User yemon choi - MathOverflow

Answer by Yemon Choi for A proof about an unconditional basis theorem

2010-03-17T20:12:42Z

I'm not a Banach-space specialist, and don't have a copy of Lindenstrauss-Tzafriri to hand (though maybe the Albiac-Kalton book would be a friendlier read, if your library has a copy?) but it seems to me quite natural to take the contrapositive and try to work with that. In other words, show that TFAE

1) X has a (closed) subspace with an unconditional basis.

2) There exists a block subspace Y of X and a real number C such that for every sequence $y_1 < y_2 < \dots < y_n$ in $Y$ we have the inequality

$$ \left\Vert \sum_{i=1}^n y_i \right\Vert \leq C \left\Vert \sum_{i=1}^n (-1)^i y_i \right\Vert $$

Certainly if we assume 2) then there is a natural candidate for the subspace desired in 1) ...

Maybe it would help if you could show us a more precise or more localised part of the problem which you're stuck on -- is it to do with the equivalent definitions of unconditional basis, or something else?

Answer by Yemon Choi for Simultaneous time-frequency concentration of orthonormal sequences?

2010-03-12T04:18:54Z

The Fourier transform on $L^2({\mathbb R})$ has an complete set of eigenvectors (that is, there is an o.n. basis of $L^2({\mathbb R})$ consisting of eigenfunctions for the FT, and they are all in the Schwartz space). Does this do what you want?

Reference needed for: every idempotent in a C*-algebra is similar to a hermitian one

2010-03-03T02:11:50Z

The result stated in the title is thoroughly standard - or that's the impression I got. I seem to remember seeing it stated somewhere in a book I was reading in the library, and then reverse-engineering a proof from the hints given.

For a preprint I'm working on, it would be preferable to give a precise citation from a "standard text", rather than spend time giving the proof "for the reader's convenience". Any suggestions?

If anyone's interested, an outline of a proof is as follows: consider an idempotent P in B(H), with H a Hilbert space, and note that we can always decompose H as an orthogonal sum with respect to which P has the block matrix form

$$ P= \left(\begin{matrix} I & R \\ 0 & 0 \end{matrix}\right) $$

Then it's not hard to see that conjugating $P$ by the invertible operator

$$ S= \left( \begin{matrix} I & R \\ 0 & I \end{matrix} \right) $$

will give

$$ E = \left(\begin{matrix} I & 0 \\\ 0 & 0 \end{matrix} \right) $$

Since $S= I+P-E$, it suffices to show that $E$ is in the C*-algebra generated by I and P (for then S will also lie in that algebra, and then we're done). This follows by messing around with various combinations of P, its adjoint, and their products.

Answer by Yemon Choi for Embeddings of Weighted Banach Spaces

2010-02-16T02:13:43Z

This is a special case of a much more general phenomenon, so I'm writing an answer which deliberately takes a slightly high-level functional-analytic POV; I think (personally) that this makes it easier to see the wood for the trees, even if it might not be the most direct proof. However, depending on your mathematical background it might not be the most helpful; so apologies in advance.

Anyway, start with a very general observation: let $E$ be a (real or complex) Banach space, and for each $n=1,2,\dots$ let $T_n:E\to E$ be a bounded linear operator which has finite rank. (In particular, each $T_n$ is a compact operator.)

Lemma: Suppose that the sequence $T_n$ converges in the operator norm to some linear operator $T:E\to E$. Then $T$ is compact.

(The proof ought to be given in functional-analytic textbooks, so for sake of space I won't repeat the argument here.)

Now we consider these specific spaces $\Omega_p$. Let $T:\Omega_p\to\Omega_{p'}$ be the embedding that you describe.

For each $n$, define $T_n: \Omega_p \to \Omega_{p'}$ by the following rule:

$$ T_n(x)_i = x_i \ {\rm if } \ |i| \leq n \ {\rm and} \ 0 \ {\rm otherwise} $$

Then each $T_n$ has finite rank (because every vector in the image is supported on the finite set $\{ i \in {\mathbb Z}^d \vert \ \vert i\vert \leq n\}$). I claim that $T_n$ converges to $T$ in the operator norm, which by our lemma would imply that $T$ is compact, as required.

We can estimate this norm quite easily (and indeed all we need is an upper bound). Let $x\in\Omega_p$ have norm $\leq 1$; that is, $$ \sum_{i\in{\mathbb Z}^d} |x_i|^R (1+ \vert i\vert)^{-p} \leq 1 $$

Then the norm of $(T-T_n)(x)$ in $\Omega_{p'}$ is going to equal $C^{1/R}$, where

$$ \eqalign{ C &:= \sum_{i\in {\mathbb Z}^d : \vert i\vert > n} |x_i|^R (1+\vert i \vert)^{-p'} \\ & = \sum_{i \in {\mathbb Z}^d : \vert i\vert > n} |x_i|^R (1+\vert i \vert)^{-p} \cdot (1+\vert i \vert)^{p-p'} \\ & \leq \sum_{i \in {\mathbb Z}^d : \vert i\vert > n} |x_i|^R (1+\vert i \vert)^{-p} \cdot (1+\vert n \vert)^{p-p'} \\ & \leq \sum_{i \in {\mathbb Z}^d } |x_i|^R (1+\vert i \vert)^{-p} \cdot (1+\vert n \vert)^{p-p'} & = (1+\vert n \vert)^{p-p'} } $$

This shows that $\Vert T-T_n\Vert \leq (1+\vert n\vert)^{(p-p')/R}$ and the right hand side can be made arbitrarily small by taking $n$ sufficiently large. That is, $T_n\to T$ in the operator norm, as claimed, and the argument is complete (provided we take the lemma on trust).

Note that we used very little about the special nature of your weights. Indeed, as Bill Johnson's answer indicates, the only important feature is that in changing weight you in effect multiply your vector $x$ by a "multiplier sequence" which lies in $c_0({\mathbb Z}^d)$, i.e. the entries "vanish at infinity".

