User jeffrey giansiracusa - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:13:23Z http://mathoverflow.net/feeds/user/4910 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/128615/a-infinity-structure-on-cohomology-and-the-weight-filtration A_infinity structure on cohomology and the weight filtration Jeffrey Giansiracusa 2013-04-24T13:32:17Z 2013-04-25T09:17:11Z <p>Let $X$ be a complex algebraic variety. The rational cohomology of $X(\mathbb{C})$ carries a canonical filtration called the weight filtration. It also carries a canonical equivalence class of $A_\infty$ structures (extending the cup product ring structure). </p> <p><strong>How do these two structures interact?</strong> Is the weight of a higher multiplication $m_n(x_1, \cdots, x_n)$ determined by the weights of the $x_i$? If the answer is no in general, then can one choose a particular nice $A_\infty$ structure within the equivalence class for which the answer is yes?</p> http://mathoverflow.net/questions/46936/is-a-map-a-homotopy-equivalence-if-its-suspension-is-so/47007#47007 Answer by Jeffrey Giansiracusa for Is a map a homotopy equivalence if its suspension is so? Jeffrey Giansiracusa 2010-11-22T22:13:51Z 2013-02-19T10:49:26Z <p>If the spaces are simply connected, or somewhat more generally, if they are simple (meaning that $\pi_1$ is abelian and acts trivially on the higher homotopy groups) then as Andreas points out, there is Whitehead's theorem that a homology isomorphism between simple spaces is a weak equivalence, and also Whitehead's other theorem that a weak equivalence between CW complexes is a homotopy equivalence.</p> <p>However, as rpotrie's example indicates, with non-simple spaces the question is more interesting, and the answer is that there are certainly examples where the suspension is a homotopy equivalence but the map itself is not.</p> <p>Here's a way to construct such a map. Let $G$ be group containing a nontrivial perfect normal subgroup $H$ (i.e., $H= [H,H]$) let $X=BG$, and let $Y=BG^+$ - Quillen's plus construction. The plus construction attaches 2 cells and 3 cells to a space to produce a new space with the same homology but with fundamental group now the quotient of the original fundamental group by $H$. The inclusion $BG \subset BG^+$ is a homology isomorphism, but the spaces have different fundamental groups so the inclusion is not a homotopy equivalence. However, if $H$ happens to be the whole commutator subgroup and you suspend the map once then $\Sigma BG$ and $\Sigma (BG^+)$ are both simply connected, and so Whitehead's theorems tell you that the map is a homotopy equivalence.</p> http://mathoverflow.net/questions/114525/the-damworld-model-of-hamilton-and-henderson The Damworld model of Hamilton and Henderson Jeffrey Giansiracusa 2012-11-26T13:10:23Z 2012-11-26T13:10:23Z <p>I've been reading some of the literature around Lovelock and Watson's famous Daisyworld earth-system model. It is a simple non-linear system of ODEs that illustrates various interesting principles in control theory and ecology/biology. In quite a few places I find references to another interesting-sounding model formulated by William Hamilton and Peter Henderson: <em>Damworld.</em> </p> <blockquote> <p>Imagine Damworld as a basin ringed by mountains, in which a single species of algae lives. The rain that falls into Damworld can leave only through one narrow outlet. Living in the outlet are creatures that feed on the algae. These creatures anchor themselves to the sediments and tend to build up a dam, like coral polyps build up a reef. The third species is one that breaks down dams for food. If the dam rises, the lake behind it swells, creating a larger supply of sunlight-warmed, nutrient-rich water in which various organisms thrive.</p> </blockquote> <p>(From Oliver Morton's 1999 article, "Is the Earth Alive?")</p> <p><strong>Question:</strong> Are there any published accounts of the Damworld model? </p> <p>I've been unable to find anything beyond simple descriptions of Damworld, such as the one above. </p> <p>(Note: I'm perfectly capable of coding my own version of Damworld - but I'd like to read about the original one.) </p> http://mathoverflow.net/questions/104796/algebraic-k-theory-with-commutative-semirings/104820#104820 Answer by Jeffrey Giansiracusa for Algebraic K-theory with commutative semirings? Jeffrey Giansiracusa 2012-08-16T08:37:23Z 2012-08-16T12:30:41Z <p>There are a few papers out there dealing with a slightly different focus - algebraic K-theory over the "field with one element" $\mathbb{F}_1$. Some of the frameworks for $\mathbb{F}_1$ algebra include semirings, so these papers might contain material that covers what you are interested in as well.</p> <p>The thesis of Nicolai Durov <a href="http://arxiv.org/abs/0704.2030" rel="nofollow">arXiv:0704.2030</a> describes a setting for algebraic geometry over a class of objects more general than rings. These objects are commutative algebraic theories. Commutative rings, commutative semirings and commutative monoids all form full subcategories of commutative algebraic theories. Among the many things Durov does in his thesis, he includes some discussion of algebraic K-theory in the final chapter.</p> <p>The "blueprints" of Oliver Lorscheid (<a href="http://arxiv.org/abs/1103.1745" rel="nofollow">arxiv.org/1103.1745</a> and its sequels) also contain semirings as a full subcategory. According to the abstract, he will eventually get to K-theory of blueprints, which will contain K-theory of semirings as a special case.</p> http://mathoverflow.net/questions/104757/homotopy-theory-of-topological-stacks-orbifolds/104767#104767 Answer by Jeffrey Giansiracusa for Homotopy theory of topological stacks/orbifolds Jeffrey Giansiracusa 2012-08-15T14:17:19Z 2012-08-16T08:46:40Z <p>Here is a simple way to talk about the homotopy type of a stack. Let $\mathfrak{X}$ be a stack and $f: U \to \mathfrak{X}$ a representable surjective submersion (an atlas) from a space $U$ (e.g., the map from the Teichmuller space to the moduli stack.) Now, form the pullback of $f$ along itself: $U\times_{\mathfrak{X}} U$. This comes with two maps to $U$ and there is a diagonal map from $U \to U\times_{\mathfrak{X}} U$. All together, these maps give a topological groupoid. The nerve of this topological groupoid is a simplicial space and the geometric realization of which is a space that one can regard as representing the homotopy type of the stack. </p> <p>Here are some easy/nice properties of the above notion of homotopy type that are easy to check.