User phil wild - MathOverflow

Question about von Neumann algebra generated by a complete algebra of projections

2010-07-23T20:55:44Z

Hi all, sorry if this is a dumb question, I don't know much about von Neumann algebras except the definition and a few relevant facts I've managed to prove by myself so I expect the answer will turn out to be well known. Anyway, let $\mathcal{H}$ be a Hilbert space, and suppose that $P$ is a commuting set of self-adjoint projections on $\mathcal{H}$, with the additional two properties:

1) $P$ is closed under complements, i.e. if $p \in P$ then so is $1 - p$.

2) $P$ is closed under suprema of arbitrary subsets, i.e. if $S \subseteq P$ then $\sup S \in P$ (here the projections on $\mathcal{H}$ are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$).

Now let $V$ denote the smallest von Neumann algebra containing $P$. Suppose that $p \in V$ is a self-adjoint projection. Is $p \in P$?

I know that $p$ is necessarily in the closure (relative to the weak operator topology) of the set of finite sums $\sum_i \lambda_i p_i$, where $p_i \in P$ and $\lambda_i \in \mathbb{R}$. It seems like it may be possible to derive a contradiction from the assumption that $q$ has a strictly smaller range than $p$, where $q \equiv \sup ${$ r \in P | r \leq p $}. But I don't know how to proceed.

Is there such a thing as the sigma-completion of a Boolean algebra?

2010-11-24T22:23:25Z

Hi all,

Suppose that $\mathcal{B}$ is a Boolean algebra. It there a way to extend $\mathcal{B}$ to a smallest Boolean algebra $\mathcal{B}'$ that contains an isomorphic copy of $\mathcal{B}$ and is countably complete, i.e. every countable subset of $\mathcal{B}'$ has a least upper bound in $\mathcal{B}'$? By "smallest" I mean that the inclusion $i: \mathcal{B} \hookrightarrow \mathcal{B}'$ has the obvious universal property, i.e. for every homomorphism $f$ from $\mathcal{B}$ to a countably complete Boolean algebra $\mathcal{C}$ there exists a unique homomorphism $g: \mathcal{B}' \to \mathcal{C}$ such that $g \circ i = f$ (it would be nice if $g$ turned out to commute with countable sups too). If no such $\mathcal{B}'$ exists, is there some other useful definition of "smallest" countably complete Boolean algebra containing $\mathcal{B}$?

If it makes any difference, I'm mostly interested in the special case where $\mathcal{B}$ is a direct limit of a sequence of finite Boolean algebras.

Edit: Thanks very much for the replies, it's a shame I can only mark one as the answer. It will take me a while to absorb the various references I've been given, so if I run into difficulty I'll bump the thread with an edit.

Edit 2: Bumping with followup question, please see my answer below.

Answer by Phil Wild for Is there such a thing as the sigma-completion of a Boolean algebra?

2011-03-17T22:27:48Z

Hi all,

I've just come up with something that's relevant to the question I asked here. I'm bumping the thread partly in case anyone else cares, and partly in case (as is more likely) I've made an error and someone can point it out. Anyway: I believe I can prove that the $\sigma$-algebra generated by a Boolean algebra $A$ (in the sense of Todd's answer, i.e. the image of a left adjoint to the forgetful functor from $\sigma$-algebras to Boolean algebras) has a rather natural representation, namely as the $\sigma$-field generated by the double dual of $A$, i.e. the smallest $\sigma$-field containing all the clopen subsets of the dual space of $A$. Here is the proof:

Let $A$ be a Boolean algebra, let $A^\star$ be its dual Boolean space and $A^{\star \star}$ the dual algebra of its dual space, i.e. the set of clopen subsets of $A^\star$. Let $\bar{A}$ be the $\sigma$-algebra of Baire sets in $A^\star$, i.e. the $\sigma$-field of subsets of $A^\star$ generated by $A^{\star \star}$. Let $\alpha: A \cong A^{\star \star}$ be the canonical isomorphism, and let $\eta: A \to \bar{A}$ be the composition of $\alpha$ with the inclusion.

Suppose given a $\sigma$-algebra $B$ and a homomorphism (of Boolean algebras) $h: A \to B$. Define $B^\star$, $B^{\star \star}$, $\bar{B}$ and $\beta: B \cong B^{\star \star}$ as before. By Theorem 41, p. 376 of [1], $B^\star$ is a $\sigma$-space, i.e. the closure of every open Baire set is open. By Theorem 42, p. 381, there is a $\sigma$-homomorphism $\phi: \bar{B} \to B^{\star \star}$ such that $\phi$ maps ever clopen set to itself.

