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0

Is a trivial consequence of: $$|T_1(x)T_2(x)|\le\|T_1\||T_2(x)|,$$ $$|T(x)|=|\overline{T(x)}|,$$ $$|T(x)\overline{T(x)}|= |T(x)|^2.$$


1

The assertion says that $X^*X\leq I$ implies $\|X\|\leq 1$. First note that $\|X\|^2=\|X^*X\|$, so it is enough to show "if $0\leq X\leq I$, then $\|X\|\leq 1$". Next you note that $0\leq X\leq I$ implies that $\sigma(X)\subset[0,1]$. And finally, using that $X$ is selfadjoint, $\|X^2\|=\|X\|^2$, which you use to show that $\|X\|$ equals the spectral ...


0

Let $f(t) = t^2$, then $f(t)$ is a non-negative operator monotone function on $[0, \infty)$. There is a theorem that says $f(||X||) \leq ||f(|X|)||$ when $f(t)$ is a non-negative operator monotone function and $||.||$ is a normalized unitarily invariant norm. Now, the statement at the beginning of $\star$ means that $|A^{1/2}B^{-1/2}|^2 \leq I$, ...


3

Let us assume that the question requires $E$ to be a locally compact Hausdorff space. It is well-known that $C(E)$ with the supremum norm is a C$^*$-algebra. Sakai proved in 1971 that a C$^*$-algebra is a dual precisely when it is a von Neumann algebra. In terms of $E$, this means that it has to be compact and extremely disconnected: the closure of every ...


2

What you do is you think of $C^*(\Gamma)$ as already represented. Then the inclusion $\Lambda\to\Gamma$ is a unitary representation, and you can apply the universality you defined.


0

This question is too broad and requires a lot of time answer it, so I suggest you to take a look at section 13 p.117 in Amenable locally compact groups. by Jean-Paul Pier


0

Recall that one definition of a faithful state is that the associated GNS representation is faithful. So if the extension is positive definite then you can do GNS to get a unitary rep of $\Gamma$ which gives a rep of $C^*(\Gamma)$. Then as the hint says the GNS rep of $C^*(\Lambda)$ will be a subrep $\pi$ composed with the rep of $C^*(\Gamma)$. But this rep ...


1

They are just saying that if $A=\bigcup_nA_n$, then $\ell^2(A)\simeq\bigoplus \ell^2(A_n)$. The direct sum has to be understood as a direct sum of Hilbert spaces, i.e. an $\ell^2$-sum. The left regular representation of $\Lambda$ leaves each direct summand invariant. So we can think of it acting that way on $\ell^2(\Gamma)$ (as a block-diagonal operator). ...


1

Yes, this is a standard notation.


1

Yes, because when you do GNS your state becomes a vector state. So, if in the direct sum you put all zeroes and the cyclic vector for your $w_\phi$ in its corresponding component, you get a vector that implements your state.


1

It is a direct consequence of the fact that any C$^*$-algebra has an approximate identity. Then Cohen's Factorization Theorem tells you directly that $A\subset A^2$. It can also be obtained directly from the following result: For any $a\in A$, there exists $b\in A$ such that $a=b(a^*a)^{1/4}\in A^2$. Proof. (from Davidson's book) Let ...


1

Here you start with $\ell^2(\Gamma)=\ell^2(\mathbb Z)$. The left regular representation $\lambda$ maps every $n\in\mathbb Z$ into $B(\ell^2(\mathbb Z))$ by $\lambda(n)f(m)=f(m-n)$. The Fourier transform $\ell^2(\mathbb Z)\to L^2(\mathbb T)$ in this context is the map $U:f\mapsto \sum_{n\in\mathbb Z}f(n)\,e^{int}$. Now, using that $\{e^{int}\}_n$ is an ...


2

Fix $F=\{s_1,\ldots,s_n\}$. Let $\bar s=(s_1,\ldots,s_n)$. Then, as a positive functional is completely positive, $$ 0\leq\phi^{(n)}(\bar s^*\bar s)=[\phi(s^{-1}t)]_{s,t\in F} $$


3

Yes. Because a finite-dimensional von Neumann algebra will always have minimal projections, and minimal projections are abelian.


1

You can just see it as an identity: the shift operator can be expressed in terms of a Taylor series, and then we just compute its closed form. There are other visualizations for this, though. You can think of $1 + \frac{d}{dx}$ as an infinitesimal shift operator, and exponentiation accumulates all of the infinitesimal shifts up into an actual shift. In ...


