User adrien - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T13:59:32Z http://mathoverflow.net/feeds/user/13552 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/126461/quantized-enveloping-algebras-at-q1/126645#126645 Answer by Adrien for Quantized Enveloping Algebras at $q=1$ Adrien 2013-04-05T16:33:11Z 2013-04-05T16:59:56Z <p>Even if strictly speaking it's not true, it's always usefull to remember that morally $$K_i=q_i^{H_i}:=\exp(H_i\log(q_i))$$ where $q_i=q^{d_i}$ for some integer $d_i$ attached to the Cartan matrix of $\mathfrak g$, and where $H_i$ is the $i$th Cartan generator of $\mathfrak g$. This explain why the classical limit of $K_i$ should be 1.</p> <p>Now set $$G_i=\frac{K_i-K_i^{-1}}{q_i-q_i^{-1}}$$</p> <p>Then using again this heuristic it's easliy seen that you have $$\lim_{q\rightarrow 1} G_i= \lim_{q\rightarrow 1}\frac{q_i^{H_i}-q_i^{- H_i}}{q_i-q_i^{-1}}= H_i$$</p> <p>It's now rather clear how to prove that it indeed work, since $G_i$ is precisely the R.H.S. of the relation you want to modify, so just replace it by $$[E_i,F_j]=\delta_{i,j} G_i$$ and add the relaiton $$(q-q^{-1})G_i=K_i$$ which is nothing but the definition of $G_i$. That you get the same algebra is obvious because you did not really change anything. The former relation does not depend on $q$ anymore so gives you the appropriate relation at the classical limit, and the latter relation just become trivial.</p> <p>Then following again the heuristic that $K_i$ should goes to 1 at the limit, take the quotient by the ideal generated by $K_i-1$, and you get the classical envelopping algebra.</p> http://mathoverflow.net/questions/122095/on-tamarkins-proof-of-etingof-kazhdan-quantization-of-lie-bialgebra On Tamarkin's proof of Etingof-Kazhdan quantization of Lie bialgebra Adrien 2013-02-17T20:06:29Z 2013-02-28T16:37:07Z <p>This question is motivated by the desire to understand better the interplay between Drinfeld associators, algebraic structures up to homotopy and related results.</p> <p>Recall that a Lie bialgebra is a Lie algebra $\mathfrak g$ with a Lie algebra structure on $\mathfrak g^*$ (cobracket) plus some compatibility conditions. These are infinitesimal versions of compatible Poisson structures on Lie group. A quantization of $\mathfrak g$ is a $\mathbb C[[\hbar]]$ deformation of the envelopping algebra of $\mathfrak g$ the coproduct of which gives back the cobracket of $\mathfrak g$ at the quasi-classical limit. Etingof-Kazhdan theorem states that every Lie bialgebra can be quantized, in a functorial way. Tamarkin's proof of this results relies on the formality of the little disc operad, and in fact so does the original proof in a slighty different language.</p> <p>The original proof use the statement "the choice of an associator leads to a quasi-triangular quasi-Hopf algebra (aka braided monidal category) from any metrizable Lie algebra (i.e. equipped with an invariant symmetric bilinear form)". Indeed, any metrizable Lie algebra leads to a representation of a certain operad in Lie algebra $\mathfrak t$ and an associator is nothing but an isomorphism from some completion of the group algebra of pure braid operad to $U(\mathfrak t)$.</p> <p>Then the proof goes a follow:</p> <ol> <li>To any Lie bialgebra $\mathfrak g$ is asociated its double $\mathfrak d$ which quasi-triangular and in particular metrizable. $\mathfrak d=\mathfrak g\oplus \mathfrak g^*$ as vector space and the pairing is just the canonical one.</li> <li>Use the above statement to turns $U(\mathfrak d)[[\hbar]]$ into a quasi-triangular quasi-Hopf algebra</li> <li>Solve the so-called "twist equation" which say that the quasi-Hopf algebra $U(\mathfrak d)[[\hbar]]$ can be turned into an honest Hopf algebra $U_{\hbar}(\mathfrak d)$. An important fact is that the product is not modified, i.e. $U_{\hbar}(\mathfrak d)$ is isomorphic to $U(\mathfrak d)[[\hbar]]$ as an algebra.</li> <li>That we have a quasi-triangular quantization means roughly that we get a quantum version of the pairing (given by the quantum R-matrix) which can be used to identify a sub-Hopf algebra $U_{\hbar}(\mathfrak g)$ of $U_{\hbar}(\mathfrak d)$ which is the quantization we were looking for. One can also show that $U_{\hbar}(\mathfrak d)$ is the quantum Drinfeld double of $U_{\hbar}(\mathfrak g)$.</li> </ol> <p>Note that thinking of $U_{\hbar}(\mathfrak g)$ as a subalgebra of $U_{\hbar}(\mathfrak d)$ is a bit misleading, and that it's somehow better to see $U(\mathfrak d)$ as an auxiliary space acting on $S(\mathfrak g)$ by differential operators. In this sense the "twist" really leads to a star product and a "star coproduct" on $S(\mathfrak g)$ whose coefficients are given by the action of some differential operators.</p> <p>Now Tamarkin's proof goes as follow:</p> <ol> <li>Any Lie bialgebra $\mathfrak g$ leads to a Gerstenhaber structure on the free graded commutative algebra $H=S(\mathfrak g_{\hbar}[-1])$ where in $\mathfrak g_{\hbar}$ the cobracket is multiplied by $\hbar$.</li> <li>The operad of Gerstenhaber algebras is quasi-isomorphic to $BU(\mathfrak t)$ (where $B$ is the bar construction).</li> <li>The little disc operad is formal, which is roughly a refinement of the above statement on the relation between braids and $\mathfrak t$, applied at the level of chain.</li> <li>The operad of chain of the little disc operad is quasi-isomorphic to the operad of brace algebras (this is a version of the so-called Deligne conjecture).</li> <li>Therefore we get a brace structure on $H$, which induces a differential graded Hopf algebra structure on the cofree coalgebra $C(H[1])$. </li> <li>Finally, so far I understand, one can show that the 0th cohomolgoy group of the dg Hopf algebra obtained this way is a Hopf algebra quantizing $\mathfrak g$.</li> </ol> <p>Admittedly I'm less confortable with this proof, though I can roughly understand how it goes, except maybe the last step. Anyway, I'm interested in understanding how thse proofs relate. Clearly the key fact is quite similar, though different: in the first case the (trivial deformation of) the category of <em>modules</em> over $\mathfrak d$ is an algebra over the operad $\mathfrak t$, while in the second case we construct a space which is an algebra over some operad of complex attached to $\mathfrak t$ through some combinatorial quasi-isomorphism. Then the same isomorphism coming from an associator is applied but in different worlds. So are related Etingof-Kahdan 2 and Tamarkin 3.</p> <p>Now I'd like to understand the relations between the other steps. The relation between the pairing on $\mathfrak d$ and the complex associated to $\mathfrak g$ is probably well known, but I didn't find a reference. The second part is probably more tricky, and I would really be happy to learn that there is indeed some relations between the Deligne conjecture and the statement that "a quasi Hopf algebra obtained from a Drinfeld associator can be turned into an Hopf algebra". Since the role of the Drinfeld double is to somehow merge the (Lie) algebra and the (Lie) coalgebra structure, it's temptating to think that there is a sort of correspondance between coalgebra deformations of $U(\mathfrak d)$ and bialgebra deformations of $U(\mathfrak g)$, and Etingof-Kahdan's proof show that it is indeed the case, in a highly non obvious way. Can this fact also be related to the Deligne conjecture ?</p> http://mathoverflow.net/questions/119354/string-diagrams-of-special-monoidal-categories-and-higher-categories/119361#119361 Answer by Adrien for String diagrams of special monoidal categories and higher categories Adrien 2013-01-19T20:48:25Z 2013-01-20T00:36:39Z <p>As you point out, in such a situation it may not be possible to compose horizontally, hence in full generality you cannot assume your underlying category to be monoidal. On the other hand, you still want to keep track somehow of the tensor product. I think it can be handled as follow: </p> <p>The structure you are looking for should be the data of a (strict, say) monoidal category $C$, an arbitrary category $D$, a functor $F:C\rightarrow D$ and a natural isomorphism $$\gamma_{-,-}:F(-\otimes -)\rightarrow F(-\otimes^{op} -)$$. It seems to me that one natural coherence condition to impose is: $$\gamma_{Z\otimes X,Y}\circ\gamma_{X\otimes Y,Z} =\gamma_{X,Y\otimes Z}^{-1}$$</p> <p>Maybe you need the "reversed" condition as well. </p> <p>Anyway, examples of such categories may be constructed as follows: let $C$ be a <strong>rigid</strong> monoidal category enriched over some category $D$, let $F$ be the functor $Hom_C(1,-)$ and $\gamma$ be given by $$Hom_C(1,X\otimes Y)\cong Hom_C(X^*,Y)\cong Hom_C(1,Y\otimes X)$$</p> <p>I'm not aware of such a definition in the litterature, but the obviously related notion of a categorical structure involving <strong>braid</strong> diagrams drawn on a surface appears in <a href="http://arxiv.org/abs/math/0702670" rel="nofollow">this paper</a> by Calaque-Enriquez-Etingof for the torus (under the name elliptic structure) and for abritrary surfaces in <a href="http://tel.archives-ouvertes.fr/tel-00762209" rel="nofollow">Philippe Humbert's thesis</a>.</p> http://mathoverflow.net/questions/95116/hamiltonians-which-commute-both-as-operators-and-as-connections/115610#115610 Answer by Adrien for Hamiltonians which commute both as operators and as connections Adrien 2012-12-06T14:02:20Z 2012-12-06T14:02:20Z <p>Hi David,</p> <p>I think there is indeed a relation, which I learned precisely from papers of Varchenko among others. All of this is rather classical and can be found e.g. in Etingof-Frenkel-Kirilov book "Lectures on Representation Theory and Knizhnik-Zamolodchikov Equations". </p> <p>The fact that the $H_i$ satisfies this stronger condition is equivalent to say that for any parameter $\kappa$ the operators $\kappa \partial_i+H_i$ alos satisfies ($\dagger$).