented $graph$ with a distin guished vertex . For, we seta funct ion on the se of vertex$\Gamma$$\star$$\Gamma^{(0)}$

$t$

$\mu$

$( \Delta\mu)(x)=\sum_{y\in\Gamma^{(O)}}\Delta(x, y)\mu(y)$,$xE\Gamma^{(0)}$denotes the incidence matrix of
whereis alled the Laplacian of F.
edges between and ).
Haar measure is, by definition, an eigen-fu nction of‘norm alized’ in the sense$\mu(\star)=1$ .

$\Gamma(i.e.,$$\{\Delta(x, y)\}_{x,y\in\Gamma^{(0)}}$

$x$

$y$

$\Delta$,$\Delta(x, y)=the$number of

$c$

$A$

$is$

Note that the eigen value of a Haar measure is positive.

$\mu$$\triangle\iota vi$

th posi $tive$ entries$wl_{1}$ich

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Lemma 30. Letbyvertices$\mathcal{G}$be a graph introdu $ced$ above and define funct ions$\mu_{0},$$\mu$on the set of$\mathcal{G}^{(0)}$$\mu_{0}(V)=dimension$ of $V$ as left module,$\mu(V)=\{\begin{array}{l}[M.\cdot N]^{1/2}\mu_{0}(V)jfV\in \mathcal{G}_{M}^{(0)_{N}}[M.\cdot N]^{-1/2}\mu_{0}(V)ifV\in \mathcal{G}_{N}^{(0)_{M}}\mu_{0}(V)otherwjse\end{array}$(Note that we are measurin dimen sions with respect to the un ique norm alized trace.).
Then restrictions of define Haar mesures for 4 graphs$g$$\mathcal{G}_{M},$

$\mu$

$)$

We consider the case for$MX_{M}\in \mathcal{G}_{M}^{(0)_{M}},$$MYN\in \mathcal{G}_{M}^{(0)_{N}},$$\mathcal{G}_{M}^{*},$$\mathcal{G}_{N},$$\mathcal{G}_{N}^{*}$. The other cases are checked in a similar way. Let$T$be a representation of an edge between $X$ and Y.
’ and$\mathcal{G}_{M}$With these notations, we have
$MY\otimes_{N}L^{2}(M)_{M}=\oplus x,\tau T^{b}(MX_{M})$$MX_{N}=\oplus_{Y,T}T(MY_{N})$,from which, taking dimensions, we deduce that$[M : N] \mu_{0}(Y)=\sum_{X,T}\mu_{0}(X)=(\Delta\mu_{0})(Y)$$\mu_{0}(X)=\sum_{Y,T}\mu_{0}(Y)=(\Delta\mu_{0})(X)$

Now the assertion follows from this.

.

$\square$Connection
Here we introduce the notion of ‘connection’ for a representation of . A sequence ofand making a closed path in the graph , is
4 edges, each taken fromcalled a cell. Given a representation of , a connection $W$ is a C-valued function on theset of cells defined by$W(C)1_{Z}=S_{2}^{*}T_{2}^{*}T_{1}S_{1}$ ,$\mathcal{G}$$\mathcal{G}_{M},$$\mathcal{G}_{M}^{*},$$\mathcal{G}_{N},$$\mathcal{G}_{N}^{*}$$\mathcal{G}$$\mathcal{G}$: $N(Y_{1})_{M}arrow NL^{2}(M)\otimes_{M}X_{M},$where, for a cell $C,$
: $M(Y_{2})_{N}arrow MX\otimes_{M}L^{2}(M)_{N}$ ,: $NZ_{N}arrow NL^{2}(M)\otimes_{M}(Y_{2})_{N}$ are intertwiners: $NZNarrow NY1\otimes_{M}L^{2}(M)_{N}$ , and$C$
representing 4 edges of . Note that $S_{2}T_{2}T_{1}^{*}S_{1}^{*}\in End(NZ_{N})$ is a scalar multiple ofthe identity because $NZ_{N}$ is irreducible. Of course, $W$ has a cohomological ambiguity
coming from the choice of vertices (realization of irreducible bimodules) and edges ofgraphs (realization of intertwiners). In other words, $W$ is unique up to cohomology-like$T_{2}$$T_{1}$$S_{1}$

gauge.

$S_{2}$

47
YAMAGAMI SHIGERU

andand suppose that there is a cell having $X$ andLet$Z$ as a part of vertices.
In the vector space $Hom(Z,X_{N})$ , we introduce a positive
definite inner product by$MX_{M}\in \mathcal{G}_{M}^{(0)_{M}}$$NZ_{N}\in \mathcal{G}_{N}^{(0)_{N}}$$(U|V)1_{Z}=V^{*}U$,$V\in Hom(Z,X_{N})$ .

$U,$

It is easy to see that the set of intertwiners$\{T_{1}S_{1}\}$indexed by left halves of cells, forms an orthonormal basis of $Hom(ZX_{N})$ . Similarly,
we have an orthonormal basis$\{T_{2}S_{2}\}$indexed by right halves of cells.
In this way, for a fixed pair of vertices $MX_{M},$ $NZN,$ $W$ can be interpreted as a unitarymatrix connecting two orthonormal bases.
Theorem 31. For a fixed air of$p$$Y_{1}\in \mathcal{G}_{N}^{(0)_{N}}$an$dY_{2}\in \mathcal{G}_{M}^{(0)_{N}}$,$\sqrt{\frac{\mu(X)\mu(Z)}{\mu(Y_{1})\mu(Y_{2})}}W(Y_{1} XZ Y_{2})$forms a unitary matrix in dexed by upper alves an lower alves of cells $cont$ aining
and .
$h$

$d$

$h$

$Y_{1}$$Y_{2}$

Before giving the proof of the theorem, we present a lemma.

be th orthogon projection onto an N-in variant closed subspace$Z\cong 1\otimes_{N}Z\subset L^{2}(M)\otimes_{N}$ Z. Then for $T\in End(ML^{2}(M)\otimes_{N}Z)$ , wea $ve$

Lemma 32. Let

$p$

$al$

$e$

$h$

$\tau_{M}’(T)=\tau_{N}’(pTp)$

$)$

If

$\{\lambda_{i}\}$is a basis for $NZ$ , then$\{1 \otimes_{N}\lambda_{i}\}$.

is a basis for $ML^{2}(M)\otimes_{N}Z$ and we have$\tau_{M}’(T)=\sum_{i}(T(1\otimes_{N}\lambda_{i})|1\otimes_{N}\lambda_{i})$$= \sum_{i}(Tp(1\otimes_{N}\lambda_{i})|p(1\otimes_{N}\lambda_{i}))$$=\tau_{N}’(pTp)$$\square$.

