On the Similarity of Boundary Triples of Symmetric Operators in Krein Spaces

Abstract

It is a classical result that the Weyl function of a simple symmetric operator in a Hilbert space determines a boundary triple uniquely up to unitary equivalence. We generalize this result to a simple symmetric operator in a Pontryagin space, where unitary equivalence is replaced by the similarity realized via a standard unitary operator.

Appendix A.

Lemma A.1

Let ${\mathfrak {H}}=({\mathfrak {H}},[\cdot ,\cdot ])$ be a Krein space with fundamental symmetry J, and consider relations G and H in ${\mathfrak {H}}$ such that $H\subseteq G^+\cap G^\bot$. Then:

• (a) If ${{\,\textrm{dom}\,}}G\bot {{\,\textrm{dom}\,}}H$ then

  $$\begin{aligned} \begin{aligned}&{{\,\textrm{ran}\,}}(JH+zI)\subseteq {\mathfrak {N}}_z(JG^+)\\&\text {or equivalently}\\&{{\,\textrm{ran}\,}}(JG+zI)\subseteq {\mathfrak {N}}_z(JH^+) \end{aligned} \end{aligned}$$

  (A.1)

  for all $z\in {\mathbb {C}}$.

• (b) If $G\,\widehat{\oplus }\,H$ is a neutral subset of ${\mathfrak {K}}$ then:

  • (i) The inclusions in (A.1) hold for both $z=\textrm{i}$ and $z=-\textrm{i}$.

  • (ii) The inclusions in (A.1) become the equalities for $z=\textrm{i}$ or $z=-\textrm{i}$ (resp. for both $z=\textrm{i}$ and $z=-\textrm{i}$) iff $G\,\widehat{\oplus }\,H$ is maximal (resp. hyper-maximal) neutral.

Remark A.2

In (b), ${{\,\textrm{dom}\,}}G\bot {{\,\textrm{dom}\,}}H$ is not assumed. As in the body of the text, the symbol $\bot$ indicates the orthogonality with respect to a Hilbert space metric $[\cdot ,J\cdot ]$, while $[\bot ]$ refers to the orthogonality with respect to $[\cdot ,\cdot ]$.

Proof

(a) ${{\,\textrm{dom}\,}}G\bot {{\,\textrm{dom}\,}}H$ implies that

$$ {{\,\textrm{dom}\,}}(JH)={{\,\textrm{dom}\,}}H\subseteq ({{\,\textrm{dom}\,}}G)^\bot = {{\,\textrm{mul}\,}}(JG^+)\,. $$

Since $G\bot H$, this implies that also

$$ {{\,\textrm{ran}\,}}(JH)\subseteq ({{\,\textrm{ran}\,}}(JG))^\bot =\ker G^+= \ker (JG^+)\,. $$

Consider $(f,f^\prime )\in JH$, so that $f^\prime +zf\in {{\,\textrm{ran}\,}}(JH+zI)$ for all $z\in {\mathbb {C}}$. Then

$$\begin{aligned} (f^\prime +zf,z(f^\prime +zf))= z(f,f^\prime )+(f^\prime ,0)+z^2(0,f). \end{aligned}$$

Since $(f,f^\prime )$, $(f^\prime ,0)$, and (0, f) are all the elements from $JG^+$, $f^\prime +zf\in {\mathfrak {N}}_z(JG^+)$.

(b)(i) Let $K\mathrel {\mathop :}=G\,\widehat{\oplus }\,H$ be neutral. Since $JH=JK\cap (JG)^\bot$, the range ${{\,\textrm{ran}\,}}(JH+zI)$, for all $z\in {\mathbb {C}}$, consists of those $k^\prime +zk$ such that $(k,k^\prime )\in JK$ and $(k^\prime ,-k)\in JG^+$. If $z=\textrm{i}$ then

$$\begin{aligned} (k^\prime +\textrm{i}k,\textrm{i}k^\prime -k)= \textrm{i}(k,k^\prime )+(k^\prime ,-k). \end{aligned}$$

Since $JK\subseteq JG^+$, this shows $k^\prime +\textrm{i}k\in {\mathfrak {N}}_\textrm{i}(JG^+)$. The case $z=-\textrm{i}$ is treated analogously.

