### Topics in Operator Theory, Lecture 3: Dilations of commuting operators

We continue in this lecture to consider dilation theory of contractions. In the theory of Sz.-Nagy and Foias, the main route proceeded from the existence of the minimal isometric and unitary dilations, to the study of how such dilations look like, and to use them to extract information about an operator from its dilations. The only application we saw until now was von Neumann’s inequality, which is not a trivial fact, but let’s admit it: somewhat rinky dinky. But after a deeper look is taken into the structure of the minimal unitary dilation, the way for more significant applications opens. One of these applications is a functional calculus (for non-selfadjoints) that extends the holomorphic functional calculus. Another application is an affirmative solution to the invariant subspace problem for certain classes of operators. The main parts of this theory are laid down in the book Harmonic Analysis of Operators On Hilbert Space.

We will not follow that route. Rather, we will see what dilation theory can help us to understand regarding tuples of commuting operators (which is also treated to some extent in the book). Surprisingly, this will lead to a truly nifty application in function theory.

#### 1. Unitary extension of families of isometries

Definition: Let $A_1, \ldots A_d$ be commuting contractions. A commuting tuple of operators $B_1, \ldots, B_d$ is said to be a commuting isometric/unitary extension/dilation of $A_1, \ldots, A_d$ if all $B_i$s are isometric/unitary and they are all either (i) extensions of the respective $A_i$s (in the case of extension); or (ii) satisfy

$A_1^{n_1} \cdots A_d^{n_d} = P_H B_1^{n_1} \cdots B_d^{n_d} \big|_H$

(in the case of dilation).

Theorem 1: Every tuple of commuting isometries has a unitary extension.

Proof: Let $V_1, \ldots, V_d$ be commuting isometries acting on $H$. Let $U_1$ be the unitary extension of $V_1$ (that we constructed in the previous post), acting on a Hilbert space $K$. We did not delve on this matter, but the $(U_1,K)$ is in fact a minimal unitary dilation, in the sense that

$K = \bigvee_{n=-\infty}^\infty U_1^n H$.

Now we shall define operators $U_2, \ldots, U_d$ on $K$ with the following properties:

1. $U_1, \ldots, U_d$ are a commuting family,
2. $U_2, \ldots, U_d$ are all isometries, and,
3. For $i=2, \ldots, d$, if $V_i$ was already an isometry, then so is $U_i$.

Once we construct the above family, the proof would almost be complete: If $U_1, \ldots, U_d$ are all unitaries, then we’ll be done. If, say, $U_2$ is not a unitary, then we repeat the above construction, first constructing the minimal unitary extension $W_2$ of $U_2$, and then extending $U_1, U_3, U_4, \ldots U_d$ so that   $W_1, W_2, \ldots, W_d$ form commuting isometries, such that whenever $U_i$ is a unitary, $W_i$ is also a unitary. In this way, we have a commuting family $W_1, \ldots, W_d$ extending $T_1, \ldots, T_d$ such that at least $W_1, W_2$ are unitaries. Continuing this way, the proof will be complete.

It remains to define the operators $U_2, \ldots, U_d : K \to K$, and to show that they have the desired properties. There is really no freedom in the definition, since we must have $U_i U_1 = U_1 U_i$, and so (keeping in mind that we require $U_i = V_i$ on $H$) we must define

(*)   $U_i U_1^n h = U_1^n V_i h$,

keeping in mind that elements of the form $U_1^n h$ span $H$. To see that this map preserves inner product (and hence well-defines an isometry), we take $g,h \in H$ and $m \leq n$, and check

$\langle U_1^n V_i h, U_1^m V_i g \rangle = \langle U_1^{n-m}V_i h, V_i g \rangle =$

$= \langle V_1^{n-m}V_i h, V_i g \rangle =\langle V_i V_1^{n-m} h, V_i g \rangle =$

$= \langle V_1^{n-m} h, g \rangle =\langle U_1^n h, U_1^m g \rangle$.

Thus, for every $i = 2, 3, \ldots, d$, the definition (*) extends to a well defined isometry on $K$, which clearly extends $V_i$ (by considering the $n=0$ case). Moreover, if $V_i$ is a unitary, then the range of $U_i$ is dense in $K$, so, being isometric with dense range, $U_i$ is a unitary.