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Proof of the theorem. The idea of the proof is ‘similar’ to the case of the left-right unitar-

ity of $W$ . Here we consider the vector space $Hom(M(Y_{2})_{N,M}L^{2}(\Lambda t)\otimes_{N}(Y_{1})_{M})$ and makeit a finite dimensional Hilbert space as described above. Let $T_{1}\in Hom(N(Y_{1})_{M,N}X_{\Lambda I})$ ,$T_{2}\in Hom(M(Y_{2})_{N,M}X_{N}),$ $S_{1}\in Hom(Z(Y_{1})_{N})$ , and $S_{2}\in Hom(NZ_{N,N}(Y_{2})_{N})$describe a cell. Since $N(Y_{1})_{M}$ and $MX_{M}$ are irreducible,$T_{1}rightarrow T_{1}^{\#}\in Hom(XL^{2}(M)\otimes_{N}(Y_{1})_{M})$is an isometry. So$T_{1}^{\#}’ s$are orthogonal to each other and$ML^{2}(M)\otimes_{N}(Y_{1})_{M}=\oplus x,\tau_{1}T_{1}^{\#}(X)$gives an irreducible decomposition. Similarly, we have an irreducible decomposition$\cdot$$MY_{2}\otimes_{N}L^{2}(\Lambda I)_{M}=\oplus x,\tau_{2}T_{2}^{b}(X)$.

forms a basis for $Hom(Y\otimes_{N}L^{2}(M)_{MM}L^{2}(M)\otimes_{N}(Y_{1})_{M})$ .
Thus
is a basis of $Hom(M(Y_{2})_{N,M}L^{2}(M)X_{N}$Then, by Frobenius reciprocity,. Since$\{T_{1}^{\#}(T_{2}^{b})^{*}\}_{X,T_{1},T_{2}}$$\{(T_{1}^{\#}(T_{2}^{b})^{*})^{r}\}$$(Y_{1})_{N})$$(T_{1}^{\#}(T_{2}^{b})^{*})^{r}=(1\otimes_{\Lambda f}T_{1}^{\#})(T_{2}^{b^{*}})^{r}=T_{1}^{\#}(T_{2}^{b^{*}})^{r}$,they form an orthogonal basis and its normalization is given by$T_{1}^{\#}T_{2}$.

Similarly, for the lower half of a cell,$(1 \otimes_{N}S_{1})S_{2}^{\#}$

forms an orthonormal basis for $Hom(M(Y_{2})_{N,M}L^{2}(M)\otimes_{N}(Y_{1})_{N})$ . Thus we have obtained
a unitary matrix$((1\otimes_{N}S_{1})S_{2}^{\#}|T_{1}^{\#}T_{2})$.

To relate this quantity with the connection $W$ , we calculate as follows: Since$((1\otimes_{N}S_{1})S_{2}^{\#}|T_{1}^{\#}T_{2})1_{Y_{2}}=T_{2}^{*}(T_{1}^{\#})^{*}(1\otimes_{N}S_{1})S_{2}^{\#}$and since $\tau_{M}’(1_{Y_{2}})=\dim_{M}Y_{2}$ (here traces on $M$ and $N$ are assumed to be normalized),$\mu_{0}(Y_{2})((1\otimes_{N}S_{1})S_{2}^{\#}|T_{1}^{\#}T_{2})=\tau_{M}’(T_{2}^{*}(T_{1}^{\#})^{*}(1\otimes_{N}S_{1})S_{2}^{\#})$$=\tau_{M}’(S_{2}^{\#}T_{2}^{*}(T_{1}^{\#})^{*}(1\otimes_{N}S_{1}))$$=\tau_{N}’(pS_{2}^{\#}T_{2}^{*}(T_{1}^{\#})^{*}(1\otimes_{N}S_{1})p)$$=\sqrt{\frac{\mu_{0}(Y_{2})}{\mu_{0}(Z)}}\sqrt{\frac{\mu_{0}(X)}{\mu_{0}(Y_{1})}}\tau_{N}’(p(S_{2}^{l})^{*}T_{2}^{*}T_{1}^{l}(1\otimes_{N}S_{1})p)$.

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YAMAGAMI SHIGERU

Since, for$\zeta_{1},$$\zeta_{2}\in Z$,$(1\otimes_{N}\zeta_{2}|(S_{2}^{l})^{*}T_{2}^{*}T_{1}^{l}(1\otimes_{N}S_{1})(1\otimes_{N}(_{1}))$$=(T_{2}S_{2}^{l}(1\otimes_{N}\zeta_{2})|T_{1}^{l}(1\otimes_{N}S_{1}(\zeta_{1})))$$=(T_{2}S_{2}(C_{2})|T_{1}S_{1}(\zeta_{1}))$,$p(S_{2}^{l})^{*}T_{2}^{*}T_{1}^{l}(1\otimes_{N}S_{1})p=S_{2}^{*}T_{2}^{*}T_{1}S_{1}=W(C)1_{Z}$and hence$\mu_{0}(Y_{2})((1\otimes_{N}S_{1})S_{2}^{\#}|T_{1}^{\#}T_{2})=\sqrt{\frac{\mu_{0}(Y_{2})}{\mu_{0}(Z)}}\sqrt{\frac{\mu_{0}(X)}{\mu_{0}(Y_{1})}}W(C)\tau_{N}’(1_{Z})$,$=\sqrt{\frac{\mu_{0}(Y_{2})}{\mu_{0}(Z)}}\sqrt{\frac{\mu_{0}(X)}{\mu_{0}(Y_{1})}}W(C)\mu_{0}(Z)$which proves the desired assertion (note that $\mu_{0}(Y_{1})\mu_{0}(Y_{2})=\mu(Y_{1})\mu(Y_{2}),$ $\mu_{0}(X)=\mu(X)$ ,and $\mu_{0}(Z)=\mu(Z))$ .
$ML^{2}(M)\otimes_{N}(Y_{1})_{M}arrow^{T_{1}^{l}}$$MX_{M}$$1\otimes_{1}\uparrow$$\uparrow T_{2}$$ML^{2}(M)\otimes_{N}Z_{N}$$arrow^{S_{2}^{l}}M(Y_{2})_{N}$$\square$Example 33.
be a finite group, and : $Garrow Aut(N)$ be an outer act,ion.
Let $N$ be a finite factor,
Let $M=Nx_{\alpha}G$ and denote by $H$ an N-M module $NM_{M}$ . Since $N’\cap M=C1$ (cf. theappendix below), $H$ is irreducible. Note that$G$$\alpha$$NM_{N}= \sum_{g\in G}N\lambda_{g}$are not equivalent to each others. It is
gives an irreducible decomposition andclear that this gives an orthogonal decomposition and each component is invariant underis equivalent to $L^{2}(N)$ as left N-module,N-N action. Since.
be an intertwiner of N-N modules. Since commutes with the left
Let
. Thusaction,is mapped intosuch that$\{N\lambda_{g}\}s$$(\overline{N\lambda_{g}}^{L^{2}})_{0}=N\lambda_{g}$$\overline{N\lambda_{g}}^{L^{2}}$$T$$T:IV\lambda_{g}arrow\overline{N\lambda_{h}}$$N\lambda_{g}=(\overline{N\lambda_{g}}^{L^{2}})_{0}$$N\lambda_{h}=(\overline{N\lambda_{h}}^{L^{2}})_{0}$$T(\lambda_{g})=a\lambda_{h}$$\exists a\in N$.