</p> <ol> <li><p>Given a space $X$ with a $G$ action, the homotopy type of $[X/G]$ is the Borel construciton, aka homotopy quotient, $EG \times_G X$. In particular, the answer to your Question 1 is affirmative, and the homotopy type of the moduli stack of curves is exactly $B\Gamma_g$.</p></li> <li><p>One can define singular and de Rham cohomology of a stack and these invariants coincide with the integral and rational cohomology of the homotopy type of the stack. This is in fact almost a tautology since, for example, the de Rham cohomology can be defined by taking a covering by a manifold, forming iterated pullbacks (to produce a simplicial manifold), taking the de Rham algebra of this simplicial manifold to get a cosimplicial dga, and then taking the totalization to get a dga.</p></li> <li><p>It follows from property 1 above that the answer to your question 2 is also affirmative.</p></li> <li><p> This notion of homotopy type is well-defined because one can check that any two atlases determine Morita equivalent topological groupoids which then have weakly equivalent nerves.</p></li> </ol> <p>If I remember correctly, Noohi uses a slightly more sophisticated notion of homotopy type. He defines a universal weak equivalence to be a representable morphism from a space $U$ to a stack $\mathfrak{X}$ such that the pullback along any morphism from a space to $\mathfrak{X}$ is a weak equivalence. $U$ can then be regarded as the homotopy type of the stack. I think this is more of less equivalent to the naive version I explained above, but it has the advantage of being a bit more functorial and there might be some other technical advantages I can't remember. David Carchedi will probably be able to give more details.</p> http://mathoverflow.net/questions/52681/weight-filtration-and-hodge-theory-for-tropical-varieties Weight filtration and Hodge theory for tropical varieties Jeffrey Giansiracusa 2011-01-20T19:41:15Z 2012-07-09T14:38:59Z <p>Many concepts is algebraic geometry have tropical analogues. </p> <p><strong>Question:</strong> Is there an analogue of the weight filtration or Hodge filtration for tropical varieties?</p> <p>A tropical curve ends up being essentially a metric graph. The tropical genus is the first Betti number of the graph. There is a period mapping (analogous to the classical Abel-Jacobi period map) from the moduli space of tropical curves of genus $g$ to the space <code>$GL_g(\mathbb{R})/O_g(\mathbb{R})$</code>. Can this period map be interpreted as a classifying map for variation of tropical Hodge structure?</p> http://mathoverflow.net/questions/91717/dual-of-idempotent-semirings/98475#98475 Answer by Jeffrey Giansiracusa for Dual of idempotent semirings Jeffrey Giansiracusa 2012-05-31T11:00:58Z 2012-05-31T11:00:58Z <p>I don't think you can ask for a characterisation of $\mathcal{A}^{op}$ that is any more concrete than the definition. </p> <p>However, $\mathcal{A}^{op}$ can be described in alternative terms via scheme theory. This won't describe it in any simpler terms, and in fact it introduces a good deal of extra complication, but it can perhaps be useful sometimes because it puts things in a more geometric setting. In standard algebraic geometry the category of commutative rings is equivalent to the opposite of the category of affine schemes (over spec $\mathbb{Z}$). Similarly, your category of idempotent commutative semirings is equivalent to the opposite of the category of affine schemes over spec $S$, where $S$ is the initial object in idempotent semirings ($S=${0,1}). This embeds as a full subcategory of the category of affine schemes over $\mathbb{N}$ (the semiring of natural numbers). For references on the scheme theory of semirings, see:</p> <ol> <li>arXiv:math/0509684, Toen-Vaquie, Under Spec Z</li> <li>arXiv:0704.2030, Durov, New Approach to Arakelov Geometry</li> <li>arXiv:1103.1745, Lorscheid, The geometry of blueprints. Part I: Algebraic background and scheme theory</li> </ol> http://mathoverflow.net/questions/84705/is-the-moduli-space-of-curves-defined-over-the-field-with-one-element Is the moduli space of curves defined over the field with one element? Jeffrey Giansiracusa 2012-01-01T22:45:49Z 2012-01-04T00:31:15Z <p>There are various frameworks around which enlarge the category of rings to include more exotic objects such as the 'field with one element,' <code>$\mathbb{F}_1$</code>. While these frameworks differ in their details, there are certain things this should be true of any object that deserves to be called <code>$\mathbb{F}_1$</code>. For example, The algebraic K-theory of <code>$\mathbb{F}_1$</code> should be sphere spectrum, and the theory of toric varieties should be defined over <code>$\mathbb{F}_1$</code>.</p> <p><strong>Question 1:</strong> Is there a moral reason why the moduli space of curves should (or should not) be defined over Spec <code>$\mathbb{F}_1$</code>?</p> <p>EDIT: For anyone who would like to be more concrete, I'm happy to take the Toen-Vaquie definition of schemes over <code>$\mathbb{F}_1$</code>. (see <a href="http://arxiv.org/abs/math/0509684v4" rel="nofollow">arXiv:math/0509684</a>). In this setup (and most of the other frameworks I know) an affine scheme over <code>$\mathbb{F}_1$</code> is just a commutative monoid $M$. After base change to $\mathbb{Z}$ this becomes the monoid ring $\mathbb{Z}[M]$. So here is a more precise question:</p> <p><strong>Question 2:</strong> Does the moduli space of curves $\mathcal{M}_{g,n}$ (over $\mathbb{Z}$, say) admit a covering by affine charts of the form spec $\mathbb{Z}[M_i]$ for commutative monoids $M_i$? If so, can this covering be chosen so that (as in the case of toric varieties) the gluing is entirely determined by maps of monoids?