Edit 17-02-10: the previous paragraph was perhaps slightly too terse. What I meant was the following: suppose that you have two weights $\omega$ and $\omega'$, such that the ration $\omega/\omega'$ lies in $c_0({\mathbb Z}^d)$. Then the same argument as above shows that the corresponding embedding will be compact. Really, this is what Bill's answer was driving at: it isn't the weights which are important, it's the fact that the factor involved in changing weight is given by something "vanishing at infinity".

Answer by Yemon Choi for Distribution of the sum of the $m$ smallest values in a sample of size $n$

2010-02-16T07:01:22Z

This is following on from Douglas Zare's answer. While not an answer in its own right,it was getting too long for a comment. Briefly, we can hit the question with brute force and look for a generating function for the desired expected values.

So, put $$f_j(y) = \sum_{k=0}^{j-1} {n \choose k} y^k (1-y)^{n-k}$$ so that using the notation of Douglas' answer, $g_j(x)=f_j(F(x))$, and the expectation of the $j$th order statistic is $$ E_j := \int_0^\infty f_j(F(x)) \ dx $$

Put $G(y,z) = \sum_{j=1}^n f_j(y) z^j$ where $z$ is a formal variable. By linearity we have $$ \sum_{j=1}^n E_j z^j = \int_0^\infty G(F(x),z) \ dx $$

We can try to write $G$ as a rational function in $y$ and $z$. Expanding out and interchanging the order of summation gives $$ \eqalign{ G(y,z) & = \sum_{j=1}^n\sum_{k=0}^{j-1} {n\choose k} y^k (1-y)^{n-k} z^j \\ & = \sum_{k=0}^{n-1} \sum_{j=k+1}^n z^j {n\choose k} y^k (1-y)^{n-k} \\ & = \sum_{k=0}^{n-1} \left( \sum_{j=1}^{n-k} z^j \right) \cdot z^k{n\choose k} y^k (1-y)^{n-k} \\ } $$

Now $$ \eqalign{ \sum_{j=1}^{M-1}jz^j = z \frac{d}{dz}\sum_{j=0}^{M-1} z^j & = z \frac{d}{dz}\left[\frac{1-z^M}{1-z}\right] \\ & = \frac{z-z^{M+1}}{(1-z)^2} - \frac{Mz^M}{1-z} & = \frac{z-z^M}{(1-z)^2} - \frac{(M-1)z^M}{1-z} } $$ so substituting this back in we get $$ \eqalign{ G(y,z) & = \sum_{k=0}^{n-1} \left[ \frac{z-z^{n-k+1}}{(1-z)^2} - \frac{(n-k)z^{n-k+1}}{1-z} \right] \cdot z^k{n\choose k} y^k (1-y)^{n-k} \\ & = \sum_{k=0}^{n-1} \frac{z}{(1-z)^2}{n\choose k} (yz)^k (1-y)^{n-k} \\ & \ \ - \sum_{k=0}^{n-1} \frac{z^{n+1}}{(1-z)^2} {n\choose k} y^k (1-y)^{n-k} \\ & \ \ - \sum_{k=0}^{n-1} \frac{nz^{n+1}}{1-z} { {n-1} \choose k } y^k(1-y)^{n-k} \\ & = \frac{z}{(1-z)^2} \left[ (1-y+yz)^n - (yz)^n \right] - \frac{z^{n+1}}{(1-z)^2} (1-y^n) - \frac{nz^{n+1}}{1-z} } $$

I guess that in theory one could plug this back in to obtain a "formula" for the generating function $\sum_{j=1}^n E_j z^j$, but I can't see how that formula might then simplify to something calculable, unless $F$ has a rather special form.

Answer by Yemon Choi for Which journals publish expository work?

2010-02-15T21:29:37Z

I'm not too familiar with Expositiones Mathematicae, but have you given them a look?

EDIT: The article I happened to have seen, which made me think that Expo Math might be along the lines Pete Clark was looking for, is this paper of T. Bühler - it modestly claims to no originality save for assembling disparate parts of the literature and writing down what's old news to connoisseurs (I'm paraphrasing here!) but of course this is, in a sense, precisely its originality & worth.

Answer by Yemon Choi for What's the space of smooth functions in L^2(R)?

2010-02-13T04:31:25Z

N.B. this answer was in response to an earlier version of the question, which only had the first two paragraphs -- hence it doesn't address what appears to have been the original poster's actual question. For that, see the answers of Leonid or Harald.

I'm not sure if this answers your question, but it might be worth noting that a measurable function $f$ on the real line is in $L^2({\mathbb R})$ if and only if its Fourier transform $\widehat{f}$ is (Plancherel theorem), while it is in $C^\infty({\mathbb R})$ if and only if we have

$$ \int_{-\infty}^\infty | \widehat{f}(x) |^2 (1+ |x|^2)^{k} < \infty \quad{\rm for }\ k=1,2,\dots $$

(this is a form of Sobolev embedding, albeit in a very special case). In particular, if I've correctly understood the notation from the wikipedia page for Sobolev spaces, the space you're after seems to be the intersection $\bigcap_{k=0}^\infty H^k({\mathbb R})$. I don't know if this goes by a particular name.

Answer by Yemon Choi for spectra of sums and products in (Banach) algebras [was: Spectrum in Banach Algebra]

2010-02-05T07:56:34Z

In general Banach algebras the spectral radius is neither subadditive nor submultiplicative; in particular, neither of the two properties you mention holds.

$2\times 2$ (real or complex) matrices should suffice to give examples, so this is not to do with any subtleties of infinite-dimensional algebras.

For example, take $ a= \left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right) $. Note that this is nilpotent, so the only point in the spectrum is zero, and hence $\sigma(a)\sigma(b)= \{0\}$ for any other matrix $b$. On the other hand, we can find $b$ for which $ab$ is not nilpotent, so that $\sigma(ab)\not\subseteq {0}$. A simple choice which works is $b=\left(\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix} \right)$, since then $ab$ is a non-zero projection (=idempotent) and so contains $1$ in its spectrum.