(b)(ii) Sufficiency: Let K be either maximal or hyper-maximal neutral; that is, either $P_+(K)={\mathfrak {K}}$ or $P_-(K)={\mathfrak {K}}$ if K is maximal and $P_\pm (K)={\mathfrak {K}}$ if K is hyper-maximal. We consider the case $P_+(K)={\mathfrak {K}}$, since the case $P_-(K)={\mathfrak {K}}$ is treated analogously. In view of (b)(i) it suffices to show that ${\mathfrak {N}}_\textrm{i}(JG^+)\subseteq {{\,\textrm{ran}\,}}(JH+\textrm{i}I)$.

If $P_+(K)={\mathfrak {K}}$ then a maximal symmetric relation JK in a Hilbert space $({\mathfrak {H}},[\cdot ,J\cdot ])$ satisfies ${{\,\textrm{ran}\,}}(JK+\textrm{i}I)={\mathfrak {H}}$. Thus, if $g\in {\mathfrak {N}}_\textrm{i}(JG^+)$ then $(\exists (k,k^\prime )\in JK)$ $g=k^\prime +\textrm{i}k$. Then

$$\begin{aligned} JG^+\ni (g,\textrm{i}g)= (k^\prime +\textrm{i}k,\textrm{i}k^\prime -k)= \textrm{i}(k,k^\prime )+(k^\prime ,-k). \end{aligned}$$

Since $JK\subseteq JG^+$, this shows $(k^\prime ,-k)\in JG^+$, i.e. $g\in {{\,\textrm{ran}\,}}(JH+\textrm{i}I)$.

Necessity: Let $K=G\,\widehat{\oplus }\,H$ be neutral and ${{\,\textrm{ran}\,}}(JH\pm \textrm{i}I)={\mathfrak {N}}_{\pm \textrm{i}}(JG^+)$. Then ${{\,\textrm{ran}\,}}(JK\pm \textrm{i}I)={\mathfrak {H}}$, thus showing $P_\pm (K)={\mathfrak {K}}$. $\square$

Lemma A.3

Let G and H be relations in a Krein space $({\mathfrak {H}},[\cdot ,\cdot ])$ with fundamental symmetry J.

• (a) The eigenspace

  $$\begin{aligned} \begin{aligned} {\mathfrak {N}}_z(G\,\widehat{+}\,H)=&{{\,\textrm{ran}\,}}((G-zI)^{-1}(zI-H)+I)\\ =&{{\,\textrm{ran}\,}}((G-zI)^{-1}-(H-zI)^{-1})\\ \supseteq&{\mathfrak {N}}_z(G)+{\mathfrak {N}}_z(H) \end{aligned} \end{aligned}$$

  (A.2)

  for all $z\in {\mathbb {C}}$.

• (b) Let $O=O(G,H)$ be the set of all those $z\in {\mathbb {C}}$ such that

  $$\begin{aligned} {{\,\textrm{ran}\,}}(G-zI)\cap {{\,\textrm{ran}\,}}(H-zI)=\{0\}. \end{aligned}$$

  (Equivalently, O is the set of all those $z\in {\mathbb {C}}$ such that $(G-zI)^{-1}(H-zI)\subseteq {\mathfrak {N}}_z(H)\times {\mathfrak {N}}_z(G)$.) Then the inclusion $\supseteq$ in (A.2) becomes the equality for all $z\in O$; hence

  $$\begin{aligned} O\cap \sigma _p(G\,\widehat{+}\,H)= O\cap (\sigma _p(G)\cup \sigma _p(H) ). \end{aligned}$$

  (A.3)