</p> http://mathoverflow.net/questions/77868/elements-of-finite-order-in-mapping-class-groups-of-high-dimensional-manifolds/77913#77913 Answer by Jeffrey Giansiracusa for Elements of finite order in mapping class groups of high dimensional manifolds Jeffrey Giansiracusa 2011-10-12T10:16:25Z 2011-10-12T10:16:25Z <p>In the case of simply connected 4-manifolds, a famous theorem of Michael Friedman asserts that $\pi_0 Homeo(M)$ is isomorphic to the group of automorphisms $Aut(Q)$ of the intersection quadratic form $Q$ on the middle homology. This is an arithmetic group, and hence it contains a finite index subgroup that is torsion free. However, $Aut(Q)$ will almost always have torsion, except possibly in some low-rank cases.</p> <p>One way of thinking about this is: For surfaces, the map from MCG to $Aut(Q)=Sp_{2g}(\mathbb{Z})$ (sending a homeomorphism to the induced automorphism of homology) has a huge kernel, namely the Torelli group. But in dimension 4 under the simply connected hypothesis this map is an isomorphism. </p> http://mathoverflow.net/questions/77425/failures-that-lead-eventually-to-new-mathematics/77488#77488 Answer by Jeffrey Giansiracusa for Failures that lead eventually to new mathematics Jeffrey Giansiracusa 2011-10-07T19:45:49Z 2011-10-07T19:45:49Z <p>This doesn't exactly fit, but I thought it might be close enough to be worth mentioning. The Fundamental Lemma in the Langlands Program was (as implied by the name) originally expected to be relatively easy result. Much of the program depended on it, and yet the Lemma remained unproven for about 2 decades until Ngô Bảo Châu's recent proof appeared.</p> http://mathoverflow.net/questions/76608/chain-complexes-of-vector-bundles/76610#76610 Answer by Jeffrey Giansiracusa for Chain complexes of vector bundles Jeffrey Giansiracusa 2011-09-28T07:55:17Z 2011-09-28T07:55:17Z <p>The space $K_n$ sits inside the space of sequences of linear maps $$L_n = \Pi_i Hom(E^i,E^{i+1}).$$ This is just a space of sequences of matrices, so it is a real vector space of dimension $\sum_i (n_i \cdot n_{i+1})$. We give it the usual euclidean topology for real vector spaces.</p> <p>The subspace $K_n$ consists of those sequences of linear maps ${f_i}$ which form a chain complex - i.e., $f_{i+1} \circ f_i = 0$. This condition is polynomial in the entries in the matrices, so $K_n$ is a real algebraic affine subvariety inside $L_n$. We give $K_n$ the subspace topology.</p> <p>The topology on the space of morphism can be described similarly by embedding the morphism set $Mor(E,F)$ into the space of sequences of linear maps $\Pi_i Hom(E^i,F^i)$, which is again a real vector space. The condition of being a morphism of chain complexes is real algebraic so the morphism space is again a real algebraic affine variety.</p> http://mathoverflow.net/questions/53306/what-can-we-learn-from-the-tropicalization-of-an-algebraic-variety/75087#75087 Answer by Jeffrey Giansiracusa for What can we learn from the tropicalization of an algebraic variety? Jeffrey Giansiracusa 2011-09-10T10:58:12Z 2011-09-10T10:58:12Z <p>In <a href="http://arxiv.org/abs/0805.1916" rel="nofollow">arxiv.org/0805.1916</a> Sam Payne shows that one can reconstruct the analytification of a quasiprojective variety over a nonarchimedean field as the inverse limit of its tropicalisations.</p> http://mathoverflow.net/questions/72613/mikhalkins-tropical-schemes-versus-durovs-tropical-schemes Mikhalkin's tropical schemes versus Durov's tropical schemes Jeffrey Giansiracusa 2011-08-10T19:19:58Z 2011-08-10T19:19:58Z <p>In Mikhalkin's unfinished draft book on tropical geometry, <a href="http://www.scribd.com/doc/47771116/Tropical-geometry-Grigory-Mikhalkin" rel="nofollow">(available here)</a> (page 26) he defines a notion of tropical schemes. It seems to me that this definition is not just a wholesale adaptation of the usual ring-theoretic scheme theory to tropical algebras, but rather a delicately chosen modification (see, especially, his definition of rational functions and his method of handling localisations).</p> <p>On the other hand, Durov's thesis <a href="http://arxiv.org/abs/0704.2030" rel="nofollow">(available here)</a> constructs a very general framework for doing algebraic geometry, and in particular, defining schemes, over things more general than commutative rings. This setting happens to include semirings such as the tropical semifield. So Durov's work gives us a theory of tropical schemes as well. (Actually, from what I can tell, the resulting theory depends on some choice of <em>localisation theory</em>.)</p> <p>Has anyone out there taken the time to compare these two theories of tropical schemes and understand how they are related?</p> http://mathoverflow.net/questions/72490/why-are-operads-useful/72515#72515 Answer by Jeffrey Giansiracusa for Why are operads useful? Jeffrey Giansiracusa 2011-08-09T20:27:40Z 2011-08-09T20:27:40Z <p>I find two different points of view useful. </p> <ol> <li><p>Further to Steve's first answer, I would say that operads put many algebraic structures into one compact and useful meta-algebraic setting. Lie, associative, commutative, Poisson, Gerstenhaber, etc. All of these fit into one nice framework which then tells us how to define cohomology theories and study the deformation theory in each setting. This universal setup also tells us how to study generators and relations, homological algebra, duality theory, and so on. Operads, somewhat like category theory, allow one to see the common structure behind many a priori different worlds.</p></li> <li><p>My other point of view is that operads, along with their siblings, the cyclic and modular operads, are all about studying structures that glue/compose along trees or graphs. Manifestations of this type of composition appear in topological field theory, infinite loop space theory, low dimensional topology, and all sorts of other places.