The same pair also works as a counter-example for the "additive question". For since $\sigma(a)=\sigma(b)=\{0\}$, we have $\sigma(a)+\sigma(b)=\{0\}$. On the other hand, $a+b$ is a reflection and hence its spectrum is $\{-1,1\}$

(edited 11-02-10 for a couple of typos/omissions)

Jonas Meyer points out in comments that one can pose the following converse question: let $A$ be a Banach algebra with identity, and suppose that we have

($*$) $\sigma(a+b)\subseteq\sigma(a)+\sigma(b)$ and $\sigma(a)\sigma(b)\subseteq\sigma(ab)$ for all $a,b\in A$.

Must $A$ be commutative?

As Jonas also pointed out in comments, the answer is in general `no': the algebra of strictly upper-triangular matrices (or, more precisely, the algebra of scalar+strictly upper triangular $m\times m$ matrices, for some fixed $m$) gives a counterexample, at least when $m\geq 3$.

A more careful version of this argument shows, I think, that if $A$ is a finite-dimensional algebra with identity, such that $A/{\rm Rad}(A)$ is commutative, then $A$ will satisfy condition $(*)$. I suspect that the same might be true for any unital Banach algebra that is commutative modulo its radical, i.e. that the finite-dimensional hypothesis is unnecessary; but at present I'm a bit too tired to check this properly.

We could therefore modify the question yet further, and ask if a Banach algebra that satisfies condition $(*)$ must be commutative modulo its radical. The answer turns out to be yes, after a wander down memory lane and a forage on MathSciNet:

MR0461139 (57 #1124) J. Zemánek. Spectral radius characterizations of commutativity in Banach algebras. Studia Math. 61 (1977), no. 3, 257--268.

The MR is short and informative enough to give in full, for those without access:

It is standard that the spectral radius is subadditive and submultiplicative on any commutative complex Banach algebra. The author proves that, for a complex Banach algebra $A$, the following three conditions are equivalent: (1) the spectral radius is sub-additive on $A$, (2) the spectral radius is submultiplicative on $A$, (3) $A$ is commutative modulo its radical. Some applications of this result to other problems in Banach algebras are given, along with references to a number of related papers.

Answer by Yemon Choi for Almost but not quite a homomorphism

2010-02-09T11:41:07Z

Related to Henry Wilton's comments: the following might not quite be what you're looking for, but seems interesting given that quasi-morphisms have been mentioned. I'm doing this from memory so if there's a gap, someone please let me know!

Let $E$ be a Hilbert space, $B(E)$ the algebra of all bounded linear operators on $E$. (Even the case $E={\mathbb R}^n$ is of interest.) Fix a small $\epsilon>0$. Then there exists $\delta>0$ with the following property:

Let $G$ be an abelian group, and let $f:G \to B(E)$ be a bounded function (i.e. $\sup_{x\in G} \| f(x) \| < \infty$) which satisfies $$ \sup_{x,y}\| f(x)f(y) - f(xy) \| \leq \delta. $$ Then there is some representation $\rho: G \to B(E)$ such that $\sup_x \| f(x)- \rho(x)\| \leq \epsilon$.

So, less formally, bounded "almost representations" of abelian groups are "near to" genuine representations.

I imagine this could be proved by an averaging argument: the way I learned of this result is as a special case of a more general one, in which the word "abelian" is replaced by the word "amenable", and the word "Hilbert" is replaced by "nice reflexive Banach". That in turn is a special case of a general result on almost multiplicative maps between Banach algebras satisfying certain conditions (due to B. E. Johnson).

Anyway, sorry this has wandered off track. The point was to say that there are contexts where things which are close to being group homomorphisms $H\to K$, might under a small perturbation be genuine homomorphisms when restricted to a specified abelian subgroup of $H$. However, in general this can't be done so as to work simultaneously for all abelian subgroups of $H$.

Answer by Yemon Choi for Mittag-Leffler condition: what's the origin of its name?

2010-02-09T00:43:21Z

The wording of your question suggests that you're familiar with the "classical" Mittag-Leffler theorem from complex analysis, which assures us that meromorphic functions can be constructed with prescribed poles (as long as the specified points don't accumulate in the region).

It turns out - or so I'm told, I must admit to never working through the details - that parts of the proof can be abstracted, and from this point of view a key ingredient (implicit or explicit in the proof, according to taste) is the vanishing of a certain $\lim_1$ group -- as assured by the "abstract" ML-theorem that you mention.

I'm not sure where this was first recorded - I hesitate to say "folklore" since that's just another way of saying "don't really know am and not a historian". One place this is discussed is in Runde's book A taste of topology: see Google Books for the relevant part.

IIRC, Runde says that the use of the "abstract" Mittag-Leffler theorem to prove the "classical" one, and to prove things like the Baire category theorem, can be found in Bourbaki. Perhaps someone better versed in the mathematical literature (or at least better versed in the works of Bourbaki) can confirm or refute this?

Answer by Yemon Choi for When can a function be recovered from a distribution?

2010-02-08T03:25:54Z

I haven't thought about this carefully enough, but it seems that there is some ambiguity in your question about what the integral $\int f\varphi$ is supposed to mean. As Ryan and Leonid have said: if you want the representing function $f$ to be locally integrable then the Radon-Nikodym theorem is what you need.

On the other hand, if you allow principal-value integrals (which is probably not what you want, I'm guessing, but I wasn't sure from your question) then I think

$$ \varphi \mapsto \int_{\rm p.v.} \frac{\varphi(t)}{t}\ dt $$

would be a tempered distribution that is in some sense `represented by a function', even though the function is not everywhere locally integrable.

