Section3The Kadison theorem and some applications

We begin by using the tools developed in Section §2 to identify the integer $a-b$ in Kadison's theorem, that is, to prove Theorem 1.3.

Proof

As a first consequence of Kadison's theorem and of the work in Section §2, we observe that if the diagonal of a projection $\projone{}$ clusters sufficiently fast around $0$ and $1$ (that is, if $a+b\lt \infty\text{,}$ or, equivalently, if $\projone{}-\projtwo{} \in \mathcal L^2$ ), then one can “read” from the diagonal the essential codimension $[\projone{}:\projtwo{}]\text{.}$ But what if $a+b= \infty\text{?}$

If $a=\infty$ and $b \lt \infty\text{,}$ from $\projtwo{}^\perp(\projone{}-\projtwo{})\projtwo{}^\perp \in \mathcal L^1$ we can deduce that $\projone{} \wedge \projtwo{}^\perp$ is finite and $s\in \mathcal L^2\text{,}$ and hence from $\projtwo{}(\projone{}-\projtwo{})\projtwo{}\not \in \mathcal L^1$ it follows that $\projone{}^\perp \wedge \projtwo{}$ is infinite. Similarly, if $a \lt \infty$ and $b=\infty$ then $\projone{} \wedge \projtwo{}^\perp$ is infinite. In either case $(\projone{},\projtwo{})$ is not a Fredholm pair and in particular, $\projone{}-\projtwo{}\not \in \K\text{.}$

Less trivial is the case when $a=b=\infty$ and $\projone{}-\projtwo{}\in \K \setminus \mathcal L^2\text{,}$ as we see from the following proposition. We first need to introduce some notation. Next, given two sequences $\xi$ and $\eta$ of non-negative numbers converging to $0\text{,}$ with $\xi^*$ and $\eta^*$ their monotone non-increasing rearrangements, we say that $\xi$ is majorized by $\eta$ ( $\xi\prec \eta$ ) if $\sum_{j=1}^n\xi^*_j\le \sum_{j=1}^n\eta^*_j$ for all $n\text{.}$

Proposition3.1

Suppose $\projone{},\projtwo{}$ are projections with $\projone{}-\projtwo{} \in \K \setminus \mathcal{L}^2\text{.}$ Then there exists a projection $\projone{}'$ such that $\projone{}'-\projtwo{} \in \K\text{,}$ $[\projone{}':\projtwo{}] \not= [\projone{}:\projtwo{}]$ and there is an orthonormal basis $\{e_n\}$ that diagonalizes $\projtwo{}$ such that $E(\projone{})=E(\projone{}')\text{.}$

Proof

By Proposition 2.7 (i), $\projone{}-\projtwo{} \in \K$ implies that $\projone{} \wedge \projtwo{}^\perp$ and $\projone{}^\perp \wedge \projtwo{}$ are both finite. Thus $\projone{}_0-\projtwo{}_0 = (\projone{}-\projtwo{}) + (\projone{} \wedge \projtwo{}^\perp - \projone{}^\perp \wedge \projtwo{}) \in \K \setminus \mathcal{L}^2\text{.}$ It suffices to prove the proposition for projections in generic position because then we simply set $\projone{}' := \projone{} \wedge \projtwo{} + \projone{} \wedge \projtwo{}^\perp + \projone{}_0'$ for the general case. So to simplify notation, assume henceforth that $\projone{},\projtwo{}$ are in generic position and have the form as in ((2.5)). In particular, $\projtwo{}\sim \projtwo{}^\perp\sim 1$ and by Lemma 2.5, $[\projone{}:\projtwo{}]=0\text{.}$

Next, choose a rank one projection $r'\le \projtwo{}^{\perp}$ and let $r:=\projtwo{}^{\perp} - r'\text{.}$ After identifying $B(\h) \simeq B(\projtwo{}\h \oplus r\h \oplus \mathbb{C})$ with $M_2(\projtwo{}B(\h)\projtwo{}) \oplus \mathbb{C}$ via a partial isometry taking $\projtwo{}\h \to r\h\text{,}$ consider the projection

\begin{equation*} \tilde{\projone{}} := \begin{pmatrix} c^2 & cs & 0 \\ cs & s^2 & 0 \\ 0 & 0 & 1 \\ \end{pmatrix}. \end{equation*}

Note that $\tilde{\projone{}} - (0 \oplus 0 \oplus 1)$ and $\projtwo{}$ are in generic position relative to their join, and hence their essential codimension is zero. This implies that $[\tilde{\projone{}},\projtwo{}] = 1 \not= 0 = [\projone{},\projtwo{}]\text{.}$