</p> <p>Hence you can take asymptotic expansion of solutions at $\kappa \rightarrow 0$ on some neighbourhood $D$ of some $z_0$, of the form $$e^{S(z)/\kappa} (f_0(z)+O(\kappa))$$</p> <p>where $S$ is a scalar valued function. Then you can show that, assuming the $H_i(z)$ are simultaneously digonalizable then $f_0$ is a common eigenvector of them, with eigenvalues $\partial_i S$. Conversly given a common eigenvector at some $z_0$ you can construct an asymptotic solution. So the usual trick, widely used in the study of the KZ equation, is to also take some asymptotic limit w.r.t. the variable $z_i$ in such a way that eigenvectors are "easy" to find. The standard example in the KZ case is the asymptotic zone </p> <p>$$|z_i-z_1| \ll |z_j -z_1|\quad if\quad i &lt; j$$</p> <p>for which, up to some change of variable, the equation can be written $$\kappa \partial_i f= \left ( \Omega_i/u_i +reg\right)f\quad i=1\dots n-1$$ where $\Omega_i=\sum_{k &lt; i} \Omega_{k,i+1}$ and $reg$ is regular at $u=0$. Then given some common eigenvector $v$ of the $\Omega_i$ with eigenvalues $\mu_i$ there exists a unique solution of the form $$(\prod u_i^{\mu_i/\kappa})(v+r(u))$$ where $r(u)$ is regular at $u=0$ and $r(0)=0$.</p> <p>I'm not very familiar with D-modules (and by the way I would be happy is someone extends on this), but you can rephrase it as follows: viewing $\kappa$ as a formal variable leads to a filtration on the algebra of differential operators on $V$ (the vector space acted on by the $H_i$) which in turn is nothing but the usual filtration by the degree of differential operators. Taking the associated graded turns the equation</p> <p>$$(\partial_i+H_i)f=0$$</p> <p>into the equation</p> <p>$$(y_i+H_i)f=0$$</p> <p>Whose solutions are clearly commons eigenvectors of $H_i$. So I'm rather confident that you can say that the spectrum of the $H_i$ for all common eigenvectors is the characteristic variety of the D-module of solutions of the differential equation you started with.</p> http://mathoverflow.net/questions/115517/what-is-the-free-monoidal-category-generated-by-a-monoid/115525#115525 Answer by Adrien for What is the free monoidal category generated by a monoid? Adrien 2012-12-05T18:14:54Z 2012-12-05T18:14:54Z <p>I think you are interpreting it as "(free monoidal category) on a monoid", while it's really "free (monoidal category on a monoid)", or rather "free (monoidal category <strong>with</strong> a monoid)".</p> <p>Now to construct it you have have to find which morphisms should exists as a consequences of the axioms, i.e. which morphisms you are sure to see whenever you have a monoid in a monoidal category. Clearly, if $X$ is an object in a monoidal category $C$, the only morphisms between $X^{\otimes n}$ and $X^{\otimes m}$ that exists whatever $(C,X)$ are, are the identity when $m=n$. So the free monoidal category is just $\mathbb{N}$ whith $\hom(m,n)=${id} if $m=n$, and is empty otherwise.</p> <p>But if $X$ is a monoid, you get by definition a morphism $X \otimes X \rightarrow X$ and a morphism $I \rightarrow X$ where $I$ is the identity object. These morphisms clearly extends to morphisms $\delta_i:X^{\otimes n+1}\rightarrow X^{\otimes n}$ and $\sigma_i: X^{\otimes n}\rightarrow X^{\otimes n+1}$ just by tensoring them with identity morphisms, and then you can compose them to get maps $X^{\otimes m}\rightarrow X^{\otimes n}$. Finally, it's easy to see that the commutation relation between these morphisms imposed by the axioms of a monoid are precisely those satisfied by the simplicial maps in $\Delta$.</p> http://mathoverflow.net/questions/108648/relations-in-a-particular-subgroup-of-the-braid-group/108650#108650 Answer by Adrien for Relations in a particular subgroup of the braid group. Adrien 2012-10-02T16:41:33Z 2012-10-02T16:41:33Z <p>You probably already noticed that, but $B_{p,q}$ is the fundamental group of $$X_n/(S_p \times S_q)$$ where $X_n$ is the configuration space of $n$ points in the complex plane. Ths may help to guess some facts about these groups.</p> <p>So far I know these group are usually called "mixed braid groups" in the litterature, though this name is sometimes used for the group of braids with the $p$ first strands fixed. Anyway, a presentation of it is obtained in S. Manfredini <strong>Some subgroups of Artin's braid group</strong> (available <a href="http://www.sciencedirect.com/science/article/pii/S0166864196001526" rel="nofollow">here</a>). Results on their relation with the representation theory of $B_n$, as well as references you may find interesting, can be found in Bellingeri, Godelle, Guaschi, <strong>Exact sequences, lower central series and representations of surface braid groups</strong> (<a href="http://arxiv.org/abs/1106.4982" rel="nofollow">arXiv</a>).</p> http://mathoverflow.net/questions/103397/categorification-of-finite-type-invariants Categorification of finite type invariants Adrien 2012-07-28T17:12:50Z 2012-08-22T14:54:27Z <p>Recall that finite type invariants are those numerical knots invariants vanishing on knots with enough singularities. The value of an invariant on a singular knots is computed using the <a href="http://ncatlab.org/nlab/show/Vassiliev+skein+relation" rel="nofollow">Vassiliev skein relation</a>.</p> <p>A universal finite type invariant can be obtained either from the Kontsevich integral, of from Chern-Simons theory. These invariants takes values in some space of Feynman diagrams (so-called chord diagrams) encoding some abstract Lie algebraic information. They are universal in the sense that every finite type invariant factors through them, up to invariants of lower order. In other words, every finite type invariant induces an invariant of chord diagrams (it is the easy part), and conversely any invariant of diagrams can be enhanced to a knot invariant (this is of course the hard part).</p> <p>Kontsevich's construction can be extended to tangles, and arguably the better way of viewing it is as a functor from the category of tangles to the category of diagrams.</p> <p>FInite type invariants are interesting because most of known numerical invariant are of finite type. In particular, setting $q=e^{\hbar}$ in any quantum invariant, one get a formal power series whose $n$th coefficient if a type $n$ invariant.</p> <p>It turns out that many of these invariants were categorified. The main example is Khovanov's categorification of the Jones polynomial: so far I understand, it's a 2-functor from the category whose morphisms are isotopy classes of tangles and 2-morphisms are cobordisms, to a 2-category whose morphisms are certain chain complexes, and 2-morphisms are morphisms of complexes up to homotopy.</p> <p>My first questions are:</p> <blockquote> <p>Is there a widely agreed definition of what the categorification of a finite type invariant should be ? What about a universal one ? Is there an interesting 2-structure on the category of chord diagrams ? Or, as hinted <a href="http://arxiv.org/abs/1106.0042" rel="nofollow">here</a>, is it better to categorify the spaces of diagrams themselves ?</p> </blockquote> <p>Indeed, since this somehow amongs to construct a "universal homological invariant", one could imagine to replace the target category of diagrams by a category which is already a category of chain complexes, categorifying the 4T relation.</p> <p>I googled some related buzzwords but did'nt find an answer so far. Of course, having the classical case in mind, a naive starting point is:</p> <blockquote> <p>Does it happen something interesting if one extends Khovanov's construction to singular tangles ? Does it leads to an "homological invariant of chord diagrams" ?</p> </blockquote> <p>Since Khovanov's construction categorifies the Kauffman bracket it's probably doable to see what happen when one flips crossings, but I'm not aware of a nice interpretation of the result.</p> <p><strong>Edit:</strong> To elaborate on my comment below. From a categorical point of view, finite type invariant are exactly the functors from the category of tangles to a "quantum" ribbon category $C$, by which I mean a graded, complete ribbon category whose braiding squares to a perturbation of the identity. Then the statement "every finite type invariant induces an invariant of chord diagrams" translates into "this functor induces a functor from the category of diagrams to the quasi-classical limit of $C$ (which is a so-called infinitesimal braided category)". It makes the relation with quantum groups rather obvious. Then Kontsevich's theorem, reformulated using results of Drinfeld, say something like "every infinitesimal braided category can be quantized". While I agree that categorifying one single finite type invariant hardly makes sense, I think that the above statements should admits nice 2-ification, which should be related to 2-Lie algebras.</p> <p>Indeed, some authors are trying to categorify quantum groups themselves, among other things in order to recover Khovanov's construction by an analog of Reshetikhin-Turaev functor.</p> <p>The paper pointed out by Chris is very interesting: it uses some notions I'm not familiar with, but gives a definition along the lines of: "an homological invariant is of finite type if its extension to a knot with enough singularity is acyclic" and prove that some perturbative expansion of Khovanov homology provide such invariants.