Answer by Yemon Choi for Theories of Noncommutative Geometry

2010-02-06T21:40:06Z

Some thoughts and links on the analysts' NCG, from someone who doesn't practice it. Caveat lector. (Some edits made to erroneous history.)

NCG a la Connes was originally non-commutative differential geometry (which is why extra structure is needed in, say, the definition of spectral triple). Having only recently looked at Connes' original long, two-part paper in the Publications of the IHES, I think this is a better place to start than the later Big-Picture-Which-Is-Really Fundable works of various people thereafter. Work of Connes & Moscovici and others on trying to generalize the Atiyah-Singer index theorem also give some indication of the original motivation. (This is where someone more expert than me should really step in and say something about the work of ~~C & M~~ Mischenko, Kasparov and their co-authors on the Novikov conjecture, or indeed the work of Higson et al. on the Baum-Connes conjecture.)

It was only afterwards that Connes started championing an NCG perspective on The Standard Model. (Although, if you want real connections with mathematical physics, there was some work of Jean Belissard on identifying gaps in spectra of certain operators in a quantum-mechanical model with K-theoretic invariants of associated C*-algebras. See this paper of Kaminker and Putnam and the references therein for more details.)

Personally, I am bit leery of the Big Picture motivation for noncommutative geometry, at least of this variety. The most useful variant of such motivation that I can think of, is that degenerate group actions on topological spaces give rise to better-behaved homotopy groupoids; thanks to a theorem of Rieffel (IIRC), when the group action on the space is nice, the commutative C*-algebra of the quotient space is Morita equivalent -- i.e. has "the same module theory" -- as the noncommutative C*-algebra of the homotopy groupoid.

I apologize for the rambly nature of this answer, but with all due respect to Anweshi I think his question, or at least the version of it which I can currently see, is so broad (as per Anton's original comments) that the only responses are either encyclopaedic - I haven't even had space to mention the historical role played by Brown-Douglas-Fillmore theory, for instance - or sales pitches. Nevertheless, if someone can persuade Nigel Higson to drop by, I'm sure he could give a much better answer ;)

Answer by Yemon Choi for Looking for an introductory textbook on algebraic geometry for an undergraduate lecture course

2010-02-06T11:03:26Z

I can't recall exactly how much background it assumes, but I found Reid's Undergraduate Algebraic Geometry quite accessible. (The forthright views in its last section can be taken either as a blemish or a bonus depending on one's POV.)

Answer by Yemon Choi for decreasing chain of subgroups in the Heisenberg group

2010-02-05T06:48:41Z

Assuming this is the discrete Heisenberg group $H=H_3({\mathbb Z})$, as in my comment above, then here is another way of looking at Mariano's answer (I think). Take any sequence of positive integers $n_1 < n_2 < \dots $ where $n_i \vert n_{i+1}$ for all $i$, and put

$$ H_i = H_3(n_i{\mathbb Z}) $$

(Mariano's answer corresponds to taking $n_i = 2^i$.)

Answer by Yemon Choi for Finite versus infinite on non-Hausdorff topologies

2010-02-05T04:51:18Z

Seeing as the comment thread to the original question is running out of control, let me just record some attempts to formulate a question which might (a) be related to what Ian Durham is asking, and (b) is more palatable to some of the people, myself included, who find the original question hard to answer meaningfully.

First of all: I guess we are taking as a working principle

... it is impossible to simultaneously have an infinite number of physical objects of non-zero size in the universe

The example given of an object in the original question is something like "a quantum channel" - now since I'm a physics ignoramus I don't know what the ontological status is of such a beast, but let's suppose for sake of discussion that it does have "size" and that therefore only finitely many of such can exist in a given physical system. This is presumably some argument about physical observables being quantized, but someone else is welcome to correct me on this.

Secondly: there are constructions in mathematical physics which seem to be of an infinite nature. The example given seems to be a "potential infinity", i.e. what are we approaching if we tensor a channel with itself repeatedly.

Now, my interpretation of what Ian may be trying to ask -- and I have to say, in my personal opinion I've not found it at all easy to discern what his underlying question is -- goes like this:

(i) are there contexts in "mainstream abstract mathematics" where an implicitly defined "object" -- such as, the solution space of some differential equation, the solution set of some algebraic equation, the set of accumulation points of some sequence -- which depends on some outside flavour (choice of ground-field for an algebraic equation; choice of topology on some ambient space which reasonably admits more than one topology; an ambient topos in which the construction is supposed to live), might have finite cardinality for one choice of flavour, but infinite cardinality for other choices?

(ii) does this have anything to do with whether we equip a given space, broadly and vaguely conceived, with a Hausdorff or a non-Hausdorff topology?

(iii) do either of these have any connection to the original subject, namely that certain mathematical constructions appear to have physical meaning yet be defined in terms of unphysical infinities?

The answer to (i) is in my view "yes, but so what?" and the answer to (ii) is in my view "I don't really think so". Moreover, I don't think (iii) is really dependent on (ii), and so my overall impression is that the "Hausdorff discussion" is a red herring.

Lastly, I am having difficulty making sense of the reasoning behind this sentence in the original question:

Now suppose that one of the various branching spacetime interpretations of quantum mechanics (MWI, MMI, etc.) is correct (personal aside: I am agnostic on this issue). The topology of the multiverse would thus be non-Hausdorff and, given these interpretations of QM, there ought to be an infinite number of branches. Given that, an infinite physical realization becomes possible.

Answer by Yemon Choi for Finite versus infinite on non-Hausdorff topologies

2010-02-04T19:05:04Z

Note: this is a response to an earlier version of the question, and so is rather speculative. I am not sure that I can edit this to be much use in response to the newer version, so I'm leaving it as it stands for now, except to strikethrough a closing remark which was wide of the mark.

I think that the original questioner may have something like this in mind (though since I'm not a telepath, Jim, corrections/comments are welcome).

