A distance formula related to a family of projections orthogonal to their symmetries

TL;DR

This paper derives an exact formula for the distance between a projection and a specific set of projections related to a hermitian involution in a Hilbert space, expanding understanding of projection distances in operator algebras.

Contribution

It introduces a precise distance formula involving the norm of eue for projections orthogonal to symmetries, within the algebra generated by e and u.

Findings

Derived an exact formula for the projection distance

Connected the distance to the norm of eue

Enhanced understanding of projection relations in operator algebras

Abstract

Let u be a hermitian involution, and e an orthogonal projection, acting on the same Hilbert space. We establish the exact formula, in terms of the norm of eue, for the distance from e to the set of all orthogonal projections q from the algebra generated by e,u, and such that quq=0.

Figures

Equations

$$\Vert eue\Vert <\xi (\approx 0.455).$$

$$\Vert e-q\Vert \le \frac{1}{2}\Vert eue\Vert +4{\Vert eue\Vert }^2.$$

$$d=\sqrt{\frac{1}{2}\left(1-\sqrt{1-{\Vert eue\Vert }^2}\right)}$$

$$d=\frac{1}{2}\Vert eue\Vert +\frac{1}{16}{\Vert eue\Vert }^3+\cdots$$

$$\mathcal{H}={\mathcal{M}}_{00}\oplus {\mathcal{M}}_{01}\oplus {\mathcal{M}}_{10}\oplus {\mathcal{M}}_{11}\oplus \left(\mathcal{M}\oplus \mathcal{M}\right),$$

$$\begin{array}{rl}u=& I\oplus I\oplus (-I)\oplus (-I)\oplus \mathrm{diag}[I,-I],\\ e=& I\oplus 0\oplus 0\oplus I\oplus \left[\begin{array}{cc}H& \sqrt{H(I-H)}\\ \sqrt{H(I-H)}& I-H\end{array}\right].\end{array}$$

$$q=a_{00}I\oplus a_{01}I\oplus a_{10}I\oplus a_{11}I\oplus \Phi (H),$$

$${\varphi }_{00}-{\varphi }_{00}^2={\varphi }_{11}-{\varphi }_{11}^2={\left|{\varphi }_{01}\right|}^2,({\varphi }_{00}+{\varphi }_{11}-1){\varphi }_{01}=0.$$

$${\varphi }_{00}^2={\varphi }_{11}^2={\varphi }_{01}{\varphi }_{10},({\varphi }_{00}-{\varphi }_{11}){\varphi }_{01}=({\varphi }_{00}-{\varphi }_{11}){\varphi }_{10}=0.$$

$${\varphi }_{00}={\varphi }_{01}=\frac{1}{2}\chi ,{\varphi }_{01}=\frac{1}{2}\chi \omega ,{\varphi }_{10}=\frac{1}{2}\chi \overline{\omega }$$

$$q=0\oplus 0\oplus 0\oplus 0\oplus \frac{1}{2}\left[\begin{array}{cc}\chi & \chi \omega \\ \chi \overline{\omega }& \chi \end{array}\right](H).$$

$$e=\left[\begin{array}{cc}H& \sqrt{H(I-H)}\\ \sqrt{H(I-H)}& I-H\end{array}\right],q=\frac{1}{2}\left[\begin{array}{cc}\chi & \chi \omega \\ \chi \overline{\omega }& \chi \end{array}\right](H).$$

$${\Phi }_{\chi ,\omega }(t)=\left[\begin{array}{cc}t-\frac{1}{2}\chi (t)& \sqrt{t(1-t)}-\frac{1}{2}\chi (t)\omega (t)\\ \sqrt{t(1-t)}-\frac{1}{2}\chi (t)\overline{\omega (t)}& 1-t-\frac{1}{2}\chi (t)\end{array}\right].$$

$$\Vert e-q\Vert ={\mathrm{esssup}}_{t\in \Delta }{\lambda }_{\chi ,\omega }(t),$$