</p> <p><strong>Edit 2</strong>: As a hint to a possible answer, let me mention this <a href="http://arxiv.org/abs/math/0005197" rel="nofollow">paper</a> by Hinich and Vaintrob which gives a very nice "propic" construction of the space of chord diagrams out of the Lie operad, making rigorous the statement that it is universal among metrizable Lie algebras. I guess that their construction applied to the operad of 2-Lie algebras should provide a good candidate for the analog of chord diagrams in a 2-ification of the theory of finite type invariants.</p> http://mathoverflow.net/questions/66084/open-problems-with-monetary-rewards/103184#103184 Answer by Adrien for Open problems with monetary rewards Adrien 2012-07-26T10:58:14Z 2012-07-26T10:58:14Z <p>It was pointed out by <a href="http://en.wikipedia.org/wiki/Randall_Munroe" rel="nofollow">Randall Munroe</a> that by proving the inconsistency of logic you can earn quite a lot of money:</p> <p><img src="http://imgs.xkcd.com/comics/applied_math.png" alt="alt text"></p> <p>(<a href="http://xkcd.com/816/" rel="nofollow">Source</a>)</p> http://mathoverflow.net/questions/102855/about-the-term-tangential-derivation-on-a-free-lie-algebra/102885#102885 Answer by Adrien for About the term "tangential derivation" on a free Lie algebra. Adrien 2012-07-22T18:05:15Z 2012-07-22T18:05:15Z <p>You may be interested in these <a href="http://katlas.math.toronto.edu/drorbn/dbnvp/wClips-120530.php" rel="nofollow">Bar Natan's lecture</a>.</p> <p>The geometric intuition is that the $n$th Lie algebra of tangential derivations can be realized into the algebra of tangential differential operators (not derivations !) on $\mathfrak g^n$ for any (say finite dimensional) Lie algebra $\mathfrak g$. If $G$ is a Lie group with Lie algebra $\mathfrak g$, then it acts on $\mathfrak g$ by the adjoint action, and the action of a tangential differential operators is a differentiation in a direction which is tangential to the orbits of the adjoint action, hence the name.</p> http://mathoverflow.net/questions/102803/software-for-numerical-constraint-satisfaction-problems/102867#102867 Answer by Adrien for software for numerical constraint satisfaction problems Adrien 2012-07-22T12:39:32Z 2012-07-22T12:39:32Z <p>It seems to me that methods based on interval analysis are very efficient for solving hard constraints satisfaction problems. In particular, SIVIA (Set Inversion Via Interval Analysis) is an algorithm which can approximate using small "boxes" a subset of $\mathbb{R}^n$ satisfying a given set of constraints. It's a branch and bound algorithm, with quite a big complexity, but if the set is actually empty it can answer rather quickly. To be precise, it find an "inner" and an "outer" approximation of this set: if the outer approximation is empty, then your set is guaranteed to be empty, if the inner approximation is non empty then your set is guaranteed to be non empty, otherwise you can't tell and have to try with a higher precision, which increase the computing time and memory usage exponentially.</p> <p>Unfortunately I'm not aware of an easy to use implementation of this algorithm. There are some links on <a href="http://www.ensta-bretagne.fr/jaulin/demo.html" rel="nofollow">Luc Jaulin</a>'s home page, and I know that there exists various interval libraries for most of programming language and numerical softwares.</p> http://mathoverflow.net/questions/101644/fiction-books-about-mathematicians/101765#101765 Answer by Adrien for Fiction books about mathematicians? Adrien 2012-07-09T12:07:53Z 2012-07-09T12:13:04Z <p>I think that many, if not all, short stories of <a href="http://en.wikipedia.org/wiki/Jorge_Luis_Borges" rel="nofollow">Jorge Luis Borges</a> qualify. Even if they're not directly about maths, they often involve some kind of strange "mathematical structure", like paradoxes, symmetries, mirrors, labyrinth, distortions of space and time. Also, the notion of infinity is a common topic of these stories: the most well known example is "<a href="http://en.wikipedia.org/wiki/The_Library_of_Babel" rel="nofollow">The library of Babel</a>", but there are many other examples like "<a href="http://en.wikipedia.org/wiki/The_Immortal_%28short_story%29" rel="nofollow">The immortal</a>" or "<a href="http://en.wikipedia.org/wiki/The_Aleph_%28short_story%29" rel="nofollow">Aleph</a>" (which is of course a reference to the standard notation for transfinite numbers).</p> http://mathoverflow.net/questions/78492/kontsevich-integral-without-associators/78499#78499 Answer by Adrien for Kontsevich Integral without associators? Adrien 2011-10-18T22:36:08Z 2012-06-24T13:53:33Z <p>As you point out, the relation between associators and the quasi-triangular structure of $U_q(\mathfrak g)$ (and the related tangles invariants) exists "only" at the Lie algebraic level, not (not yet) at the universal one. Roughly speaking, this is because the twisting which absorb the associator is not $\mathfrak g$-invariant, ie modifies the coalgebra structure, which a priori doesn't make sense at the level of chord diagrams.</p> <p>So far I know, there is no combinatorial construction of a universal finite type invariant which can avoid associators. But of course it's something people are looking for.</p> <p>But it turns out that the theory of quantum $R$-matrices is more related to the theory of <strong>virtual</strong> knotted objects (see this <a href="http://mathoverflow.net/questions/47899/utility-of-virtual-knot-theory/52442#52442" rel="nofollow">answer</a> of Greg Kuperberg). This is more or less "Bar Natan's dream" that a universal finite type invariant for virtual knotted objects should corresponds somehow to Etingof--Kazhdan quantization of Lie bialgebras.</p> <p>There is also a baby version of this, which is Alekseev-Enriquez-Torrossian solution of the Kashiwara Vergne conjecture based on associators. It turns out that they constructs a kind of universal twist which can "kill" the associator, and a kind of universal solution of the quantum Yang-Baxter equation, living in a bigger algebra than the algebra of horizontal chords diagrams. According again to Bar Natan, this corresponds more or less to a universal finite type invariant for "welded knots". See: <a href="http://www.math.toronto.edu/drorbn/papers/WKO/" rel="nofollow">http://www.math.toronto.edu/drorbn/papers/WKO/</a></p> <p>You may also find this paper interesting : <a href="http://arxiv.org/abs/math/0211031" rel="nofollow">Towards a Diagrammatic Analogue of the Reshetikhin-Turaev Link Invariants</a></p> <p><strong>Edit:</strong> Some details about the relation between the Alekseev-(Enriquez)-Torossian construction and Vassiliev invariants.</p> <p>They start from the Lie algebra $\mathfrak{tder}_n$ of "tangential derivations" of the free Lie algebra $\mathfrak{f}_n$ on $n$ generators, that is the Lie algebra of endomorphism sending each generator $x_i$ to $[x_i,a_i]$ for some $a_i \in \mathfrak f_n$.</p> <p>Let $r^{i,j}$ be the element mapping $x_i$ to $[x_i,x_j]$, and $x_k$ to 0 for $k\neq i$. Then it leads to a solution of the "classical Yang-Baxter equation whose second leg lives in a commutative subalgebra", i.e.: $$[r^{1,3},r^{2,3}]=0$$ and $$[r^{1,2},r^{1,3}+r^{2,3}]=0$$ as does, for example, the $r$-matrice of the Drinfeld double of a cocommutative Lie bialgebra. They also prove that $R=\exp(r)$ ($r:=r^{1,2}$) satisfies the Quantum Yang Baxter equation. Therefore, one get this way a representation of the (pure) braid group in $\exp(\mathfrak{tder}_n)\rtimes S_n$. The $r^{i,j}$ can be thought of as "arrow diagrams" and are related to the welded braid group.</p> <p>On the other hand, exactly like in the usual theory of the Yang-Baxter equation, we have that $t^{i,j}=r^{i,j}+r^{j,i}$ satisfies the infinitesimal braid relations, also called 4t relation for horizontal chord diagrams. We thus get an <em>injective</em> morphism from the algebra of Horizontal chord diagram into $U(\mathfrak{tder}_n)$. Therefore we can put an associator and get another representation of the braid group in $\exp(\mathfrak{tder}_n)\rtimes S_n$ which is precisely the "Kontsevich integral for braids".</p> <p>Then the main result of [AT] is the following identity: let $\Phi$ be (the image in $\exp(\mathfrak{tder}_3)$ of) an associator, then there exists some $F \in \exp(\mathfrak{tder}_2)$ such that $$F^{2,3}F^{1,23}=\Phi F^{1,2} F^{12,3}$$</p> <p>where the indices correspond to some maps $\mathfrak{tder}_2 \rightarrow \mathfrak{tder}_3$ modelled on the coproduct of an envelopping algebra. They also show that:</p> <p>$$R=F e^{t/2} (F^{2,1})^{-1}$$</p> <p>Hence $F$ is a universal version of a Drinfeld twist for Quasi-Hopf algebra, which is able to "kill" the associator, and according to the theory it implies that the corresponding representations of the braid group are the same.</p> <p><strong>Edit 2:</strong> This was actually the situation for braids, let me add a few word about knots. It turns out that while usual braids embeds into welded one, this is far from being true at the level of knots. So the twist also intertwines between the restriction of the invariant for welded knots to usual knots, and the image of the Kontsevich integral in the space of arrow diagrams, but the resulting invariant is much weaker.</p> <p>Welded braids can be identified with the group of basis conjugating automorphisms of a free group, and the map it gets from usual braids is nothing but the Artin representation. This representation is faithfull, but its extension to string links (and in particular to long knots) has a huge kernel. At the level of diagrams, the algebra corresponding to knots is free commutative with two generators in degree one, and one generator in each degree greater than one. Hence it has a rather simple structure, while its analog for usual knots is very complicated.