When we explain the definition of Hausdorff, we talk about separating points from other points using open neighbourhoods. Now if your intuition comes from metric spaces, these neighbourhoods are balls. So one might think that Hausdorffness is to do with separating things by balls, Furthermore, one might be thinking of some notion of minimal ball size -- this isn't what Hausdorffness is about, but bear with me! I'm trying to recreate a train of thought, not recapitulate the correct definitions -- and so get the idea that in "contexts which are Hausdorff" certain postulated objects -- the collection of all Widgets that satisfy the Sveshnikov-Pelikan equation - are forced to be finite because of "the need to separate constituent parts with balls". (See the original post's 4th para.)

Pursuing this train of thought, one might then wonder that if this postulated object is not finite, this is something to do with the failure of Hausdorffness. Again, I think this comes from a misapprehension about the Hausdorff separation condition; but at least this interpretation makes some sense of the original post's 1st para.

For the record: of course, things are more likely to be Hausdorff when you have more open sets, and of course a completely discrete space is Hausdorff for the trivial reason that every set is open. At the other extreme, a space with the indiscrete topology and more than two points has no chance of being Hausdorff. However, this has absolutely nothing to do with constraining the underlying set to be finite or infinite, contra the apparent guess of the original question. (If it helps: in the definition of Hausdorff, we don't constrain "the size of our open balls" before picking our two points; given any two points in, say, a metric space, we then have the freedom to choose a mesh size which will distinguish between them.) In particular, the sentence which starts "The topology of the multiverse would therefore be non-Hausdorff..." is in my view based on a misunderstanding, either of the word "Hausdorff" or the word "therefore"...

A comment for Ian, if I may: the reaction you got was because you started with a misuse of mathematical terminology, inserted into a question with words like "context" which are by nature philosophical/cultural rather than mathematical. It's that clash of tones which I think confused/irked some of the commenters.

Answer by Yemon Choi for On operator ranges in Hilbert & Banach spaces

2010-02-04T20:53:40Z

My initial impression is that for what you want, you're going to need a notion of $A^*: E\to E$ when $A:E\to E$ is an operator on a Banach space. I don't know much about this, but some years ago did see this short paper

MR2053349 (2005a:46045)
Gill, Tepper L.(1-HWRD-EE); Basu, Sudeshna(1-HWRD); Zachary, Woodford W.(1-HWRD-EE); Steadman, V.(1-DC)
Adjoint for operators in Banach spaces. Proc. Amer. Math. Soc. 132 (2004), no. 5, 1429--1434

which requires a choice of Hilbert space rigging $H_1 \hookrightarrow E \hookrightarrow H_2$.

One thing that might go wrong with $(1) \implies (3)$ in general Banach spaces is the non-existence, in general, of projections from $E$ onto a closed subspace. However, that doesn't rule out the possiblity that something like $(1)\implies(3)$ does indeed hold; I'd need to think about this a bit more.

Edit: ah, I see that in your setting the operators go from one Banach space to another, rather than from the space to itself. That might make a difference: and indeed, since you're mapping into a $C(K)$-space and not just an arbitrary one, more tools might be available.

Answer by Yemon Choi for How exactly is Hochschild homology a monad homology?

2009-10-28T23:12:26Z

This is partly in response to Reid, but also intended as general clarification.

As I understand it, Peter's original question was:

-- here is the Hochschild chain complex for an algebra $A$ and bimodule $M$, as defined in Hochschild's original papers; -- it is the chain complex associated to a certain simplicial object as defined on the Wikipedia page; -- one is told that this object comes from the bar construction (or standard resolution) associated to some monad; -- where/what is the monad?

The last one seems to be Reid's underlying point/question. Tyler says you can get it, up to a dimension-shift, from the adjunction between k-modules and k-algebras (at least when $A=M$). My earlier recollection was that this more naturally leads to cyclic homology a.k.a. additive K-theory as defined by Feigin and Tsygan, but I have yet to check this against a copy of their paper. (The point is that in characteristic zero, the cyclic homology of a free tensor algebra on a given k-module, coincides with the cyclic homology of the ground field, so one can take free resolutions of a given $k$-algebra and then use spectral sequence arguments.) On reflecting a bit more, because the Hochschild homology of a free (=tensor) algebra is confined to degrees 0 and 1, perhaps one can also obtain $H_n(A,M)$ as Tyler suggests, by taking the free algebra resolution of A (in the category of k-algebras) and then hitting the resulting simplicial object with a suitable functor - but this seems trickier than in the commutative case (Andre-Quillen) and I can't get hold of a copy of Quillen's paper at the moment.

Alors. As I understand it, following Weibel's book (and the papers of Barr & Beck et al), the simplicial object (in the category of $k$-modules) that yields the Hochschild chain complex, arises by applying a certain Hom-functor (namely ${}_A{\rm Hom}_A(\ \cdot\ ,X)$ ) to another simplicial object, say $\beta(A)$, in the category of $A$-bimodules.

Now $\beta(A)$ is not contractible in the category of $A$-bimodules, in general, and doesn't come from a (co)monad on that category. However, $\beta(A)$ can be identified with another simplicial object $F(A)$, which lives in the category of $A$-modules.

What is $F(A)$?

Well, take a step back and consider the adjunction between $k$-modules and $A$-modules (maybe you need $k$ to be a field at this point, maybe not). That gives rise to a bar construction in $A$-mod, namely for any given $M$ in $A$-mod one obtains a simplicial object $F(M)$ which is given in each degree by

$$ F_{-1}(M)=M\quad,\quad F_n(M) = M \otimes A^{\otimes n+1} \ {\rm for }\ n \geq 0. $$

Note that this is contractible in $A$-mod by the general machinery of the bar resolution associated to a monad. There was nothing to stop us taking $M=A$, that's a perfectly good $A$-module; and on doing so, lo and behold, we get the same simplicial object $F(A)$.