• (c) Suppose $G\bot H$ as linear subsets of ${\mathfrak {K}}$. Let O be as in (b). Then

  $$\begin{aligned} \sigma _p(G\,\widehat{\oplus }\,H)= ( O\cap (\sigma _p(G)\cup \sigma _p( H) ) )\amalg ({\mathbb {C}}\smallsetminus O). \end{aligned}$$

  (A.4)

  (The symbol $\amalg$ denotes the union of disjoint sets.) In particular:

  • (i) $\sigma _p(G\,\widehat{\oplus }\,H)=\emptyset$ (resp. $\sigma ^0_p(G\,\widehat{\oplus }\,H)=\emptyset$) iff $O={\mathbb {C}}$ and $\sigma _p(G)=\sigma _p(H)=\emptyset$ (resp. $O\supseteq {\mathbb {C}}_*$ and $\sigma ^0_p(G)=\sigma ^0_p(H)=\emptyset$).

  • (ii) $\sigma _p(G\,\widehat{\oplus }\,H)={\mathbb {C}}$ (resp. $\sigma ^0_p(G\,\widehat{\oplus }\,H)={\mathbb {C}}_*$) iff $O\subseteq \sigma _p(G)\cup \sigma _p(H)$ (resp. ${\mathbb {C}}_*\cap O\subseteq \sigma ^0_p(G)\cup \sigma ^0_p(H)$).

Proof

(a) Let $f\in {\mathfrak {N}}_z(G\,\widehat{+}\,H)$, i.e. $(f,zf)\in G\,\widehat{+}\,H$. Then $f=g+h$ with $(g,g^\prime )\in G$ and $(h,h^\prime )\in H$ such that $g^\prime +h^\prime =z(g+h)$, i.e. $(\exists u)$ $(g,u)\in G-zI$ and $(h,u)\in zI- H$, i.e. $(h,g)\in (G-zI)^{-1}(zI-H)$. Since the arguments are reversible, this proves (A.2).

(b) First we show that $z\in O$ iff $z\in {\mathbb {C}}$ satisfies $(G-zI)^{-1}(H-zI)\subseteq {\mathfrak {N}}_z(H)\times {\mathfrak {N}}_z(G)$. This will follow from a general claim: If X and Y are relations in ${\mathfrak {H}}$ then ${{\,\textrm{ran}\,}}X\cap {{\,\textrm{ran}\,}}Y$ is trivial iff $X^{-1}Y\subseteq \ker Y\times \ker X$. (Notice that always $X^{-1}Y\supseteq \ker Y\times \ker X$.) Indeed, ${{\,\textrm{ran}\,}}X\cap {{\,\textrm{ran}\,}}Y$ consists of those u such that $(\exists x)$ $(\exists y)$ $(x,u)\in X$ and $(y,u)\in Y$, while $X^{-1}Y$ is the set of those (y, x) such that $(\exists u)$ $(x,u)\in X$ and $(y,u)\in Y$. Therefore, if ${{\,\textrm{ran}\,}}X\cap {{\,\textrm{ran}\,}}Y$ is trivial then $(y,x)\in X^{-1}Y$ implies $(y,x)\in \ker Y\times \ker X$. And conversely, if $X^{-1}Y=\ker Y\times \ker X$ then $u\in {{\,\textrm{ran}\,}}X\cap {{\,\textrm{ran}\,}}Y$ implies $u=0$.

This shows in particular that the inclusion $\supseteq$ in (A.2) becomes the equality for all $z\in O$.

(c) Let

$$\begin{aligned} L_z\mathrel {\mathop :}=(G-zI)^{-1}(zI-H)+I. \end{aligned}$$

By (a), ${\mathfrak {N}}_z(G\,\widehat{\oplus }\,H)$ is trivial iff so is ${{\,\textrm{ran}\,}}L_z$, i.e.iff $L_z\subseteq {\mathfrak {H}}\times \{0\}$. We show that $\ker L_z$ is trivial, meaning that the last inclusion is equivalent to $L_z=\{0\}$, which in turn is equivalent to $(G-zI)^{-1}(zI-H)=\{0\}$.

