Φιλομαθὴς εἰμί.semper studēnsTumblr (3.0; @studeo)http://studeo.tumblr.com/Gamers discover protein structure that could help in war on HIV<a href="http://arstechnica.com/science/news/2011/09/gamers-discover-protein-structure-relevant-to-hiv-drugs.ars">Gamers discover protein structure that could help in war on HIV</a>: <blockquote>
<p>Minimization algorithms only work so well, however. The idea behind Foldit is that human pattern recognition and problem solving skills can succeed where the algorithms fail. The group’s first publication described this success; now, Foldit’s players have solved an important problem related to AIDS research.</p>
<p>[…]</p>
<p>The Foldit players were given ten different, incorrect models as a starting place. Over the course of sixteen days, players tweaked the designs, coming up with tens of thousands of variations. Then, the breakthroughs: a player named “spvincent” came up with a structure much better than any previous iteration, and two more players (“grabhorn” and “mimi”) quickly improved on it. With a few more days of work, the researchers completed the solution. Achievement unlocked, indeed.</p>
</blockquote>
<p>Human computing grid!</p>http://studeo.tumblr.com/post/10506218608http://studeo.tumblr.com/post/10506218608Thu, 22 Sep 2011 03:34:06 +0100computationoptimization"A logician is someone who thinks that a formula is a mathematical object."“A logician is someone who thinks that a formula is a mathematical object.”<br/><br/> - <em>Adrian Mathias</em>http://studeo.tumblr.com/post/10363243111http://studeo.tumblr.com/post/10363243111Sun, 18 Sep 2011 17:45:21 +0100logicphilosophymathematicsAn intuitionistic axiom of choice...?<ol><li>The law of excluded middle is invalid intuitionistically. [Brouwer; Heyting]</li>
<li>The axiom of choice is a theorem of intuitionistic type theory. [Martin-Löf, 1984, p. 50]</li>
<li>The axiom of choice implies the law of excluded middle in the internal logic of a topos, and so in particular in any intuitionistic set theory. [<a href="http://dx.doi.org/10.2307/2039868">Diaconescu, 1975</a>]</li>
</ol><p>We <em>appear</em> to have a contradiction. But this is an illusion, since the ‘axiom of choice’ in (2) and (3) are actually subtly different. [<a href="http://plato.stanford.edu/entries/axiom-choice/choice-and-type-theory.html">1</a>] Indeed, the axiom of choice of (3) follows from what <a href="http://dx.doi.org/10.1007/978-1-4020-8926-8_10">Martin-Löf [2009]</a> calls the ‘<em>extensional</em> axiom of choice’, but not the <em>intensional</em> axiom of choice of (2), which, according to Bishop [1967], is evident ‘because a choice is <em>implied by the very meaning of existence</em>.’</p>
<!-- more -->
<h3>References</h3>
<ul><li>Bishop, Errett (1967). <em>Foundations of constructive analysis</em>. </li>
<li>Diaconescu, Radu (1975). Axiom of choice and complementation. In: <em>Proceedings of the American Mathematical Society</em> 51.1, pp. 176–177. </li>
<li>Martin-Löf, Per (1984). <em>Intuitionistic type theory</em>.</li>
<li>Martin-Löf, Per (2009). 100 years of Zermelo’s axiom of choice: What was the problem with it? In: <em>Synthese Library</em> 341, pp. 209–219.</li>
</ul>http://studeo.tumblr.com/post/9169061350http://studeo.tumblr.com/post/9169061350Sat, 20 Aug 2011 17:32:00 +0100intuitionismaxiom of choicetopos theoryThe sum of human knowledge is not deductively closed<p>From <a href="http://www.scottaaronson.com/blog/">Scott Aaronson</a>'s recent essay, <em><a href="http://eccc.hpi-web.de/report/2011/108/">Why philosophers should care about computational complexity</a></em>:</p>
<blockquote>
<p>While the details differ, what most formal accounts of knowledge have in common is that they treat an agent’s knowledge as <em>closed</em> under the application of various deduction rules like the ones above. In other words, agents are considered <em>logically omniscient</em>: if they know certain facts, then they also know all possible logical consequences of those facts.</p>
<p>Sadly and obviously, no mortal being has ever attained or even approximated this sort of omniscience (recall the Turing quote from the beginning of Section 1). So for example, I can know the rules of arithmetic without knowing Fermat’s Last Theorem, and I can know the rules of chess without knowing whether White has a forced win.</p>