Thus, Hochschild homology, regardless of the choice of coefficients, can be thought of as "coming from" a comonad - namely, that induced on $A$-mod by the forgetful functor from $A$-mod to $k$-mod. In my opinion, that is probably the (co)monad they are talking about.

It so happens that, since $F(A)$ is contractible in $A$-mod and hence a fotiori in $k$-mod, the "chain-complex-ification" of $\beta(A)$ is, as a chain complex in $R$-bimod, a resolution of $R$ by $k$-relatively projective $R$-bimodules -- and hence applying ${}_R{\rm Hom}_R(\ \cdot \ ,X)$ to it and taking homology coincides with taking $k$-relative Tor of $R$ and $X$ as R-bimodules. Hence the point of view that Hochschild homology is a special case of relative Tor.

Finally, I actually agree with Reid that this is not the best example to motivate (co)monad (co)homology. Group cohomology with coefficients in the ground field; or indeed André-Quillen cohomology, which is given by a "free algebra" adjunction but only for commutative algebras, or sheaf cohomology, would be better. (No originality in my choices; I've cribbed them out of Weibel Section 8.6).

(Apologies for the length and the tediousness, by the way.)

Answer by Yemon Choi for Coloring Points in the Plane

2010-02-03T06:28:22Z

I remember seeing this while grazing online back in the 20th century. The relevant phrase to google would be something like "chromatic number of the plane"; see also this wikipedia page.

Answer by Yemon Choi for Which R-algebras are the group ring of some group over a ring R?

2010-01-30T20:31:18Z

It doesn't answer your question, but might be worth noting en passant: depending on the choice of $R$, two different groups might give rise to the same group ring. The standard example is that when ${\mathbb C}$ is the field of complex numbers, and $G$ and $H$ are two finite abelian groups with the same cardinality, then the group algebras ${\mathbb C}G$ and ${\mathbb C}H$ are isomorphic. (In fact, even if you equip them with their natural $C^*$-algebra norms, the corresponding $C^*$-algebras will be *-isomorphic.)

If we work over a field $k$ of characteristic zero, then an old result of Kaplansky tells us that for any group $G$ (not necessarily finite) the group algebra $kG$ is "directly finite", in the sense that every left invertible element is right invertible. So any $k$-algebra which fails to have this property cannot arise as $kG$ for some group $G$.

There may perhaps also be homological obstructions of various sorts, but I don't know too much about such things in this setting.

Answer by Yemon Choi for Decoupling lemma for the Lambda(p) problem

2009-11-03T06:47:47Z

Having had a quick look, does the following work? Put $x= \sum_i x_i/3$ and put

$$ y(t) = \sum\nolimits_{i \in {R^1}_t} x_i = \sum_i \eta_i(t)x_i $$

and try to substitute these into (3.2).

Observe that

$$ \begin{aligned} |x| + |y(t)| = | \frac13 \sum_i x_i | + | \sum_i \eta_i x_i | & \leq | \frac13 \sum_i x_i | + | \sum_i x_i / 3 | + | \sum_i (\eta_i - 1/3)x_i | \\ &\leq | \sum_i x_i | + | \sum_i (\eta_i - 1/3)x_i | \end{aligned} $$

and this should give what we want on the RHS of the formula you're asking about.

Answer by Yemon Choi for Is Thompson's group F residually finite?

2010-01-31T23:26:46Z

The answer is "no" -- the quickest way to see this appeals to the following nontrivial fact: the commutator subgroup of $F$, denoted by $F'$ as usual, is infinite and simple.

Armed with this, we argue as follows. Let $N$ be a normal subgroup in $F$ of finite-index; then $N\cap F'$ is going to be normal in $F'$ and of finite index in $F'$. Hence $N\cap F'=F'$, that is, $N$ contains $F'$. So the intersection of all finite-index normal subgroups of $F$ must contain $F'$. But if $F$ were residually finite then this intersection would only contain the identity element, and the result follows.

If you don't mind me asking: is this a question out of curiosity, or one that you've run into during your studies or research?

Translation of "le nilradicalisé de g"

2010-01-31T21:43:01Z

I apologize for asking something that might well be found in a mathematical dictionary, but the similarity of the French word to an English one is frustrating my attempts to Google the answer (and the library is shut at time of typing). I suspect the answer should be obvious to those who, unlike me, know some basic Lie group/Lie algebra terminology.

Some context: I am reading an old paper of Dixmier from 1969, which has the following construction/definition. Let $\mathfrak g$ be a Lie algebra (characteristic zero, finite-dimensional), let $\mathfrak n$ be its largest nilpotent ideal -- the nilradical -- and put ${\mathfrak h}=[{\mathfrak g},{\mathfrak g}]+{\mathfrak n}$. Dixmier calls ${\mathfrak h}$ "le nilradicalisé de ${\mathfrak g}$".

Literal translation would surely be "the nilradicalised", but that sounds more like a mopey university indie band than a mathematical object. So what is the usual name for this object in English?

Answer by Yemon Choi for Quantum channels as categories: question 1.

2010-01-31T22:13:55Z

I'm not sure if it answers your question, but if ${\mathcal C}$ is a finite-dimensional $C^*$-algebra, and if $f$ and $g$ are completely positive, trace-preserving, linear maps from ${\mathcal C}\to {\mathcal C}$, then the composite map $g\circ f$ is going to be completely positive, trace-preserving, and linear. So one can indeed define a certain category to have ${\mathcal C}$ as its sole object, and have as its set of morphisms the collection of all CP-TP linear maps from ${\mathcal C}$ to itself. The associativity rule comes for free just because composition of functions is an associative operation.

If, on the other hand, you define quantum channels to be maps of a certain concrete form (rather than as being maps which preserve certain structure) then probably one needs to do a direct calculation similar to yours.

This might not be quite what you were asking, but I hope it helps.

