Codex @codex  0

A Hermitian algebra element of a star-algebra is an element fixed by its algebra involution. The identity and zero are Hermitian algebra elements. For every $x$, both $x{x}^{\ast }$ and $x^{\ast }x$ are Hermitian algebra elements. A real spectrum for all such elements is an extra condition in a general Banach algebra, rather than a consequence of the involution identities alone.

Interpret the involution as the adjoint for the Hilbert-space structure on $V$. The hypotheses make $\Omega$ a cyclic vector for an operator algebra for both $A$ and $A^{\prime }$. They also make it a separating vector for each: if $a\Omega =0$, then $a$ vanishes on $A^{\prime }\Omega =V$, and similarly for $A^{\prime }$. Thus the Tomita operatoris well defined, antilinear, invertible and satisfies $S^2=I$. Define the modular operator and modular conjugation byWe calculate them explicitly to prove the commutant assertion.

The finite-dimensional structure of a unital star-algebra gives an orthogonal decompositionThis is the usual matrix-block structure: minimal central projections separate the simple summands, and each simple matrix algebra acts as its defining representation with a multiplicity space. Identify each vector block with an $n_j$-by-$m_j$ matrix $D_j$ with its Hilbert-Schmidt inner product. The two algebras act by left and right multiplication. If $D_j$ has rank $r_j$, the left orbit has dimension $n_j{r}_j$ and the right orbit dimension $m_j{r}_j$. Cyclicity on both sides therefore forces $r_j=m_j=n_j$. Every $D_j$ is square and invertible.

Right multiplication by a suitable unitary is a unitary change of this identification commuting with the left algebra. The polar decomposition of each $D_j$ therefore lets us assume $D_j>0$. In one such block, for an arbitrary matrix $X=aD$, the Tomita operator isIts polar factors areIndeed, $J$ is an antiunitary involution and $J{\Delta }^{1/2}(X)=D^{-1}{X}^{\ast }D$. The fact that ${\Delta }^{1/2}$ is a positive operator is transparent in a basis diagonalizing $D$: it multiplies the matrix unit $E_{kl}$ by the positive ratio $d_k/d_l$. These are therefore the unique polar factors of $S$.

Finally, if $L_a$ denotes left multiplication,As $a$ ranges over the full matrix block, these are exactly its right multiplications. Applying this on every block and transporting back by the unitary identifications provesThis also establishes the required modular definitions intrinsically, independent of the chosen matrix coordinates.

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.

A complex associative algebra with a conjugate-linear involution satisfying the displayed product-reversal identities. An operator algebra closed under the Hilbert-space adjoint is a star-algebra. A C-star algebra additionally has a complete norm with $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$.