User aaron tikuisis - MathOverflow

Answer by Aaron Tikuisis for Homomorphisms preserving constant functions

2013-05-23T17:48:13Z

This is perhaps just a partial answer; but I feel that Will's answer is incomplete, and this is a more interesting question that he is suggesting.

Given a *-homomorphism $\phi$, composing with point-evaluation at any point $x \in S^1$ gives us a representation $$\phi_x: C(S^1, M_n) \to M_m, $$ which is therefore of the form $$ \phi_x(f) = \alpha_x(\text{diag}(f(y(x,1)), \dots, f(y(x,k)))), $$ where $\alpha_x$ is an automorphism of $M_m$ and $y(x,1),\dots,y(x,k) \in S^1$ (and $k=m/n$).

Now, $\phi$ sends constant functions to constant functions if and only if $\alpha_x \circ \alpha_y^{-1}$ acts as the identity on the $M_m \otimes 1_k \subset M_n$.

I'm not sure if there is a more explicit characterization. Some remarks:

Certainly, $\phi$ preserving constant functions doesn't imply that $\alpha_x = \alpha_y$.
$\alpha_x$ isn't uniquely determined (when there are repeats among $y(x,1),\dots,y(x,k)$, and related to this, needn't be continuous.

Answer by Aaron Tikuisis for Inductive limit of mapping tori

2013-05-11T07:57:12Z

Your algebras $A_n$ and $B_n$ are isomorphic (and thus isomorphic to $C(S^1,M_n)$). Due to a classification theorem of Elliott, this means that, if your limits are simple, then they are not isomorphic iff their Elliott invariant (ordered $K_0$-group, $K_1$-group, trace simplex, and pairing between traces and $K_0$) differs.

I'm not sure about the other constraints you have (and I'm not sure if you really want the fibres of $A_n$ to be $M_n$ instead of say $M_{k_n}$ for some integer $k_n$), but probably the $K_0$-groups are forced to be isomorphic (eg. if the connecting maps are unital). But, you should probably be able to get different $K_1$-groups.

Answer by Aaron Tikuisis for Inductive limit of C*-algebras

2013-05-06T20:16:19Z

The converse holds if the inductive limits involve semiprojective building blocks. That is: suppose $$\varinjlim (A_i,\phi_i^{i'}) \cong \varinjlim (B_j,\psi_j^{j'})$$ (here my notation is $\phi_i^{i'}:A_i \to A_{i'}$, etc.), and each $A_i$ and $B_j$ are separable and semiprojective, then there exists subsequences $(A_{i_k}), (B_{j_k})$ and an approximate intertwining between these.

Subgroups of $\mathbb{Z}^n$

2013-02-16T17:18:36Z

I hope that the following problem isn't actually elementary (at least, for the sake of the fact that I'm posting it here), and I apologize if it is. I did try hard to solve it first.

Let $V$ be a $\mathbb{Q}$-vector subspace of $\mathbb{Q}^n$, and let $G = V \cap \mathbb{Z}^n$. Does there exist a linearly independent generating set for $G$ (i.e. a subset of $G$ such that every element of $G$ can be expressed uniquely as as $\mathbb{Z}$-linear combination of elements of this subset)? Is there an algorithm to find it (given a basis for $V$)?

Answer by Aaron Tikuisis for General recipe for building C*-algebras out of combinatorial object

2013-02-04T06:29:31Z

Benjamin's answer nicely describes where the relations for graph $C^\ast$-algebras come from. Here is an attempt to answer the question suggested by the title: what is the general recipe for constructing $C^\ast$-algebras from other mathematical objects. This may not work for every type of $C^\ast$-construction, but it seems to be the general formula behind at least group $C^\ast$-algebras, crossed products, graph $C^\ast$-algebras (and more generally, $C^\ast$-algebras of inverse semigroups), semigroup $C^\ast$-algebras, and ring $C^\ast$-algebras.