For $b\in N$ ,$T(\lambda_{g}b)=T(\alpha_{g}(b)\lambda_{g})=\alpha_{g}(b)T(\lambda_{g})=\alpha_{g}(b)a\lambda_{h}$$T(\lambda_{g}b)=T(\lambda_{g})=a\lambda_{h}b=a\alpha_{h}(b)\lambda_{h}$

,

,

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A REPORT ON OCNEANU’S LECTURE

which means that$\alpha_{g}(b)a=a\alpha_{h}(b)$,$\forall b\in N$.

From the outerness of the action,$a=\{\begin{array}{l}scalarifg=h0otherwise\end{array}$proving the assertion.
Now we consider the decomposition of $M\otimes_{N}M=H^{*}\otimes_{N}H$ . For$\rho\in\hat{G}$

, set

.

$\xi_{ij}^{\rho}=\sum_{g}\rho_{i}i(g^{-1})\rho_{g}^{*}\otimes_{N}\lambda_{g}\in H^{*}\otimes_{N}H$

Since$(a\lambda_{g}\otimes\lambda_{h}|a’\lambda_{g’}\otimes\lambda_{h’})=\tau([a’\lambda_{g’}, a\lambda_{g}]_{NN}[\lambda_{h}, \lambda_{h’}])$$=\tau(\alpha_{g^{l-1}}(a’)^{*}[\lambda_{*}g’’\lambda_{g}]_{N}\alpha_{g^{-1}}(a)_{N}[\lambda_{h}, \lambda_{h’}])$$=\delta_{g,g’}\delta_{h,h’}\tau(aa)$

,

$a,$

$a\in N$ ,we have$(a \lambda_{h}\xi_{ij}^{\rho}|b\lambda_{k}\xi_{kl}^{\sigma})=\delta_{h,k}\tau(ab^{*})\sum_{g}\rho_{i}i(g^{-1})\overline{\sigma_{kl}(g-1)}$$=\delta_{h,k}\tau(ab^{*})\delta_{\rho,\sigma}\delta_{i,k}\delta_{j,l}|G|$.

Thus$\sum_{\rho,i,j}M\xi_{i,j}^{\rho}$

is an orthogonal sum in. In particular,
basis in$H^{*}\otimes_{N}H$$\ell^{2}(G)$. By the Peter-Weyl’s theorem,
we can find a sequence$\forall h\in G$$\{\rho_{ij}(g^{-1})\}_{\rho,i,j}$$\{c_{ij}^{\rho}\in C\}$

forms asuch that$\sum_{\rho,i,j}c_{ij}^{\rho}\rho_{ij(g^{-1})=\delta_{g,h}}$contains. Sincewhich means that$h\in G$ , which shows thatM-invariant, it contains. Thus we have obtained an orthogonal decomposition
in$\sum_{\rho,i,j}M\xi_{i,j}^{\rho}$$\lambda_{h}^{*}\otimes_{N}\lambda_{h}$$N\lambda_{a}\lambda_{h}^{*}\otimes_{N}\lambda_{h},$$\sum_{\rho,i,j}\Lambda t\xi_{i,j}^{\rho}$$a,$$H^{*}\otimes_{N}H$$H^{*} \otimes_{N}H=\sum_{\rho^{1},j}M\xi_{i,j}^{\rho}$.

Since, by an easy computation,$\xi_{ij}^{\rho}x=x\xi_{ij}^{\rho}$

,

$xEN$

$\xi_{ij}^{\rho}\lambda_{h}=\lambda_{h}\sum_{k}\rho_{ik}(h)\xi_{kj}^{\rho}$,$h\in G$,is apparently left
is dense$\sum_{\rho,i,j}M\xi_{i,j}^{\rho}$

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YAMAGAMI SHIGERU

is an M-M sub-module.
To see that these are irreducible, letbe an intertwiner of M-M modules.
is isomorphic to $L^{2}(M)$ as left M-module,
and
Since

$X_{j}^{\rho} \equiv\sum_{k}M\xi_{kj}^{\rho}$$T:X_{j^{\rho}}arrow X_{k}^{\sigma}$$(\overline{IM\xi_{ij}^{\rho}})_{0}=M\xi_{ij}^{\rho}$$\overline{M\xi_{ij}^{\rho}}$.

$(X_{j}^{\rho})_{0}=( \sum_{i}\overline{M\xi_{ij}^{\rho}})_{0}=\sum_{i}(\overline{M\xi_{ij}^{\rho}})_{0}$

Since

$T$

maps

$(X_{j}^{\rho})_{0}$into$(X_{k}^{\sigma})_{0},$$\exists a_{il}\in M$

such that$T( \xi_{ij}^{\rho})=\sum_{l}a_{il}\xi_{lk}^{\sigma}$

Futhermore, since$T$.

commutes with left action of $N$ , we have$a_{i1}\in N’\cap M=C1$ .

If we consider the right action of onand . Thus, by Schur’s lemma,$G$$\rho$$\xi_{ij}^{\rho}$, we can show that the matrix $A=(a_{il})$ intertwines$\sigma$$a_{il}=\{\begin{array}{l}a\delta_{il}if\rho=\sigma 0otherwise\end{array}$In other words,
if$X_{j}^{p}\not\cong X_{l}^{\sigma}$$X_{j}^{\rho}\cong X_{l}^{\rho}$,$\rho\neq\sigma$$\forall j,$$k$.

Since $N’\cap M=C1,$ $H=NL^{2}(M)_{M}$ and $H^{*}=ML^{2}(M)_{N}$ are irreducible modules.
Next, we show that all the irreducible components in $L^{2}(M)\otimes_{N}\cdots\otimes_{N}L^{2}(M)$ arecontained in the ones already listed. To see this, we first consider the decomposition of$NL^{2}(M)\otimes_{N}L^{2}(M)_{N}$ . Using, we have an orthogonal decomposition$M= \sum_{g\in G}N\lambda_{g}$$L^{2}(M) \otimes_{N}L^{2}(M)=\sum_{g,h\in G}\overline{INV\lambda_{g}\otimes_{N}\lambda_{h}}$

.