2

So you have suggested $e^{t \frac d{dx}} f(x) = f(x+t)$. This is to be expected, because you would formally expect (i) $e^{0 \frac d{dx}}$ to be the identity operator, (ii) $\frac d{dt} [e^{t \frac d{dx}} f(x)] \big|_{t=0} = f'(x)$, and (iii) $e^{t \frac d{dx}} e^{s \frac d{dx}}e^{(s+t) \frac d{dx}}$. And look, the formula you propose works. Look here for ...


2

As Daniel mentioned, we have to assume that $\Gamma$ is infinite. Also, that the constant diagonal is not zero. There is probably a more elegant way, but here is an argument. If you have such a compact $T$, you can assume that the constant diagonal is the main diagonal (otherwise you replace $T$ with $ts^{-1}T$ which is again compact, and $\langle ...


2

Each $x$ is a $\textit{finite}$ linear combination of elements from $\Gamma$. So since elements of $\Gamma$ get mapped to unitaries you have that $x=\sum a_gu_g\Rightarrow\|x\|\leq \sum |a_g|\cdot\|u_g\|=\sum|a_g|<\infty $ since the sum if finite.


2

If $H$ is a Hilbert space, given a unitary representation $\gamma:\Gamma\rightarrow\mathcal{U}(H)$, how do you induce a (at first, simply linear) function $\widetilde{\gamma}:\mathbb{C}(\Gamma)\rightarrow\mathcal{B}(H)$? What about the converse: if $c:\mathbb{C}(\Gamma)\rightarrow\mathcal{B}(H)$ is a (unital) *-representation, how do you obtain a unitary ...


2

You are probably used to think of $\ell^\infty(\Gamma)$ as the set of bounded sequences indexed by $\Gamma$. That's nothing but $$\{f:\Gamma\to\mathbb C:\ \sup\{|f(t)|:\ t\in \Gamma\}<\infty\}. $$ If you think about it $f_t$ is nothing but notation for $f(t)$.


1

For any $a\in A$, $b\in \mathcal J$, we have $$ \|a\|'=\|a\|'-\|b\|'\leq\|a+b\|'\leq\|a\|'+\|b\|'=\|a\|'. $$ So $\|a+\mathcal J\|'=\|a\|'$. Then $$ \|a^*a+\mathcal J\|'=\|a^*a\|'=\|a\|'^2=\|a+\mathcal J\|'^2. $$


2

Well, since $\{\delta_t\}$ is a orthonormal basis for $\ell^2(\Gamma)$, to find $\lambda_s^\ast$ and $\rho_s^\ast$, it suffices to look at $\langle \lambda_s^\ast \delta_t, \delta_u \rangle$ and $\langle \rho_s^\ast \delta_t, \delta_u \rangle$. For example, let's find $\lambda_s^\ast$. Now, $$ \langle \lambda_s^\ast \delta_t, \delta_u \rangle := \langle ...


2

Suppose that $\pi_1,\pi_2:C^*(\Gamma)\to B(H)$ are $*$-homomorphisms that satisfy $\pi_1(s)=u_s=\pi_2(s)$ for all $s\in \Gamma$. Then $$ \pi_1(\sum_{s\in \Gamma}a_ss)=\sum_sa_2\pi_1(s)=\sum_sa_s\pi_2(s)=\pi_2(\sum_sa_ss). $$ Also, $$ \|\pi_j(x)\|\leq\sup\{\|\pi(x)\|:\ \pi\}=\|x\|_u, $$ so both $\pi_1,\pi_2$ are contractions (in particular, continuous). As ...


3

Suppose that $\overline{aA_+a}\subsetneq A_+$. Use Hanh-Banach to construct a functional with $f(b)=1$ and $f(aca)=0$ for all $c\in A_+$. This functional, extended to all of $A$, is a linear combination of four states (see II.6.3.4 in Blackadar): at least one of them, say $f_1$, will satisfy that $f_1(b)>0$; by the Jordan Decomposition, we also have that ...


0

In your setting, $f$ is just a bounded function on the group with values in real or complex numbers. $f(t)$ is its value at $t$.


1

In the polar decomposition $x=v|x|$, the partial isometry $v$ maps the range of $x^*$ into the range of $x$. The former is below $(1-p)$, while the latter is below $p$.