To the mathematical object in question is associated a natural and concrete Hilbert space and a natural collection of operators on that Hilbert space (often called the regular representation). Take the obvious $*$-algebraic relations that hold between these operators; then these are the relations used to define the universal $C^\ast$-algebra.

Here is this formula worked out for a few examples:

Group $C^\ast$-algebras: For a discrete group $G$, the natural concrete Hilbert space is $\ell^2(G)$ (with canonical ONB $(\xi_g)_{g \in G}$), and to each group element $g \in G$ is associated the operator $\lambda_g$ defined by $$ \lambda_g(\xi_h) = \xi_{gh}. $$ The obvious relations here are, that $\lambda_g$ is a unitary and $\lambda_g\lambda_h = \lambda_{gh}$; that is to say, that $\lambda$ is a group homomorphism between $G$ and the unitary group of $\mathcal{B}(\ell^2(G))$.

Of course, the $C^*$-algebra generated by $\{\lambda_g\}$ is called the reduced $C^\ast$-algebra of $G$, while the universal one (generated by $\{u_g: g \in G\}$ satisfying the relations that they are unitary and $u_gu_h = u_{gh}$) is called the (full) group $C^*$-algebra of $G$.

Dynamical systems: Though this can be done more generally, let's stick to a discrete group $G$ acting by homeomorphisms $\alpha_g$ on a compact metric space $X$. The natural concrete Hilbert space is $\ell^2(G \times X)$ (with ONB $(\xi_{g,x})_{g \in G, x \in X}$). For each $g \in G$, we may associate the operator $\lambda_g$ defined by $$ \lambda_g(\xi_{h,x}) = \xi_{gh,x}. $$ For each $f \in C(X)$, we may also associate the operator $D_f$ defined by $$ D_f(\xi_{g,x}) = f(g^{-1}x)\xi_{g,x}. $$ The obvious relations are: that $g \mapsto \lambda_g$ is a group homomorphism from $G$ to the unitary group; that $f \mapsto D_f$ is a $*$-homomorphism; and that $$ \lambda_g D_f \lambda_g^* = f \circ \alpha_g. $$

Graph $C^\ast$-algebras: The details of this construction are more or less explained in Benjamin's answer. The concrete Hilbert space has an orthonormal basis $(\xi_{\pi})$ indexed by the nonempty paths in the graph. An edge $e$ gives rise to the operator $v_e$ given by $$ v_e(\xi_{\pi}) = \begin{cases} \xi_{e\pi},\ &\text{if $e\pi$ is a path,} \\ 0,\ &\text{otherwise.} \end{cases} $$ A vertex $v$ gives rise to the operator $p_v$ which is the projection onto the span of $$ \{\xi_{\pi}: \pi \text{ is a non-empty path beginning at $v$}\}. $$

Answer by Aaron Tikuisis for Realizing universal C-algebras as concrete C-algebras

2013-02-01T23:14:58Z

Here is a further supplement: a tip for how to check if a $C^\ast$-algebra $A$ is the universal $C^\ast$-algebra for a given presentation. (I probably learned this from Terry Loring's book "Lifting solutions to perturbing problems in $C^\ast$-algebras".)

First, check that $A$ really is generated by a set of elements satisfying the given relations.

Second, check that every irreducible representation of the universal $C^\ast$-algebra is a representation of $A$. Say your generators are $x_1,\dots,x_n$. Then an irreducible representation would be generated by elements $X_1,\dots,X_n$. Since the centre of an irreducible representation is trivial, anything built out of the $X_i$'s that $*$-commutes with all the $X_i$'s is a scalar - so this approach works well if your relations entail a certain amount of commutativity, since commuting elements.

For example, to show that the universal $C^*$-algebra on a self-adjoint element of norm at most $1$ is $C_0([-1,1] \setminus {0})$, the second part above would go as follows. Let $X$ be the generator in an irreducible representation. Then $X$ is a scalar, which is self-adjoint (i.e. real) and has norm at most $1$. So $X = t \in [-1,1]$, and this representation corresponds to evaluating at this point $t$.

von Neumann automorphisms: does convergence on a dense algebra imply $u$-convergence?