$${\Phi }_{1,\omega }(t)=\left[\begin{array}{cc}t-\frac{1}{2}& \sqrt{t(1-t)}-\frac{1}{2}\omega (t)\\ \sqrt{t(1-t)}-\frac{1}{2}\overline{\omega (t)}& \frac{1}{2}-t\end{array}\right],$$

$${\lambda }_{1,\omega }(t)=\sqrt{\frac{1}{2}-\sqrt{t(1-t)}\mathrm{Re}\omega (t)}.$$

$$q_0=\frac{1}{2}\left[\begin{array}{cc}I& I\\ I& I\end{array}\right],$$

$$\Vert e-q_0\Vert =\sqrt{\frac{1}{2}-\underset{t\in \Delta }{\min }\sqrt{t(1-t)}}=\sqrt{\frac{1}{2}\left(1-\sqrt{1-\underset{t\in \Delta }{\max }{\left|2t-1\right|}^2}\right)}.$$

$$\underset{t\in \Delta }{\max }\left|2t-1\right|=\Vert eue\Vert .$$

$$\Phi (t)=(2t-1)\left[\begin{array}{cc}t& \sqrt{t(1-1)}\\ \sqrt{t(1-1)}& 1-t\end{array}\right],$$

$$z:=\frac{1}{2}(e-ue{u}^{\ast })=\left[\begin{array}{cc}0& \sqrt{H(I-H)}\\ \sqrt{H(I-H)}& 0\end{array}\right]\in \mathcal{C}(e,ue{u}^{\ast }),$$

$$z(z^2{)}^{-1/2}=\left[\begin{array}{cc}0& I\\ I& 0\end{array}\right].$$

$$q_{x,y}=\frac{1}{2}\left[\begin{array}{ccc}{\left|x\right|}^2& x\overline{y}& x\\ \overline{x}y& {\left|y\right|}^2& y\\ \overline{x}& \overline{y}& 1\end{array}\right],$$

$$\Vert e-q_{x,y}\Vert =\sqrt{(1+{\left|y\right|}^2)/2}.$$

$$\frac{1}{2}\left[\begin{array}{ccc}1& 0& \omega \\ 0& 0& 0\\ \overline{\omega }& 0& 1\end{array}\right].$$

Full text

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1 Introduction

Let $\mathcal{H}$ be a Hilbert space and let $\mathcal{B}(\mathcal{H})$ stand for the $C^{*}$-algebra of all bounded linear operators acting on $\mathcal{H}$. Given a hermitian involution $u\in\mathcal{B}(\mathcal{H})$, denote by ${\mathcal{Q}}_{u}$ the set of all orthogonal projections $q\in\mathcal{B}(\mathcal{H})$ for which $quq=0$.

Theorem 1.2 of [1] can be stated as follows:

Theorem 1

Let $e\in\mathcal{B}(\mathcal{H})$ be an orthogonal projection such that

[TABLE]

Then there exists $q\in{\mathcal{Q}}_{u}$ for which

[TABLE]

Further, $q$ is in the $C^{*}$-subalgebra of $\mathcal{B}(\mathcal{H})$ generated by $e,ueu^{*}$.

Note that the distance between any two orthogonal projections does not exceed one. So, estimate (1) is useful only when $\left\|eue\right\|$ is smaller than the positive root of $8x^{2}+x-2$, that is, approximately $0.441$.

We will provide an explicit formula for the distance from $e$ to the intersection of ${\mathcal{Q}}_{u}$ with the $W^{*}$-algebra ${\mathcal{W}}(e,u)$ generated by $e,u$, as well as for the element on which this distance is attained. No a priori restriction on $\left\|eue\right\|$ is needed, and the respective $q$ indeed lies in the $C^{*}$-algebra ${\mathcal{C}}(e,ueu^{*})$ generated by $e,ueu^{*}$ whenever $\left\|eue\right\|<1$.

Theorem 2

Let $e,u\in\mathcal{B}(\mathcal{H})$ be, respectively, an orthogonal projection and a hermitian involution. Denote by ${\mathcal{H}}_{\pm}$ the eigenspace of $u$ corresponding to its eigenvalue $\pm 1$. Then the distance $d$ from $e$ to ${\mathcal{Q}}_{u}\cap{\mathcal{W}}(e,u)$ is one if the range of $e$ has a non-trivial intersection with ${\mathcal{H}}_{+}$ or ${\mathcal{H}}_{-}$, and is given by the formula

[TABLE]

otherwise.