Since
$\overline{INV\lambda_{g}\otimes_{M}\lambda_{h}}\ni a\lambda_{g}\otimes_{N}\lambda_{h}rightarrow a\lambda_{gh}\in\overline{N\lambda_{gh}}$gives an isomorphism as N-N module, there appears no new irreducible component in$L^{2}(M)\otimes_{N}L^{2}(M)$ . Repeating these arguments, we find thatdoes
not generate new irreducible modules.
In this way, we have described the vertices. Let us now determine the edges. Since$L^{2}(\Lambda f)\otimes_{N}\cdots\otimes_{N}L^{2}(M)$${}_{N}H_{N}= \sum_{g\in G}IV\lambda_{g}$,52
A REPORT ON OCNEANU’S LECTURE

there is one edge between (the vertex determined by) $H$ and (the vertex of) each irreducible
N-N module. On the other hand, the orthogonal decomposition$N(X_{j}^{\rho})_{M}= \sum_{i}N(\overline{\xi_{ij}^{\rho}M})_{M}$and the fact

$N(\overline{\xi_{ij}^{\rho}M})_{M}\cong {}_{N}H_{M}$shows that there are $n=\dim\rho$ edges between$X_{j^{\rho}}$and $H$ .

Markov IYace
In tbis part, von Neumann algebras are assumed to be finite sums of finite factors.
be the set of minimal central
For a von Neumann algebra of this type, let. In the following,projections in . Then a trace of is determined by the values on$(\tau(p_{1}),\ldots,\tau(p_{m}))$.
to express the row vector
we use the notation$AX_{B}$(resp.
) be the set of minimal
be a bimodule of finite type. LetLetandcentral projections in (resp. ). Define a bipartite graph having vertices$AX_{B}$is called connected if the graph obtainedwith edges if $p_{i}Xq_{j}\neq 0$ . The bimodulein this way is connected.
$A$

$A$

$\{p_{i}\}_{1\leq i\leq m}$$\tau$$A$$\{p_{i}\}$$\tau(p)$$\{p_{i}\}$$\{q_{j}\}$

$B$

$A$

$\{p_{i}\}$$\{q_{j}\}$be traces on $A,$
Definition 34. Let $AX_{B}$ be bim $odu$ le of finite type andare called Markov traces (or alanced traces) if $\exists a>0,$ $b>0$
andrespectively.
such th at$a$$\tau_{A}$$\tau_{A},$$B$$\tau_{B}$$b$$\tau_{B}$$\tau_{A}=a\tau_{B}’|_{A}$,$\tau_{B}=b\tau_{A}’|_{B}$.

is connected, there is a
Proposition 35. Let $AX_{B}$ be a bimodule of finite type. If$bal$$ique$,
positive multiplica ion.
on
which
is
un
up
tracesonandtoair of anced$AX_{B}$$p$$A$$\tau_{A}$$B$$\tau_{B}$$t$

Before the proof of the proposition, we introduce some useful notations.
the se of minimal central projections in $von$ Neum $ann$
Definition 36. Let, define matrices $L(X)=$
algebras $A,$ , respecti vely. For a $bim$ odule of finite type$(L_{ij}(X))$ and $R(X)=(R_{ij}(X))$ by$\{p_{i}\},$$\{q_{j}\}$

$t$

$B$

$AX_{B}$$L_{ij}(X)=\dim_{Ap;}(p_{i}Xq_{j})$,$R_{ij}(X)=\dim(p;Xq_{j})_{Bq_{j}}$,where imensions are meas red with respect to the norm alized races. (Note thatare factors.)$d$$Bq_{j}$$u$$t$$Ap_{i}$

an

$d$

53
YAMAGAMI SHIGERU

Lemma 37.
(i) Letbe traces on $A,$$\tau_{A},$$\tau_{B}$$B$

respectively. Then we have$\tau_{A}’(q)=\tau_{A}(p)L(X)$,$\tau_{B}’(p)=\tau_{B}(q)^{t}R(X)$.

(ii) $If_{A}X_{B}$ an $d_{B}Y_{C}$ are bimodules of finite $type$ , then$R(X\otimes_{B}Y)=R(X)R(Y)$ .

$L(X\otimes_{B}Y)=L(X)L(Y)$ ,$)$(i) From Lemma 1 (iii),$\tau_{A}’(q_{j})=\dim_{A^{A}}^{\tau}X_{q_{j}}$$= \sum_{i}\dim_{A^{A}}^{\tau}(p_{i}Xq_{j})$$= \sum_{i}\tau_{A}(p_{i})L_{ij}(X)$.
Similarly for(ii) is an immediate consequence of Corollary 21 (i).

.

$\tau_{B}’(p_{i})$$\square$Going back to the proof of Proposition 35, for a trace,
Then is a balanced trace if and only if$\tau$of , let$A$$\tau‘=\tau_{A}’x$

and$\tau_{B}=\tau’|_{B}$.

$\exists\alpha>0$$\tau$$(\tau_{B})’|_{A}=\alpha\tau$.

Since $\tau_{B}’(p)=\tau_{B}(q)^{t}R(X)$ and $\tau_{B}(q)=\tau(p)L(X)$ from the above lemma, this conditionis equivalent to$\tau(p)L(X)^{t}R(X)=\alpha\tau(p)$.

Now we can apply the Perron-Frobenius theory because the positve matrix $L(X)^{t}R(X)$ isirreducible by the connectedness assumption on X.
$\square$Remark. If$\tau_{A}$and$\tau_{B}$are balanced through $AX_{B}$ and$\tau_{A}=a\tau_{B}’|_{A},$$\tau_{B}=b\tau_{A}’|_{B}$, then$ab=||L(X)^{t}R(X)||$ .

Proposition 38. Let $AX_{BB}Y_{C}$ be bimodules of fi nite type abe traces on$A,$ $B,$.
andare $bal$ anced through $BYC$ , then ,(i) Ifare balanced through.
are $bal$ anced throughor $BYC$ is connected, thenis connected.
(ii) If either$I1d\tau_{A},$$\tau_{B},$$\tau_{C}$$C$$\tau_{A},$$AX_{B}$$\tau_{B}$$\tau_{B},$$\tau_{C}$$\tau_{A}$$AX\otimes_{B}Y_{C}$$\tau_{C}$$AX_{B}$$AX\otimes_{B}Y_{C}$(i) We may suppose thatwithforbe the set of minimal central projections insuitable $a>0$ and $c>0$ . Let$A,$ $B,$ $C$ , respectively. Then by a repetition of Lemma 36 (i), we have$)$$\tau_{A}=\tau_{B}’|_{A},$$\{p_{i}\},$$\{q_{j}\},$$\tau_{C}=\tau_{B}’|_{C}$$\tau_{B}=a\tau_{A}’|_{B},$$\{r_{k}\}$$\tau_{A}(p)=ac\tau_{A}(p)L(X)L(Y)^{t}R(Y)^{t}R(X)$$\tau_{C}(r)=ac\tau_{C}(r)^{t}R(Y)^{t}R(X)L(X)L(Y)$ .

$\tau_{B}=c\tau_{C}’|_{B}$54
A REPORT ON OCNEANU’S LECTURE

andare balanced (cf. the proof of Proposition 35).
Now Lemma 37 (ii) shows thatis not connected, i.e., minimal central projections in and(ii) Suppose thatin split into two parts. Then the set of minimal central projections in splits into twoparts and hence both $AX_{B}$ and $BYC$ are not connected.
$\tau_{A}$$\tau_{C}$$A$$AX\otimes_{B}Y_{C}$

$B$

$C$

$\square$Now we extend Proposition 20.