2012-07-11T11:23:49Z

Let $M$ be a separable von Neumann algebra and let $A$ be a (von Neumann-)dense *-subalgebra. Suppose that $\alpha,\alpha_1,\alpha_2,\dots$ are automorphisms of $M$, such that for every $a \in A$, $$ \alpha_n(a) = \alpha(a) $$ for all $n$ sufficiently large. Does it follow that $\alpha_n$ converges to $\alpha$ in the $u$-topology?

(Also: what about if we weaken the hypothesis to just assuming that $\alpha_n(a)$ converges in norm to $\alpha(a)$, for all $a \in A$?)

This question is inspired by a related MO question, where an explicit example was requested of a sequence of inner automorphisms on the hyperfinite $III_1$- (or $II_1$-) factor which converge to an outer automorphism. An answer involved a sequence of inner automorphisms satisfying, in particular, the hypotheses of my question, although the proof of convergence uses more information. Perhaps other examples answering that question would be available if the answer to my question is yes.

Answer by Aaron Tikuisis for Pointwise limit at Lebesgue's point

2012-07-10T09:54:34Z

Since $L^1(\mathbb{R}^d)$ really means equivalence classes of integrable functions, I am interpreting the question as follows. Given $f \in L^1(\mathbb{R}^d)$, does there exist $g$ which is equal to $f$ almost everywhere and such that for almost every $x \in \mathbb{R}^d$, $$\lim_{x' \to x} g(x') = g(x)? $$

Here is a counterexample, even with $d=1$. Let $A \subset [0,1]$ be a measurable set with measure strictly less than $1$, such that for every open subset $U$ of $[0,1]$, the measure of $U \cap A$ is nonzero, and set $f=\chi_A$, the characteristic function of $A$.

(Such $A$ can be constructed, for example, taking an enumeration $\mathbb{Q} \cap (0,1) = {x_n}$ and setting $A = \bigcup_{n=1}^\infty \left(x_n-2^{-(n+2)}, x_n + 2^{-(n+2)}\right)$.)

Suppose that $g$ is equal to $f$ almost everywhere. Since the measure of $A$ is less than $1$, $B := g^{-1}(0)$ has positive measure, and we shall now show that $g$ is discontinuous at each $x \in B$.

Fix $x \in B$. For every $\epsilon > 0$, we have that $(x-\epsilon,x+\epsilon) \cap A$ has nonzero measure, and therefore, there exists a point $x'\in (x-\epsilon,x+\epsilon)$ such that $g(x') = 1$. Hence, $g$ is not continuous at $x$.

Answer by Aaron Tikuisis for Examples of conjectures that were widely believed to be true but later proved false

2012-05-04T12:10:08Z

This can perhaps be considered more of a meta-conjecture than a conjecture: Hilbert's program, http://en.wikipedia.org/wiki/Hilbert's_program. The conjecture would be: that set theory (or some set of axioms suitable for doing math) can be proven consistent. Gödel's Incompleteness Theorem disproved this conjecture.

I don't have a reference, but I have the impression that, at the time, Hilbert's program seemed attainable, and Gödel's result came as a surprise.

Answer by Aaron Tikuisis for Cesaro means and Banach limits

2012-04-09T06:10:16Z

We can characterize Banach limits as continuous functionals on $\ell^\infty$ which vanish on $$ X := \{(x_n - x_{n+1}): (x_n) \in \ell^\infty\} $$ and which send the constant sequence $(1,1,\dots)$ to $1$.

Note that $X$ is a subspace. The Hahn-Banach Theorem tells us that we are asking: if $(y_n) \in \ell^\infty$ has Cesaro mean $0$, is it in the closure of $X$? (And the converse question is: does every element of $X$ have Cesaro mean $0$? Yes; since the $n^\text{th}$ Cesaro mean of $(x_n-x_{n+1})$ is $(x_1-x_{n+1})/n$, which converges to $0$ since $(x_n)$ is uniformly bounded.)