For small values of $\left\|eue\right\|$, it is instructive to compare (1) with the Taylor expansion of (2):

[TABLE]

2 Proof of the main result

Using the canonical representation [2] (see also [3] or a more recent survey [4]) of the pair $e,(u+I)/2$ of orthogonal projections, we can find an orthogonal decomposition of $\mathcal{H}$ into six summands,

[TABLE]

with respect to which

[TABLE]

(Here and in what follows we use the notation $\operatorname{diag}[X_{1},\ldots,X_{k}]$ for block diagonal matrices with $X_{1},\ldots X_{k}$ as their diagonal blocks.) Note that in (3) the subspaces ${\mathcal{M}}_{00}$ and ${\mathcal{M}}_{11}$ (resp, ${\mathcal{M}}_{01}$ and ${\mathcal{M}}_{10}$) are the intersections of the range (resp, the kernel) of $e$ with ${\mathcal{H}}_{+}$ and ${\mathcal{H}}_{-}$. The (hermitian) operator $H$ is the compression of $e$ onto ${\mathcal{M}}:={\mathcal{H}}_{+}\ominus({\mathcal{M}}_{00}\oplus{\mathcal{M}}_{01})$. By construction, $H$ has its spectrum $\Delta$ lying in $[0,1]$ and $0,1$ are not its eigenvalues.

Elements of ${\mathcal{W}}(e,u)$ with respect to the same decomposition (3) look as

[TABLE]

where $\Phi=\begin{bmatrix}\phi_{00}&\phi_{01}\\ \phi_{10}&\phi_{11}\end{bmatrix}$, $a_{ij}\in\mathbb{C}$, and the functions $\phi_{ij}$ are Borel-measurable and essentially bounded on $\Delta$, in the sense of the spectral measure of $H$ ([5], see also [3, 4]). Consequently, $q\in{\mathcal{W}}(e,u)$ is an orthogonal projection if and only if $a_{ij}\in\{0,1\}$, the functions $\phi_{00},\phi_{11}$ are real-valued, while $\phi_{01},\phi_{10}$ are complex conjugate, and

[TABLE]

On the other hand, direct computations immediately reveal that condition $quq=0$ is equivalent to

[TABLE]

Solving the system of equations (6)–(7) yields

[TABLE]

with $\chi$ being a characteristic function of some subset of $\Delta$ and unimodular $\omega$.

So, elements of ${\mathcal{Q}}_{u}\cap{\mathcal{W}}(e,u)$ have the form

[TABLE]

The rest of the reasoning depends on whether or not the subspaces ${\mathcal{M}}_{00}$, ${\mathcal{M}}_{11}$ are actually present in the decomposition (3).

Case 1. At least one of the subspaces ${\mathcal{M}}_{00}$, ${\mathcal{M}}_{11}$ is different from zero, that is, the range of $e$ contains some eigenvectors of $u$.

Since for any $q$ of the form (8) the restriction of $e-q$ on ${\mathcal{M}}_{00}\oplus{\mathcal{M}}_{11}$ is the identity, we then have $\left\|e-q\right\|=1$. Consequently, $d=1$. Note that in this case also $\left\|eue\right\|=1$.

Case 2. ${\mathcal{M}}_{00}={\mathcal{M}}_{11}=\{0\}$. Since both $e$ given by (4) and $q$ given by (8) have zero restrictions onto ${\mathcal{M}}_{01}\oplus{\mathcal{M}}_{11}$, we may without loss of generality suppose that in place of (3) simply ${\mathcal{H}}={\mathcal{M}}\oplus{\mathcal{M}}$, and respectively

[TABLE]

So, $e-q=\Phi_{\chi,\omega}(H)$, where

[TABLE]

Consequently,

[TABLE]

where $\lambda_{\chi,\omega}(t)$ is the positive eigenvalue of $\Phi_{\chi,\omega}(t)$, and ess is understood in the sense of the spectral measure of $H$.