</p></li> </ol> http://mathoverflow.net/questions/57589/heuristic-behind-a-infty-algebras/57604#57604 Answer by Jeffrey Giansiracusa for Heuristic behind $A_{\infty}$ - algebras Jeffrey Giansiracusa 2011-03-06T21:54:15Z 2011-03-06T21:54:15Z <p>I like to think of definition of $A_\infty$ algebras as what you get when you take a space with an action of Stasheff's $A_\infty$ operad of associahedra and you take cellular chains. The algebraic definition follows directly from the cellular structure of the associahedra - they are convex polytopes with poset of faces isomorphic to the opposite of the poset of planar rooted trees and edge contractions. </p> http://mathoverflow.net/questions/55018/tropical-homological-algebra Tropical homological algebra Jeffrey Giansiracusa 2011-02-10T11:18:25Z 2011-02-11T04:44:19Z <p>Has anyone out there thought about homological algebra over the tropical semifield $\mathbb{T}$? For example, I'm interested in the Hochschild homology and cyclic homology of tropical algebras, if this can be made to make sense. </p> <p> Perhaps I should add some motivation. Topologists and geometric group theorists have been interested in the moduli space of metric graphs for at least 25 years now, mainly because of its appearance as a classifying space for automorphism groups of free groups. This space turns out to have a second identity as the moduli space of tropical curves, and people in tropical geometry tell me that it should probably in fact carry the structure of a tropical orbifold. </p> <p>This means that, in addition to homological invariants built from (sheaves of) continuous of PL functions on the space (containing essentially the information of rational homotopy), one can try to build and study homological invariants made from the tropical structure sheaf. I'm interested in what sort of geometric information these invariants might carry. Is there perhaps some new information hiding in here? </p> <p>It is always exciting when one finds that an object one has known for many years turns out to have a hidden new structure.</p> <p>Following Zoran's comment, it looks like Durov has constructed, among many other things, a model category structure on complexes of modules over $\mathbb{T}$. This means that, in principle, something like homological algebra can be done. But it's a different matter to explicitly develop homological algebra.</p> <p>So, to expand on my original question, here are some explicit questions. </p> <ol> <li><p>What is a tropical chain complex?</p></li> <li><p>Is there an explicit tropical analogue of the usual Hochschild chain complex and does it compute the correct derived functor?</p></li> <li><p>Ordinary Hochschild homology carries a Gerstenhaber algebra structure. Is there an analogous structure on a tropical Hochschild homology?</p></li> <li><p>Same questions for cyclic homology.</p></li> </ol> http://mathoverflow.net/questions/50238/poincare-duality-and-the-a-infty-structure-on-cohomology Poincare duality and the $A_\infty$ structure on cohomology Jeffrey Giansiracusa 2010-12-23T11:32:48Z 2011-01-23T10:46:52Z <p>If $X$ is a topological space then the rational cohomology of $X$ carries a canonical $A_\infty$ structure (in fact $C_\infty$) with differential <code>$m_1: H^*(X) \to H^{*+1}(X)$</code> vanishing and product <code>$m_2: H^*(X)\otimes H^*(X) \to H^*(X)$</code> coinciding with the cup product.</p> <p>If $X$ is a closed manifold then its cohomology algebra satisfies Poincare duality. This is a condition that refers only to the $m_2$ part of the $A_\infty$ structure. There are things more general than manifolds that satisfy rational Poincare duality, such as rational homology manifolds. In general, a space that satisfies rational Poincare duality is called \emph{rational Poincare duality space}. </p> <p>Obviously, the statement that $X$ is a rational Poincare duality space places no additional restrictions on the $A_\infty$ structure beyond the condition on the product. </p> <p><strong>Question 1</strong> Can one construct from the higher multiplications obstructions for a rational Poincare duality space to be rationally equivalent to a rational homology manifold, or a topological or smooth manifold?</p> <p>Here is a second and somewhat related question. Poincare duality says that $H^*(X)$ is self-dual (with an appropriate degree shift). Thus the adjoints of the higher multiplication maps make the cohomology into an $A_\infty$ coalgebra (with appropriate adjustment of the grading). </p> <p><strong>Question 2</strong> How does this $A_\infty$ coalgebra structure interact with the $A_\infty$ algebra structure on the cohomology?</p> http://mathoverflow.net/questions/52189/higher-homotopy-algebraic-structure-on-the-homology-of-an-operad Higher homotopy algebraic structure on the homology of an operad Jeffrey Giansiracusa 2011-01-15T22:13:32Z 2011-01-20T13:07:47Z <p>Given a DGA $A$, then by standard techniques such as homological perturbation theory, the ring structure on the homology $H(A)$ extends to a minimal <code>$A_\infty$</code>-algebra structure such that $H(A)$ is quasi-isomorphic to $A$, and moreover, this structure on the homology is unique up to isomorphism. The <code>$A_\infty$</code> algebra structure is describe explicitly, as usual, by the collection of higher multiplications <code>$m_n: H(A)^{\otimes n} \to H(A)$</code> (which are closely related to Massey products).</p> <p>An operad in the category of chain complexes can be thought of as a generalisation of a DGA - we now have a sequence of complexes $P(n)$, and the <code>$\circ_i$</code> compositions are associative products between these. The homology $H(P(n))$ is again an operad in chain complexes (with zero differential), and the above fact for DGAs should generalise to say that $H(P)$ carries the additional structure of a strongly homotopy operad for which it is quasi-isomorphic to $P$. Has this structure been describes explicitly in terms of higher $n$-ary analogues of the <code>$\circ_i$</code> compositions? I would be very grateful if someone could point me towards an appropriate reference.</p> http://mathoverflow.net/questions/51978/why-does-ribbon-graph-cohomology-compute-cohomology-of-mcg/52051#52051 Answer by Jeffrey Giansiracusa for Why does (Ribbon) Graph (co)Homology Compute (co)Homology of MCG? Jeffrey Giansiracusa 2011-01-14T07:41:51Z 2011-01-14T07:41:51Z <p>My favorite way to 'see' the connection between ribbon graphs and mapping class groups is to use the contractibility of the complex of arcs in a surface. Given a surface $S$, the arc complex $\mathcal{A}(S)$ has vertices given by isotopy classes of arcs with endpoints lying on the boundary of the surface, and simplices given by disjoints collections of such arcs. Hatcher has a rather attractive way to construct a contraction of this complex down to a single vertex.