Now, choose an orthonormal basis $\{e_n\}$ that diagonalizes $\projtwo{}\text{,}$ $E$ its corresponding conditional expectation, and let $\xi$ be the diagonal sequence of $s^2\text{.}$ Then under natural notations we have

\begin{equation*} E(\projone{})= E_\projtwo{}(\projone{})\oplus E_{\projtwo{}^\perp}(\projone{}) \end{equation*}

and furthermore, $E_\projtwo{}(\projone{})=E_\projtwo{}(\tilde{\projone{}})= E_\projtwo{}(c^2)$ and $E_{\projtwo{}^\perp}(\projone{})=E_{\projtwo{}}(s^2)= \diag \xi\text{.}$

Since $\projone{}-\projtwo{} \in \K \setminus \mathcal{L}^2\text{,}$ by Proposition 2.7 we have that $s^2 \in \K \setminus \mathcal{L}^1\text{,}$ that is $\xi\to 0$ but $\xi\not \in \ell^1\text{.}$ By the Schur–Horn theorem for compact operators [20] $\xi$ is majorized by the eigenvalues sequence $\lambda(s^2)$ of the operator $s^2$ and hence,

\begin{equation*} \xi\prec \lambda(s^2)\prec \lambda \begin{pmatrix} s^2 & 0 \\ 0 & 1 \\ \end{pmatrix}. \end{equation*}

Now $ \begin{pmatrix} s^2 & 0 \\ 0 & 1 \\ \end{pmatrix} $ is a positive compact operator with zero kernel belonging to $\B(\projtwo{}^\perp \h)\text{.}$ Hence by [20] there is a unitary $u\in \B(\projtwo{}^\perp \h)$ such that

\begin{equation*} \diag \xi= E_{\projtwo{}^\perp\h}\big (u \begin{pmatrix} s^2 & 0 \\ 0 & 1 \\ \end{pmatrix} u^*\big). \end{equation*}

Let $u':= 1\mid _{\projtwo{}\h}\oplus u$ and $\projone{}':= u'(\tilde \projone{})u'^*\text{.}$ Then

\begin{equation*} E(\projone{}')= E_\projtwo{}(c^2)\oplus \diag \xi = E(\projone{}). \end{equation*}

Since $u'\projtwo{}u'^{*} = \projtwo{}$ we have

\begin{equation*} [\projone{}':\projtwo{}] = [u\tilde{\projone{}}u^{*}:u\projtwo{}u^{*}] = [\tilde{\projone{}}:\projtwo{}] \not= [\projone{},\projtwo{}]. \qedhere \end{equation*}

As a second application of Theorem 1.3 and of the techniques used to prove it, we will consider a recent work by Bownik and Jasper [6]. Based on Kadison's characterization of diagonals of projections, Bownik and Jasper characterized the diagonals of selfadjoint operators with finite spectrum and in a key part of their analysis they too encountered an index obstruction similar to the one in Theorem 1.1 (ii). Following their notations, if $z\in \B(\h)$ is a selfadjoint operator with finite spectrum we let $\sigma(z)= \{a_j\}_{j=-m}^{n+r}$ and $\projone{}_j= \chi_{\{a_j\}}(z)$ be the spectral projection corresponding to the eigenvalue $a_j\text{,}$ so that

$\begin{equation*} z= \sum _{j=-m}^{n+r}a_j \projone{}_j. \end{equation*}$

For ease of notations perform if necessary a transformation so to have

$\begin{equation*} \tr(\projone{}_j)\lt \infty \text{ for }j\lt 0 \text { and } j>n+1, \quad a_0=0, \text{ and } a_{n+1}=1. \end{equation*}$

Let $\{e_n\}$ be an orthonormal basis, $\{d_n\}$ be the diagonal of $z$ with respect to that basis and let as in Theorem 1.1,

$\begin{equation*} a = \sum_{d_n \le 1/2 } d_n \qquad\text{and}\qquad b = \sum_{d_n > 1/2 } (1-d_n), \end{equation*}$

Then their Theorem 4.1, which is a key component of the necessity part of their characterization, states that

Theorem3.2[6]

If $a+b\lt \infty$ then

$\tr(\projone{}_j) \lt \infty$ for $0\lt j\lt n+1\text{;}$
$a-b - \sum_{j\ne n+1} a_j\tr(\projone{}_j) \in \mathbb Z.$

Here of course we use the convention that $0\cdot \infty=0$ and so $a_0\tr(\projone{}_0)=0$ whether $\tr(\projone{}_0)$ is finite or not.