</p> <p>In fact, one of the main claim of Bar Natan's paper is that the universal invariant for welded knots is roughly the Alexander polynomial.</p> <p>However, let me mention that usual knots does embed into virtual one, and that the above story is an important step towards something similar in the virtual case. </p> http://mathoverflow.net/questions/94860/lie-algebra-for-a-general-group/95133#95133 Answer by Adrien for "Lie algebra" for a general group ? Adrien 2012-04-25T08:52:17Z 2012-06-22T00:58:22Z <p>If $G$ is a finitely generated group which is torsion free nilpotent of class $n$, then $G$ is the Lie group of some $\mathbb{Z}$-Lie algebra $\mathfrak g$ which is also nilpotent of class $n$. Hence you can define an algebraic group $G(k)$ for any field $k$ by taking the exponential of $\mathfrak g\otimes_{\mathbb{Z}} k$.</p> <p>Now if $G$ is a discrete group, define the rational serie as $D_i(G)=${$x \in G, x^r \in \Gamma_i(G)$ for some $r>0$ } where $\Gamma_i(G)$ is the $i$th term of the lower central serie. Therefore, by construction $G/D_i(G)$ is torsion free nilpotent of class $i$, hence you can associate to it a Lie algebra $\mathfrak g_i(k)$. Now define $\mathfrak g(k)$ as the inverse limit of the $\mathfrak g_i(k)$. it is called the Malcev Lie algebra of $G$. It is a complete, separated pro-nilpotent Lie algebra. Set $G(k)=\exp(\mathfrak g(k))$, it is a pro-unipotent group coming with a morphism $G \rightarrow G(k)$ which is universal for this property. Note that if $\bigcap_{i\geq 0} D_i(G)$ = {1} (such a group is called residually torsion free nilpotent) this morphism is injective, so it may happen that $G(k)$ capture a lot of things about $G$.</p> <p>Indeed every representation of $\mathfrak g(k)$ extends to a representation of $G(k)$ just by taking the exponential, and therefore to a representation of $G$. Conversly, every $k$-(pro-)unipotent representation of $G$ induces a representation of $\mathfrak g(k)$.</p> <p>Note that $\mathfrak g(k)$ is a rather complicated object, so it doesn't seems to help a lot. But in many interesting case, $\mathfrak g(k)$ is isomorphic as a filtered Lie algebra to an "easy to handle" graded Lie algebra (namely to the associated graded of $G$, see Ralph's answer). In that case you really get something like the relation between a Lie group and its easier to handle Lie algebra.</p> http://mathoverflow.net/questions/100033/interesting-mathematical-documentaries/100048#100048 Answer by Adrien for Interesting mathematical documentaries Adrien 2012-06-19T20:56:23Z 2012-06-20T09:33:55Z <p>The soundtrack is still in progress, so it's not yet fully available but <a href="http://science.unitn.it/~dalvit/" rel="nofollow">Ester Dalvit</a> made a very interesting movie about braids and knots. At the end it will be distributed under a Creative commons license.</p> <p>A trailer and parts of the movie are available <a href="http://matematita.science.unitn.it/braids/index.html" rel="nofollow">here</a>. Up to now three (old versions of) chapters are available, explaining respectively through computer generated animations:</p> <ul> <li>the group structure of the braid group</li> <li>"combing" of braids and handle reduction (i.e. solutions to the world problem)</li> <li>Alexander and Markov theorems (i.e. the relation with knots).</li> </ul> http://mathoverflow.net/questions/97703/list-of-hall-basis/97725#97725 Answer by Adrien for list of Hall basis Adrien 2012-05-23T08:05:31Z 2012-05-23T14:36:51Z <p>These are easily obtained with <a href="http://www.sagemath.org/" rel="nofollow">SAGE</a>:</p> <pre><code>for i in range(1,6): for w in StandardBracketedLyndonWords(2, i): print w </code></pre> <p><strong>Edit:</strong> And for the graded case, since the function which generates Lyndon words knows what a <a href="http://en.wikipedia.org/wiki/Composition_%28number_theory%29" rel="nofollow">composition</a> is, you can use the function </p> <pre><code>WeightedIntegerVectors(d,[d1,..,dk]) </code></pre> <p>which find all positive solutions of $$\sum \lambda_i d_i=d$$ for a given $d$. Then for any given solution $L=[\lambda_1,\dots,\lambda_n]$ in the form of a Python list,</p> <pre><code>LyndonWords(L): </code></pre> <p>will return all the Lyndon words on $n$ letters containing exactly $\lambda_i$ times the $i$th letter. You'll get this way all Lyndon words of degree $d$. <strong><em>Warning</em></strong>: there is just a small issue: the LyndonWords function seems to have trouble with lists beginning by 0, so the code below use a modified function, see the end of this post... </p> <p>Example:</p> <pre><code>for i in range(1,6): print "degree "+str(i) L=WeightedIntegerVectors(i,[1,2]) for l in L: for w in MyLyndon(list(l)): print sage.combinat.lyndon_word.standard_bracketing(w) </code></pre> <p>gives</p> <pre><code>degree 1 1 degree 2 2 degree 3 [1, 2] degree 4 [1, [1, 2]] degree 5 [[1, 2], 2] [1, [1, [1, 2]]] </code></pre> <p>Since Omar pointed this out, let me recall that standard bracketing of Lyndon words provides a Hall basis, maybe not "the" Hall basis you have in mind.</p> <hr> <p>If I'm not wrong, a Lyndon word o composition $(0,\dots,0,k_{j+1},\dots,k_n)$ with $j$ 0's at the beginning is the same as a Lyndon word of composition $(k_{j+1},\dots,k_n)$ with letters shifted by $j$ (since it has to be a Lyndon basis of the sub-Lie algebra generated by $x_{j+1},\dots,x_n$. So hopefully the following code will do the trick:</p> <pre><code>def myLyndon(e): if e == []: return k=0 while (e[k]==0): k=k+1 for z in sage.combinat.necklace._sfc(e[k:], equality=True): yield LyndonWord([i+k+1 for i in z], check=False) </code></pre> http://mathoverflow.net/questions/97556/noncommutative-localization-of-a-ring-complete-construction/97561#97561 Answer by Adrien for Noncommutative Localization of a Ring : Complete Construction Adrien 2012-05-21T14:11:50Z 2012-05-22T01:13:22Z <p>If I remember well, the second chapter of </p> <blockquote> <p>J. C. McConnell, J. C. Robson. <strong>Noncommutative Noetherian rings</strong>, vol. 30 of Graduate Studies in Mathematics (American Mathematical Society, Providence, RI, 2001)</p> </blockquote> <p>contains a rather detailled proof of the Ore's theorem.</p> <p><strong>Edit:</strong> I just checked it on <a href="http://books.google.ch/books?hl=en&amp;lr=&amp;id=c3fWk8LoSvgC&amp;oi=fnd&amp;pg=PR13&amp;dq=Noncommutative+Noetherian+rings&amp;ots=8UtVO6F-wQ&amp;sig=9f4RvubqAdPJboT3B-aqeTJPJ30&amp;redir_esc=y#v=onepage&amp;q=Noncommutative%2520Noetherian%2520rings&amp;f=false" rel="nofollow">Google books</a> and they allow zero divisors as well. The point is that if $S$ satisfies the Ore's condition (which is nothing but your first condition), then the set {$r\in R, rs=0$ for some $s\in S$} is an ideal in $R$ which is precisely the kernel of the natural map $R\rightarrow RS^{-1}$.</p> http://mathoverflow.net/questions/85305/orbifold-fundamental-group-and-configuration-space Orbifold fundamental group and configuration space Adrien 2012-01-10T00:14:50Z 2012-05-16T15:06:53Z <p>Hi,</p> <p>I'm not very familiar with (even simple examples of) orbifolds, so my first question is:</p> <blockquote> <p>Let $C_2$ be $\mathbb{C}$ with one cone singularity at 0 of index 2. What is the fundamental group of $C_2$ minus $k$ points ?</p> </blockquote> <p>My naive answer is: take $\mathbb{C}^*$ minus the same $k$ points. Its fundamental group is freely generated by the $k+1$ loops around the punctures. Now decide that you don't have a "hole" in 0 anymore, but a cone singularity, meaning that the generators corresponding to a loop around 0 is now of order 2. Then I would say that the fundamental group of $C_2$ minus $k$ points is $\langle a_0,\dots,a_k | a_0^2=1\rangle$, ie $Z_2\ltimes F_{2k}$, where $F_{2k}$ is generated by {$a_i,a_0a_ia_0, i\geq 1$}.</p> <p>Now recall the following construction: take the pure braid group $P_n$ with its standard generators $x_{i,j}, 1\leq i &lt; j\leq n$ given by taking the $j$th strand, letting it go behind all other strand, loop around the $i$th one and going back. Then it's quite easy to see that the subgroup generated by the $x_{i,n}$ is free: it is the subgroup of pure braids for which all but the last strand are fixed straight lines. In fact, it leads to a semidirect product decomposition $P_n=P_{n-1}\ltimes F_{n-1}$. This decomposition is actually a so called "almost direct" product, which is quite an important fact.</p> <p>This construction has a nice geometric interpretation: let $X_n$ be the configuration space of $n$ points in $\mathbb{C}$, and recall that $P_n=\pi_1(X_n)$. Then the map $X_n \rightarrow X_{n-1}$ which forget the last coordinate is a locally trivial fibration with fiber $\mathbb{C}$ minus $n-1$ points. Then it induces a (split) short exact sequence of fundamental groups</p> <p>$$1\rightarrow F_{n-1} \rightarrow P_n \rightarrow P_{n-1}\rightarrow 1$$</p> <p>Let's try to do something similar with the "orbifold braid group" of $C_2$, that is the fundamental group $P_n(C_2)$ of $O_n=${$z_1,\dots,z_n \in C_2, z_i \neq z_j$}.</p> <p>It seems to me that $P_n(C_2)=P_{n+1}/ \langle x_{1,i}^2=1,i=2 \dots n+1 \rangle$.</p> <p>The above construction seems to work "at the algebraic level": let $G_n$ be the subgroup of $P_n(C_2)$ generated by (the images of) $x_{i,n+1}$. What is stated in <a href="http://www.springerlink.com/content/p17178r57191kg88/" rel="nofollow">this paper</a> (in a slighty different form) is that $P_n(C_2)=P_{n-1}(C_2) \ltimes G_n$, and that it is an almost direct product too. </p> <p>But $G_n$ satisfies some relations, for example $x_{i,n+1}$ and $x_{0,n+1}x_{i,n+1}x_{0,n+1}$ commute for a given $i$, hence it is not isomorphic to the fundamental group of $C_2$ minus $n-1$ points (at least if my first naive try is not wrong). While this construction strongly looks like to and shares many algebraic properties with the construction for $P_n$, it does not seems to come from a natural geometric construction. So my real question is:</p> <blockquote> <p>Am I wrong somewhere ? Is there a natural interpretation of $G_n$ ?