The answer is no. Consider the sequence $(y_n)$ that has $1$ once, followed by $-1$ three times, then $1$ five times, and so on. One can compute the Cesaro mean, and see that it approaches $0$ in the limit. But $(y_n)$ is not in the closure of $X$.

Surely, if it were, then let $(x_n) \in \ell^\infty$ be such that $$ \|(y_n) - (x_n-x_{n+1})\|_\infty < 1/2. $$ Let $M$ be a natural number, $M \geq \|(x_n)\|$. Let $n$ be an index such that $$ y_n = \cdots = y_{n+4M} = 1. $$ Then for $i=1,\dots,4M$, $$ x_{n+i} < x_{n + i-1} - y_{n + i - 1} + 1/2 = x_{n + i - 1} - 1/2, $$ and summing these up, we find $$ x_{n+4M} < x_n - 4M/2. $$ This contradicts the assumption that $\|(x_n)\| \leq M$.

Answer by Aaron Tikuisis for Affine Homeomorphism between a compact set K and the state space on A(K)

2012-04-06T13:29:06Z

I believe that this is Theorem 7.1 in Ken Goodearl's book "Partially ordered abelian groups with interpolation."

Properties of orthogonality-preserving c.p. maps between $C^*$-algebras

2012-03-20T09:49:41Z

Suppose that $A,C$ are $C^*$-algebras and $\phi:A \to C$ is a completely positive, orthogonality-preserving linear map. (Orthogonality preserving means: if $a,b \in A$ satisfy $ab=0$ then $\phi(a)\phi(b) = 0$.) Then:

(i) For any $a,b,c \in A$, $$ \phi\left(ab\right)\phi\left(c\right) = \phi\left(a\right)\phi\left(bc\right) $$ (in the special case that $A$ is unital, this is equivalent to $\phi\left(a\right)\phi\left(b\right) = \phi\left(1\right)\phi\left(ab\right)$ for any $a,b \in A$);

(ii) For any $a,b \in A$, $$ \left\| \phi\left(ab\right) \right\| \leq \|a\|\cdot\left\|\phi\left(b\right)\right\|; $$

(iii) If $A$ is unital and simple, then for any $a \in A$, $$ \left\| \phi\left(a\right) \right\| = \|a\|\cdot\left\|\phi\left(1\right)\right\|. $$

In fact, there is a rich structure theorem about completely positive, orthogonality-preserving maps (in the literature, they are called "order zero" instead of "orthogonality-preserving" ), Theorem 2.3 of Winter, Zacharias, "Completely positive maps of order zero," Münster J. Math, 2009 (see also Corollary 3.1); and I can prove these statements easily using the structure theorem. But, my question is: can we prove any of the facts above directly (without appealing to this structure theorem)?

(I am intentionally not restating the structure theorem here because my question is about not using it.)

Answer by Aaron Tikuisis for Bounding 2nd Eigenvalue of a Pseudo-Rotation-ish matrix

2012-03-17T15:39:20Z

With reference to the tensor product formulation that I gave in my comment, we notice that $S$ is unitarily equivalent to $$ diag(\alpha, \alpha^2, \dots, \alpha^q) $$ where $\alpha = exp(2\pi i/q)$, and likewise $T$ is unitarily equivalent to $$ diag(\beta, \dots, \beta^p) $$ where $\beta = exp(2\pi i/p)$. Therefore, $A$ is unitarily equivalent to $$ 1/2 \begin{pmatrix} I_{pq} & diag(\gamma,\dots,\gamma^{pq}) \\ diag(\gamma,\dots,\gamma^{pq}) & I_{pq} \end{pmatrix}, $$ where $\gamma=\alpha\beta$. Subtracting $1/2I_{2pq}$ from this gives $$ 1/2 \begin{pmatrix} 0_{pq} & diag(\gamma,\dots,\gamma^{pq}) \\ diag(\gamma,\dots,\gamma^{pq}) & 0_{pq} \end{pmatrix} = 1/2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \otimes diag(\gamma,\dots,\gamma^{pq}). $$

