Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 1 / 8

A bounded self-adjoint operator $T$ is a positive operator when $⟨T\xi ,\xi ⟩\ge 0$ for every $\xi$. Equivalently its quadratic form is nonnegative. We construct its positive square root of an operator directly and prove uniqueness.

If $T=0$ its positive square root is zero. Otherwise set $r=\Vert T\Vert$, $B=T/r$, and $C=I-B$. Then $0\le B\le I$ and $C$ is positive. It is also a contraction. Indeed, the Cauchy-Schwarz inequality for the positive form $⟨B\xi ,\eta ⟩$ gives $\Vert B\xi {\Vert }^2\le ⟨B\xi ,\xi ⟩$, and henceThe scalar binomial series has the formThe last equality follows by taking $z\uparrow 1$ in the nonnegative series. Thereforeconverges in operator norm. Every $C^n$ is a positive contraction: its even powers are squares, its odd powers have form $C^{k}C{C}^k$, and their norms are at most one. Each partial sum for $S/\sqrt{r}$ is consequently positive, since it is at least $(1-{\sum }_{n\le N}{c}_n)I$. The limit is positive. Multiplying the absolutely convergent series and using the scalar coefficient identity gives $S^2=r(I-C)=T$. This construction also shows that $S$ is a norm limit of polynomials in $T$ and thus commutes with every operator commuting with $T$.

Now suppose $R$ is another positive square root. It commutes with $T=R^2$, hence with $S$. ThusThe difference vanishes on the range of $R+S$ and on its closure. It also vanishes on its kernel: if $(R+S)\xi =0$, the fact that $R$ and $S$ are positive operators gives $⟨R\xi ,\xi ⟩=⟨S\xi ,\xi ⟩=0$, and the Cauchy-Schwarz inequality for their positive forms implies $R\xi =S\xi =0$. Since $R+S$ is self-adjoint, the orthogonal sum of its kernel and closed range is all of $H$. Therefore $R=S$, proving

The polar decomposition of a bounded operator states that every bounded $T:H\to H$ is uniquelywhere $V$ is a partial isometry with initial space $\overline{\mathrm{Ran}|T|}=(\ker T{)}^{\perp }$ and final space $\overline{\mathrm{Ran}T}$. Its initial and final orthogonal projections are $V^{\ast }V$ and $V{V}^{\ast }$.

Existence of $|T|$ was proved in (a). For every $\xi$,Thus the map $|T|\xi \mapsto T\xi$ is well defined and isometric on $\mathrm{Ran}|T|$. It extends uniquely to an isometry between its closed range and $\overline{\mathrm{Ran}T}$. Extend it by zero on $\ker |T|=\ker T$ to obtain $V$. The equation and the projection assertions follow immediately. Any other normalized partial isometry with the same product must agree on the dense range of $|T|$ and vanish on its orthogonal complement, giving uniqueness. The kernel normalization is essential; without it an action on the kernel could be added.

A useful consequence for a Von Neumann algebra $M$ is that the polar factors of $T\in M$ also belong to $M$. The polynomial construction in (a) puts $|T|$ in the norm-closed algebra $M$. For $\epsilon >0$, the inverse of $|T|+\epsilon I$ belongs to $M$ by a convergent Neumann series, andThese operators are contractions, since $\Vert |T|\xi \Vert \le \Vert (|T|+\epsilon I)\xi \Vert$. On vectors $|T|\xi$ the difference from $V$ has norm at most $\epsilon \Vert \xi \Vert$, and on the kernel it is zero. Density and the uniform bound prove the strong limit. Strong closedness of $M$ then gives $V\in M$. This justifies the use of algebra-valued polar factors in Question 7(c).

A closed subspace invariant under a star-algebra is reducing, since it is invariant under every adjoint as well. Thus the orthogonal projections $P_1,P_2$ onto $H_1,H_2$ commute with $M$. Consider the bounded intertwinerIts kernel is $H_1\cap H_2$, so its initial polar space isIts closed range isFor the inclusion from left to right, $(I-P_2)h_1=h_1-P_2{h}_1$ lies in $H_1+H_2$ and is orthogonal to $H_2$. Conversely, approximate any $z$ in the displayed $K_2$ by $h_{1,n}+h_{2,n}$ and apply $I-P_2$. The images converge to $z$ and equal $T{h}_{1,n}$, proving the reverse inclusion after closure. The overline on the sum is present in the original PDF and is essential for potentially nonclosed sums.

Take the polar decomposition $T=V|T|$ between these Hilbert spaces. Because $T$ intertwines the two actions of $M$, its adjoint does also, so $T^{\ast }T$ commutes with the action on $H_1$. Its positive square root therefore commutes as well. The identity $V|T|\xi =T\xi$ shows on its dense initial range that $V$ intertwines, and continuity extends this to $K_1$. It is an isometry onto $K_2$. ConsequentlyThe unitary here is a unitary between the indicated Hilbert space modules, not an assertion that the unclosed algebraic sum is a Hilbert space.

One implication is immediate: a unitary module equivalence makes each space an isometric copy of the entire other space, which is closed and invariant. For the converse, choose isometric intertwiners $u:H_1\to H_2$ and $v:H_2\to H_1$ supplied by the two embeddings. Their ranges are closed reducing subspaces. We construct the unitary, proving the Schröder-Bernstein theorem for Hilbert space modules in this setting.

Put $w=vu$, an isometric intertwiner of $H_1$ into itself, and let $D=H_1\ominus v{H}_2$. The spaces $w^{n}D$, $n\ge 0$, are mutually orthogonal. For $m>n$, applying the isometry relation reduces their inner product to one between $D$ and $w^{m-n}D\subseteq w{H}_1\subseteq v{H}_2$, which is zero. Henceis a closed reducing submodule of $H_1$, and $K=D\oplus wK$. Its orthogonal complement lies in $v{H}_2$, because it is orthogonal to $D$. Combining these decompositions givesApply the inverse isometry $v^{\ast }$ on $v{H}_2$ to obtainThe operatoris therefore an isometry onto $H_2$: it is isometric on each orthogonal summand and their images are the two orthogonal summands just exhibited. Both pieces intertwine the action of $M$, as do the projections onto $K$ and $K^{\perp }$, so $W$ is a unitary module equivalence. This proves the reverse implication with no finiteness assumption on either module.