</p> </blockquote> <p><strong>Edit</strong>: Here is roughly what happen: assuming that $n=2$ for the sake of simplicity, it doesn't make sense to "freeze" the first strand (and its negative) and to make the second one loop around because the following relation holds:</p> <p><img src="http://img690.imageshack.us/img690/5962/braid.png" alt="braid"></p> <p>now pushing the red loop (seen as a loop in the 2-punctured plane) to the bottom plane, we see that it has to be identified with its conjugate by a loop around the two strands at once, ie by the product of the generators of $F_2$. Therefore, this product has to be central, leading to the relation holding in $G_n$ above. So one can ask:</p> <blockquote> <p>Is there a topological space modelled on this situation, i.e. which looks like to the "complementary in $\mathbb{C}\times[0,1]$ of two strands modulo homotopy". Or at least, is there a way to prove that there are no other relations than declaring that the big loop is central ?</p> </blockquote> http://mathoverflow.net/questions/94954/connes-kreimer-hopf-algebra-and-cosmic-galois-group Connes-Kreimer Hopf algebra and cosmic Galois group Adrien 2012-04-23T15:07:33Z 2012-04-24T09:49:46Z <p>Hi,</p> <p>I'm interested in the relation between the two following constructions motivated by renormalization:</p> <ul> <li>Connes-Kreimer gives an interpretation of the renormalization procedure in the framework of an Hopf algebra $H$ of rooted trees. By Cartier-Milnor-Moore theorem, $H$ is the algebra of regular functions on some pro-algebraic group, which turns out to be the so-called <a href="http://en.wikipedia.org/wiki/Butcher_group" rel="nofollow">Butcher group</a> $B$. </li> <li>Connes-Marcolli then considered a differential equation satisfied by divergences appearing in the above work, which leads to the introduction of a category of "equisingular flat connection" which, so far I understand, are more or less specific $B$-valued principal bundles equipped with a flat connection up to some equivalence relation. It turns out that this category is tannakian, meaning that it is equivalent to the category of modules over some pro-algebraic group $G$. They observe that $G$ acts on any renormalizable theory in a nice way, hence the name "cosmic Galois group" which was coined by Cartier. It is more than an analogy, since $G$ is (non-canonically) isomorphic to some motivic Galois group.</li> </ul> <p>I do not claim to understand these things at all, especially the "physical" part, so my question is maybe naive. The vague question is:</p> <blockquote> <p>Is there a "direct relation" between the group $G$ and the Hopf algebra $H$ ? </p> </blockquote> <p>Of course there are some relations between the two, so maybe a more precise question is: </p> <blockquote> <p>Is there a definition of $G$ purely in terms of combinatorics of Feynman graphs, or as the automorphism group of something (something else than a fiber functor) ?</p> </blockquote> <p>In fact, if I understand Marcolli's survey correctly, $G$ action on physical theories factors through an action of $B$ (but it seems rather surprizing, so it's very likely that I misunderstood something). So for the sake of curiosity, another naive question is: </p> <blockquote> <p>"Why" is it $G$, and not $B$ which plays the role of a cosmic Galois group ?</p> </blockquote> <p>My motivation for this question comes from the (highly non-trivial) fact that (some group which is almost) $G$ embeds into the graded Grothendieck-Teichmuller group $GRT$, which can be defined as the automorphism group of a certain operad of Feynman graphs. Since $G$ is more or less by definition related to the combinatorics of these graphs, I'm wondering if there is a concrete "combinatorial" definition of it.</p> <p><strong>Edit:</strong> To be precise they also consider a more general Hopf algebra of Feynman diagrams (not only rooted trees). They call the corresponding pro-algebraic group the group of "diffeographism". My claim about the action of $G$ is related to this group and not to $B$, I guess.. So let's assume that I'm asking my question in this setting too.</p> <p><strong>Edit 2:</strong> I should have mentionned that there are in fact several Hopf algebras of Feynman graphs whose choice depends on the particular physical theory you're considering (i.e. you have to choose a specific "type" of Feynman graph), so there are in fact plenty of diffeographism groups, and $G$ is universal among them (which in fact answer my last question, although the Butcher group still seems to play a distinguished role). On the other hand it seems to me that mathematically it still make sense to consider the Hopf algebra of all Feynman graphs, but maybe what you get is precisely $\mathcal H _{\mathbb U}$ (see Gjergji's answer) ?</p> <p>And just to expand what I'm saying in my comment to Gjergji's answer, it's rather clear for example that $\mathcal H _{\mathbb U}$ is isomorphic to the dual of the <a href="http://arxiv.org/abs/hep-th/9407124" rel="nofollow">algebra of noncommutative symmetric functions</a>, since the latter is also by definition a free graded noncommutative algebra with one primitive generator in each degree (which are analogs of <a href="http://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial" rel="nofollow">power sum symmetric polynomials</a>). In fact the abstract definition of $\mathcal H _{\mathbb U}$ itself is rather explicit and combinatorial, hence I'm really looking for a more conceptual link.</p> http://mathoverflow.net/questions/91927/compatibility-of-the-kz-connection-with-operadic-composition/91964#91964 Answer by Adrien for Compatibility of the KZ connection with operadic composition Adrien 2012-03-23T00:52:35Z 2012-04-08T19:38:12Z <p><strong>Edit</strong>: Answer reformulated</p> <p>Hi Pavol,</p> <p>While it's not true that you can extend the KZ connection on the compactified space, there is a well defined procedure to "specialize" it to get a connection on each boundary component. My claim is that these specializations are precisely obtained through application of operadic composition.</p> <p>The precise story is as follows: De Concini--Procesi introduced a compactification $Y_n$ of the configuration space satisfying the following properties:</p> <ul> <li>The complementary $K=Y_n\backslash C_n$ is a divisor with normal crossings</li> <li>The KZ connection induces a meromorphic connection $\nabla_n$ on $Y_n$ with logarithmic singularities on $K$.</li> </ul> <p>These properties means the following: let $D$ be a smooth irreducible divisor. Then there exists some local coordinates $z_1,\dots,z_{n-1},t$ such that $D$ is defined by $t=0$, and such that $$\nabla_n=d-\omega_D(z_1,\dots,z_{n-1},t)+A \frac{dt}t$$ where $\omega_D$ is holomorphic at $t=0$ and $A$ is a constant (the residue of the defining 1-form of $\nabla_n$). Now for any choice of a tangential basepoint inside $D$ the monodromy of $\nabla_n$ around $D$ is then given by $e^A$.</p> <p>Now you can specialize $\nabla_n$ to a flat, meromorphic connection on $D$ by setting $$\nabla_D:=d-\omega_D(z_1,\dots,z_{n-1},0)$$</p> <p>It turns out that there is a natural identification $D=Y_{n-1}$, and that $D$ is the image of some operadic composition $$\mu:Y_{n-1}\times Y_2 \rightarrow Y_n$$ Let $$\mu:\mathfrak t_{n-1}\times \mathfrak t_{2}\rightarrow \mathfrak t_{n}$$ be the "same" operadic composition.</p> <p>Now my precise attempt to answer your question is the following</p> <blockquote> <p><strong>Claim:</strong> $$\nabla_D=\mu(\nabla_{n-1})$$</p> </blockquote> <p>Now you can repeat this process on $D=Y_{n-1}$ and get the compatibility with other operadic maps.</p> <h2>Example</h2> <p>Consider the KZ connection on $Y_4$ (obtained from the original KZ connection by setting $z_1=0,z_2=x,z_3=y,z_4=1$):</p> <p>$$\nabla_4=t_{12}d\log x+t_{23}d\log(x-y)+t_{24}d\log(x-1)+t_{13}d\log y+t_{34}d\log(y-1)$$ Let $D \subset Y_4$ be the divisor associated to the hyperplane $x=y$ (that is, to the hyperplane $z_2=z_3$ in $C_n$). Setting $t=x-y$ one gets: $$\nabla_4=t_{12}d\log x+t_{13}d\log(x-t)+t_{24}d\log(x-1)+t_{34}d\log(x-t-1)+t_{23}d \log t$$</p> <p>Then we can define a connection on $D$ it by setting $t=0$ in the holomorphic part of $\nabla_4$: $$\nabla_D= t_{12}d\log x+t_{13}d\log x+t_{24}d\log(x-1)+t_{34}d\log(x-1)$$</p> <p>i.e. $$\nabla_D= (t_{12}+t_{13})d\log x+(t_{24}+t_{34})d\log(x-1)=\nabla_3^{1,23,4}$$</p> http://mathoverflow.net/questions/92327/hopf-structure-of-uqsl2/92331#92331 Answer by Adrien for Hopf structure of Uq(sl(2)) Adrien 2012-03-27T00:31:15Z 2012-03-27T01:38:57Z <p>Hi Ryan,</p> <p>Let me elaborate on my answer to your <a href="http://mathoverflow.net/questions/92046/quantum-group-uqsl2" rel="nofollow">previous question</a>. Somehow, deforming only the algebra structure is easy, in the sense that if you give some generators and some relations you're sure to get... an algebra. So just take the same generators as for $U(\mathfrak{sl}_2)$, some random relations whose quasi-classical limit gives you the defining relations of $U(\mathfrak{sl}_2)$ and you're done. Of course it's not a very interesting approach, but at least you are sure to get something matching your requirements.</p> <p>On the other hand, deforming the Hopf structure is hard: constructing a well defined algebra map $\Delta$ is easy, but coassociativity is a hard condition which is of course not guaranted at all if you pick something at random.