</p> <p>By an easy inductive argument one can deduce from the contractibility of the arc complex that the poset $\mathcal{A}_0(S)$ of simplices for which the arcs cut the surface into pieces that are all discs is also contractible. Such a collection of arcs has a dual ribbon graph.</p> <p>The mapping class group $MCG(S)$ acts on the poset $\mathcal{A}_0(S)$, and the homotopy quotient is equivalent to the classifying space of the mapping class group since the poset is contractible. On the other hand, one can build a model for the homotopy quotient by taking the category $MCG(S) \int \mathcal{A}_0(S)$ in which objects are objects of $\mathcal{A}_0(S)$ and a morphism $x \to y$ is a mapping class group element $\alpha$ such that $\alpha\cdot x \subset y$ (it is a theorem of Thomason that this models the homotopy quotient). By sending a collection of arcs to its dual ribbon graph one sees that this category is equivalent to the category of ribbon graphs of type $S$.</p> http://mathoverflow.net/questions/35925/natural-examples-of-finite-dimensional-spaces-with-interesting-2-type/50883#50883 Answer by Jeffrey Giansiracusa for Natural examples of finite dimensional spaces with interesting 2-type Jeffrey Giansiracusa 2011-01-01T21:24:40Z 2011-01-01T21:24:40Z <p>In low dimensional topology there are host of examples along the lines of the curve complex of a surface. Someone like Andy Putman will be able to provide a more detailed explanation, but here is a summary. You build a simplicial complex in which vertices are isotopy classes of closed curves in a surface, and n-simplices are disjoint (n+1)-tuples of curves. If you require that no curve bounds a disc (or an annulus if your surface has boundary) then it is fairly easy to see that the complex is finite dimensional. The mapping class group of the underlying surface acts on this complex.</p> <p>Now, the homotopy types of this complex and its variations, and of the quotient by the mapping class group, are very interesting objects. In particular, studying the low dimensional homotopy groups allow you to do things like construct presentations for the mapping class group. </p> <p>This sort of construction is really a small industry in low dim topology/geometric group theory. Braid groups, automorphism groups of free groups, surface mapping class groups, 3-manifold mapping class groups, and various subgroups of these, can all be studied via complexes of this type, and often we only understand the complex as far as its 2-type.</p> http://mathoverflow.net/questions/50473/why-does-the-group-act-on-the-right-on-the-principal-bundle/50637#50637 Answer by Jeffrey Giansiracusa for Why does the group act on the right on the principal bundle? Jeffrey Giansiracusa 2010-12-29T09:23:46Z 2010-12-29T09:23:46Z <p>There is another possible reason for this convention: to make the notation for the Borel construction $EG \times_G X$ a little nicer. Writing $Y\times_G X$ generally implies that $Y$ has a right action and $X$ has a left action (although of course there really isn't any difference between right and left $G$-spaces). If we habitually work with left $G$-spaces $X$, then we end up wanting the universal principal bundle $EG \to BG$ to be defined by a right action.</p> http://mathoverflow.net/questions/49800/are-there-graph-models-for-other-moduli-spaces/49873#49873 Answer by Jeffrey Giansiracusa for Are there graph models for other moduli spaces? Jeffrey Giansiracusa 2010-12-19T11:08:33Z 2010-12-19T11:08:33Z <p>Yes, graphs come up in various moduli contexts quite a bit! As the previous two answers indicated, $Out(F_n)$ is closely related to graphs. This goes back to the original paper of Vogtmann and Culler where outer space was first introduced. $BOut(F_n)$ can be modelled as the moduli space of metric graphs with first Betti number equal to $n$. If you consider the moduli space of graphs as an orbifold then this has the right integral homotopy type (but if you take the coarse quotient space then it is just a rational classifying space because $Out(F_n)$ acts on outer space with finite stabilisers).</p> <p>Given a cyclic operad $P$ in the category of topological spaces one can talk about the space of graphs with vertices labelled by $P$. An ordinary graph is the same as a $Comm$-labelled graph since the commutative cyclic operad is just a point in each arity. A ribbon graph is the same as an $Assoc$-labelled graph since in cyclic arity $n$ the associative cyclic operad is the set of cyclic oderings on $n$ letters.</p> <p>So the Culler-Vogtmann work can be interpreted as saying that $BOut(F_n)$ is the moduli space of rank $n$ $Comm$-graphs. Moduli spaces of Riemann surfaces with marked points are are homotopy equivalent to spaces of $Assoc$-graphs. There are some other results of this type. A Mobius graph is like a ribbon graph but with edges possibly given a half-twist; these are the same as graphs labelled by the cyclic operad $InvAss$ for associative algebras with an involution (perhaps could be called the hermitian associative operad). The spaces of Mobius graphs are homotopy equivalent to the moduli spaces of surfaces with a Klein structure (an unoriented version of complex structure), or equivalently, the classifying spaces of the mapping class groups of unorientable surfaces. Another result of this type is that the space of graphs labelled by the framed little 2-discs cyclic operad is homotopy equivalent to the moduli space of 3-dimensional oriented handlebodies.