Proposition 39. Let $A_{2}B$ fnite sums of finite factors with norm alized traces$wh$ ich arealanced through a bimodule $AX_{B}$ of finite type. $If_{B}Y$ is a left B-module, then$\tau_{A},$$\tau_{B}$$b$.

$\dim_{A^{A}}^{\tau}X\otimes_{B}Y=\dim_{A^{A}}^{\tau}X\dim_{B^{B}}^{\tau}Y$$)$

A careful reading of the proof of Proposition 20 gives the result.

$\square$

The index of a bimodule introduced before is now extended to non-factor case. Let$AX_{B}$ be a connected bimodule. The index $[X]=[X]$ is defined to be$[X]=\dim_{A^{A}}^{\tau}(X)\dim(X)_{B^{B}}^{\tau}$

where$\tau_{A}$and$\tau_{B}$,denote unique normalized balanced traces.

Remark. $[X]=||L(X)^{t}R(X)||$ .
Corollary 40. Let $AX_{B},$ $BYC$ be $bimodu$ les of finite type and suppose that there exist$trac$ es
andare balanced thro $ughX$ an $dY$
of $A,$ $B,$ $Csuch$ th atrespectively. Then$[AX\otimes_{B}Y_{C}]=[X][BY_{C}]$ .
$\tau_{A},$$\tau_{B},$$\tau_{A},$$\tau_{C}$$\tau_{B}$$\tau_{B},$$\tau_{C}$Jones Construction

In this part, as an application of (the existence of) Markov traces, we shall construct socalled Jones projections via Frobenius transform.
traces on $A,$
Lemma 41. Let $AX_{B}$ be a bimodule of finite type with. If
be a basis forfor $somea>0$ , thenan$\tau_{A},$$d\{\mu_{j}\}$$(X)_{B^{B}}^{\tau}$$\tau_{B}’|_{A}=a\tau_{A}$$\sum_{j}A[\mu i, \mu i]=\dim(X)_{B^{B}}^{\tau}1_{A}$.

$\tau_{B}$$B$

respectively,

55
YAMAGAMI SHIGERU
$)$

Let

$\{p_{i}\}$be the set of minimal central projections in$A$. Since$a( \sum_{j}A[\mu i, \mu;])=\sum_{j}A[a\mu_{j}, \mu_{i}]$$= \sum_{j}A[\sum_{k}[\mu_{k}, a\mu_{j}]_{B},$$\mu J$]$= \sum_{j,k}A[\mu_{k}, \mu i[a\mu_{j}, \mu_{k}]_{B}]$$= \sum_{j,k}A[\mu_{k}, \mu i[\mu i, a^{*}\mu_{k}]_{B}]$$= \sum_{k}A[\mu_{k}, a_{l}^{*}\iota_{k}]$$=( \sum_{k}A[\mu_{k}, \mu_{k}])a$$\forall$$a$

E $A$ ,

is a scalar multiple of . To determine the scalar, we first remark that, which implies thatis a basis for$p_{i} \sum_{i^{A}}[\mu_{j}, \mu_{j}]$$\{p_{i}\mu;\}_{j}$,$p_{i}$$(p_{i}X)_{B}$$\tau_{A}(p_{i}\sum_{j}A[\mu;, \mu i])=\tau_{A}(\sum_{j}A\lceil y;\mu_{j}, p;\mu i])=\sum_{j}(p_{i}\mu_{j}|p_{i}\mu_{j})$$=\dim(p;X)_{B^{B}}^{\tau}=\tau_{B}’(p_{i})$and hence$p;( \sum_{j}A[\mu j, \mu j])=\frac{\tau_{B}’(p_{i})}{\tau_{A}(p_{i})}p_{i}=ap_{i}$by the assumptiontion.

$\tau_{B}’|_{A}=a\tau_{A}$,. Summing up these relations over , we get the asser$i$$\square$be the normalized balanced traces.
Now suppose that $AX_{B}$ is connected and let
be a vector corresponding to $1_{X}\in End(AX_{B})$ by Lemma 24, andLetcall it the standard vector.
$\tau_{A},$$\tau_{B}$$\epsilon_{X}\in(X\otimes_{B}X^{*})^{A}$

Lemma 42.
(i)(ii) For a basis$(\epsilon_{X}’|\xi’\otimes_{B}\xi^{*})=(\xi|\xi’)$$\{\mu j\}$

of

for

$(X)_{B^{B}}^{\tau}$$\xi,$$\xi’\in X_{0}$.

,$\epsilon_{X}=\sum_{j}\mu j^{\otimes}B\mu_{j}^{*}$(iii)$\forall\varphi\in(X\otimes_{B}Y)_{0}$,$A[\varphi, \epsilon_{X}]=[\epsilon x, \varphi]_{A}$

(iv)

.

$A[\xi’\otimes_{B}\xi^{*}, \epsilon x]=A[\xi’, \xi]$, for $\xi_{)}\xi’EX_{0}$ .

.

56
A REPORT ON OCNEANU’S LECTURE(v)$A[\epsilon_{X}, \epsilon_{X}]=\dim X_{B}1_{A}$.

(i) and (ii) follow from the correspondence
given in Lemma 24. (iii)
is a consequence of the A-invariance of . (iv) results from (i) and the definition ofoperator-valued inner product. (v) is a combination of (ii) and Lemma 41.
$)$$\Leftrightarrow$$\epsilon_{X}$$1_{X}$$\epsilon_{X}$$\square$Remark. From (ii) and the proof of Lemma 23 (ii),$\epsilon_{X}\in(X\otimes_{B}X^{*})_{0}$Again, by Lemma 24,. Its concrete form is given by.

defines an intertwiner$\epsilon_{X}\otimes_{A}\epsilon_{X}\in((X\otimes_{B}X^{*})\otimes_{A}(X\otimes_{B}X^{*}))^{A}$$e_{X}^{0}\in\cdot End(AX\otimes_{B}X_{A}^{*})$$e_{X}^{0}(\xi’\otimes_{B}\xi^{*})=\epsilon_{X}[\epsilon_{X}, \xi’\otimes_{B}\xi^{*}]_{A}=\epsilon_{XA}[\xi’, \xi]$,$\xi,$$\xi’\in X_{0}$.

is associated withSince the adjoint ofLemma 24),is self-adjoint. Now the calculation$(\epsilon_{X}\otimes_{A}\epsilon_{X})^{*}=\epsilon_{X}^{*}\otimes_{A}\epsilon_{X}^{*}=\epsilon_{X}\otimes_{A}\epsilon_{X}$$e_{X}^{0}$$e_{X}^{0}$$e_{X}^{0}e_{X}^{0}(\xi’\otimes_{B}\xi^{*})=e_{X}^{0}(\epsilon_{XA}[\xi’, \xi])=e_{X}^{0}(\epsilon_{X})_{A}[\xi’, \xi]$$= \sum_{j}e_{X}^{0}(\mu_{j}\otimes_{B}\mu_{j}^{*})_{A}[\xi’, \xi]$$= \epsilon x\sum_{j}A[\mu_{i}, \mu_{i}]_{A}[\xi’, \xi]$(by Lemma 41)$=\dim(X)_{B^{B}}^{\tau}\epsilon_{XA}[\xi’, \xi]$$=\dim(X)_{B^{B}}^{\tau}e_{X}^{0}(\xi’\otimes_{B}\xi^{*})$,

shows that$e_{X}= \frac{1}{\dim(X)_{B^{B}}^{\tau}}e_{X}^{0}$is a projection in End$(AX\otimes_{B}X_{A}^{*})$, which is called the Jones projection.