Answer by Yemon Choi for How can I calculate the characteristic function of these distributions? [previously: difficult integral]

2010-01-31T21:26:30Z

If $q$ is a positive integer, then I think one can find this in any one of several undergraduate textbooks on complex analysis, where it's usually one of the standard examples to show the power of contour integration. I dimly remember something like this in Priestley's little OUP book, for instance. For arbitrary positive real values of $q$, I can't remember how this works I'm afraid.

(This is probably the sort of question which you could try out on fellow colleagues/students first, in my view.)

Answer by Yemon Choi for Distribution of 1-norm for Gaussian Unitary Ensemble

2009-10-26T19:43:09Z

I'm not at all expert on random matrix stuff, but until someone more qualified pops up, would you be interested in some crude estimates on Z in the $n \to \infty$ asymptotic? Or do you really want something sharper in each dimension?

EDIT/UPDATE: OK, here's my hand-wavy argument, it's been a while since I did any probability theory, so caveat lector and all that.

I'm going to use the GOE just because that's the one I know better and to save me worrying about stray scaling factors.

The idea is that for any $n \times n$ matrices $S$ and $T$ we always have

$\Vert S\Vert_1 \Vert T \Vert_{\rm op} \geq |{\rm tr}(ST)|$

where the subscript 1 denotes Ky-Fan/Schatten 1-norm and the subscript "op" denotes usual operator norm. In particular, if $S=T$ is self-adjoint then

$\Vert S\Vert_1 \geq || S ||_2^2\, /\, || S ||_{\rm op}$

Now when S is GOE(n,$\sigma^2$) then $n^{-2} \Vert S\Vert_2^2$ is strongly concentrated round its mean (which is $\sigma^2$) -- it's the average of a bunch of independent random variables so we could use variance estimates and Chebyshev, or probably some stronger exponential tail estimates.

Also, when S is GOE(n,$\sigma^2$) then $n^{-1/2} \Vert S\Vert_{\rm op}$ is strongly concentrated round $2\sigma$ - one can get exponential tail estimates, at least for an upper bound of $(2+\epsilon)\sigma$ for any positive $\epsilon$. I think this is folklore or a special case of Big Machinery, but as I said I have a more elementary proof, albeit one which is probably not original.

So, there is going to be a high probability (for $n$ large) that || S ||₂² is bigger than $(1-\epsilon)\sigma^2n^2$, and there is going to be a high probability (for $n$ large) that $\Vert S\Vert_{\rm op}$ is less than $(2+\epsilon)\sigma n^{1/2}$. On the intersection of these two events you're going to find that

$\Vert S\Vert_1 \geq (1-\epsilon)n^2\sigma^2 / (2+\epsilon)\sigma n^{1/2}$

which gives the lower bound I was claiming.

Answer by Yemon Choi for Geometric Interpretation of Trace

2010-01-31T02:05:24Z

If you are just working in a finite-dimensional Euclidean space, then by using the fact that we can calculate the trace of $A$ as $\sum_{j=1}^n \langle Ae_j, e_j\rangle$ for $any$ choice of orthonormal basis $e_1,\dots, e_n$, one obtains

${\rm Tr}(A) = \int_{x\in B} \langle Ax, x\rangle \,dm(x)$

where $B$ is the Euclidean unit sphere, and $m$ is the uniform measure on $B$ normalised to have total mass $1$. This is perhaps not quite as geometric as you want, but perhaps seems less dependent on a choice of coordinates.

Also, the wikipedia page refers to the trace as being (related to) the derivative of the determinant -- does that not seem `geometric'?

Answer by Yemon Choi for Orthogonal matrices with small entries

2010-01-29T10:01:17Z

Here's an idea which I think might be expandable to a solution once some details are filled in. (I am rather tired at the moment, though, so apologies if there is a cretinous error in what follows.)

We'll do the case $n=4m-1$ where $m$ is an integer; the case $n=4m-3$ is similar.

Let $C$ be a $2m\times 2m$ matrix which has the required form. Let $A$ be the $n\times n$ matrix with $C$ in the top left corner, $1$ on the remaning $2m-1$ diagonal entries, and zero elsewhere. Let $B$ be the $n\times n$ matrix with $C$ in the bottom right corner, $1$ on the remaining $2m-1$ diagonal entries, and zero elsewhere.

$A=\left[\begin{matrix} C & 0 \\ 0 & I_{2m-1} \end{matrix} \right]\quad,\quad B= \left[\begin{matrix} I_{2m-1} & 0 \\ 0 & C \end{matrix} \right]$

Both $A$ and $B$ will be real orthogonal since $C$ is. Consider the matrix $AB$, which being the product of real orthogonal matrices will also be orthogonal. I claim that the entries will all be $O(\sqrt{n})$ as required.

In more detail:

-- If both $i$ and $j$ are $\leq 2m-1$, then $(AB)_{ij}=A_{ij}=C_{ij}$ which is small by our choice of $C$; by symmetry, we can dispose of the case where both $i$ and $j$ are $\geq 2m+1$ in a similar way.

-- If $i\leq 2m-1$ and $j\geq 2m+1$, then on considering $\sum_r A_{ir}B_{rj}$ we see that the only nonzero contribution comes when $r\leq 2m$ and $r\geq 2m$, i.e. when $r=2m$ and so $(AB)_{ij}=A_{i,2m}B_{2m,j}$ is small.

-- If $i=2m$ or $j=2m$ then a similar analysis shows that $(AB)_{ij}$ can't be bigger than the entries of $C$ (at least up to some constant independent of $m$).

-- If $i\geq 2m+1$ and $j\leq 2m-1$ then $(AB)_{ij}=0$.

That should handle the case $n=4m-1$. The case $n=4m-3$ can be done in a similar fashion, but this time we will have extra factors of $3$ floating around since we have $3\times n$ and $n\times 3$ regions to consider, rather than just $1\times n$ and $n\times 1$ regions.

