Functional Models and Minimal Contractive Liftings

Abstract

Based on a careful analysis of functional models for contractive multi-analytic operators we establish a one-to-one correspondence between unitary equivalence classes of minimal contractive liftings of a row contraction and injective symbols of contractive multi-analytic operators. This allows an effective construction and classification of all such liftings with given defects. Popescu’s theory of characteristic functions of completely non-coisometric row contractions is obtained as a special case satisfying a Szegö condition. In another special case of single contractions and defects equal to 1 all non-zero Schur functions on the unit disk appear in the classification. It is also shown that the process of constructing liftings iteratively reflects itself in a factorization of the corresponding symbols.

Appendix

In our analysis of functional models in Sect. 2 we need a few technical results about how the kernel of the adjoint changes if we go to a restriction of an operator. We provide them in the following lemma. These are quite general observations, useful in particular to describe the geometry of invariant subspaces.

Lemma 6.1

Let $T \in \mathcal B ( \mathcal H_1, \mathcal H_2)$ where $\mathcal H_1, \mathcal H_2$ are Hilbert spaces and $\mathcal K_i \subset \mathcal H_i,\; i=1,2,$ be subspaces such that $T \mathcal K_1 \subset \mathcal K_2$. Define $\tilde{T}:=T|_{\mathcal K_1} :\mathcal K_1 \mapsto \mathcal K_2$ and $\mathcal N_i := \mathcal H_i\ominus \mathcal K_i,\; i=1,2$. Then

• (i) $\mathrm{ker}\,{\tilde{T}}^*= \{\xi \in \mathcal K_2: T^* \xi \in \mathcal N_1 \}$.

• (ii) $P_{\mathcal K_2}\, \mathrm{ker}\,T^* \subset \mathrm{ker}\,{\tilde{T}}^*$. Assume in addition that $T$ is an isometry. Set $\mathcal L := \overline{\mathrm{span}}\, \{\mathcal N_2,\, T \mathcal N_1\} \ominus \mathcal N_2$. Then

• (iii) $\overline{\mathrm{span}}\,\{\mathcal L, P_{\mathcal K_2}\, \mathrm{ker}\,T^*\} = \mathrm{ker}\,{\tilde{T}}^* = \mathcal L \oplus (\mathrm{ker}\,T^*\cap \ \mathcal K_2)$.

Proof

• (i) For $\xi \in \mathcal K_2$

  $$\begin{aligned} \xi \in \mathrm{ker}\,{\tilde{T}}^*&\Leftrightarrow \langle \xi , {\tilde{T}} \eta \rangle = 0 \; ~\mathrm{for~ all }~ \; \eta \in \mathcal K_1\\&\Leftrightarrow \langle \xi , T\eta \rangle = 0 \; ~\mathrm{for~ all }~\; \eta \in \mathcal K_1\\&\Leftrightarrow \langle T^*\xi , \eta \rangle = 0 \; ~\mathrm{for~ all }~ \; \eta \in \mathcal K_1\\&\Leftrightarrow T^*\xi \in \mathcal N_1. \end{aligned}$$

• (ii) Decompose $\xi \in \mathrm{ker}\,T^*$ as $\xi = \xi _{\mathcal K_2} \oplus \xi _{\mathcal N_2}$ such that $\xi _{\mathcal K_2} \in \mathcal K_2$ and $\xi _{\mathcal N_2} \in \mathcal N_2$. Since $T^* (\mathcal N_2) \subset \mathcal N_1$ it follows that $T^*\xi _{\mathcal N_2} \in \mathcal N_1$. Thus $T^*\xi _{\mathcal K_2} = T^* \xi - T^*\xi _{\mathcal N_2} =0- T^*\xi _{\mathcal N_2} \in \mathcal N_1$. We conclude that $\xi _{\mathcal K_2} \in \mathrm{ker}\,{\tilde{T}}^*$ by (i). Finally because $P_{\mathcal K_2}\, \xi = \xi _{\mathcal K_2}$ we obtain $P_{\mathcal K_2} \; \mathrm{ker}\,T^* \subset \mathrm{ker}\,{\tilde{T}}^*$ which is (ii).

• (iii) First note that if $\xi \in \mathrm{ker}\,T^* \cap \ \mathcal K_2$ then $\xi \perp \overline{\mathrm{range}\, T} \supset T \mathcal N_1$ and $\xi \perp \mathcal N_2$. Therefore $\mathrm{ker}\,T^* \cap \ \mathcal K_2 \perp \mathcal L$.

Further $\mathcal L \subset \mathcal N_2^\perp = \mathcal K_2$ and $T^* \mathcal L \subset T^*\big [\overline{\mathrm{span}}\,\{\mathcal N_2, T \mathcal N_1\} \big ] = \mathcal N_1$, by the assumption that $T$ is an isometry. So $\mathcal L \subset \mathrm{ker}\,{\tilde{T}}^*$ by (i). Clearly $\mathrm{ker}\,T^* \cap \ \mathcal K_2 \subset \mathrm{ker}\,{\tilde{T}}^*$ by (i). Together we have $\mathcal L \oplus (\mathrm{ker}\,T^*\cap \ \mathcal K_2) \subset \mathrm{ker}\,{\tilde{T}}^*$.

For the opposite inclusion let $\xi \in \mathrm{ker}\,{\tilde{T}}^* \ominus \mathcal L$. Because $\xi \in \mathrm{ker}\,{\tilde{T}}^*$ we have $\xi \in \mathcal K_2$ and $T^* \xi \in \mathcal N_1$ by (i). But from $\xi \perp \mathcal L$ and $\xi \perp \mathcal N_2$ we also get, using the definition of $\mathcal L$, that $\xi \perp T \mathcal N_1, i.e., T^* \xi \perp \mathcal N_1$. Hence $T^* \xi = 0$ and $\xi \in \mathrm{ker}\,T^*\cap \ \mathcal K_2$. We have now established the second equality in (iii).

Finally it is clear that $\mathrm{ker}\,T^* \cap \mathcal K_2 \subset P_{\mathcal K_2}\,\mathrm{ker}\,T^*$, hence $\mathcal L \oplus (\mathrm{ker}\,T^*\cap \ \mathcal K_2) \subset \overline{\mathrm{span}}\,\{\mathcal L, P_{\mathcal K_2}\, \mathrm{ker}\,T^*\}$. On the other hand we have seen above that $\mathcal L \subset \mathrm{ker}\,{\tilde{T}}^*$ and we have $P_{\mathcal K_2}\,\mathrm{ker}\,T^* \subset \mathrm{ker}\,{\tilde{T}}^*$ by (ii). Hence we also have $\overline{\mathrm{span}}\,\{\mathcal L, P_{\mathcal K_2}\, \mathrm{ker}\,T^*\} \subset \mathrm{ker}\,{\tilde{T}}^* = \mathcal L \oplus (\mathrm{ker}\,T^*\cap \ \mathcal K_2)$ and we have proved the first equality in (iii). $\square$