2

This is more a question about functional analysis than algebraic topology: A discrete group $\Gamma$ is a group equipped with the discrete topology. If $\Gamma$ has no topology (that is, we are in a purely algebraic setting), we can always equip it with the discrete topology, and we say that $\Gamma$ is discrete nonetheless. 2./3. Given a (discrete) set ...


2

A group with the discrete topology (one may say that a discrete group is a group with no topology at all). $\ell_2(\Gamma)$ is the Hilbert space of square-summable functions on $\Gamma$, $\ell_\infty(\Gamma)$ the space of all bounded functions on $\Gamma$ with the supremum norm, $B(\ell_2(\Gamma))$ the algebra of all bounded operators on the Hilbert space ...


0

1) The point here is to construct finie-dimensional representations. The restrictions $A_n$, $B_n$ are finite-dimensional but not unitaries, so the trick is to use the fact that any contraction is a corner of a unitary. Indeed, given any contraction $T$ n $B(H)$, the operator $$ \begin{bmatrix}T&(I-TT^*)^{1/2}\\ (I-T^*T)^{1/2}&-T^*\end{bmatrix}\in ...


0

That states are of the form you mention is true for $B(H)$ for any Hilbert space. In this case, you are asking about the dimension 2 case, and in such case one can give a very explicit answer. Let $f:M_2(\mathbb C)\to\mathbb C$ be a state. For any $x\in M_2(\mathbb C)$, we have $x=\sum_{k,j}x_{kj}e_{kj}$ for the matrix units $e_{11},e_{12},e_{21},e_{22}$. ...


3

Here is a simple example. Let $X$ be a non-compact LCH space. Every multiplicative linear functional has the form $\phi_x(f) := f(x)$ for $f \in C_0(X)$, but if you choose a sequence (or net) $x_n$ which goes off to $\infty$ then $\phi_{x_n}(f) = f(x_n) \to 0$ for every $f$ which means $\phi_{x_n} \to 0$ in the weak*-topology.


2

Consider the Banach algebra $C_0(Z)$ and the evaluation functionals $l_n(f)=f(n)$. They all have norm 1 but converge to 0 in weak-star. The fact that this can't happen in the unital case is because all the functionals send the identity to 1, hence so must their limit, so the limit can't be 0.


2

Gelfand spectrum is a set of non-zero characters. For non-unital algebra you can not prove that the limit of the net of characters is non-zero character, that is the place where you argument fails. In fact, Alexandrov's compactification of the spectrum of non-ubital Banach algebra is just an addition of zero character.


0

As you say, one needs to test that $\omega\circ\varphi$ is normal for all normal states $\omega$. I see two immediate ways to achieve the proof from here: 1) Use that $\omega=\text{Tr}(H\cdot)$ for some positive trace-class $H$. Then, if $T_j\nearrow T$, $$ ...


1

Yes it does. As $M_n$ is simple, both homorphisms are monomorphisms. If we replace $A$, $B$ with $\pi_1(M_n)$, $\pi_2(M_n)$ respectively, we may assume that $\pi_1$ and $\pi_2$ are isomorphisms. In particular, from $\pi_1(U)\in A_+$ we get that $\sigma(U)=\sigma(\pi_1(U))\subset[0,\infty)$. But then $\sigma(U)=\{1\}$, and $U=I$. With a similar reasoning ...


1

The answer is no, in general. Let $n=m=2$. Define, for $x\in M_2(\mathbb C)$, $$ f(x)=x_{11}\,\begin{bmatrix}2&0\\0&1\end{bmatrix}+x_{22}\,\begin{bmatrix}1&0\\0&0\end{bmatrix}. $$ This map is linear. It is also positive, because if $x\geq0$ then $x_{11},x_{22}\geq0$ and the two matrices are positive. And it is completely positive, because ...


0

The definition of the essential spectrum does not ask for any specific property of the operator. It is just the spectrum of the image of the operator in the Calkin algebra. More explicitly, $\sigma_{\rm ess}(T)$ consists of those $\lambda\in\sigma(T)$ such that there exists a sequence $\{x_n\}\subset H$ with no convergent subsequence and such that ...


3

Yes. You can factorise the identity map on $A\oplus B$ in the point-ultraweak topology taking direct sums of the respective approximations for $A$ and $B$. Then you pretend that $M_{k_A(n)}$ and $M_{k_B(n)}$ live in some bigger full matrix algebra and you are done. In other words, you can replace the matrix algebras $M_{k(n)}$ in the definition of ...