<p>[…]</p>
<p>Intuitively, we want to say that your “knowledge” consists of various non-logical facts (“grass is green”), together with <em>some</em> simple consequences of those facts (“grass is not pink”), but not necessarily <em>all</em> the consequences, and certainly not all consequences that involve difficult mathematical reasoning. Unfortunately, as soon as we try to formalize this idea, we run into problems.</p>
<p>The most obvious problem is the lack of a sharp boundary between the facts you know right away, and those you “could” know, but only after significant thought. (Recall the discussion of “known primes” from Section 3.3.) A related problem is the lack of a sharp boundary between the facts you know “only if asked about them,” and those you know even if you’re <em>not</em> asked.</p>
</blockquote>
<p>Of course, the same is <em>obviously</em> true of belief systems. If one holds beliefs which can be shown to imply a logical contradiction, if beliefs were deductively closed (and logic classical, or at least intuitionistic), then one would believe anything and everything, by the principle that <em>ex falso quodlibet</em>. But then again we have such things as <a href="http://en.wikipedia.org/wiki/Paraconsistent_logic">paraconsistent logic</a> to deal with this particular problem.</p>http://studeo.tumblr.com/post/8717822075http://studeo.tumblr.com/post/8717822075Wed, 10 Aug 2011 03:51:30 +0100philosophylogiccomplexityCracking nuts<blockquote>
<p>Je pourrais illustrer la deuxième approche, en gardant l’image de la noix qu’il s’agit d’ouvrir. La première parabole qui m’est venue à l’esprit tantôt, c’est qu’on plonge la noix dans un liquide émollient, de l’eau simplement pourquoi pas, de temps en temps on frotte pour qu’elle pénètre mieux, pour le reste on laisse faire le temps. La coque s’assouplit au fil des semaines et des mois – quand le temps est mûr, une pression de la main suffit, la coque s’ouvre comme celle d’un avocat mûr à point !</p>
</blockquote>
<p class="quote_source">— Alexandre Grothendieck, <em>Récoltes et semailles</em></p>
<blockquote>
<p>I can illustrate the second approach with the same image of a nut to be opened. The first analogy that came to my mind is of immersing the nut in some softening liquid, and why not simply water? From time to time you rub so the liquid penetrates better, and otherwise you let time pass. The shell becomes more flexible through weeks and months—when the time is ripe, hand pressure is enough, the shell opens like a perfectly ripened avocado!</p>
</blockquote>
<p class="quote_source">— as translated by Colin McLarty, <em>The rising sea: Grothendieck on simplicity and generality </em></p>http://studeo.tumblr.com/post/8630447619http://studeo.tumblr.com/post/8630447619Mon, 08 Aug 2011 04:45:00 +0100Logicians on safari<a href="http://www.scottaaronson.com/blog/?p=152">Logicians on safari</a>: <p>Some cute examples of computability theory applied to specific problems.</p>
<blockquote><ol><li>We now know that, if an alien with enormous computational powers came to Earth, it could prove to us whether White or Black has the winning strategy in chess. To be convinced of the proof, we would <em>not</em> have to trust the alien or its exotic technology, and we would <em>not</em> have to spend billions of years analyzing one move sequence after another. We’d simply have to engage in a short conversation with the alien about the sums of certain polynomials over finite fields.</li>
<li>There’s a finite (and not unimaginably-large) set of boxes, such that if we knew how to pack those boxes into the trunk of your car, then we’d also know a proof of the Riemann Hypothesis. Indeed, <em>every</em> formal proof of the Riemann Hypothesis with at most (say) a million symbols corresponds to some way of packing the boxes into your trunk, and vice versa. Furthermore, a list of the boxes and their dimensions can be feasibly written down.</li>
<li>Supposing you do prove the Riemann Hypothesis, it’s possible to convince someone of that fact, <em>without revealing anything other than the fact that you proved it</em>. It’s also possible to write the proof down in such a way that someone else could verify it, with very high confidence, having only seen 10 or 20 bits of the proof.</li>