Finally,

$U_i U_j U_1^n h = U_1^n V_i V_j h = U_1^n V_j V_i h = U_j U_i U_1^n h$,

and since elements of the form $U_1^n h$ are total in $K$, it follows that $U_i U_j = U_j U_i$. This completes the proof.

Exercise A: Show that the unitary dilation constructed in the above proof is minimal.

Exercise B: Prove that a unitary dilation of an isometry (or isometries) is an extension.

#### 2. Ando’s theorem

If $T_1, \ldots, T_d$ are any $d$ contractions, then we know from Exercise C in the previous post that there are isometric dilations $V_1, \ldots, V_d$ that simultaneously dilate any noncommutative monomial. However, if the $T_1, \ldots, T_d$ are assumed to be commuting, there is no reason that the isometries that you constructed in your solution to Exercise C will also commute (recall the construction and try to understand where this fails). In fact, we will soon see that in general, when $d \geq 3$, a $d$-tuple of commuting contractions does not have an isometric (nor a unitary) dilation.

The case $d=2$ is special.

Theorem 2 (Ando’s isometric dilation theorem): Every pair of commuting contractions has an isometric dilation (in fact, an isometric co-extension).

Proof: The proof is “not deep”, and just boils down to finding a sufficently clever construction. Let $A_1, A_2 \in B(H)$ be two commuting contractions. We define the isometric dilation as follows. We begin by defining $G = H \oplus H \oplus H \oplus H$, and

$K = H \oplus G \oplus G \oplus G \oplus \cdots$

We begin by defining isometric dilation $V_1$ and $V_2$ by

$V_i (h_0, (h_1, h_2, h_3, h_4), (h_5, h_6, h_7, h_8), \ldots) = (A_ih_0, (D_{A_i} h_0, 0 , h_1, h_2), (h_3, h_4, h_5, h_6), \ldots)$,

where $D_{A_i} = (I - A_i^* A_i)^{1/2}$. Each $V_i$ is an isometric dilation – in fact, a co-extension – of $A_i$, but they do not necessarily commute:

$V_1 V_2 (h_0, (h_1, h_2, h_3, h_4), (h_5, h_6, h_7, h_8), \ldots) = (A_1 A_2 h_0 (D_{A_1} A_2 h_0, 0, D_{A_2}h_0, 0), (h_1, h_2, h_3, h_4), \ldots)$

while

$V_2 V_1 (h_0, (h_1, h_2, h_3, h_4), (h_5, h_6, h_7, h_8), \ldots) = (A_2 A_1 h_0 (D_{A_2} A_1 h_0, 0, D_{A_1}h_0, 0), (h_1, h_2, h_3, h_4), \ldots)$.

For $V_1 V_2$ to be equal to $V_2 V_1$ we need that $(D_{A_1} A_2 h_0, 0, D_{A_2}h_0, 0) = (D_{A_2} A_1 h_0, 0, D_{A_1}h_0, 0)$. There is really no reason for equality to hold here, the fact that $A_1$ and $A_2$ commute does not help at all (think of a simple example where equality fails).

However, we have the following (writing $h$ for $h_0$):

$\|(D_{A_1} A_2 h, 0, D_{A_2}h, 0)\|^2 = \langle (I - A_1^*A_1)A_2 h, A_2 h \rangle + \langle (I - A_2^*A_2) h, h \rangle = \|h\|^2 - \|A_1 A_2\|^2$

and by commutativity and symmetry, this clearly equals $\|(D_{A_2} A_1 h, 0, D_{A_1}h, 0)\|^2$. We can therefore find a unitary $U : G \to G$ such that

$U(D_{A_1} A_2 h, 0, D_{A_2}h, 0) = (D_{A_2} A_1 h, 0, D_{A_1}h, 0)$

for all $h \in H$ (this unitary is guaranteed to exist thanks to the zeros that we stuffed in the definition; think about it). Letting $W = I \oplus U \oplus U \oplus \cdots$, we now define $U_1 = W V_1$ and $V_2 W^*$. It is now easy to check that $U_i$ is still a co-extension of $A_i$ and also that $U_1$ and $U_2$ commute. This concludes the proof.

Corollary (Ando’s unitary dilation theorem): Every pair of commuting contractions has a unitary dilation.