$\beta h \alpha^{-1}$ is a homomorphism $A^{\star \star} \to B^{\star \star}$, so by duality there is a unique continuous function $f: B^\star \to A^\star$ such that $f^{-1} P = \beta h \alpha^{-1} (P)$ for every $P \in A^{\star \star}$. It is easy to see that $f^{-1} S$ is a Baire set whenever $S$ is, so define

$f^\star : \bar{A} \to \bar{B}$; $S \mapsto f^{-1} S$.

$f^\star$ is clearly a $\sigma$-homomorphism. Let

$\bar{h} \equiv \beta^{-1} \phi f^\star: \bar{A} \to B$.

Then one may check, using the defining property of $f$ and the fact that $\phi$ maps clopen sets to themselves, that $\bar{h} \eta = h$. The uniqueness of $\bar{h}$ with this property follows from the fact that the range of $\eta$ generates $\bar{A}$.

So there's the alleged proof; I can't see anything wrong with it but the result strikes me as being "too good to be true", and if it is true then I'm surprised I didn't see any reference to it online before I started this thread. So I'll be grateful if anyone can spot a mistake.

[1] Steven Givant and Paul Halmos, Introduction to Boolean Algebras, Springer 2009

Answer by Phil Wild for Examples of non-rigorous but efficient mathematical methods in physics

2010-12-09T00:46:49Z

Another example from theoretical high-energy physics I've encountered: sometimes when physicists have some equation of motion for an arbitrary number $N$ of particles with positions $x_i$, e.g. something of the form $\frac{1}{N}\sum_i f(x_i) + \frac{1}{N^2}\sum_{ij} g(x_i, x_j) = 0$, they wish to know what the solutions to this equation look like for large $N$. A technique they use is to replace the variables $x_i$ with a probability measure $\mu$ on the space of their possible values, which is supposed to represent the number of $x_i$'s in a given region in the large $N$ limit, and instead of solving the original equation they solve the analogous equation in $\mu$, e.g. $\int f(x) \mathrm{d}\mu(x) + \int g(x, y) \mathrm{d}(\mu \times \mu) (x, y) = 0$. In fact it's not hard to come up with a toy example where the original equation can be solved exactly for all $N$ and the solutions "look like" a particular probability distribution in the large $N$ limit, but that probability distribution fails to satisfy the corresponding equation, and for that reason I have some doubt that this method can be turned into something rigorous.

Do functions defined on global elements give rise to arrows in a well-pointed topos?

2010-10-14T22:14:03Z

Hi all,

Sorry if this question is not the right level for mathoverflow, but I already tried math.stackexchange and received no answers.

Suppose that $\mathcal{E}$ is a well-pointed elementary topos, that $X$ and $Y$ are objects of $\mathcal{E}$, and that $F$ is a function which maps global elements $p: 1 \to X$ to global elements $F(p): 1 \to Y$ (here $1$ is the terminal object of $\mathcal{E}$). Does there exist a (necessarily unique) arrow $f: X \to Y$ in $\mathcal{E}$ such that $fp = F(p)$ for all $p$? Equivalently, is any object in a well-pointed topos the coproduct over its global elements of $1$? It's easy to show that the answer is "yes" if the coproduct exists since the induced map $\coprod_{p \in \Gamma X} 1 \to X$ is iso. But I don't know whether the coproduct exists in general.

Question about equivalence relation defining integers in an elementary topos

2010-10-01T20:29:48Z

Hi all,

Let $\mathcal{E}$ be an elementary topos with natural number object $N$, and let $+: N \times N \to N$ be the the addition arrow; I expect that the nature of $N$ and $+$ will turn out to be irrelevant to my question, but if so they should at least make its motivation clear. Let $E$ be the pullback of $+$ along itself, with projections $p, q: E \to N \times N$; for example if $\mathcal{E}$ is the topos of sets then $E$ may simply be taken to be the set of quadruples $(n, m, n', m') \in N^4$ such that $n + m' = n' + m$, with $a(n, m, n', m') = (n, m')$, $b(n, m, n', m') = (n', m)$. Let $f_1, f_2: E \to N \times N$ be given by

$f_1 \equiv \left< p_1 a, p_2 b \right>$

$f_2 \equiv \left< p_1 b, p_2 a \right>$

(here $p_1, p_2: N \times N \to N$ are the projections and $\left< f, g \right>$ denotes the product arrow $X \to N \times N$ of arrows $f, g: X \to N$). For example in the topos of sets again, $f_1 (n, m, n', m') = (n, m)$ etc.. Let $c: N \times N \to Z$ be the coequaliser of $f_1$ and $f_2$, so $Z$ is the integer object in $\mathcal{E}$.

My question is: if $g, h, g', h': X \to N$ are such that $c \left< g, h \right> = c \left< g', h' \right>$, is it always the case that $+ \left< g, h' \right> = + \left< g', h \right>$? Equivalently, is $E$ with the arrows $f_1$, $f_2$ the pullback of $c$ along itself?