</ol></blockquote>http://studeo.tumblr.com/post/8517374979http://studeo.tumblr.com/post/8517374979Fri, 05 Aug 2011 17:21:18 +0100computer sciencecomputabilityAlgebraic real analysis<a href="http://www.tac.mta.ca/tac/volumes/20/10/20-10abs.html">Algebraic real analysis</a>: <p>An interesting paper from Peter Freyd, of adjoint functor theorem fame.</p>
<blockquote>
<p>The title is wishful thinking; there ought to be a subject that deserves the name “algebraic real analysis.”</p>
<p>Herein is a possible beginning.</p>
</blockquote>
<p>Not a short paper, nor easy! However, his final appendix makes for interesting reading. A quote:</p>
<blockquote>
<p>I was appalled by the gap between applied mathematical experience and what we could even imagine proving. How does one integrate over all continuous functions to arrive at the expected error of a particular method?</p>
<p>Of course one can carve out finite dimensional vector spaces of continuous functions and compute an expected error thereon. But all continuous functions? It’s easy to prove that there is no measure—not even a finitely additive measure—on the set of all continuous functions assuming at least that we ask for even a few of the most innocuous of invariance properties. Yet experience said that there was, indeed, such a measure on the set of functions one actually encounters.</p>
<p>But it wasn’t just a problem in mathematics: I learned from physicists that they succeed in coming to verifiable conclusions by pretending to integrate over the set of all paths between two points. Again it is not hard to prove that no such “Feynman integral” is possible once one insists on a few invariance properties.</p>
<p>Even later I learned (from the work of David Mumford) about “Bayesian vision”: in this case one wants to integrate over all possible “scenes” in order to deduce the most probable interpretations of what is being seen. A scene is taken to be a function from, say, a square to shades of gray. It would be a mistake to restrict to continuous functions—sharp contour boundaries surely want to exist. Quite remarkable “robotic vision” machines had been constructed for specific purposes by judiciously cutting down to appropriate finite-dimensional vector spaces of scenes. But once again, there is no possible measure on sets of all scenes which enjoy even the simplest of invariance conditions.</p>
<p>Thus three examples coming with quite disparate origins—math, science, engineering—were shouting that we need a new approach to measure theory.</p>
<p>One line of hope arose from the observation that the non-existence proofs all require a very classical foundation. There’s the enticing possibility that a more computationally realistic setting—as offered, say, by effective topoi—could resolve the difficulties. A wonderful dream presents itself: the role of foundations in mathematics—and its applications—could undergo a transformation similar to the last two centuries’ transformation of geometry.</p>
</blockquote>http://studeo.tumblr.com/post/8292080482http://studeo.tumblr.com/post/8292080482Sun, 31 Jul 2011 09:24:21 +0100mathematicscategory theoryalgebrareal analysisWhat is the intersection of π and the Cantor set?<p>An amusing observation from <a href="http://nlab.mathforge.org/nlab/show/Trimble+on+ETCS+I">Todd Trimble’s essay on the elementary theory of the category of sets</a>:</p>
<blockquote>
<p>My own reaction is that ZFC is perhaps way too powerful! For example, the fact that ∈ is an endo-relation makes possible the kind of feedback which can result in things like Russell’s paradox, if one is not careful. Even if one is free from the paradoxes, though, the point remains that ZFC pumps out not only all of mathematics, but all sorts of dross and weird by-products that are of no conceivable interest or relevance to mathematics. One might think, for example, that to understand a model of ZFC, we have to be able to understand which definable pairs (<em>a</em>, <em>b</em>) satisfy <em>a</em> ∈<em> b</em>. So, in principle, we can ask ourselves such otherwise meaningless gibberish as “what in our model and implementation is the set-theoretic intersection of the real number π and Cantor space?” and expect to get a well-defined answer. When you get right down to it, the idea that everything in mathematics (like say the number <em>e</em>) is a “set” is just plain bizarre, and actually very far removed from the way mathematicians normally think. And yet this is how we are encouraged to think, if we are asked to take ZFC seriously as a foundations.</p>