If $\chi(t)=0$ for some $t\in\Delta$, then the respective $\lambda_{\chi,\omega}(t)$ equals one, guaranteeing $\left\|e-q\right\|=1$. We should concentrate therefore on elements $q$ with $\chi(t)\equiv 1$. Then we have

[TABLE]

and

[TABLE]

Since $\omega$ is unimodular, to minimize $\lambda_{1,\omega}(t)$ for any given $t$ we should take $\omega(t)=1$. The respective element $q$ is simply

[TABLE]

$\lambda_{1,1}(t)=\sqrt{\frac{1}{2}-\sqrt{t(1-t)}}$, and

[TABLE]

In order to justify (2), it remains only to observe that

[TABLE]

But this is indeed the case, since $eue=\Phi(H)$ with the matrix

[TABLE]

the eigenvalues of which are zero and $2t-1$.

3 Additional comments

1

Recall [6] that elements of $C^{*}$-algebra ${\mathcal{C}}(e,u)$ generated by $e$ and $u$ are those of the form (5) for which the functions $\phi_{ij}$ are continuous on $\Delta$ and such that $\phi_{01}(j)=\phi_{10}(j)=0$, $a_{ij}=\phi_{ii}(j)$ if $j\in\Delta$ ($i,j=0,1$). From (10) we therefore conclude that the element $q_{0}\in{\mathcal{W}}(e,u)$ on which the distance from $e$ to ${\mathcal{Q}}_{u}$ is attained does not lie in ${\mathcal{C}}(e,u)$ if the spectrum of $H$ contains [math] or $1$.

On the other hand, due to (11) condition $\left\|eue\right\|<1$ guarantees that $0,1\notin\Delta$, and thus the invertibility of the operator $H(I-H)$. Moreover, ${\mathcal{M}}_{00}={\mathcal{M}}_{11}=\{0\}$, as was observed in Section 2. So, without loss of generality $e$ is given by the first formula in (9), while $u=\operatorname{diag}[I,-I]$. From here:

[TABLE]

$z^{2}=\operatorname{diag}[H(I-H),H(I-H)]$ is positive definite and also lies in ${\mathcal{C}}(e,ueu^{*})$, and therefore so does $(z^{2})^{-1/2}=\operatorname{diag}[\left((H(I-H)\right)^{-1/2},\left(H(I-H)\right)^{-1/2}]$. Along with $z$ and $(z^{2})^{-1/2}$, the algebra ${\mathcal{C}}(e,ueu^{*})$ contains their product

[TABLE]

We conclude from (10) that $q_{0}=\frac{1}{2}(I+z(z^{2})^{-1/2})\in{\mathcal{C}}(e,ueu^{*})$.

2

The distances from $e$ to the sets ${\mathcal{Q}}_{u}$ and ${\mathcal{Q}}_{u}\cap{\mathcal{W}}(e,u)$ may not coincide. To illustrate, consider ${\mathcal{H}}={\mathbb{C}}^{3}$, $u=\operatorname{diag}[1,1,-1]$ and $e=\operatorname{diag}[1,0,0]$. Then ${\mathcal{Q}}_{u}$ consists of zero and all matrices of the form

[TABLE]

with the parameters $x,y\in\mathbb{C}$ satisfying $\left|x\right|^{2}+\left|y\right|^{2}=1$. An easy computation shows that

[TABLE]

So, the distance from $e$ to ${\mathcal{Q}}_{u}$ equals $1/\sqrt{2}$ and is attained on all the matrices $q_{\omega,0}$ with $\left|\omega\right|=1$, that is, having the form

[TABLE]

On the other hand, the algebra generated by $e$ and $u$ consists simply of all $3$-by-$3$ diagonal matrices. The only diagonal matrix lying in ${\mathcal{Q}}_{u}$ is [math], and $d=1$ in full agreement with Theorem 2.

4 Acknowledgments

The author was supported in part by Faculty Research funding from the Division of Science and Mathematics, New York University Abu Dhabi.