</p> <p>To expand on some of Jim's comments a bit further: The connection to graph homology is as follows. First you need to know that there is a duality functor for cyclic operads in chain complexes. Sometimes it is called dg-duality, and sometimes it is called the Bar construction or Koszul duality. (Strictly speaking, the duality is given by a bar construction that produces a cyclic cooperad, followed by applying linear duality to turn in back into a cyclic operad; this construction agrees up to quasi-isomorphism with the koszul duality construction for cyclic operads that are Koszul.) The associative cyclic operad is self-dual. The commutative cyclic operad is dual to the Lie cyclic operad. </p> <p>The general theorem is that if $P$ is a cyclic operad in topological spaces, then $C_*P$ is a cyclic operad in chain complexes with dual $D(C_*P)$, and $D(C_*P)$-graph homology computes the cohomology of the space of $P$-labelled graphs. This is why $Lie$ graph homology computes the cohomology of $Out(F_n)$ and $Ass$ graph homology computes the cohomology of moduli spaces of Riemann surfaces. </p> http://mathoverflow.net/questions/46830/hodge-star-and-harmonic-simplicial-differential-forms Hodge star and harmonic simplicial differential forms Jeffrey Giansiracusa 2010-11-21T16:39:39Z 2010-11-21T16:39:39Z <p>Is there a notion of harmonic forms and Hodge theory for Sullivan's piecewise smooth differential forms on a simplicial set?</p> <p>Let me recall some background.</p> <p><strong>Hodge Theory on a Riemannian manifold</strong> A Riemannian metric $g$ an $n$-dimension closed manifold $M$ gives a Hodge star operator on the smooth differential forms $*: \Omega^k(M) \to \Omega^{n-k}(M)$, a nondegenerate inner product on $\Omega^k(M)$ given by $\langle \alpha, \beta \rangle = \int_M \alpha \wedge * \beta$, and a codifferential $\delta$ that is the adjoint of the usual exterior differential $d$. The Laplacian is $\Delta = \delta d + d \delta$, and the harmonic forms are those which are in the kernel of the Laplacian.</p> <p>Hodge theory asserts that the space of harmonic forms is isomorphic to the real cohomology of $M$. I.e., every harmonic form is closed, and each cohomology class contains a unique harmonic representative.</p> <p><strong>Sullivan's piecewise smooth differential forms on a simplicial complex</strong> Let $K$ be a simplicial set. A differential form on $K$ is essentially a smooth differential form on each simplex of $K$ subject to compatibility conditions given by the face and degeneracy maps. In detail, $\Omega^*(\Delta^\bullet)$ is a simplicial object in commutative differential graded algebras, and the algebra of piecewise smooth forms on $K$ is<br> <code>$A_{C^\infty}(K) = Hom_{\mathrm{SSet}}(K_\bullet, \Omega^*(\Delta^\bullet))$</code></p> <p><strong>The question</strong> Suppose now that $K$ is a finite simplicial set (i.e., a simplicial set with finitely many nondegenerate simplices) with an appropriate version of a Riemannian metric. Is there a notion of harmonic forms and Hodge theory for $A_{C^\infty}(K)$? </p> http://mathoverflow.net/questions/45036/spectral-sequences-opening-the-black-box-slowly-with-an-example/45084#45084 Answer by Jeffrey Giansiracusa for Spectral sequences: opening the black box slowly with an example Jeffrey Giansiracusa 2010-11-06T18:12:58Z 2010-11-06T18:12:58Z <p>In Jim Stasheff's original papers on $A_\infty$-algebras he generalises the bar construction and its spectral sequence to $A_\infty$-algebras, and then he identifies all of the differentials explicitly in terms of things that look like (duals of) Massey products - he calls them Yessam products. </p> <p>I realise that this isn't exactly an explicit compution, but it is at least a general explicit description of the higher differentials that is not formal, and to get it requires a bit of the guts of the spectral sequence.</p> http://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/40965#40965 Answer by Jeffrey Giansiracusa for Compelling evidence that two basepoints are better than one Jeffrey Giansiracusa 2010-10-03T22:04:00Z 2010-10-03T22:04:00Z <p>Here is an interesting example where groupoids are useful. The mapping class group <code>$\Gamma_{g,n}$</code> is the group of isotopy classes of orientation preserving diffeomorphisms of a surface of genus $g$ with $n$ distinct marked points (labelled 1 through n). The classifying space <code>$B\Gamma_{g,n}$</code> is rational homology equivalent to the (coarse) moduli space <code>$\mathcal{M}_{g,n}$</code> of complex curves of genus $g$ with $n$ marked points (and if you are willing to talk about the moduli orbifold or stack, then it is actually a homotopy equivalence)</p> <p>The symmetric group <code>$\Sigma_n$</code> acts on <code>$\mathcal{M}_{g,n}$</code> by permuting the labels of the marked points.</p> <p><strong>Question:</strong> How do we describe the corresponding action of the symmetric group on the classifying space <code>$B\Gamma_{g,n}$</code>?</p> <p>It is possible to see <code>$\Sigma_n$</code> as acting by outer automorphisms on the mapping class group. I suppose that one could probably build an action on the classifying space directly from this, but here is a much nicer way to handle the problem.</p> <p>The group $\Gamma_{g,n}$ can be identified with the orbifold fundamental group of the moduli space. Let's replace it with a fundamental groupoid. Fix a surface $S$ with $n$ distinguished points, and take the groupoid where objects are labellings of the distinguished points by 1 through n, and morphisms are isotopy classes of diffeomorphisms that respect the labellings (i.e., sending the point labelled $i$ in the first labelling to the point labelled $i$ in the second labelling). </p> <p>Clearly this groupoid is equivalent to the original mapping class group, so its classifying space is homotopy equivalent. But now we have an honest action of the symmetric group by permuting the labels on the distinguished points of $S$.</p> http://mathoverflow.net/questions/38846/a-infty-structure-of-cohomology-of-a-space/38873#38873 Answer by Jeffrey Giansiracusa for $A_{\infty}$ structure of (co)homology of a space Jeffrey Giansiracusa 2010-09-15T21:10:34Z 2010-09-15T21:10:34Z <p>Your category $X^X$ is just the group of homeomorphisms of $X$. This group certainly acts on the homology and cohomology, making them strict modules. But since the group of homeomorphisms is actually acting on $X$, it gives automorphisms of the rational homotopy type. The rational homotopy type of $X$ can be encoded in an $A_\infty$ algebra structure on the rational cohomology ring (technically, it is a $C_\infty$ structure, which is a special kind of $A_\infty$ structure). Thus the group of homeomorphisms of $X$ gives homotopy self-equivalences of the $A_\infty$ algebra $H^*(X)$. That is, a homeomorphism $\phi: X \to X$ gives an $A_\infty$ map <code>$H^*(X) \to H^*(X)$</code> that is an equivalence. Note that such a map contains potentially more information than simply an automorphism of $H^*(X)$ as an ordinary ring.