</blockquote>
<p>This has been bugging me ever since I understood that the ontology of ZFC contains sets, and only sets. A bit too barren for my taste. I expect the objects of my universe to have types other than just the one!</p>http://studeo.tumblr.com/post/7807429670http://studeo.tumblr.com/post/7807429670Tue, 19 Jul 2011 18:12:31 +0100mathematicsset theoryA logical interpretation of some bits of topology<a href="http://xorshammer.com/2011/07/09/a-logical-interpretation-of-some-bits-of-topology/">A logical interpretation of some bits of topology</a>: <blockquote>
<p>Although topology is usually motivated as a study of spatial structures, you can interpret topological spaces as being a particular type of logic, and give a purely logical, non-spatial interpretation to a number of bits of topology.</p>
</blockquote>
<p>I should note that the ‘semi-decidable logic’ he describes in this post is in fact the propositional fragment of geometric logic. A model of propositional geometric logic would be precisely the frame of open sets of a locale (a ‘pointless’ topological space), or equivalently, a complete Heyting algebra.</p>http://studeo.tumblr.com/post/7476532973http://studeo.tumblr.com/post/7476532973Mon, 11 Jul 2011 03:02:57 +0100mathematicslogictopologyExponential objects in a category with zero objects<p>To be precise, in a category with a pair of objects with at least two distinct arrows between them, there cannot be exponential objects if there is a zero object. (Proof below.) This shows that, for example, the set of all homomorphisms from one group to another is not “naturally” a group. On the other hand, the set of all homomorphisms from one abelian group to another <em>is</em> naturally an abelian group, but it is not an exponential object in the strict sense.</p>
<!-- more -->
<p>Indeed, let $\mathcal{C}$ be such a category, and $f, g : A \to B$ two distinct arrows. Recall that a <strong>zero object</strong> is an object $0$ which is <em>simultaneously</em> initial and terminal, i.e. for every object $C$, there is a <em>unique</em> arrow $0 \to C$ (so $0$ is initial) and a <em>unique</em> arrow $C \to 0$ (so $0$ is terminal). An <strong>exponential object</strong> is an object $B^A$ such that for every object $C$, there is a natural bijection between arrows $C \times A \to B$ and $C \to B^A$, natural in the sense that for any arrow $h : C \to C’$, the diagram $$\begin{matrix} \text{Hom}(C \times A, B) & \cong & \text{Hom}(C, B^A) \newline \uparrow & & \uparrow \newline \text{Hom}(C’ \times A, B) & \cong & \text{Hom}(C’, B^A) \end{matrix}$$ commutes, where the left vertical arrow is the precomposition map $(h \times \text{id})^*$ and the right vertical arrow is the precomposition map $h^*$, and similarly for any arrow $k : B \to B’$ (except covariantly instead of contravariantly). In succinct terms, we could say that the functor $- \times A : \mathcal{C} \to \mathcal{C}$ is left-adjoint to the functor $(-)^A : \mathcal{C} \to \mathcal{C}$.</p>
<p>In any category with a terminal object $1$ and binary products, $1 \times A \cong A$, so if there were an exponential object $B^A$, we would have a natural bijection between arrows $A \to B$ and $1 \to B^A$; but if $0 = 1$ (i.e. the terminal object is also initial), then there can only be one arrow $1 \to B^A$, hence, only one arrow $A \to B$. But this contradicts our hypothesis that we have two distinct arrows $A \to B$—so there cannot be an exponential object $B^A$.</p>