I've spent a while trying to prove it is but I just keep going round in circles, so any hints will be much appreciated.

Need help understanding a topos theory proof (any topos generated by subobjects of 1 in whose subobject lattices are complete and Boolean satisfies AC)

2010-09-12T14:12:57Z

Hi all, I'm reading Mac Lane & Moerdijk's book "Sheaves in Geometry and Logic" and I don't understand a proof; sorry if this is the wrong place to ask, if there's somewhere better please let me know.

The proof in question is of Proposition VI.1.8 (page 276 in the paperback), which states:

Let $\mathcal{E}$ be a topos which is generated by subobjeccts of $1$, and moreover has the property that for each object $E$, $\mathrm{Sub}(E)$ is a complete Boolean algebra. Then $\mathcal{E}$ satisfies the axiom of choice.

Here the axiom of choice is the statement that any epimorphism $p: X \to I$ has a section $s: I \to X$, i.e. $ps = 1_I$. The proof starts as follows:

Let $p: X \to I$ be an epimorphism in $\mathcal{E}$. By completeness of $\mathrm{Sub}(I)$, we can apply Zorn's lemma and find a maximal subobject $m: M \to I$ such that $p$ has a section $s: M \to X$, i.e., $ps = m$.

I don't see how to show that Zorn's lemma applies here. Presumably I need to show that, given a linearly ordered set of subobjects $m_i: M_i \to I$ of $I$, $m_i = m_j k_{ij}$ for some $k_{ij}: M_i \to M_j$ whenever $m_i \leq m_j$, together with a set of sections $s_i: M_i \to X$ such that $p s_i = m_i$, $s_j k_{ij} = s_k$, there exists a section $s: M \to X$ such that $ps= m$, where $m: M \to I$ is the least upper bound in $\mathrm{Sub}(I)$ of the $m_i$'s, and such that the restriction of $s$ to each $M_i$ is $s_i$. It seems plausible that $m: M \to I$ will turn out to be a colimit in $\mathcal{E}/I$ of the collection $m_i: M_i \to I$, from which the required result would follow easily, but I don't know how to prove this. Can anybody help?

Question about projections on a Hilbert space

2010-07-25T23:29:42Z

Sorry for the vague title, I can't think of a better one that isn't overly long.

Suppose that $S$ is a commuting set of projection operators on a Hilbert space. I'll introduce the following notation: if $p \in S$, let $p^+ \equiv p$ and $p^- \equiv 1 - p$. Let $I \equiv ${$+, -$}. The projections are ordered by defining $p \leq q$ whenever the range of $p$ is contained in the range of $q$; this makes the set of all projections into a complete lattice. Is the following identity true?

$\sup_{f \in I^S} \inf_{p \in S} p^{f(p)} = 1$

In the case where $S$ is finite with elements $p_1, p_2, \ldots p_n$, the left hand side of this equation is simply the product over $i$ of $p_i + (1 - p_i)$, so I'm interested in whether this can be generalised to the infinite case. It's easy to see that the following two statements are equivalent to the above:

If $\inf_p p^{f(p)} x = 0$ for all $f$, then $x = 0$

If $\sup_p p^{f(p)} x = x$ for all $f$, then $x = 0$

but I have no idea how to prove either of these.

My reason for asking is that I'm trying to show that, if $\mathcal{H}_1$ and $\mathcal{H}_2$ are Hilbert spaces, then if a projection on $\mathcal{H}_1 \otimes \mathcal{H}_2$ is of the form $\sup_i p_i \otimes q_i$, with $p_i$ and $q_i$ drawn from some complete Boolean algebras of projections on $\mathcal{H}_1$ and $\mathcal{H}_2$ respectively, then the $q_i$ may be chosen to satisfy $q_i q_j = 0$ when ever $i \neq j$. So if anybody knows of an alternative way to prove that, or knows that it's false, then by all means say so.

Comment by Phil Wild

2010-12-13T23:22:53Z

The formula is not of the form given by the example in my original post; rather than a single equation it gives an equation for each $x_i$ in terms of the other $x_j$, which are to be solved simultaneously. The analogous problem involving the measure $\mu$ would involve simultaneously solving an equation in $\mu$ with a free variable $x$, for every $x$. Sorry if this wasn't clear in my answer, but the example was just intended to be illustrative rather than completely general (and I should have written "system of equations" rather than "equation").

Comment by Phil Wild

2010-12-13T18:09:29Z

Oh, upon rereading your first comment I should add that the solutions $\mu_N$ in the example I came up with were far from unique.

Comment by Phil Wild

2010-12-13T17:53:07Z

I'm afraid I don't remember the details, though IIRC it was something simple like $\partial/\partial x_i \left(\sum_j |x_j|^2 - a \sum_{jk} |x_j - x_k|^4 \right) = 0$.