We will present an independent proof of this result and at the same time identify the integer in (ii) proving that if we set $\projtwo{}$ as in Theorem 1.1 to be the projection on $\overline{\spans}\{e_j\mid\ d_j> 1/2 \}\text{,}$ then

$\begin{equation} a-b - \sum_{j\ne n+1} a_j\tr(\projone{}_j)= [\projone{}_{n+1}:\projtwo{}]. \tag{3.1} \end{equation}$

First we need an extension to positive elements of the equivalence of (i) and (ii) in Proposition 2.8.

Lemma3.3

Let $\mathcal J$ be a proper ideal, $x\in \B(\h)_+$ a positive contraction, and $\projtwo{}\in \B(\h)$ a projection.

If $\projtwo{}-\projtwo{}x\projtwo{}\in \mathcal J$ and $\projtwo{}^\perp x \projtwo{}^\perp\in \mathcal J\text{,}$ then $x-\projtwo{}\in \mathcal J^{ 1/2 }$ and $x\chi_{[0, \eps]}(x)\in \mathcal J$ for every $0\lt \eps \lt 1\text{.}$
Assume that $x$ is a projection or that $\mathcal J$ is idempotent (i.e., $\mathcal J= \mathcal J^2$ ). If $x-\projtwo{}\in \mathcal J^{ 1/2 }$ and $x\chi_{[0, \eps]}(x)\in \mathcal J$ for some $0\lt \eps \lt 1\text{,}$ then $\projtwo{}-\projtwo{}x\projtwo{}\in \mathcal J$ and $\projtwo{}^\perp x \projtwo{}^\perp\in \mathcal J\text{.}$

Proof

(i). Since $\projtwo{}^\perp x \projtwo{} x \projtwo{}^\perp\le \projtwo{}^\perp x^2 \projtwo{}^\perp\le \projtwo{}^\perp x \projtwo{}^\perp\in \mathcal J\text{,}$ it follows that $\projtwo{} x \projtwo{}^\perp$ and $\projtwo{}^\perp x \projtwo{}$ belong to $\mathcal J^{ 1/2 }\text{.}$ But then

\begin{equation*} x-\projtwo{} = (\projtwo{}x\projtwo{}- \projtwo{}) + \projtwo{}^\perp x \projtwo{}^\perp + \projtwo{} x \projtwo{}^\perp+ \projtwo{}^\perp x \projtwo{}\in \mathcal J^{ 1/2 }. \end{equation*}

Let $\eps>0$ and let $x_\eps:= x\chi_{[0, \eps]}(x)\text{.}$ Then $0\le \projtwo{}^\perp x_\eps \projtwo{}^\perp\le \projtwo{}^\perp x \projtwo{}^\perp\in \mathcal J\text{,}$ whence $\projtwo{}^\perp x_\eps \projtwo{}^\perp\in \mathcal J\text{.}$ Furthermore,

\begin{equation*} 1-x\ge (1-x) \chi_{[0, \eps]}(x)\ge (1-\eps) \chi_{[0, \eps]}(x)\ge (1-\eps)x_\eps \end{equation*}

and hence $\projtwo{}-\projtwo{}x\projtwo{}\ge (1-\eps)\projtwo{}x_\eps \projtwo{}.$ Thus $\projtwo{}x_\eps \projtwo{}\in \mathcal J$ and since

\begin{equation*} 0\le x_\eps \le 2\big(\projtwo{}x_\eps \projtwo{} + \projtwo{}^\perp x_\eps \projtwo{}^\perp)\in \mathcal J \end{equation*}

it follows that $x_\eps\in \mathcal J\text{.}$

(ii). The case when $x$ is a projection is given by ((2.12)). Assume then that $\mathcal J$ is idempotent. Then $x-\projtwo{}\in \mathcal J^{ 1/2 }= \mathcal J$ implies $\projtwo{}-\projtwo{}x\projtwo{}=-\projtwo{}(x-\projtwo{})\projtwo{}\in \mathcal J\text{.}$ Furthermore, $(x-\projtwo{})^2= x^2-x\projtwo{}-\projtwo{}x+\projtwo{}$ hence $\projtwo{}^\perp x^2 \projtwo{}^\perp= \projtwo{}^\perp (x-\projtwo{})^2 \projtwo{}^\perp\in \mathcal J\text{.}$ Then

\begin{equation*} \projtwo{}^\perp (x-x_\eps) \projtwo{}^\perp\le \frac{1}{\eps}\projtwo{}^\perp (x-x_\eps)^2 \projtwo{}^\perp\le \frac{1}{\eps}\projtwo{}^\perp x^2 \projtwo{}^\perp \in \mathcal J \end{equation*}

and hence

\begin{equation*} \projtwo{}^\perp x \projtwo{}^\perp = \projtwo{}^\perp (x-x_\eps) \projtwo{}^\perp + \projtwo{}^\perp x_\eps \projtwo{}^\perp \in \mathcal J. \qedhere \end{equation*}