</p> http://mathoverflow.net/questions/38865/homology-of-bundles-over-a-triangulated-base-and-a-infty-algebras/38871#38871 Answer by Jeffrey Giansiracusa for Homology of bundles over a triangulated base and $A_\infty$-algebras Jeffrey Giansiracusa 2010-09-15T20:58:46Z 2010-09-15T20:58:46Z <p>I don't remember the details, but this is pretty much what K. Igusa does as the first step in his construction of higher Franz-Reidemeister torsion invariants. Have a look at his book and some of the papers on the arXiv (arXiv:math/0212383 and arXiv:math/0303047).</p> http://mathoverflow.net/questions/36222/what-are-natural-questions-to-ask-about-an-operad/36227#36227 Answer by Jeffrey Giansiracusa for What are natural questions to ask about an operad? Jeffrey Giansiracusa 2010-08-20T21:32:40Z 2010-08-20T21:32:40Z <p>Here are a few of my favorite questions about operads at the moment:</p> <ol> <li><p>Is your operad actually a cyclic operad (in the sense of Getzler-Kapranov)?</p></li> <li><p>If so, what is the modular operad that it generates?</p></li> <li><p>Is it a quadratic operad? Is it Koszul?</p></li> <li><p>Is it cofibrant? If not, what does a nice cofibrant replacement look like? E.g., the associative operad is not cofibrant, but the A-infinity operad made from the associahedra is an interesting cofibrant replacement.</p></li> <li><p>Does it have interesting morphisms to or from other operads? If so, then people are often interested in the deformation theory of such a morphism.</p></li> </ol> http://mathoverflow.net/questions/33833/what-is-a-principal-refinement-of-a-postnikov-system/33845#33845 Answer by Jeffrey Giansiracusa for What is a principal refinement of a Postnikov system? Jeffrey Giansiracusa 2010-07-29T20:32:22Z 2010-07-29T20:32:22Z <p>The key idea is that not all fibrations $E \to B$ with fibre an Eilenberg-MacLane space $K(\pi,n)$ can be constructed by pulling the principal path fibration $K(\pi,n) \to PK(\pi,n+1) \to K(\pi,n+1)$ along a classifying map $B \to K(\pi,n+1)$. If you can construct the fibration in this way then the classifying map is the Postnikov $k$-invariant. Clearly this at least requires that the group $\pi$ is abelian. </p> <p>Now, it is a nice little exercise to check that existence of a principal refinement of the Postnikov tower is equivalent to $\pi_1$ being nilpotent and acting nilpotently on all of the higher homotopy groups.</p> <p>(Recall that a group $G$ acts nilpotently on a group $H$ if $H$ has a finite sequence of $G$-invariant subgroups <code>$H \supset H_1 \supset H_2 \supset \cdots H_k = 1$</code> such that <code>$H_i/H_{i+1}$</code> is abelian and the action of $G$ on it is trivial.)</p> <p>The general idea of Hilton, Mislin, and Roitberg is that is it obvious how to localise abelian groups, and nilpotent groups are those which can be assembled from abelian groups one layer at a time. So we can localise nilpotent groups by working one layer at a time, and then we can localise nilpotent spaces by working up the refined Postnikov tower one stage at a time.</p> http://mathoverflow.net/questions/32387/computation-of-homology-groups-of-m-g-n/32410#32410 Answer by Jeffrey Giansiracusa for Computation of homology groups of $M_{g,n}$ Jeffrey Giansiracusa 2010-07-18T21:51:31Z 2010-07-18T21:51:31Z <p>There is some general machinery that is perfectly suited to this question.</p> <p>Consider the following setup: let $U$ be the complement of a normal crossing divisor $D$ in a compact complex manifold (or orbifold) $X$. (In the special case at hand, <code>$U = \mathcal{M}_{g,n}$</code>, and $X$ is the Deligne-Mumford compactification.) With a bit of work one can see that the Leray spectral sequence for the inclusion $U\hookrightarrow X$ has $E_2$ page given by <code>$E^{p,q}_2 = \oplus_S H^{q-2p}(D_S;\mathbb{C})$</code> where the sum runs over the closed boundary strata of codimension $p$ and the differential on this page is given by <code>$da = \Sigma_T \pm (i_{S,T})_! a$</code> where <code>$a\in H^{q-2p}(D_S;\mathbb{C})$</code> and $T$ runs over codimension $p-1$ boundary strata that contain <code>$D_S$</code>, and <code>$(i_{S,T})_!$</code> is the pushforward along the inclusion <code>$D_S \hookrightarrow D_T$</code>. One can also get this spectral sequence from the weight filtration on the complex of forms with logarithmic poles along $D$.</p> <p>Deligne proved that this spectral sequence degenerates at the <code>$E_2$</code> page. So the cohomology of this page is the associated graded for $H^*(U;\mathbb{C})$.</p> <p>In the case of the moduli space of curves and its Deligne-Mumford-Knudsen compactification, the $E_2$ page is described in terms of the cohomology of the various strata of the boundary, which are isomorphic to smaller compactified moduli spaces. There have been quite a few papers that used this spectral sequence to prove interesting things. For instance, there is an old paper of Voronov (alg-geom/9708019) (from just a couple of years before the proof of the Madsen-Weiss theorem) in which he analyzes this SS to show that the rational homotopy type of <code>$\mathcal{M}_{g,n}$</code> is stable in the Harer-Ivanov stable range and moreover, it is formal in this range. </p> <p>As an aside, very shortly after Voronov's paper, there was Tillmann's paper in which she showed that <code>$\mathcal{M}_{g,n}$</code> has the homology of an infinite loop space in the stable range, which also implies rational homotopy formality in the stable range.