Comment by Phil Wild

2010-12-09T17:39:52Z

Well, coming up with an appropriate definition of "converges to" would be one of the difficulties in making the technique rigorous, but in the toy example the solution for a given $N$ consisted of the $N^{\mbox{th}}$ roots of unity in the complex plane, and the probability distribution they "look like" was the measure uniformly concentrated on the unit circle. I don't know if there's any notion of convergence that works, but the real examples I saw were of the same form (i.e. sets of points lying at regular intervals on submanifolds of $\mathbb{R}^n$ being approximated by uniform measures).

Comment by Phil Wild

2010-11-25T18:08:13Z

Sorry if I wasn't sufficiently clear, but Todd's interpretation of my OP is correct - I have no guarantee that the f's in which I'm interested preserve any infinite joins that may exist in B. I appreciate all the comments nonetheless, they are very interesting.

Comment by Phil Wild

2010-10-20T22:58:03Z

(sorry for the slow reply, only just noticed your comment) Well the question arose in the course of trying to understand a book that everyone working in the field has probably already read and understood, so I didn't think it would qualify as "research level". But thanks, it's good to know I'm not wasting MO's time.

Comment by Phil Wild

2010-10-14T23:01:16Z

In proving that the so-defined sets satisfy extensionality, the authors suppose that every point (i.e. node covered by the root) of a tree $T_1$ is isomorphic to some point of $T_2$, and then state that this gives a morphism from the subobject of points of $T_1$ to that of $T_2$. Do you know if the proof is wrong, or can the relevant morphisms be shown to exist in this case? If it's wrong, can it be fixed?

Comment by Phil Wild

2010-10-14T22:52:23Z

Very nice, thank you. Are you familiar with Mac Lane and Moerdijk's book? My question arose in trying to understand Section VI.10 in which they claim to prove that a well-pointed topos with NNO which satisfies the axiom of choice gives rise to a model of restricted Zermelo set theory with the axiom of choice. They do this by identifying sets with internal trees which satisfy certain additional properties, and identify two trees if there is an internal isomorphism of trees between them. [contd.]

Comment by Phil Wild

2010-10-14T22:43:29Z

I considered trying to use Yoneda, but I think it just takes me back where I started - constructing such a natural transformation $\sigma$ would in particular involve constructing the image $\sigma_X (1_X) \in \mathcal{R}(X, Y)$, which is what I need.

Comment by Phil Wild

2010-10-02T05:31:26Z

Excellent, thanks.

Comment by Phil Wild

2010-10-02T02:35:27Z

OK, thanks. Is the fact you mention easy to prove, and if so, can you give me a hint? (I don't have access to a university library so checking the reference you give won't be so easy.) One of the reasons for my original question was that Mac Lane and Moerdijk state without proof that the natural map $c \left< 1_N, 0! \right>: N \to Z$ is monic, which would follow easily from the positive answer. Do you know if there's another way to see that it's monic?

Comment by Phil Wild

2010-10-02T00:34:53Z

Thanks for your reply and apologies if I'm missing something, but I don't see how that helps with my question. I need to show that $f_1, f_2$ is the kernel pair of some arrow. Does the fact that $\left< f_1, f_2 \right>$ is an equivalence relation imply that it's also a kernel pair in an arbitrary (not necessarily Grothendieck) topos? Mac Lane and Moerdijk mention that the implication (equivalence relation \Rightarrow kernel pair) does not hold in an arbitrary category.

Comment by Phil Wild

2010-09-12T23:10:02Z

Yes, I think the proposition is supposed to apply to elementary topoi. I don't understand your comment that you "don't know why $\mathrm{Sub}(I) should be complete [...]" - I don't think it's supposed to be true in general, it's just one of the hypotheses of the proposition.

Comment by Phil Wild

2010-09-12T17:46:59Z

Sorry if I'm being thick, but I don't follow your reply. Are the colimits you refer to those in the original topos $\mathcal{E}$? If so then how do I know that they exist? The completeness of the subobject lattices shows that the corresponding colimits exist in $\mathrm{Sub}(I)$; I don't see how it follows that $\mathcal{E}$ itself has all relevant colimits.

Comment by Phil Wild

2010-09-12T17:39:46Z

Thanks for your reply. I can see how showing that filtered colimits in $\mathrm{Sub}(I)$ coincide with those in $\mathcal{C}$ gives me the result I want. However I do not see how Charles's answer shows this, can you explain? Regarding your last paragraph, the version of Zorn's lemma I know says that a non-empty poset in which every chain has an upper bound has a maximal element; since any element is an upper bound for the empty chain, this case is automatic for any non-empty poset.