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## New answers tagged operator-algebras

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Let $\pi,H_1$ be a Strinespring dilation of $\phi$ on $A$. Now let $K$ be some Hilbert space with $\dim K>\dim H_1$ and $T\in B(H,K)$ some operator. Let $H_2=H_1\oplus K$, $V_2:H\to H_2$ be given by $V_2:\xi\mapsto V\xi\oplus T\xi$, $\pi_2:A\to B(H_2)$ by $\pi_2(a)=\pi(a)\oplus 0$. Then $$V_2^*\pi_2(a)V_2\xi=(V^*\oplus T^*)(\pi(a)\oplus 0)(V\xi\oplus ... 1 If you only consider elementary tensors, then what would their sum be? Remember that you want A\odot H to be a vector space. 2 If you do not need any control over the norm of the vector \alpha, then, yes, such a vector exists. Take any vector with norm less than \min (\frac{\varepsilon}{2}, \frac{ \varepsilon}{ 2 \Vert T \Vert}). If you add the assumption that \Vert \alpha \Vert=1, such a vector \alpha need not exist. Consider the operator T(x)=-x. Its restriction to any ... 3 You ask what is the definition of a positive element in E; the element a \in E is positive if a is positive when we think of it as an element in A. But an operator system is more than that. It also allows us to define an order structure on M_n(E) for each n and to say whether each matrix [a_{ij}]\in M_n(E) is positive or not. The statement "an ... 0 The case you mention is one of the very few where the commutant can be expressed explicitly. Namely,  L(G)'=R(G),  where R(G)  is the von Neumann algebra of the right regular representation R_g:e_h\mapsto e_{hg} . 1 Any finite dimensional Banach algebra of dimension not a square is not isomorphic to B(X) for any Banach space X. 1 It follows just from the definition and the fact that \varphi is *-preserving: if a\in A_\varphi, then$$ \varphi((a^*)^*a^*)=\varphi(aa^*)=\varphi(a)\varphi(a^*)=\varphi(a^*)^*\varphi(a^*). $$1 You have that A is a maximal ideal in its unitization. Then you can construct in \tilde A a quasicentral approximate unit \{b_j'\}\subset A. Note also that a state on A has a unique extension to a state in \tilde A; let us still call this extension \phi. We also have \limsup \phi(b_j')=1: if \phi(b_j')<1-\delta for all j, then ... 2 To see that L\subset\ker\phi, if a\in L then by Kadison's inequality$$ 0\leq|\phi(a)|^2=\phi(a)^*\phi(a)\leq\phi(a^*a)=0, $$so \phi(a)=0. And now for b\in L^*, then b^*\in L, so \phi(b)=\overline{\phi(b^*)}=0 (this last part is just the fact that the kernel of a selfadjoint functional contains adjoints; it is an ideal, in fact). 1 Note that V^*V is a positive element with \|V^*V\|=1. This means that \sigma(V^*V)\subset[0,1]. If you consider the function f(x)=x(1+h(x)). If h is chosen appropriately, then |f(x)|\leq1 for all x (because the term 1+h(x) is positive only on the small neighbourhood of t). Now functional calculus gives you ... 1 Let _AX_B be the bimodule, then by definition, \widetilde{X} is the same space with right A action as x\cdot a=a^{*}x and similarly for left B action. The inner product on \widetilde{X} is still <x,y>_A=_A<x,y> Now the map from _AX_B\otimes \widetilde{_BX_A} to _AA_A is given on elementary tensors by x\otimes y \rightarrow ... 3 You have a typo at the end of your question: it should say xp=yp. This is similar to Lusin's theorem in the sense that in the original theorem you have the continuous functions hanging aroudn betweent the measurable functions, and off a set of small measure, you can get your measurable function equal to the continuous function. Here you have to think of ... 1 Indeed we also have T_F \to T in the strong operator topology. The strong operator topology is defined by the seminorms$$p_F(S) = \sup \{\lVert S(x)\rVert : x \in F\},$$where F traverses the finite subsets of X. The construction immediately yields$$p_F(T_F - T) = 0$$for any linearly independent finite F\subset X, and it is straightforward to ... 1 I wouldn't really call the idea of using rank-one projections a "trick". Here is what I consider an even more straightforward approach, which uses the same idea in a more explicit way: Let \theta:B(H)\to B(H) be a *-automorphism. Fix an orthonormal basis \{\xi_j\} of H, and write E_{jj} for the corresponding rank-one projections, i.e. ... 1 If you have access to Davidson's "C^*-Algebras by Example", it is Theorem I.9.16 there. If you don't, I'll try to post the proof later, but it is a one-page+ affair. 1 There's a classic trick to produce the unitary. I was told once who this is due to, but I've unfortunately forgotten. Recall that, if \xi,\eta \in H, then \eta \otimes \overline \xi denotes the rank-1 operator which sends \zeta \mapsto \langle \xi,\zeta \rangle \eta. Here, the inner-product is conjugate linear in the 1st slot. It's easy to check that ... 1 Just for fun, the matrix trick mentioned in the answer by @Tom Cooney: Assume A is unital. If not, extend f to unitization of A. For any a \in A,$$ \left[ \begin{matrix} 1 & a^*\\ a & aa^* \end{matrix} \right] $$is positive in M_2(A). The 2-positivity of f then tells you$$ f(a) f(a^*) \leq f(aa^*). $$1 Your question raises the point of what is meant by an "isomorphism of von Neumann algebras". Any C^*-isomorphism will of course preserve order, and then it will be normal: if \varphi:A\to B is a *-isomorphism and \{a_j\}\subset A^{\rm sa} is a bounded increasing net with least upper bound a, then \varphi(a) is the least upper bound for ... 0 If T_gh=0, then g=0 on [1/2,1]; any polynomial that is zero on an interval is equally zero, so g=0. That is, h is separating. The commutant of A agrees with the commutant of the weak closure of A. This is the set of multiplication operators by all essentially bounded functions (i.e., L^\infty[0,1]). In particular T_k, with k the ... 0 Let H=L^{2}[0,2\pi]. Let f \in \mathcal{D}(A) iff f is equal a.e. to an absolutely continuous periodic function \tilde{f} on [0,2\pi] such that \tilde{f}'\in L^{2}[0,2\pi]. Define A : \mathcal{D}(A)\subset H\rightarrow H by Af=-i\tilde{f}' for all f\in\mathcal{D}(A). A is closed, and the domain of A is dense in H. (A is, in fact, ... -1 Proof: For all h \in H, \langle A^* BA h, h \rangle = \langle BA h, A h \rangle \leq \|B\| \cdot \langle A h, A h \rangle = \langle \|B\|A^*A h, h \rangle. 2 In general it is not the identity of H what you get, but the range projection of h. For example if$$ h=\begin{bmatrix}1/2&0\\0&0\end{bmatrix}, $$then$$ (h+1/n ...

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One of the most important results about completely positive maps is Stinespring's Dilation Theorem. Suppose that $f:A \to B$ is a completely positive map, where $A$ and $B$ are $C^*$-algebras. Then we can find a Hilbert space $H$ such that $B \subseteq B(H)$. Stinespring's Theorem then states that there exists a Hilbert space $K$, $*$-homomorphism \$\pi: A ...

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