</p> <p>First of all, it's strictly speaking not completely true that $K=q^H$ since $H \not\in U_q(\mathfrak{sl}_2)$, but it is still the right intuition, justified by the "formal" setting below. Then the same argument as in my previous answer apply: </p> <p>$$\Delta(K)=q^{\Delta(H)}=q^{H\otimes 1+1\otimes H}=(q^H \otimes 1)(1 \otimes q^H)=K\otimes K$$ </p> <p>All of this is more natural if you take the formal version $U_{h}(\mathfrak{sl}_2)$ which is a topological $\mathbb{C}[[h]]$-Hopf algebra. In that case $H$ is a true generator and there are many reasons for which you want to let the "Cartan part" really undeformed. It is, indeed, a general fact: for a semi simple Lie algebra $\mathfrak{g}$, the sub-Hopf algebra of $U_h(\mathfrak{g})$ generated by the $H_i$'s is isomorphic as an Hopf algebra to $U(\mathfrak{h})[[h]]$, and the isomorphism is just the identity $H_i\mapsto H_i$. In fact, it is known that $U_{h}(\mathfrak{g})$ is isomorphic <strong>as an algebra</strong> to $U(\mathfrak{g})[[h]]$, but now the isomorphism is far from being trivial. Still, it clearly show that the important things is the whole Hopf algebra structure.</p> <p>Therefore we know the coproduct for $K$, or at least we expect that there exists an Hopf structure such that $\Delta(K)=K\otimes K$. Now you can do some computation, write down the constraints imposed by co-associativity, still trying to get something "familiar" (ie s.t. $E$ and $F$ are close from being primitive). I may be wrong but I'm not sure that there is another way of finding something explicit, so what you did was right.</p> <p>Now for general $\mathfrak g$ you can try to "glue together" copies of $U_q(\mathfrak{sl}_2)$. Again there is nothing obvious here, and finding these formulas was a great achievement. It seems to me, for example, that no similar deformation for non semisimple Lie algebras are known except in a few cases.</p> <p>As I said in the abovementionned answer, the very much existence of a non-trivial deformation of the Hopf structure is a highly non-trivial fact, which "could have been false". The deep reason for which such object exists is really related to the theory of Drinfeld associators.</p> http://mathoverflow.net/questions/92046/quantum-group-uqsl2/92049#92049 Answer by Adrien for Quantum group Uq(sl(2)) Adrien 2012-03-24T00:31:06Z 2012-03-24T02:57:10Z <p>Hi Ryan,</p> <p>You can prove that if $a,b$ are some elements in an algebra such that $[a,b]=\lambda b$ for $\lambda$ a scalar, then (in a context where this expression makes sense) $q^a b q^{-a}=q^{\lambda} b$: rewrite the relation as $$ab=b(a+\lambda)$$ then $$a^nb=b(a+\lambda)^n$$ therefore $$q^ab=\sum \frac{\log(q)^na^n}{n!}b=b\sum \frac{\log(q)^n(a+\lambda)^n}{n!}= bq^{(a+\lambda)}=q^{\lambda}bq^a$$ since $\lambda$ commutes with everything.</p> <p>Therefore it's in fact not a deformation but the same relation with $q^H$ instead of $H$.</p> <p>In my opinion, what is more miraculous is the existence of a non-trivial deformation of the Hopf structure. Although it does not leads to explicit formulas, the conceptual explanation for that comes from the Kohno-Drinfeld theorem, or more generally from the Etingof-Kazhdan quantization functor.</p> http://mathoverflow.net/questions/91425/the-word-problem-in-braid-groups/91435#91435 Answer by Adrien for The word problem in braid groups Adrien 2012-03-17T00:34:21Z 2012-03-20T10:26:27Z <p>Here "reduced" refer to the so-called handle reduction algorithm introduced by Dehornoy.</p> <p>So far I remember, this algorithm does not provide a normal form, therefore I agree that the statement is false: two reduced word may represent the same braid even if they are different. Indeed, it can be checked with the following Java applet: <a href="http://www.math.unicaen.fr/~tressapp/index.html" rel="nofollow">http://www.math.unicaen.fr/~tressapp/index.html</a> (it applies handle reduction the other way as you does, i.e. it pushes handles to the left, so in that case the reduction will produce the same word. But if you swap $b_1$ and $b_3$ you still get isotopic braid, and now the applet will tell you that the words are already reduced).</p> <p>The true statement is that a word represent the trivial braid if and only if its reduction is the empty word. Of course it still solve the word problem: checking if $w_1,w_2$ represent the same braid is the same as checking if $w_1w_2^{-1}$ represents the trivial braid. </p> <p><strong>Edit.</strong> For the sake of completeness let me recall why not leading to a normal form is not a negative point against this algorithm.</p> <ol> <li>When you use normal forms, testing equality is trivial, but multiplication and inversion are hard because you have to normalize the result each time. So it's somehow better to apply algebraic operation only at the word level, and to apply the algorithm only when you want to test equality.</li> <li>Of course you can still do it with a normal form algorithm. But since you ask less, you can hope to find a more efficient algorithm. Indeed, handle reduction is strongly believed to be of quadratic complexity, and is without any doubt the most efficient solution to the word problem in the braid group.</li> <li>Maybe the most important things is that handle reduction actually leads to a left (or right) ordering on the braid group, and to a bi-ordering on the pure braid group, which has quite a lot of group-theoretic consequences.</li> </ol> http://mathoverflow.net/questions/89976/q-analog-of-the-matrix-exponential/89982#89982 Answer by Adrien for q-analog of the matrix exponential Adrien 2012-03-01T18:22:37Z 2012-03-02T11:33:18Z <p>It's more an example than a general answer. Details can be found here : <a href="http://arxiv.org/abs/math/0512500" rel="nofollow">http://arxiv.org/abs/math/0512500</a></p> <p>It is convenient to replace $q$ by $q^2$ in the formula that you gave. Doing so we have the following desirable identities:</p> <ul> <li>$\exp_q(x)\exp_{-q}(-x)=1$</li> <li>if $xy=q^2yx$, then $\exp_q(y)\exp_q(x)=\exp_q(x+y)$</li> </ul> <p>Now if $\mathfrak g$ is a simple Lie algebra, $\Phi^+$ a choice of positive roots, $h_i,e_{\alpha},f_{\alpha}$ the generators of $U_q(\mathfrak g)$ associated to the Chevalley basis of $\mathfrak g$, $>$ a normal ordering on $\Phi^+$, and $q_{\alpha}=q^{(\alpha,\alpha)/2}$, then the R-matrix of $U_q(\mathfrak g)$ is given by</p> <p>$$R=K \prod_{\alpha \in \Phi^+}^> R_{\alpha}$$ where $$K=q^{\sum h_i \otimes h^i}$$ and $$R_{\alpha}=\exp_{q_{\alpha}^{-1}}((q_{\alpha}-q_{\alpha}^{-1})e_{\alpha} \otimes f_{\alpha})$$</p> <p>Here $q$ is either a generic complex number, or a variable, in which case we work over the field $\mathbb{Q}(q^{\frac12})$.</p> <p>It is a universal formula, but of course you can specialize it to an element of $End(V\otimes V)$ for any $U_q(\mathfrak g)$-module $V$.</p> http://mathoverflow.net/questions/86874/is-every-finite-group-a-quotient-of-the-grothendieck-teichmuller-group/86875#86875 Answer by Adrien for Is every finite group a quotient of the Grothendieck-Teichmuller group? Adrien 2012-01-28T01:53:54Z 2012-01-28T10:50:19Z <p>It's only a partial answer since this <a href="http://www.math.jussieu.fr/~leila/farb.pdf" rel="nofollow">survey</a> is already 5 years old, but it suggest that almost nothing is known about (non-abelian) finite quotients of $\widehat{GT}$ (question 1.7).</p> <p><strong>Edit:</strong> I should maybe recall what happen in the abelian case, and why it's encouraging: elements of $\widehat{GT}$ are pairs $(f,\lambda)$ where $f$ is in the derived subgroup of $\hat{F}_2$, and $\lambda \in \hat{\mathbb{Z}}^{\times}$, satisfying some complicated equations. It turns out that the set theoretic map $(f,\lambda) \mapsto \lambda$ induces a <em>surjective</em> group morphism $\widehat{GT}\rightarrow \hat{\mathbb{Z}}^{\times}$. And the good news is that the composite</p> <p>$$G_{\mathbb{Q}} \hookrightarrow \widehat{GT} \rightarrow \hat{\mathbb{Z}}^{\times}$$</p> <p>is nothing but the <a href="http://en.wikipedia.org/wiki/Cyclotomic_character" rel="nofollow">cyclotomic character</a>.</p> http://mathoverflow.net/questions/85985/symmetric-polynoms-are-hopf-algebra-what-for-one-needs-co-product/86008#86008 Answer by Adrien for Symmetric polynoms are Hopf algebra ? What for one needs co-product ? Adrien 2012-01-18T16:03:05Z 2012-01-20T13:38:08Z <p>In fact there are several Hopf algebra structures on this algebra, mainly because it's one of the many occurence of the free commutative graded algebra with exactly one generator in each degree $A=\mathbb C[h_1,\dots,h_n,\dots]$, where $h_i$ is of degree $i$. Here the $h_i$'s can be either the elementary or the complete symmetric polynomials. Of course that's the grading which makes things interesting, otherwise it would just be the usual polynomial algebra on infinitely many variables. </p> <p>Indeed, let $G$ be the group of formal power series with constant terme equal to 1. Then the algebra $O(G)$ of polynomial function on $G$ is generated by the linear maps $\lambda_k$ defined by $$\langle \lambda_k,1+\sum a_nX^n \rangle=a_k/k!