40.2 $B(H)$
Definition 40.2.1 ($B(H)$).label Let $H$ be a complex Hilbert space, then $B(H) = L(H; H)$ is the algebra of all bounded linear operators on $H$.
Lemma 40.2.2.label Let $H$ be a complex Hilbert space, then the strong and weak operator topologies on $U(B(H))$ coincide.
Proof. Let $S, T \in U(B(H))$ and $\xi \in H$, then
which tends to $0$ as $S \to T$ in the weak operator topology.$\square$
Definition 40.2.3 (Partial Isometry).label Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if $T|_{\ker(T)^\perp}$ is an isometry. In which case, $\ker(T)^{\perp}$ is the initial space of $T$, and $T(H)$ is the final space of $T$.
Proposition 40.2.4.label Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^{*}T$ is a projection. In which case, $T^{*}T$ is a projection onto $\ker(T)^{\perp}$.
Proof. ($\Rightarrow$): Suppose that $T$ is a partial isometry. Let $x \in \ker(T)^{\perp}$, then $\dpn{Tx, Tx}{H}= \norm{x}_{H}^{2}$ and $\dpn{T^*Tx, x}{H}= \norm{x}_{H}^{2}$. By polarisation, for each $x, y \in \ker(T)^{\perp}$,
Therefore $T^{*}T$ is idempotent. As $T^{*}T$ is self-adjoint, it is a projection onto $\ker(T)^{\perp}$.
($\Leftarrow$): Suppose that $T^{*}T$ is a projection, then for each $x \in \ker(T)^{\perp}$, $\dpn{Tx, Tx}{H}= \dpn{T^*Tx, x}{H}= \norm{x}_{H}^{2}$.$\square$
Corollary 40.2.5.label Let $H$ be a complex Hilbert space and $T \in B(H)$, then $T$ is a partial isometry if and only if $T^{*}$ is a partial isometry.
Proof, [Corollary 12.7, Zhu93]. Suppose that $T$ is a partial isometry, then $P = T^{*}T$ is a projection onto $\ker(T)^{\perp}$ by Proposition 40.2.4. In which case, $T(T^{*}T) = T$ and $(TT^{*})^{2} = T(T^{*}T)T^{*} = TT^{*}$, so $TT^{*}$ is a projection, and $T^{*}$ is a partial isometry by Proposition 40.2.4.$\square$
Theorem 40.2.6 (Polar Decomposition).label Let $H$ be a complex Hilbert space and $T \in B(H)$, then there exists a unique pair $(P, V) \in B(H)^{2}$ such that:
- (1)
$P$ is positive.
- (2)
$V$ is a partial isometry.
- (3)
$T = VP$.
- (4)
$\ker P = \ker V$.
The pair $(P, V)$ is the polar decomposition of $T$, and
- (5)
$V$ is a partial isometry from $\ker(T)^{\perp}$ to $\ol{T(H)}$.
- (6)
$P$ and $V$ are contained in the von Neumann algebra generated by $T$.
Proof, [Theorem 12.8, Theorem 18.9, Zhu93]. (1): Let $P = |T| = \sqrt{T^{*}T}$, then $P$ is positive (1).
(2): For each $x \in H$,
Let $V_{0}: P(H) \to H$ be defined by $V(Px) = Tx$, then $V_{0}$ extends to a well-defined isometry $\ol{P(H)}\to H$. Further extend $V_{0}$ to $V$ by setting its value to $0$ on $P(H)^{\perp}$, then $V$ is a partial isometry.
(3): For any $x \in H$, $Tx = V_{0}Px = VPx$.
(4), (5): Since the initial space of $V$ is $\ol{P(H)}= \ker(T)^{\perp}$, $\ker(V) = P(H)^{\perp} = \ker(P)$.
(Uniqueness): Let $T = WQ$ be a polar decomposition of $T$ satisfying (1)-(4). By Proposition 40.2.4, $W^{*}W$ is a projection onto $\ker(W)^{\perp} = \ker(Q)^{\perp} = \ol{Q(H)}$. Thus $P^{2} = T^{*}T = QW^{*}WQ = Q^{2}$, and $P = Q$ by uniqueness of the positive square root.
Since $VP = WP$ and $\ker(V) = \ker(W) = P(H)^{\perp}$, $V = W$ on $H$, and the polar decomposition is unique.
(6): Let $A$ be the von Neumann algebra generated by $T$. By Theorem 39.3.1, $A$ is a unital $C^{*}$-algebra, so $P = \sqrt{T^{*}T}\in A$. To see that $V \in A$, it is sufficient to apply the Bicommutant Theorem.
To this end, let $S \in A'$, then $TS = ST = SVP$ and $VSP = VPS = TS$, so $SV$ and $VS$ agree on $\ol{P(H)}$. Since $\ker(V) = \ker(P) = \ol{P(H)}^{\perp}$, $SV|_{\ker(P)}= 0$. On the other hand, as $SP = PS$, $S(\ker(P)) \subset \ker(P) = \ker(V)$, so $VS|_{\ker(P)}= 0$ as well. Therefore $V \in A'' = A$.$\square$
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