The eigenvalues of $\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ are $\pm 1$, while the eigenvalues of $diag(\gamma,\dots,\gamma^{pq})$ are $\gamma,\dots,\gamma^{pq}$, so the eigenvalues of $$ 1/2 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \otimes diag(\gamma,\dots,\gamma^{pq}) $$ are $\pm\gamma/2,\dots,\pm\gamma^{pq}/2$. The eigenvalues of $A$ are therefore $(1\pm \gamma)/2,\dots,(1\pm\gamma^{pq})/2$. Your desired bound follows.

Answer by Aaron Tikuisis for A point in the weak closure but not in the weak sequential closure

2012-03-15T13:10:13Z

I don't believe that $0$ is a weak cluster point of this set. For example, consider $y \in \ell^2$ defined by $$y(k) = 1/k.$$ Then we have, for any $m,n$ that $$\langle x_{m,n}, y \rangle = m/n + n/m \geq 2.$$ Therefore, the weak neighbourhood $$ \{x \in \ell^2: |\langle x, y\rangle| < 1\} $$ of $0$ does not intersect $S$.

Answer by Aaron Tikuisis for Lattice-ordered group

2012-03-13T13:55:35Z

This is not a lattice-ordered group. As mentioned by boumol, it is (partially-)ordered. A simple characterization of $(G,G_+)$ being lattice-ordered (where $G$ is an ordered group with positive cone $G_+$) is the following: every intersection of two translates of $G_+$ is itself a translate of $G_+$, i.e. for any $x,y \in G$, there exists $z \in G$ such that

$(x + G_+) \cap (y + G_+) = z + G_+.$

(This is equivalent to being lattice-ordered, since this says that $z$ is the supremum of $x$ and $y$.)

In your case, use $x=(0,1)$ and $y=(1,0)$ to show that your ordered group is not lattice-ordered.

Answer by Aaron Tikuisis for when does a $C^*$-algebra have no nonzero unital quotient?

2012-03-10T13:59:52Z

It seems that "$A$ has a nonzero unital quotient" should mean: there exists a proper ideal $J$ of $A$ (possibly $J=0$) such that $A/J$ is unital. You would be correct to say that, if $A$ is nonunital and simple, then it has no nonzero unital quotient.

Elliott-Handelman's result does not contradict the second result that you cited, because in that case, $qAq$ is unital.

Comment by Aaron Tikuisis

2013-05-12T06:43:54Z

I'm very sorry - I totally read over the part about real numbers. But this means that you have 8 K-groups to try to use to show that the limits are non-isomorphic. Do your limits have the same K-theory?

Comment by Aaron Tikuisis

2013-05-07T15:18:43Z

The sub-unital hypothesis is a simplifying assumption (a "WLOG" if you will), since any positive linear map can be rescaled to produce a sub-unital map.

Comment by Aaron Tikuisis

2013-02-17T17:52:16Z

I might mention that, although I have some regrets about posting the question in this form (mainly that it makes me look bad), I wouldn't feel the same had I done the internet research and then asked simply about an algorithm. (I can only speculate that I wouldn't have found Smith Normal Form on my own in that circumstance.) Insofar as I got the answer I wanted, I am happy that I asked, though I suppose that people who sneak in questions from their math homework can often say the same.

Comment by Aaron Tikuisis

2013-02-17T17:46:51Z

I was surprised to log in to MO to be told that I've earned a "Nice Question" badge. My reason for feeling, at the time of posting, that this question might be unreasonable is that I had thought seriously about it for a day or so, and it felt like the sort of question that is either nontrivial or could be solved in that amount of time. In retrospect, I really should have found the answer to my first question using Wikipedia. (Part of what mislead me was thinking that the special form of $G$ could be important - it isn't an arbitrary subgroup of $\mathbb{Z}^n$.)