Lemma 43. In $X\otimes_{B}X^{*}\otimes_{A}X$ , we a $ve$ the $rel$ a tion$h$$(1\otimes e_{X}\cdot)(e_{X}\otimes 1)(1\otimes e_{X}\cdot)=[X]^{-1}(1\otimes e_{X}\cdot)$

.

(see

57
YAMAGAMI SHIGERU$)$

Take bases$\{\lambda_{i}\}$for $AX$ and$\{\mu_{j}\}$for$X_{B}$. For$\xi_{1},$$\xi_{2},$$\xi_{3}\in X_{0}$,$(\dim X_{B})(\dim_{A}X)^{2}(1\otimes e_{X}*)(e_{X}\otimes 1)(1\otimes e_{X}\cdot)(\xi_{1}\otimes_{B}\xi_{2}^{*}\otimes_{A}\xi_{3})$$=(1\otimes e_{X}^{0}.)(e_{X}^{0}\otimes 1)(1\otimes e_{X}^{0}.)(\xi_{1}\otimes_{B}\xi_{2}^{*}\otimes_{A}\xi_{3})$$= \sum_{i}(1\otimes e_{X}^{0}.)(e_{X}^{0}\otimes 1)(\xi_{1}\otimes\lambda_{i}^{*}\otimes\lambda_{i}[\xi_{2}, \xi_{3}]_{B})$$= \sum_{i,j}(1\otimes e_{X}^{0}.)(\mu_{j}\otimes\mu_{jA}^{*}[\xi_{1}, \lambda_{i}]\otimes\lambda_{i}[\xi_{2}, \xi_{3}]_{B})$$\lambda_{i}[\xi_{2}, \xi_{3}]_{B}]_{B}$$= \sum_{i,j}\mu_{i}\otimes\epsilon_{X}\cdot[A[\lambda_{i}, \xi_{1}]\mu_{j},$$= \sum_{i,j}\mu_{j}\otimes\epsilon_{X}\cdot[\mu_{j,A}[\xi_{1}, \lambda_{i}]\lambda_{i}[\xi_{2}, \xi_{3}]_{B}]_{B}$$= \sum_{j}\mu_{j}\otimes\epsilon_{X^{*}}[\mu_{j}, \xi_{1}[\xi_{2}, \xi_{3}]_{B}]_{B}$$= \sum_{j}\mu_{j}[\mu_{j}, \xi_{1}[\xi_{2}, \xi_{3}]_{B}]_{B}\otimes\epsilon_{X}$$=\xi_{1}[\xi_{2}, \xi_{3}]_{B}\otimes\epsilon_{X}$.

.

$=(1\otimes e_{X}^{0}.)(\xi_{1}\otimes\xi_{2}^{*}\otimes\xi_{3})$$=\dim_{A}X((1\otimes e_{X})(\xi_{1}\otimes\xi_{2}^{*}\otimes\xi_{3})$.

$\square$Now we define towers of algebras.

Definition 44. Define two sequences of algebras by$A_{1}=End(X_{B})\subset A_{2}=End(X\otimes_{B}X_{A}^{*})\subset A_{3}=End(X\otimes_{B}X^{*}\otimes_{A}X_{B})\subset\ldots$,$B_{1}=End(AX)^{opp}\subset B_{2}=End(BX^{*}\otimes_{A}X)^{opp}\subset B_{3}=End(AX\otimes_{B}X^{*}\otimes_{A}X)^{opp}\subset\ldots$,

which are called the towers of algebras associated with a bimodule $AX_{B}$ .
and, we introdu ce normalized traces
andas the normaliz ation of$al$canonic
traces associ ated with defin $ing$ modules of $each$ algebras.

On.

$A_{i}$$B_{j}$$\tau_{A;}$$\tau_{B_{j}}$In(a tensor product of finite number of bimodules),mark oneand split the tensor product into two parts, and then consider the algebras
of intertwiners in the left and right sides of the splitting: For example, if we split at a point
in which acts, then the left and right algebras in the tensor product are defined by$V\equiv\cdots\otimes_{A}X\otimes_{B}X^{*}\otimes_{A}X\otimes_{B}\ldots$$of\otimes s$

$A$

End

$(\cdots\otimes_{B}X_{A}^{*})$,

End $(AX\otimes_{B}\ldots)$ .

They are one of algebras in the towers of algebras and, by Lemma 19, commutants ofeach other. In particular, the original tensor productadmits a structure of bimodule.
$V$58
A REPORT ON OCNEANU’S LECTURE

Similarly, if we mark two points in the tensoring operations and divide into three parts,
then the left and the right parts give rise to algebras onwhich commute each other. In
this way, we have many kinds of bimodule structures in . Say, $V=A_{i}V_{B_{J}}$ in this fashion.
Then, by Proposition 38 (i),are balanced through .
, and
$V$

$V$

$V$

$\tau_{A}$$V$$\tau_{B_{J}}$Remark. The sets of left-(or $right-$ ) $bounded$ elements are the same for any bimodule
structures described above (cf. Lemma 23 $(i)$ ).
act on $X\otimes_{B}X^{*}from$ left and right and $regard$ it anbimodule.
.
, for
.
$ds$
to an isomet $ric$ isomorphism $L^{2}(A_{1})arrow X\otimes_{B}X^{*}$ , and
exten(here the conjuga $tion*stands$ for
gives a stan ard represen tation of
the J-involution in the standard representation).