Answer by Yemon Choi for Category theory and model theory as "natural" counterparts

2010-01-29T10:25:27Z

I find this a difficult question to answer, but let me try for your Q1. It could be that some people don't feel comfortable promoting vague analogies, or indeed spending time discussing them. Signal-noise ratio, to be blunt. In particular, your point 9 is not really the sort of thing we want to spend time belabouring. Point 7 does not say much about either model theory or category theory; the fact I can't eat rocks or wood says little about the common material composition of either. Point 5 is again an observation that both lions and tables have legs.

There is, I think common ground between ideas from model theory and categorical frameworks; but this is something where the devil is in the detail and not in the blue sky. 'Tis very like a whale, one might say.

In my Philistine opinion, of course.

Answer by Yemon Choi for Conditional probabilities are measurable functions - when are they continuous?

2010-01-29T03:03:11Z

Even if your probability measure is absolutely continuous with respect to Lebesgue measure on $\Omega={\mathbb R}^2$, I don't think this suffices for the function $f$ you have defined to be continuous (just take $X$ to be independent from $Y$, i.e. your probability measure is just the product of two probability measures, "one on each axis", and choose the one for $X$ to be something in $L^1({\mathbb R}, {\mathcal B}, dx)$ which is discontinuous.

On the other hand, if ${\mathbb P}$ is not just absolutely continuous with respect to Lebesgue measure on $\Omega$, but has a continuous density function wrt said measure, then your function $f$ will be continuous -- just because integrating over a ball of radius $\eta$ with centre $\eta$ can only smooth things out, so that continuity of the original density function goes over to continuity of your conditional probability. [This is a fairly straightforward observation using basic properties of usual integration in the plane.]

So in your example, the Gaussian structure isn't really relevant as far as I can see. Also, if the original density function is strictly positive on the cylinder ${(x,y) : |y-y_0|<\eta }$, then the conditional probability you've defined will also be strictly positive; this condition is evidently not necessary, but I suspect in the examples you're interested in something like it should hold.

In between these two extremes, I'm not sure what else one can say. Perhaps, from your point of view, it's more important to go up to infinite-dimensional $\Omega$ but place restrictions on the kind of probability measure which you wish to consider.

Comment by Yemon Choi

2010-03-17T23:22:31Z

Is there any reason to expect there to be a nice answer? Also, why do you want to know the actual distribution, rather than (say) estimates on the moments or on the tails?

Comment by Yemon Choi

2010-03-17T23:20:53Z

As Jacques says: without more motivation on the OP's part, it's difficult to know what more to say on this problem. At present it just seems like an idle question; and like many people here I have my own idle questions to worry about...

Comment by Yemon Choi

2010-03-17T23:17:15Z

To try and put a more positive spin on what Zev has said: if you're having trouble working out where to start (so that the reason you can't give more details or working is because you yourself aren't certain what to try) then I think one of the sites listed in the FAQ might be of more help to you. It's natural to get stuck on basic problems if they're slightly beyond one's level; it's just that MO is not really addressed to people in that situation

Comment by Yemon Choi

2010-03-17T21:00:56Z

A bound in terms of what?

Comment by Yemon Choi

2010-03-17T20:16:42Z

It is worth noting that (AFAIK) there is no group which is known to be non-sofic. Also, in more or less the same breath, Kaplansky observes that the group von Neumann algebra of a discrete group, like any finite von Neumann algebra, is Dedekind finite (also sometimes called directly finite). A slightly lower tech proof can be extracted from M. S. Montgomery, Left and right inverses in group algebras, Bull. Amer. Math. Soc. 75 (1969) 539--540. (Might also shamelessly plug arxiv.org/abs/1003.1650 while I'm typing...)

Comment by Yemon Choi

2010-03-16T23:57:54Z

Ah, I see that according to Wikipedia, meme now means the same thing as trope. Well, as someone who still thinks that "alternate" should not be used as a synonym for "alternative", I'd best shut up now :)

Comment by Yemon Choi

2010-03-16T23:56:31Z

meme = abstract unit of cultural replication, no? I remain to be convinced that the word brings real benefit over "idea" or "folklore".

Comment by Yemon Choi

2010-03-16T09:59:14Z

I presume that you meant independent gaussian RVs?

Comment by Yemon Choi

2010-03-16T01:33:29Z

I get the impression Dissertationes is not meant for expository work per se. "Utility for a broad readership" just means that one is allowed a more leisurely style, I think.

Comment by Yemon Choi

2010-03-15T08:52:30Z

Thanks everyone for the comments on why my initial impression was naive.

Comment by Yemon Choi

2010-03-15T06:40:13Z

With the proviso that much of the older "folk history" is, I'm told, not accurate, or is misleading. I went to several entertaining history of mathematics lectures where this was pointed out vehemently. Also, the history of ideas is really tricky, because we have to try and understand how e.g. the Greeks thought, not how we would think about what they appear to describe.

Comment by Yemon Choi

2010-03-15T06:37:10Z

What have you worked out about the analogous non-discretized question? That is, if you put $h(x) = (x+2)^{3/2} - 2(x+1)^{3/2} + x^{3/2}$, can you show that $h$ is bounded by small constants? This would be my first attempt to get a handle on the problem; if it works, then you can examine more closely what happens upon taking "floors". (I don't think the problem has anything really to do with number theory per se.)

Comment by Yemon Choi

2010-03-14T11:44:37Z

Further to Scott's comment: please consult the FAQ for other sites which might be more appropriate for this kind of question.

Comment by Yemon Choi

2010-03-14T09:06:35Z

If you don't mind me asking: (a) why do you want to know? mere curiosity, or as a precursor to some other question? (b) what goes wrong with a direct attempt to use the definition of (sequential) compactness?

Comment by Yemon Choi

2010-03-13T20:22:49Z

OK, I was misinterpreting the tensor product of two modules as being a bimodule. If we're only taking the module action on, say, the first factor, then yes I agree that projectives are preserved. (I notice that Ben has immediately gone to finite-dimensional algebras; which is where I may have been talking at cross-purposes to everyone else.)