The relation $L_z$ consists of the pairs $(h,g+h)$ such that $(\exists u)$ $(g,zg+u)\in G$ and $(h,zh-u)\in H$. Since $G\bot H$

$$\begin{aligned} \Vert u\Vert ^2-(1+|z|^2)\left\langle g,h\right\rangle = z\left\langle u,h\right\rangle -\overline{z}\left\langle g,u\right\rangle \end{aligned}$$

where the scalar product $\left\langle x,y\right\rangle \mathrel {\mathop :}=[x,Jy]$ and the norm $\Vert x\Vert ^2\mathrel {\mathop :}=\left\langle x,x\right\rangle$ for all x, $y\in {\mathfrak {H}}$. If $g+h=0$ then by the above

$$\begin{aligned} \Vert u\Vert ^2+(1+|z|^2)\Vert g\Vert ^2 +2\Re (z\left\langle u,g\right\rangle )=0. \end{aligned}$$

But by Cauchy–Schwarz

$$\begin{aligned} \Vert u\Vert ^2+(1+|z|^2)\Vert g\Vert ^2 +2\Re (z\left\langle u,g\right\rangle )\geqslant&(\Vert u\Vert -|z|\,\Vert g\Vert )^2+ \Vert g\Vert ^2\\ \geqslant&\Vert g\Vert ^2\geqslant 0 \end{aligned}$$

so that then $g=0=u$ and $h=0$.

Suppose now $(G-zI)^{-1}(zI-H)=\{0\}$. Multiplying both sides by $G-zI$ from the left implies that

$$\begin{aligned} R\mathrel {\mathop :}=({\mathfrak {H}}\times {{\,\textrm{ran}\,}}(G-zI))\cap (zI-H)\subseteq \{0\}\times {{\,\textrm{mul}\,}}G. \end{aligned}$$

Then

$$\begin{aligned} {{\,\textrm{mul}\,}}R={{\,\textrm{ran}\,}}(G-zI)\cap {{\,\textrm{mul}\,}}H\subseteq {{\,\textrm{mul}\,}}G \end{aligned}$$

and then $R=\{0\}$, since

$$\begin{aligned} {{\,\textrm{mul}\,}}G\cap {{\,\textrm{mul}\,}}H={{\,\textrm{mul}\,}}(G\cap H)=\{0\} \end{aligned}$$

by $G\bot H$. In particular

$$\begin{aligned} {{\,\textrm{ran}\,}}R={{\,\textrm{ran}\,}}(G-zI)\cap {{\,\textrm{ran}\,}}(H-zI)=\{0\}. \end{aligned}$$

This shows that ${\mathbb {C}}\smallsetminus \sigma _p(G\,\widehat{\oplus }\,H)\subseteq O$. The latter combined with (A.3) yields (A.4), which in turn yields (i) and (ii). $\square$

Remark A.4

For closed relations G and H and for $z\in \rho (G)\cap \rho (H)$, the second equality in (A.2) is given in [7, Lemma 1.7.2].

Lemma A.5

Let G be a relation in a Krein space ${\mathfrak {H}}$. In order that the eigenspaces corresponding to distinct eigenvalues of the adjoint $G^+$ should be mutually disjoint, that is, $(\forall z,z_0\in {\mathbb {C}})$

$$\begin{aligned} {\mathfrak {N}}_z(G^+)\cap {\mathfrak {N}}_{z_0}(G^+)=\{0\}\quad \text {if}\quad z\ne z_0 \end{aligned}$$

it is necessary and sufficient that ${{\,\textrm{dom}\,}}G+{{\,\textrm{ran}\,}}G$ should be dense in ${\mathfrak {H}}$.