Lemma 45. Let
(i)
(ii)
(iii)

$A_{1}- A_{1}$$A_{1}$$\epsilon_{X}\in(X\otimes_{B}X^{*})^{A_{1}}$$(a_{1}\epsilon x|\epsilon_{X})=\dim X_{B}\tau_{A_{1}}(a_{1})$$a_{1}\in A_{1}$$A_{1}\ni a_{1}\vdasharrow a_{1}\epsilon_{X}$$d$$(X\otimes_{B}X^{*}, \epsilon_{X})$$)$(i) For $a_{1}\in A_{1}=End(X_{B})$ , and$A_{1}$$\xi,$$\xi’\in X_{0}$$((a_{1}\otimes_{B}1)\epsilon_{X}|\xi’\otimes_{B}\xi^{*})=(a_{1}\xi|\xi’)$,(by Lemma 42 $(i)$ )$=(\epsilon_{X}|\xi’\otimes_{B}\xi^{*}a_{1}^{*})$$=$((by Lemma 42 $(i)$ )$\epsilon_{X}|(\xi’\otimes_{B}\xi^{*})(1\otimes_{B}$a$1)^{*}$$=(\epsilon_{X}(1\otimes_{B}a_{1})|\xi’\otimes_{B}\xi^{*})$(ii) For a basis$\{\mu i\}$of

$X_{B}$).

,$(a_{1} \epsilon_{X}|\epsilon_{X})=\sum_{j}(a_{1}\mu;\otimes_{B}\mu_{j}^{*}|\epsilon_{X})$(by Lemma 42 $(i)$ )$= \sum_{j}(a_{1}\mu_{j}|\mu i)$$=\tau_{X_{B}}’(a_{1})$(by Lemma 42 (ii))(by Corollary 13 $(i)$ )$=\tau_{X_{B}}’(1)\tau_{A_{1}}(a_{1})$$=\dim X_{B}\tau_{A_{1}}(a_{1})$.

is cyclic for(iii) By (ii) just proved, we need to show the following two facts;
and
.
for
as $L^{2}(B)_{B}\oplus\cdots\oplus L^{2}(B)_{B}\oplus pL^{2}(B)_{B}$ and
For the proof of cyclicity, we express
use the matrix form of End $(X_{B})$ . Then it is easy to see thatshould be written in terms of the basis associated with the above expression) is a dense
linear subspace of$\epsilon_{X}$$(a_{1}\epsilon_{X})^{*}=\epsilon_{X}a_{1}^{*}$$A_{1}$$a_{1}\in A_{1}$$X_{B}$$\{(a_{1}\otimes_{B}1)\epsilon_{X} ; a_{1}\in A_{1}\}(\epsilon_{X}$$X\otimes_{B}X^{*}\cong(L^{2}(B)_{B}\oplus\cdots\oplus L^{2}(B)_{B}\oplus pL^{2}(B)_{B})\otimes_{B}(L^{2}(B)_{B}\oplus\cdots\oplus L^{2}(B)_{B}\oplus pL^{2}(B)_{B})$.

59
YAMAGAMI SHIGERU

The property related with the $involution*follows$ from$((a_{1}\epsilon_{X})^{*}|\xi’\otimes_{B}\xi^{*})=(\xi\otimes_{B}\xi’*|a_{1}\epsilon_{X})$$=(a_{1}^{*}\xi\otimes_{B}\xi’*|\epsilon_{X})$$=(a_{1}^{*}\xi|\xi’)$$=(\epsilon_{X}|\xi’\otimes_{B}\xi^{*}a_{1})$$=(\epsilon_{X}a_{1}^{*}|\xi’\otimes_{B}\xi^{*})$.

口Corollary 46.

$A_{1}[\varphi, \epsilon_{Z}]\epsilon_{X}=(\dim X_{B})\varphi$

This follows from the fact that
Example 10). 口
$)$

for

$\varphi\in(X\otimes_{B}X^{*})_{0}$$(\dim X_{B})^{-1/2}\epsilon_{X}$.

corresponds toin$1_{A_{1}}$Proposition 47. The representation of ona an dard representat ion $wi$ th *the an onical in volution.

$(X\otimes_{B}X^{*})\otimes_{A}\cdots\otimes_{A}(X\otimes_{B}X^{*})$$A_{i}$

$is$

$st$

$)$

(iii).

$L^{2}(A_{1})$(cf.

(i-times)$c$Instead of $AXB$ , use $A:(X\otimes_{B}X^{*}\otimes_{A}X\otimes_{B}\cdots\otimes (X^{*} or X))_{(A}$ or$B$)

in Lemma 45$\square$Definition 48. Define a sequence of projections$e_{1},$ $e_{2},$in$\ldots$$X\otimes_{B}X^{*}\otimes_{A}X\otimes_{B}\ldots$(finitely many tensor product) by$e_{1}=e_{X}\otimes_{A}1_{X}\otimes_{B}1_{X}*\otimes_{A}$.

...

$e_{2}=1_{X}\otimes_{B}e_{X}\cdot\otimes_{B}1_{X}\cdot\otimes_{A}$$e_{3}=1_{X}\otimes_{B}1_{X}$...

$\otimes_{A}e_{X}\otimes_{A}\ldots$

Lemma 49. We have$(1_{X}\otimes_{B}End(X_{B})^{opp})\cap\{e_{X}\}’=1_{X}\otimes_{B}$$A$

in End $(AX\otimes_{B}X^{*})$ .

Let $S\in End(X_{B})$ and suppose that the right action of
From
$)$

$1_{X}\otimes_{B}S$

commutes with$(e_{X}(\xi’\otimes_{B}\xi^{*}))(1\otimes_{B}S)=(_{A}[\xi’, \xi]\epsilon_{X})(1\otimes_{B}S)=A[\xi’, \xi]\epsilon_{X}(1\otimes_{B}S)$and$e_{X}((\xi’\otimes_{B}\xi^{*})(1\otimes_{B}S))=e_{X}(\xi’\otimes_{B}(S^{*}\xi)^{*})=A[\xi’, S^{*}\xi]\epsilon_{X}$we have$A[\xi’, \xi]\epsilon_{X}(1\otimes_{B}S)=A[\xi’, S^{*}\xi]\epsilon x$.

,$e_{X}$.

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A REPORT ON OCNEANU’S LECTURE

in this relation (note that
Taking, we haveand then applying$A[\epsilon_{X)}\cdot]$$\epsilon_{X}(1\otimes_{B}S)\in(X\otimes_{B}X^{*})_{0}$since$\epsilon_{X}E(X\otimes_{B}X^{*})_{0}$)$\tau_{A}(\cdot)$$(A[\epsilon_{X}, \epsilon x(1\otimes_{B}S)]\xi|\xi’)=(A[\epsilon_{X}, \epsilon_{X}]S^{*}\xi|\xi’)$

Since$A[\epsilon_{X}, \epsilon_{X}]=\dim X_{B}1_{A}$(Lemma 42 $(v)$ ), this implies$S(\xi’)=(\dim X_{B})_{A}^{-1}[\epsilon_{X}(1\otimes_{B}S), \epsilon_{X}]\xi’$and hence$S$.

is expressed as a left action of an element in A.

,

口Lemma 50. Let $AX_{B,B}Y_{C}$ be bimodules with $bal$ anced traces an set $Az_{c}=AX\otimes_{B}Y_{C}$ .
If $A=End(X_{B})$ , then$d$$e_{Z}=1_{X}\otimes_{B}e_{Y}\otimes_{B}1_{X^{*}}$

$)$

in

$X\otimes_{B}Y\otimes_{C}Y^{*}\otimes_{B}X^{*}$.