$$</p> <p>Letting $\lambda_k$ having degree $k$, then the map $h_k \rightarrow \lambda_k$ is an isomorphism of graded algebra. But being an algebra of function on a group, $O(G)$ has a natural Hopf algebra structure given by</p> <p>$$\Delta(f)(a \otimes b)=f(ab)$$ and $$S(f)(a)=f(a^{-1})$$</p> <p>If you think as the $h_i$'s as being the elementary symmetric polynomials, then this coproduct is the same as the coproduct of Dan's answer (if I'm not wrong). This is not just an abstract isomorphism, however, but if I remenber well it is reminiscent of the fact that coefficients of a polynomial are elementary symmetric functions of its roots. </p> <p>But if you take instead the group of formal power series of the form $$X+\sum_{n\geq 1} a_nX^{n+1}$$ , whose multiplication is given by the composition of formal power series, then you get again the same graded algebra but the above formula leads to a non-commutative coproduct (leading to the so-called <a href="http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula" rel="nofollow">Faa di Bruno</a> Hopf algebra). </p> <p><strong>Edit:</strong> Let me add a few words about the motivations. Once you already know the fundamental theorem of symmetric functions, the above isomorphism may seems tautological and not very interesting. In fact, the existence of an (actually several) very explicit isomorphism(s) from the algebra of symmetric functions to $A$ is nothing but a reformulation of this theorem. On the other hand, the above definition is arguably one of the most natural definition of $A$, and you get the Hopf structure for free.</p> <p>Somehow, the fundamental theorem tells you that the algebra structure of symmetric functions is not that interesting. But it turns out that many interesting combinatorial identities can be deduced from the Hopf structure, and especially from the fact that it's self dual. Hence the pull pack of the coproduct and antipode to the algebra of symmetric functions itself has many interesting combinatorial applications. The same is true for the other Hopf algebra structure, since it identifies combinatorial identities between symmetric functions, and the computation of the composition inverse of a formal power serie. </p> http://mathoverflow.net/questions/84977/graph-of-dependencies-from-a-latex-file/84985#84985 Answer by Adrien for Graph of dependencies from a Latex file Adrien 2012-01-05T18:30:22Z 2012-01-05T21:48:29Z <p>Well, it's probably time to learn <a href="http://en.wikipedia.org/wiki/Regular_expression" rel="nofollow">regular expressions</a></p> <p><img src="http://imgs.xkcd.com/comics/regular_expressions.png" alt="alt text"> (<a href="http://xkcd.com/208/" rel="nofollow">http://xkcd.com/208/</a>)</p> <p>More seriously, I don't know if a graphical tools already exists, but writing a small say Python script which parse every \ref, \eqref and \cite in each proof environment using regular expressions is rather easy. Maybe the only ambiguous things is to associate each proof environment with the corresponding theorem/lemma. Then it would again be easy to generate <a href="http://en.wikipedia.org/wiki/Graphviz" rel="nofollow">Graphviz</a> code from that, and Graphviz will automatically produce a nice oriented graph showing all the dependencies. </p> http://mathoverflow.net/questions/75500/algorithm-denominators-of-elements-of-a-rational-affine-space Algorithm/denominators of elements of a rational affine space Adrien 2011-09-15T09:30:32Z 2011-12-14T19:11:47Z <p>Hi,</p> <p>I hope it's not a trivial question... Suppose I have a finite dimensional vector space $V$ over $\mathbb{Q}$ with a distinguished basis (in my case it's the $k$th graded piece of the free associative algebra on two generators), and a one dimensional affine space</p> <p>$$S=\lbrace a+\lambda b,\lambda \in \mathbb{Q} \rbrace$$</p> <p>where $a,b \in V$ are explicitly given. As you may have guessed, $a$ and $b$ are respectively a particular solution of a linear equation and a basis of the kernel of the associated linear map, whcih were computed using a computer algebra system (SAGE).</p> <p>Let's call denominator of an element $x$ of $S$ the LCM of all denominators of the coordinates of $x$ expressed in the distinguished basis. The context doesn't really matter, but there is a general result on my particular problem stating that there exists an element of $S$ whose denominator is smaller than a given constant $D$. It turns out that the solution $a$ returned by SAGE match exactly this bound, i.e., its denominator is precisely $D$, which actually I don't find so surprising after all. </p> <p>But now, of course, I'm curious to know:</p> <ol> <li>if this bound is optimal, i.e., if there exists solution with a smaller denominator</li> <li>how many such solutions there are, i.e., if this bound on the denominators is sufficient to select a particular (maybe up to some trivial modifications) solution.</li> </ol> <p>So my question can be formulated as follows:</p> <blockquote> <p>Given two vector $a,b \in \mathbb{Q}^n$, is there an algorithm which can find a $\lambda \in \mathbb{Q}$ such that the denominator of $a+\lambda b$ is minimal ?</p> </blockquote> <p><strong>Edit:</strong> In order to give a motivation to the question: I wrote down a small SAGE program which computed a rational, even, Drinfeld associator (very roughly a formal power serie in two non-commuting variables satisfying some complicated equation) up to and including degree 8. It can be computed recursively by solving linear equations. It turns out that such an element is uniquely determined up to degree 6 (and hence, up to degree 7 because it's even), but in degree 8 the kernel is one dimensional. The result I was mentionning is a <a href="http://arxiv.org/abs/1002.2331" rel="nofollow">paper</a> by Alekseev, Podkopaeva and Severa proving the existence of associators whose denominators satisfies a specific upper bound.</p> <p>It seems to me that the question is already hard when $n=1$. Still, in my case $n$ is not that big, and even a "brute force" algorithm would be fine. I'm also wondering is there is, at least, an algorithm which can approximate the result (something like LLL does for some minimzation problem in lattices).</p> http://mathoverflow.net/questions/80405/presentation-of-the-pure-artin-groups/80472#80472 Answer by Adrien for Presentation of the pure Artin groups Adrien 2011-11-09T12:25:00Z 2011-11-09T13:14:17Z <p>I don't know any reference where such a presentation is written down for any Artin group. But:</p> <ul> <li>You can find In this paper of Enriquez a presentation for the pure Artin group of type B: <a href="http://arxiv.org/abs/math/0408035" rel="nofollow">http://arxiv.org/abs/math/0408035</a> Proposition 1.1 (by setting $N=2$ is the formulaes)</li> <li>For all the infinite families, the corresponding pure braid groups are iterated semi-direct products of free groups. (For all the families but the $D_n$ one, it follows from the fact that the corresponding hyperplane arrangements are of fiber type, hence these are even almost-direct product). It should leads quite easily to a nice presentation of these groups.</li> </ul> <p><strong>Edit:</strong> You may also be interested by this paper of Crips and Paris: <a href="http://arxiv.org/abs/math/0210438" rel="nofollow">http://arxiv.org/abs/math/0210438</a>. Recall that $W_{B_n}=(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$ and that $W_{D_n}=(\mathbb{Z}/2\mathbb{Z})^{n-1}\rtimes S_n$. It is proved in this paper that $B_{B_n}=F_n \rtimes B_n$ and that $B_{D_n}=F_{n-1}\rtimes B_n$ where $B_n$ is the braid group of type $A$ and $F_n$ is a free group, and that the canonical projection from the braid group to the Coxeter group is compatible with these decomposition (in the sense that in both cases it restricts to the obvious projections $B_n \rightarrow S_n$ and $F_k \rightarrow (\mathbb{Z}/2\mathbb{Z})^k$).</p> http://mathoverflow.net/questions/79272/reshetikhin-turaev-and-links-with-a-distinguished-component Reshetikhin-Turaev and links with a distinguished component Adrien 2011-10-27T14:39:31Z 2011-10-28T10:25:52Z <p>Hi,</p> <p>This question came up to me when reading the paper of Cartier on Vassiliev invariants, but it can probably be turned into a more general question.</p> <p>Let $T$ be the category whose objects are finite sequences of ${+,-}$ (including the empty one) and whose morphisms are framed, oriented tangles. In particular, $End(\emptyset)$ is the set of framed oriented links. It's quite well known that $T$ is the free ribbon category on one object, hence if $C$ is a ribbon category and $V \in C$ then there is a functor $F:T\rightarrow C$ mapping $+$ to $V$ and $-$ to $V^*$. In particular, $F$ associate to any link viewed as an element of $End(\emptyset)$ an automorphism of the unit object of $C$ (with respect to the monoidal structure of $C$).</p> <p>Cartier applies this construction in the case $C$ is the ribbon category having the same objects as $T$, whose morphisms are $k[[\hbar]]$-linear combinations of chords diagrams, and whose ribbon structure comes essentially from a Drinfeld associator. The point is that he doesn't apply the above construction to $End(\emptyset)$ but rather to $End(+)$, and claims that its restriction to knots is a universal finite type invariant.</p> <p>It's quite clear that the closure operation identifies $End(+)$ with the set of links with a distinguished component. In particular it also contains the set of knots, which explains that his claim makes sense. But still, I'm wondering</p> <blockquote> <p>Why did he make this particular choice ?</p> </blockquote> <p>There is, of course, an obivous answer. The product on $End(+)$ can be identified with the connected sum of links along the distinguished component. Hence, it contains the set of knots as a sub-monoid and not only as a subset. As a nice consequence, the invariant he constructs is compatible with the connected sum of knots, which is of course a desirable feature.