Notice that if $\mathcal J$ is not idempotent and $k\in \mathcal J^{ 1/2 }_+\setminus \mathcal J$ is a positive contraction, then $x:=1-k$ and $\projtwo{}:=1$ satisfy both hypotheses of Lemma 3.3 (ii) but $k=\projtwo{}-\projtwo{}x\projtwo{}\not \in \mathcal J\text{.}$

Now we can proceed with the proof of Theorem 3.2 and ((3.1)).

Proof

Set $x = \sum_{j=1}^{n+1} a_j \projone{}_j\text{.}$ Then $0 \le x \le 1$ and

\begin{equation*} z-x = \sum_{j=-m}^{-1}a_j \projone{}_j+\sum_{j=n+2}^{n+r}a_j \projone{}_j \quad\text{has finite rank.} \end{equation*}

As in ((1.2)) we have that

\begin{equation*} a= \tr(\projtwo{}^{\perp} x \projtwo{}^{\perp}) + \tr(\projtwo{}^\perp (z-x) \projtwo{}^\perp) = \tr(\projtwo{}^\perp z \projtwo{}^\perp) \end{equation*}

and

\begin{equation*} b= \tr(\projtwo{}-\projtwo{}x\projtwo{}) - \tr(\projtwo{}(z-x)\projtwo{}) = \tr( \projtwo{}- \projtwo{}z\projtwo{}), \end{equation*}

hence $\projtwo{} (z-\projtwo{}) \projtwo{} \in \mathcal L^1$ , $\projtwo{}^\perp(z-\projtwo{})\projtwo{}^\perp \in \mathcal L^1\text{,}$ and

\begin{equation} a-b = \tr\big( \projtwo{}(z-\projtwo{})\projtwo{}+ \projtwo{}^\perp (z-\projtwo{})\projtwo{}^\perp \big). \tag{3.2} \end{equation}

We also have $\projtwo{}(x-\projtwo{})\projtwo{}\in \mathcal L^1$ and $\projtwo{}^\perp (x-\projtwo{}) \projtwo{}^\perp \in \mathcal L^1\text{,}$ hence by Lemma 3.3, it follows that $x-\projtwo{}\in \mathcal L^2$ and

\begin{equation*} \sum_{j=1}^n a_j \projone{}_j= x\chi_{[0, a_n]}(x) \in \mathcal L^1. \end{equation*}

But then $x-\projone{}_{n+1}= \sum_{j=1}^n a_j \projone{}_j$ has finite rank and in particular, $\tr(\projone{}_j)\lt \infty$ for $0\lt j \lt n+1\text{,}$ thus proving (i). As a consequence,

\begin{equation*} \projone{}_{n+1}-\projtwo{}= \projone{}_{n+1}-x+ x-\projtwo{}\in \mathcal L^2 \end{equation*}

and hence by Proposition 2.8,

\begin{equation} [\projone{}_{n+1}:\projtwo{}]= \tr\big(\projtwo{}(\projone{}_{n+1}-\projtwo{})\projtwo{}+ \projtwo{}^\perp(\projone{}_{n+1}-\projtwo{})\projtwo{}^\perp\big). \tag{3.3} \end{equation}

Furthermore, $y:= z-\projone{}_{n+1}= \sum_{j\ne n+1} a_j \projone{}_j $ has finite rank and in particular is in $\mathcal L^1\text{,}$ so that

\begin{equation} \sum_{j\ne n+1} a_j\tr(\projone{}_j)= \tr(y)= \tr(\projtwo{}y\projtwo{}+\projtwo{}^\perp y \projtwo{}^\perp). \tag{3.4} \end{equation}

Finally from ((3.2)) and ((3.3)),

\begin{align*} a-b & = \tr\big( \projtwo{}(\projone{}_{n+1}-\projtwo{})\projtwo{}+ \projtwo{}^\perp(\projone{}_{n+1}-\projtwo{})\projtwo{}^\perp +\projtwo{}y\projtwo{}+ \projtwo{}^\perp y \projtwo{}^\perp \big)\\ & =[\projone{}_{n+1}:\projtwo{}]+ \tr(y). \end{align*}

Thus by ((3.4)), $a-b- \sum_{j\ne n+1} a_j\tr(\projone{}_j)= [\projone{}_{n+1}:\projtwo{}]\in \mathbb Z\text{.}$