Comment by Aaron Tikuisis

2013-02-17T17:09:22Z

Please don't upvote this question anymore.

Comment by Aaron Tikuisis

2013-02-16T18:09:53Z

I hope my question didn't offend you Fernando. I was tentative about asking it, because I wasn't sure if it was research-level. Perhaps, sometimes you need to see the answer before you realize how easy a question is. Thank you for pointing out Smith Normal Form, since I didn't learn about it in my undergrad or graduate career

Comment by Aaron Tikuisis

2013-02-03T12:57:09Z

Perhaps this is a way to view it, although I am unfamiliar with the Yoneda Lemma (and I'd be grateful if you elaborated). However, note that what's useful about the method I mentioned is that one needn't check that for *any* $C^\ast$-algebra $B$ with generators satisfying the given relations, there is a $\ast$-homomorphism $A \to B$ sending generators to respective generators. Rather, one only needs to check this when $B$ is irreducible. This is a consequence of the fact that, for any $C^\ast$-algebra, the kernels of all irreducible representations have trivial intersection.

Comment by Aaron Tikuisis

2012-10-20T22:07:34Z

This seems to be asked (question 2), and answered, here: mathoverflow.net/questions/50302/…

Comment by Aaron Tikuisis

2012-08-20T12:49:44Z

What is $f_i$ supposed to be?

Comment by Aaron Tikuisis

2012-08-04T18:10:07Z

Greg, surely a modification of this works, if instead of insisting on rotating the matrix 180 degrees, you allow yourself a more general permutation of the Jordan blocks. This should take care of the algebraically closed case, and I don't see how to do the general case.

Comment by Aaron Tikuisis

2012-07-13T09:26:08Z

Yes, this is what I meant by separable.

Comment by Aaron Tikuisis

2012-07-11T18:03:45Z

Dense in the strong topology, or equivalently, the weak one or any of those other von Neumann algebra topologies except the norm topology. Also equivalently, $A''=M$.

Comment by Aaron Tikuisis

2012-07-11T11:12:20Z

One more question: how can we prove that the flip isn't inner?

Comment by Aaron Tikuisis

2012-07-11T11:04:07Z

Thanks! I take it from your proof that $u$-convergence does mean point-norm convergence on the predual?

Comment by Aaron Tikuisis

2012-07-10T10:46:13Z

I was thinking of the same example, but couldn't see how to prove the convergence. Can you explain this part?

User aaron tikuisis - MathOverflow

Answer by Aaron Tikuisis for Homomorphisms preserving constant functions

Answer by Aaron Tikuisis for Inductive limit of mapping tori

Answer by Aaron Tikuisis for Inductive limit of C*-algebras

Subgroups of $\mathbb{Z}^n$

Answer by Aaron Tikuisis for General recipe for building C*-algebras out of combinatorial object

Answer by Aaron Tikuisis for Realizing universal C*-algebras as concrete C*-algebras

von Neumann automorphisms: does convergence on a dense algebra imply $u$-convergence?

Answer by Aaron Tikuisis for Pointwise limit at Lebesgue's point

Answer by Aaron Tikuisis for Examples of conjectures that were widely believed to be true but later proved false

Answer by Aaron Tikuisis for Cesaro means and Banach limits

Answer by Aaron Tikuisis for Affine Homeomorphism between a compact set K and the state space on A(K)

Properties of orthogonality-preserving c.p. maps between $C^*$-algebras

Answer by Aaron Tikuisis for Bounding 2nd Eigenvalue of a Pseudo-Rotation-ish matrix

Answer by Aaron Tikuisis for A point in the weak closure but not in the weak sequential closure

Answer by Aaron Tikuisis for Lattice-ordered group

Answer by Aaron Tikuisis for when does a $C^*$-algebra have no nonzero unital quotient?

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Answer by Aaron Tikuisis for Realizing universal C-algebras as concrete C-algebras