</p> <p>It seems quite obvious to me that this construction applied to $End(\emptyset)$ also gives a universal finite type invariant, just with a different normalization. <s>I would also say that it's a general fact that for an arbitrary ribbon category these invariants are more or less the same. </s></p> <p>There is still a problem if you want to realize this invariant. For example if $\mathfrak g$ is a simple Lie algebra, a choice of $t \in S^2(\mathfrak g)^{\mathfrak g}$ leads to an infinitesimal symmetric monoidal category structure on $\mathfrak g$-mod. Putting again a Drinfeld associator you get a ribbon category structure on $\mathfrak g[[\hbar]]$-mod, hence a functor from $T$ for each $V \in \mathfrak g$-mod which factor through Cartier's category of chord diagramm.</p> <p>Now the usual RT construction leads to a numerical invariant, while Cartier's one leads to an endomorphism of $V$. It's not a big deal, one can take its quantum trace in order to get something numerical, but it's somehow artificial since this endomorphism is already an invariant<s>, although I'm pretty convinced that no information is lost in doing that, for the reason explained above.</s> But taking the usual trace instead of the "quantum" one, for example, also leads to a (less natural, but still) invariant in that case...</p> <p>Well, it's probably not a real subtle problem but rather a technical detail, but I'm wondering if there is a natural way to handle/understand this, or a "philosophical" picture explaining the relations between these invariants.</p> <p><strong>Edit:</strong> This <a href="http://arxiv.org/abs/0711.4229" rel="nofollow">paper</a> of Nathan Geer deals with the situations mentionned by Theo and explain that the cut open invariant may be non trivial even if the quantum trace vanish and that's an interesting problem to see how it can be turned into an invariant which doesn't depends on the choice of the distinguished component. </p> http://mathoverflow.net/questions/77904/is-there-a-reasonable-definition-of-the-height-of-a-transcendental-number/78056#78056 Answer by Adrien for Is there a reasonable definition of the height of a transcendental number Adrien 2011-10-13T19:58:11Z 2011-10-13T19:58:11Z <p>It does'nt quite answer your question, but maybe there is a resaonable definition of the height of a <a href="http://en.wikipedia.org/wiki/Ring_of_periods" rel="nofollow">period</a>. The ring of period is a countable over-ring of the field of algebraic numbers, sharing (at least conjecturally) many properties with them. This ring contains most of interesting transcendental numbers.</p> http://mathoverflow.net/questions/124772/h-adic-completion-of-u-q-fraksl-2 Comment by Adrien Adrien 2013-03-17T16:20:37Z 2013-03-17T16:20:37Z Alexander&gt; But $U_q$ is not a deformation of $U$, even if $q$ is a variable. Abtan&gt; I'm confused about the objects you're considering: your $U_{\hbar}(\mathfrak g)$ is just the trivial deformation of $U(\mathfrak g)$, and not what people usually mean when they write $U_{\hbar}(\mathfrak g)$ (though they are indeed non canonically isomorphic as algebras). And if I understand correctly the second algebra you look at is $U_q(\mathfrak g)[[\hbar]]$ for some complex number $q$, and in particular without any relation between $q$ and $\hbar$, is it really what you meant ? http://mathoverflow.net/questions/122095/on-tamarkins-proof-of-etingof-kazhdan-quantization-of-lie-bialgebra/122917#122917 Comment by Adrien Adrien 2013-02-28T16:30:40Z 2013-02-28T16:30:40Z That's amazing, thanks ! I really don't know much about Koszul duality, but if I undertsand well you are claiming that (whenever it make senses and under some assumption), if $A$ is a P-algebra over some operad P, then the <i>category of modules</i> over the Koszul dual of A is a P-algebra in Cat ?I'm a bit confused about your last statement: the category of modules over a quasi-triangular (quasi-)Hopf algebra H is braided monoidal, hence an algebra over $E_2$ in Cat, so it's temptating to thinks that H is koszul dual to an E_2 algebra, not necessarily E_3, but I'm probably missing something. http://mathoverflow.net/questions/122095/on-tamarkins-proof-of-etingof-kazhdan-quantization-of-lie-bialgebra Comment by Adrien Adrien 2013-02-28T16:20:50Z 2013-02-28T16:20:50Z Theo&gt; I would be happy to discuss this ! I'll send you an e-mail soon. http://mathoverflow.net/questions/122065/an-element-g-in-a-group-such-that-neither-g1-nor-g-ne-1-can-be-proved Comment by Adrien Adrien 2013-02-17T15:12:55Z 2013-02-17T15:12:55Z Two words $u,v$ in the generators represents the same element of $G$ iff $uv^{-1}$ represent 1. Hence it seems to me your question reduces to the existence of finitely presented group whose word problem is unsolvable, of which there are many known examples. http://mathoverflow.net/questions/119991/is-the-quantum-algebra-unique-up-to-isomorphism-in-deformation-quantization/120089#120089 Comment by Adrien Adrien 2013-01-28T16:04:12Z 2013-01-28T16:04:12Z @Damien&gt; That's what I meant, your point 2) is definitely true at the universal level, but that being true or not for actual Poisson structures is open and probably a hard problem, isn'it ? http://mathoverflow.net/questions/119991/is-the-quantum-algebra-unique-up-to-isomorphism-in-deformation-quantization/120089#120089 Comment by Adrien Adrien 2013-01-28T12:44:36Z 2013-01-28T12:44:36Z Damien&gt; Isn't your point 2 the content of Dolgushev's paper arXiv:1109.6031, at least at the universal level ? It states that the action of grt on homotopy classes of stable formality isomorphisms is free and transitive (hence highly non trivial). However, as noticed in arXiv:1211.4230 (and if I understand correctly, which is far from being granted) there is no known example of an actual Poisson structure for which the realization of this action is non trivial. http://mathoverflow.net/questions/118511/is-the-universal-enveloping-algebra-of-a-finite-dimensional-lie-algebra-left-no Comment by Adrien Adrien 2013-01-10T10:47:05Z 2013-01-10T10:47:05Z The universal enveloping algebra of a finite dimensional Lie algebra is a so-called G-alegbra, hence is left and right Noetherian (see e.g. <a href="http://www.singular.uni-kl.de/Manual/3-1-5/sing_510.htm" rel="nofollow">singular.uni-kl.de/Manual/3-1-5/sing_510.htm</a>). Note that this includes quantized enveloping algebra as well. http://mathoverflow.net/questions/107424/knot-security-when-to-trust-your-life-with-a-knot Comment by Adrien Adrien 2012-09-17T23:04:42Z 2012-09-17T23:04:42Z Maybe this paper can help: Alexander Coward, Joel Hass, Topological and physical knot theory are distinct (<a href="http://arxiv.org/abs/1203.4019" rel="nofollow">arxiv.org/abs/1203.4019</a>) http://mathoverflow.net/questions/106442/braidings-and-isomorphism-classes-in-a-monoidal-category Comment by Adrien Adrien 2012-09-05T19:28:24Z 2012-09-05T19:28:24Z When you say &quot;satisfies the hexagonal identities&quot;, how do you define $\gamma_{X,X\otimes X}$ ? Defining a braiding needs some kind of naturality, I think (at least on &quot;one leg&quot;, and you can take the hexagon for the other leg as an actual definition, like in the construction of the Drinfeld center). http://mathoverflow.net/questions/103397/categorification-of-finite-type-invariants Comment by Adrien Adrien 2012-07-30T11:06:30Z 2012-07-30T11:06:30Z Looks very interesting, thanks ! http://mathoverflow.net/questions/103397/categorification-of-finite-type-invariants Comment by Adrien Adrien 2012-07-28T20:08:08Z 2012-07-28T20:08:08Z Somehow, we can rephrase your objection by saying that categorification is clearly related to &quot;non-perturbative&quot; things (rational quantum groups, Jones polynomial, and Chern-Simons theory according to Witten) while finite type invariants are by nature perturbative things, related to perturbative expansion of CS theory, formal quantum groups and formal expansion of the Jones polynomial. http://mathoverflow.net/questions/103397/categorification-of-finite-type-invariants Comment by Adrien Adrien 2012-07-28T19:42:26Z 2012-07-28T19:42:26Z I agree that there is a point here, as it makes hardly senses to categorify one single numerical invariant. On the other hand, these numbers are in fact endomorphisms of the base field in some category, or better images of endomorphisms of the unit object in the monoidal category of diagrams (which are not numbers anymore) through some functor. http://mathoverflow.net/questions/66084/open-problems-with-monetary-rewards/103184#103184 Comment by Adrien Adrien 2012-07-26T12:21:15Z 2012-07-26T12:21:15Z I should have mentioned that the hidden text is : &quot;Dear Reader: Enclosed is a check for ninety-eight cents. Using your work, I have proven that this equals the amount you requested.&quot; http://mathoverflow.net/questions/102803/software-for-numerical-constraint-satisfaction-problems/102867#102867 Comment by Adrien Adrien 2012-07-22T15:28:36Z 2012-07-22T15:28:36Z As the name suggest, the general goal of SIVIA is to approximate the inverse image of some subset of $\mathbb{R}^k$ by some function, and when the precision goes to 0, the abovementionned conditions implies that bot the inner and the outer approximation goes to the exact set you're looking for. http://mathoverflow.net/questions/102803/software-for-numerical-constraint-satisfaction-problems/102867#102867 Comment by Adrien Adrien 2012-07-22T15:26:53Z 2012-07-22T15:26:53Z Yes, it is theoretically convergent. The point is that if $f$ is an elementary function/operation, then $f$ applied to an interval is not generally an interval. So it's replaced by an &quot;interval valued&quot; function which satisfy some conditions which guarantee that you recover the true function when the size of the interval is small enough. Then the precision of the algorithm is roughly related to the size of boxes you allows at the boundary, i.e. between the inner and outer approximation.