<p>Now, let me return to my earlier comment about the category of abelian groups, $\textbf{Ab}$. It is true that the set $\text{Hom}(A, B)$ is naturally an abelian group $\text{hom}(A, B)$ if $A, B$ are abelian groups: we say that $\textbf{Ab}$ has <strong>internal hom objects</strong>. The internal hom object satisfies a different universal property, however: we have a natural bijection between arrows $C \to \text{hom}(A, B)$ and $C \otimes A \to B$, where $\otimes$ is the <strong>tensor product</strong>. The reason the above argument breaks down here is because $0 \otimes A \cong 0$ for every abelian group $A$—in other words, $0$ is <em>not</em> an identity element for $\otimes$. On the other hand, it is true that $\mathbb{Z} \otimes A \cong A$ for every $A$, and there certainly is a natural bijection between the arrows $A \to B$ and $\mathbb{Z} \to \text{hom}(A, B)$, so all is well again.</p>http://studeo.tumblr.com/post/7076065181http://studeo.tumblr.com/post/7076065181Thu, 30 Jun 2011 09:06:00 +0100category-theorymathematicsIntuitionistic logic and Kuratowski's closure–complement problem<p>Suppose I give you a subset of a <a title="topological space" href="http://en.wikipedia.org/wiki/Topological_space">topological space</a>. Using only the operations of closure and complementation, how many subsets can you form using this?</p>
<p>As it turns out, the answer is a finite number: 14. The proof is somewhat intricate, and one of the steps is to prove that the interior of the complement of the interior of the complement of the interior of the complement of the interior is the same thing as the interior of the complement of the interior. When I first did this exercise a year ago, I found this identity to be somewhat mysterious, but I just realised that it can be explained using the fact that the internal logic of a topological space is <a title="intuitionistic" href="http://en.wikipedia.org/wiki/Intuitionistic_logic">intuitionistic</a>!</p>
<!-- more -->
<p>Since it’s a bit unwieldy to keep saying “interior of the complement”, let’s write ${\sim}A$ for the interior of the complement of a subset $A$ of the topological space $X$, and say that ${\sim}A$ is the <strong>pseudocomplement</strong> of $A$. The claim is that, for all subsets $A \subseteq X$, $${\sim}{\sim}{\sim} (A^{\circ}) = {\sim} (A^{\circ})$$ If you know something about intuitionistic logic, this should immediately ring bells in your head, and in fact, by the time you finish reading this sentence, you probably don’t need to read the rest of this post.</p>
<p>But for the benefit of readers who don’t know intuitionistic logic, let me just briefly explain what it is. In classical logic, we have the <a href="http://en.wikipedia.org/wiki/Law_of_excluded_middle">law of excluded middle</a>, or <em>tertium non datur</em>, meaning that for any proposition $p$, $p \lor \lnot p$ is a tautology. This is equivalent to the rule of <a title="double negation elimination" href="http://en.wikipedia.org/wiki/Double_negation_elimination">double negation elimination</a>, which states that $p$ is deducible from $\lnot \lnot p$. In intuitionistic logic, we reject the law of excluded middle, and it is no longer valid to claim that $p$ and $\lnot \lnot p$ are the same thing. Nonetheless, something weaker is true: you can deduce $\lnot p$ from $\lnot \lnot \lnot p$, and vice versa.</p>
<p>It turns out that it is possible to interpret intuitionistic logic in terms of reasoning about open subsets of a topological space, and under this interpretation, logical negation corresponds exactly with pseudocomplementation. So, when $A$ is an open subset, it is no longer surprising that $${\sim}{\sim}{\sim} A = {\sim} A$$ That’s not quite what we need, since $A$ may or may not be open; but it’s easy enough to fix — we just take the interior of $A$ and we obtain the identity $${\sim}{\sim}{\sim} (A^{\circ}) = {\sim} (A^{\circ})$$ which is what was claimed.</p>http://studeo.tumblr.com/post/6179821857http://studeo.tumblr.com/post/6179821857Sat, 04 Jun 2011 17:45:00 +0100mathematicstopologylogicArrows in mathematics<p>A well-known point of view holds that category theory is really about the morphisms between objects, rather than about the objects themselves. I like the idea, but it just occurred to me that there’s another kind of arrow which is quite fundamental in mathematics: the implication arrow.</p>