2

For von Neumann algebras in general it is false, there is two much leeway in the choice of the traces. For instance let $M=N=\mathbb C\oplus\mathbb C$, with $f$ the identity map. Let the traces be $$ \tau_M(x,y)=\frac x4+\frac{3y}4, \ \ \tau_n=\frac{3x}4+\frac y4. $$ Then, for $a=(1,0)$, $$ \tau_M(a)=\tau_M(1,0)=\frac14<\frac34=\tau_N(1,0)=\tau_N(f(a)) ...


1

The difference between nuclearity and exactness is that in the range of the maps $\psi_n$: when $\psi_n:M_{k(n)}(\mathbb C)\to \pi(A)$, the algebra is nuclear. When the range of $\psi_n$ is allowed to be bigger than $\pi(A)$, then $A$ is exact.


1

Likely means that for every $a\in F$ there exists $b\in B$ with $\|a-b\|<\varepsilon$. You fix $F=\{a_1,\ldots,a_n\}\subset A$ finite and $\varepsilon >0$. By hypothesis there exists $B$ nuclear, and $\{b_1,\ldots,b_n\}\subset B$ with $\|a_j-b_j\|<\varepsilon/3$. As $B$ is nuclear there exist $k\in \mathbb N$, $\varphi:B\to M_k(\mathbb C)$, and ...


1

It is a basic consequence of the Hahn-Banach theorem that the weak and norm closures of a convex set agree. This is the key result used in the argument. Here they want to use that fact, but they have the problem that they need to be able to choose a net of operators in $C$. That's the role of the finite sets $F$: they will be the indices of the net: for ...


1

If $A$ is a C$^*$-algebra and $p\in A$ is a projection, then $pAp$ is hereditary. For this, assume that $A\subset B(H)$, and suppose that $a\leq b$, with $a\in A$, $b\in pAp$. Then, as $p$ is a projection, $b=pbp$. From $pbp-a\geq0$, we have $$ 0\leq(I-p)(pbp-a)(I-p)=-(I-p)a(I-p). $$ This implies that $(I-p)a(I-p)=0$. Writing $a=c^*c$, we have ...


3

The key concept here is that of annihilator. For a subset $X\subset A$, the annihilator of $X$ is the subset of $A^*$ given by $$ X^o=\{f\in A^*: f(x)=0\,\forall x\in X\}. $$ (note that almost all equal signs below mean isomorphism) It is not hard to prove that $X^{oo}=X^{**}$. Now if $A=X\oplus Y$, then $A^*=X^*\oplus Y^*$. Also $X=A/Y$, $Y=A/X$. And it is ...


4

This is a proof by induction. The base case is $[A, B] = I = 1 B^{1-1}$. For the induction step, assuming it is true for $n=k$, i.e. $[A,B^k] = k B^{k-1}$, we compute: $$ [A, B^{k+1}] = A B^{k+1} - B^{k+1} A = A B^{k+1} - B^k A B + B^k A B - B^{k+1} A$$ Now $$A B^{k+1} - B^{k} A B = (A B^k - B^k A) B = k B^{k-1} B = k B^k$$ and $$B^kAB - B^{k+1} A = B^k ...


2

Note that Murray-von Neumann equivalence depends on the algebra where you consider it. Usually stably equivalence of projections is considered on $(A\otimes K(H))^{**} $, and you require $P\otimes1\sim Q\otimes1$. So let $A=M_3 (\mathbb C) $, $P=E_{22}+E_{33}$, $Q=E_{33}$ (or any other rank-one projection for that matter), which are non equivalent as they ...


1

Since you already know that $M_n(A^{**})=M_n(A)^{**}$, the matrices are playing no role here. So all you need to show is that $A_+$ is ultraweakly dense in $A^{**}_+$.


2

Every von Neumann algebra $M$ has unique predual $M_*$. Second dual $A^{**}$ of $C^*$ algebra $A$ in fact is a von Neumann algebra which is called enveloping von Neumann algebra. For $M=A^{**}$ we have unique predual which is clearly (now) is $A^*$.


2

The relevant consequence of the Hahn-Banach theorem(s) here is that in a normed space $X$, for every convex subset $M$, the weak closure and the norm-closure coincide. Since the norm topology is finer than the weak topology, we always have $\operatorname{cl}_{\lVert\cdot\rVert}(M) \subset \operatorname{cl}_w(M)$, and the Hahn-Banach theorem tells us that we ...



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