Corollary (Ando’s inequality): $\|p(A,B)\| \leq \sup_{|z|=|w|=1}|p(z,w)|$ for every polynomial $p \in \mathbb{C}[z,w]$ and every pair of commuting contractions $A,B$

Proof: For the proof, one proceeds as usual, with the difference that now one needs the spectral theorem for commuting normals, rather than the spectral theorem for single operators. Alternatively, one can use the theory of commutative C*-algebras.

Exercise C: Define the notion of “minimal dilation” for isometric and unitary dilations of commuting tuples. Prove that if a tuple has a dilation, then it has a minimal dilation. Prove that the minimal isometric/unitary dilation of a pair of contractions is not unique.

Theorem 3 (commutant lifting theorem): Let $A\in B(H)$ be a contraction, and let $(V,K)$ be the minimal isometric dilation of $A$ (which, we know, is a co-extension). For every $B \in B(H)$ commuting with $A$, there exists $R \in B(K)$ that commutes with $V$, is a co-extension of $B$, and satisfies $\|R\| = \|B\|$

Proof: WLOG, $\|B\| = 1$. Let $U, W \in B(L)$ be an isometric co-extension of $A, B$, given by Ando’s theorem. Then $U$ is an isometric co-extension of $A$, so by Exercise C in the previous post, $L = K \oplus K'$ and $U = V \oplus V'$, where $H \subseteq K$ and $(V,K)$ is the minimal isometric co-extension of $A$. We can therefore write, with respect to the decomposition $L = K \oplus K'$,

$U = \begin{pmatrix} V & 0 \\ 0 & V' \end{pmatrix}$

and

$W = \begin{pmatrix} R & X \\ Y & Z \end{pmatrix}$.

We claim that this $R = P_K W \big|_K$ does the job. It is a co-extension of $B$ because $W$ is, and so $\|B\|\leq \|R\| \leq \|W\| = 1 = \|B\|$, whence $\|B\| = \|W\|$. All that remains to prove is that $R$ commutes with $V$. One can see this simply by multiplying the above matrices and looking at the $(1,1)$ entry. Alternatively, being a direct sum, we have $P_K W = P_K W P_K = W P_K$. So

$RV = P_K W P_K U P_K = P_K W U P_K = P_K U W P_K = P_K U P_K W P_K = VR$.

The proof is done.

#### 3. Counter example

The following example is due to Kaijser and Varopoulos.

Example: Let $A_1, A_2, A_3 \in M_5$ be given by

$A_1 = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 1/\sqrt{3} & -1/\sqrt{3} & -1/\sqrt{3} & 0\end{pmatrix}$

$A_2 = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & -1/\sqrt{3} & 1/\sqrt{3} & -1/\sqrt{3} & 0\end{pmatrix}$

$A_3 =\begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & -1/\sqrt{3} & -1/\sqrt{3} & 1/\sqrt{3} & 0 \end{pmatrix}$

Let $p(z) = z_1^2 + z_2^2 + z_3^2 - 2z_1 z_2 - 2 z_2 z_3 - 2 z_1 z_3$.

Using your either your brain or your favourite computer algebra software (or both), you should check the following facts:

1. $\|A_i\|\leq 1$ for all $i$,
2. $A_i A_j = A_j A_i$ for all $i,j$,
3. $\|p\|_\infty := \sup_{\|z\|_1=1}|p(z)| = 5$,
4. and finally:

$p(A) = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 3 \sqrt{3}& 0 & 0 & 0 & 0 \end{pmatrix}$

and so $\|p(A)\| = 3 \sqrt{3} > \|p\|_\infty$. It follows that $A_1, A_2, A_3$ do not have a unitary dilation (otherwise, they would satisfy a von Neumann inequality), and therefore they also cannot have an isometric dilation.

This raises the question: let $A = (A_1, \ldots, A_d)$ be a tuple of commuting contractions, and suppose that they satisfy a von Neumann type inequality:

$\|p(A)\| \leq \sup_{\|z\|_1}|p(z)|$.

Does it follow that $A$ must have a unitary dilation? In other words, is the failure of a von Neumann type inequality the only obstruction to the existence of dilations? We leave this question for now. In the next lecture, we will apply the dilation theory that we have developed thus far to function theory.