<p>In fact, the Curry–Howard correspondence even makes a connection between the two kinds of arrows, so perhaps it is really is true that mathematics is actually about arrows between objects, rather than about numbers or shapes or sets…</p>http://studeo.tumblr.com/post/5906357614http://studeo.tumblr.com/post/5906357614Fri, 27 May 2011 21:41:00 +0100mathematicscategory theoryRational numbers as functions: a first look at Spec ℤ<p>In Kempf’s <u>Algebraic Varieties</u>, he says something to the effect that a scheme is what you get when you replace the algebras of functions that arise naturally in algebraic geometry by more general rings, which may or may not be rings of functions. It turns out that we can use this mirror between algebra and geometry to interpret the rational numbers as functions on a topological space, albeit with some complications.</p>
<!-- more -->
<p>I recently spent some time thinking about $\operatorname{Spec} \mathbb{Z}$, in my attempt to understand how ramification of primes in the ring of integers of a number field corresponds to ramification in algebraic geometry and Riemann surfaces. Although I eventually figured it out by translating the geometric picture into algebra, I thought I would reverse the process and try to figure out the geometry of $\operatorname{Spec} \mathbb{Z}$, and the first step in doing that is understanding its structure sheaf, and to do that we need to look at the underlying topological space.</p>
<p>Recall that the spectrum of a ring is the set of all its prime ideals, endowed with the topology where the closed sets are exactly the sets of prime ideals contained in <em>any</em> ideal of the ring. In the case of the integers, this means that $\operatorname{Spec} \mathbb{Z}$ can be identified with the set of all prime numbers (since $\mathbb{Z}$ is a principal ideal domain), plus $0$ (since $(0)$ is a prime ideal too), and the closed sets are precisely all the finite sets of prime numbers (and the set of all prime numbers, of course). Hence, all the non-empty open sets of $\mathbb{Z}$ open are cofinite and dense.</p>
<p>The local ring at the prime $p$ is the localisation $\mathbb{Z}_{(p)}$ of $\mathbb{Z}$ at the prime ideal $(p)$, which is (isomorphic to) the subring of rational numbers $a/b$ where $p$ does <em>not</em> divide $b$. Note that this means that the local ring at $0$ — which is the generic point of this space — is just the whole ring $\mathbb{Z}$. The residue field at $p \ne 0$ is the finite field $\mathbb{F}_p$, and the residue field at $0$ is $\mathbb{Q}$. Since $0$ is the generic point of $\operatorname{Spec} \mathbb{Z}$, we may interpret $\mathbb{Q}$ as the field of rational functions on $\operatorname{Spec} \mathbb{Z}$. But in what sense are rational numbers functions?</p>
<p>In order to make the analogy clear, we need to understand the notion of evaluating a rational function at a point on an affine variety. Recall that the (closed) points of an affine variety $X$ are in natural bijection with the maximal ideals of the global function ring $\mathcal{O}(X)$; the local ring $\mathcal{O}_P$ at a point $P \in X$ is $\mathcal{O}(X)$ localised at that maximal ideal $\mathfrak{m}_P$, and the residue field is (isomorphic to) the base field $k$. The evaluation map is then just the natural quotient map $\text{ev}_P : \mathcal{O}_P \to \mathcal{O}_P / \mathfrak{m}_P$.</p>
<p>With that in mind, we can now “evaluate” rational numbers $a/b$ at a prime $p$. Recall that if $p$ does not divide $b$, $a/b \in \mathbb{Z}_{(p)}$, so we may evaluate $a/b$ at $p$ in that case, and its value is just $\bar{a} \bar{b}^{-1}$, where $\bar{a}$ and $\bar{b}$ are the residue classes of $a$ and $b$ (respectively) in $\mathbb{F}_p$. For example, $6/11$ evaluated at $7$ is $5 \pmod 7$. Note that $a/b$ is regular at $0$ if and only if it is an integer, and in that case it evaluates to itself. Note also that, unlike the case of algebraic varieties, the residue field varies from point to point, so we can’t really identify $a/b$ with some function $f : \operatorname{Spec} \mathbb{Z} \to A$ where $A$ is some fixed ring. Nonetheless, we have something analogous to a function: $a/b$ can be recovered from its values at all primes; $a/b$ can have multiple poles and multiple zeros, just like a rational function on a variety; and not every assignment of values gives rise to a rational number, just as not every continuous function gives rise to a rational function.</p>http://studeo.tumblr.com/post/5777431761http://studeo.tumblr.com/post/5777431761Mon, 23 May 2011 22:09:00 +0100mathematicsschemesalgebraic-geometry