Since $A=B’$ , if we use Lemma 19 in $Z\otimes_{C}Z^{*}=.X\otimes_{B}Y\otimes cY^{*}\otimes_{B}X^{*}$ , we have
End $(AZ\otimes_{C}Z^{*})=1_{X}\otimes xEnd(BY\otimes_{C}Y^{*}\otimes_{B}X^{*})$$End(Z\otimes_{C}Z_{A}^{*})=End(X\otimes_{B}Y\otimes_{C}Y_{B}^{*})\otimes_{B}1_{X}\cdot$,

and then
End $(AZ\otimes_{C}Z_{A}^{*})=1_{X}\otimes_{B}End(BY\otimes_{C}Y_{B}^{*})\otimes_{B}1_{X}\cdot$ .

Since $e_{Z}\in End(AZ\otimes_{C}Z_{A}^{*})$ , we can set$e_{Z}^{0}\cdot=1_{X}\otimes_{B}T\otimes_{B}1_{X}\cdot$with $T\in End(BY\otimes_{C}Y_{B^{*}})$ . Then, for a basis$\{\lambda_{i}\}$,of $AX$ and$\eta_{1},$$\eta_{2}\in Y_{0}$,$\sum_{i,j}(\lambda_{i}\otimes_{B}\eta_{1}\otimes c\eta_{1}^{*}\otimes_{B}\lambda_{j}^{*}|e_{Z}^{0}(\lambda_{i}\otimes_{B}\eta_{2}\otimes c\eta_{2}^{*}\otimes_{B}\lambda_{j}^{*}))$$= \sum_{i,j}(\lambda_{i}\otimes_{B}\eta_{1}\otimes c\eta_{1}^{*}\otimes_{B}\lambda_{j}^{*}|\lambda_{i}\otimes_{B}T(\eta_{2}\otimes c\eta_{2}^{*})\otimes_{B}\lambda_{j}^{*})$.

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YAMAGAMI SHIGERU

Since$e_{Z}^{0}(\zeta’\otimes_{C}\zeta^{*})=\epsilon_{ZA}[\zeta’, \zeta]$, the left hand side is calculated as$\sum_{i,j}(\lambda;\otimes_{B}\eta_{1}\otimes c\eta_{1}^{*}\otimes_{B}\lambda_{j}^{*}|\epsilon_{ZA}[\lambda_{i}\otimes_{B}\eta_{2}, \lambda_{j}\otimes_{B}\eta_{2}])$$= \sum_{i,j}(\lambda_{i}\otimes_{B}\eta_{1}|_{A}[\lambda_{i}\otimes_{B}\eta_{2}, \lambda_{j}\otimes_{B}\eta_{2}]\lambda_{j}\otimes_{B}\eta_{1})$$= \sum_{i,j}(\lambda_{i}\otimes_{B}\eta_{1}|_{A}[\lambda_{i}, \lambda_{jB}[\eta_{2}, \eta_{2}]]\lambda_{j}\otimes_{B}\eta_{1})$$= \sum_{j}((\sum_{i}A[\lambda_{jB}[\eta_{2}, \eta_{2}], \lambda_{i}]\lambda_{\{})\otimes_{B}\eta_{1}|\lambda_{J}\otimes_{B}\eta_{1})$$= \sum_{j}(\lambda_{jB}[\eta_{2}, \eta_{2}]\otimes_{B}\eta_{1}|\lambda_{j}\otimes_{B}\eta_{1})$$= \sum_{j}(\lambda_{jB}[\eta_{2}, \eta_{2}]_{B}[\eta_{1}, \eta_{1}]|\lambda_{j})$$= \sum_{j}\tau_{B}(B[\eta_{2}, \eta_{2}]_{B}[\eta_{1}, \eta_{1}][\lambda_{i}, \lambda_{i}]_{B})$(Lemma 41)$=\dim_{A}X\tau_{B}(B[\eta_{2}, \eta_{2}]_{B}[\eta_{1}, \eta_{1}])$$=\dim_{A}X(\eta_{1}\otimes_{C}\eta_{1}^{*}|e_{Y}^{0}(\eta_{2}\otimes_{C}\eta_{2}^{*}))$.

On the other hand, the right hand side is calculated as$\sum_{i,j}(\lambda_{i}\otimes_{B}\eta_{1}\otimes c\eta_{1}^{*}\otimes_{B}\lambda_{j}^{*}|\lambda_{i}\otimes_{B}T(\eta_{2}\otimes_{C}\eta_{2}^{*})\otimes_{B}\lambda_{j}^{*})$$= \sum_{i,j}([\lambda_{i}, \lambda_{i}]_{B}\eta_{1}\otimes_{C}\eta_{1}^{*}|T(\eta_{2}\otimes_{C}\eta_{2}^{*})[\lambda_{j}, \lambda_{j}]_{B})$.

$=(\dim_{A}X)^{2}(\eta_{1}\otimes c\eta_{1}^{*}|T(\eta_{2}\otimes c\eta_{2}^{*}))$

Comparing the results (note that $\dim Z_{C}=\dim X_{B}\dim Y_{C}$ , and $\dim_{A}X\dim X_{B}=1$ since$A=B’)$ , we have$e_{Y}=(\dim Z_{C})^{-1}T$$i.e.$,,$ez=1_{X}\otimes_{B}e_{Y}\otimes_{B}1_{X}$..

口Proposition 51.
satisfies the following $rel$ a tion :(i)$\{e_{i}\}_{i\geq 1}$$s$$e_{i\pm 1}e_{i}e_{i\pm 1}=[X]^{-1}e_{i\pm 1}$$e_{i}e_{j}=e_{j}e_{i}$,,$|i-j|\geq 2$ .

62
A REPORT ON OCNEANU’S LECTURE(ii) $A_{i}=\{A_{1}, e_{1}, \ldots, e_{i-1}\}(i\geq 2)$ .
(i) is a consequence of Lemma 43.
(ii) We shall prove by the induction on . When $i=2$ , this is a consequence of Lemma 49.
Suppose that this is valid for some . To be specific, consider the case $i=2k$ . Applying
Lemma 49 to
$)$

$i$

$i$

$Z=A_{2k-1}( \frac{2k-1time.s}{X\otimes_{B}X^{*}\otimes_{A}\cdot\cdot\otimes_{A}X})\otimes_{B}X_{A}^{*}$,We have$\{A_{i}\otimes_{B}1_{Z}\cdot, e_{Z}\}=A_{i+1}\otimes_{A}1$

Since

by Lemma 50, the induction proceeds.

$e_{Z}=e_{i}\otimes_{A}1$.
口

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Hilbert, J. Operator Theory 9 (1983),
7. J.-L. Sauvageot, Sur le Prodiut Ten’ oriel Relativ d’Ezpacez
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$h$$d\epsilon$