Proof

Let $f\in {\mathfrak {N}}_z(G^+)\cap {\mathfrak {N}}_{z_0}(G^+)$, $z\ne z_0$. Then $(f,zf)\in G^+$ and $(f,z_0f)\in G^+$ implies $f\in {{\,\textrm{mul}\,}}G^+\cap \ker G^+$. Conversely, if $f\in {{\,\textrm{mul}\,}}G^+\cap \ker G^+$ then $(\forall w\in {\mathbb {C}})$ $(f,wf)\in G^+$; hence in particular $f\in {\mathfrak {N}}_z(G^+)\cap {\mathfrak {N}}_{z_0}(G^+)$. Therefore, ${\mathfrak {N}}_z(G^+)\cap {\mathfrak {N}}_{z_0}(G^+)$ for $z\ne z_0$ is trivial iff ${{\,\textrm{mul}\,}}G^+\cap \ker G^+= ({{\,\textrm{dom}\,}}G+{{\,\textrm{ran}\,}}G)^{[\bot ]}$ is trivial. $\square$

Lemma A.6

A closed symmetric operator in a Hilbert space has property (P), see Definition 6.10, iff it is densely.

Proof

Step 1. Let T be a closed symmetric relation in a Hilbert space ${\mathfrak {H}}$. Then $(\forall z\in {\mathbb {C}}_*)$ ${\mathfrak {N}}_z(T^*)\cap {{\,\textrm{dom}\,}}T=\{0\}$. For, if $f\in {\mathfrak {N}}_z(T^*)$ then $(f,zf)\in T^*$ and if also $f\in {{\,\textrm{dom}\,}}T$ then $(\exists f^\prime )$ $(f,f^\prime )\in T$, i.e. $f^\prime -zf\in {{\,\textrm{ran}\,}}(T-zI)\cap {{\,\textrm{mul}\,}}T^*$; but the latter set is trivial, see e.g.[28, Eq. (2.4)].

Step 2. We use the first von Neumann formula

$$\begin{aligned} T^*=T\,\widehat{\oplus }\,\widehat{{\mathfrak {N}}}_\textrm{i}(T^*)\,\widehat{\oplus }\,\widehat{{\mathfrak {N}}}_{-\textrm{i}}(T^*). \end{aligned}$$

Since ${\mathfrak {N}}_{\pm \textrm{i}}(T^*)\cap {{\,\textrm{dom}\,}}T=\{0\}$, ${{\,\textrm{mul}\,}}T^*$ consists of the elements $f^\prime +\textrm{i}f_\textrm{i}-\textrm{i}f_{-\textrm{i}}$, where $(f,f^\prime )\in T$, $f_{\pm \textrm{i}}\in {\mathfrak {N}}_{\pm \textrm{i}}(T^*)$, and $f+f_\textrm{i}+f_{-\textrm{i}}=0$; hence $f=0$, $f^\prime \in {{\,\textrm{mul}\,}}T=\{0\}$, and $f_{-\textrm{i}}=-f_{\textrm{i}}\in {\mathfrak {N}}_{\textrm{i}}(T^*)\cap {\mathfrak {N}}_{-\textrm{i}}(T^*)$. By Lemma A.5${\mathfrak {N}}_{\textrm{i}}(T^*)\cap {\mathfrak {N}}_{-\textrm{i}}(T^*)= {{\,\textrm{mul}\,}}T^*\cap \ker T^*$, so ${{\,\textrm{mul}\,}}T^*\subseteq {{\,\textrm{mul}\,}}T^*\cap \ker T^*$ implies that ${{\,\textrm{mul}\,}}T^*\cap \ker T^*=\{0\}$ iff $T^*$ is an operator. $\square$

Remark A.7

In Step 1 one could instead use that $(\forall z\in {\mathbb {C}})$

$$\begin{aligned} {\mathfrak {N}}_z(T^*)\cap {{\,\textrm{dom}\,}}T={\mathfrak {N}}_z (T^*\cap ({{\,\textrm{dom}\,}}T)^2) \end{aligned}$$

and that $T^*\cap ({{\,\textrm{dom}\,}}T)^2$ is a symmetric relation in a Hilbert space $\overline{{{\,\textrm{dom}\,}}}T$.