</p> http://mathoverflow.net/questions/128615/a-infinity-structure-on-cohomology-and-the-weight-filtration Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2013-04-24T17:37:32Z 2013-04-24T17:37:32Z Dan, thanks for reminding me about that question - I had forgotten about it. Unfortunately nobody ever gave an answer to that one. http://mathoverflow.net/questions/128615/a-infinity-structure-on-cohomology-and-the-weight-filtration Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2013-04-24T17:36:50Z 2013-04-24T17:36:50Z Jan, even without knowing that smooth projective implies formal, smooth projective implies immediately that weight equals degree. http://mathoverflow.net/questions/44021/which-manifolds-are-homeomorphic-to-simplicial-complexes/44039#44039 Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2013-03-12T10:19:37Z 2013-03-12T10:19:37Z Ciprian Manolescu has just posted a paper in which he claims to prove that no such homology 3-sphere exists. arXiv:1303.2354 Pin(2)-equivariant Seiberg-Witten Floer homology and the Triangulation Conjecture http://mathoverflow.net/questions/104757/homotopy-theory-of-topological-stacks-orbifolds/104767#104767 Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2012-08-26T02:35:24Z 2012-08-26T02:35:24Z If you build the Borel construction in the right way then the map from the homotopy quotient to the coarse quotient will have as its fibres the classifying spaces of the isotropy groups. If the isotropy groups are all finite then their classifying spaces have trivial rational cohomology and a version of the Vietoris Mapping Theorem gives the rational equivalence asked for in Q2. Alternatively, one could take singular simplicial sets and use a spectral sequence argument. http://mathoverflow.net/questions/104757/homotopy-theory-of-topological-stacks-orbifolds/104767#104767 Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2012-08-16T08:43:25Z 2012-08-16T08:43:25Z @Jacob, I should have been a little more careful. I was starting with a stack in the category of smooth manifolds (or topological spaces). You're correct in that if we start with a complex algebraic stack then we should first pass to the associated analytic stack. If it is nonsingular then forgetting the complex analytic structure gives a stack in manifolds. If it is singular then you still get a stack in topological spaces, which still has a homotopy type as I described The only part where you need manifolds is for de Rham forms. http://mathoverflow.net/questions/84878/spec-mathbbz-in-absolute-geometry Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2012-01-04T15:06:43Z 2012-01-04T15:06:43Z The problem, as I understand it, is not in defining spec Z as an object over <code>$F&#95;1$</code>. There are actually quite a few definitions of $F_1$ and the first property that any putative definition must have is that it must map to Z. The difficulty is in finding a setup where spec Z has various desired properties, such as admitting a nice compactification that behaves like a curve over $F_1$ (I think this is called the Deninger program). http://mathoverflow.net/questions/84705/is-the-moduli-space-of-curves-defined-over-the-field-with-one-element/84785#84785 Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2012-01-03T08:59:35Z 2012-01-03T08:59:35Z Thanks for the very clear explanation. http://mathoverflow.net/questions/84705/is-the-moduli-space-of-curves-defined-over-the-field-with-one-element Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2012-01-02T18:44:20Z 2012-01-02T18:44:20Z @Chris - in low dimensional topology there is an abundance of interesting sets on which the mapping class group acts, such as the n-simplices of the curve complex and all kinds of related objects. In the examples that spring to mind there isn't a fixed basepoint, but one can of course always add a fixed basepoint. So of course, a family of interesting questions is whether any of these sorts of objects admit algebraic descriptions. This wasn't my original motivation for asking the question, but it could have been. http://mathoverflow.net/questions/84705/is-the-moduli-space-of-curves-defined-over-the-field-with-one-element Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2012-01-02T12:28:00Z 2012-01-02T12:28:00Z I think what you mean is that spec $\mathbb{F}_1$ should be a final object, so everything maps to it. This is different from being defined over it. E.g. $\mathbb{Z}$ is initial among rings and so any scheme maps to spec $\mathbb{Z}$, but there are certainly things that are not defined over $\mathbb{Z}$ - i.e., there exists schemes over spec $k$ that are not the pullback (along spec $k$ $\to$ spec $\mathbb{Z}$) of a scheme over $\mathbb{Z}$. http://mathoverflow.net/questions/82376/twisted-de-rham-cohomology-and-eilenberg-mac-lane-spaces/82394#82394 Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2011-12-01T21:33:10Z 2011-12-01T21:33:10Z Fabio, you are on the right. Fernando is only talking about Z-graded cohomology. But if you instead ask about 2-periodic cohomology (Z/2-graded) then I believe you can have more interesting twists. http://mathoverflow.net/questions/82376/twisted-de-rham-cohomology-and-eilenberg-mac-lane-spaces Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2011-12-01T16:13:01Z 2011-12-01T16:13:01Z I believe the answer to this question (and probably far more than you ever wanted to know) is contained in the paper &quot;Periodic twisted cohomology and T-duality&quot; by Bunke, Schick and Spitzweck, in Asterisque 337 (2011) - <a href="http://arxiv.org/abs/0805.1459" rel="nofollow">arxiv.org/abs/0805.1459</a> http://mathoverflow.net/questions/77425/failures-that-lead-eventually-to-new-mathematics/77489#77489 Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2011-10-10T08:40:12Z 2011-10-10T08:40:12Z Algebraic geometers tend to use the word 'space' quite flexibly too. And as a group, I'd have to say from my experience that Russians are the most flexible in this regard. I've heard the work space used in talks to refer to any of: topological space, simplicial set, spectra, manifold, variety, algebraic space, stack, vector space, Lie algebra, A_infty algebra, L_infty algebra, noncommutative ring, C^* algebra, and a category! http://mathoverflow.net/questions/73772/concrete-example-of-infty-categories Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2011-08-26T16:23:04Z 2011-08-26T16:23:04Z Is Lurie's paper sketching the classification of topological field theories not considered up-to-date enough now? http://mathoverflow.net/questions/72490/why-are-operads-useful/72515#72515 Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2011-08-10T12:04:21Z 2011-08-10T12:04:21Z For example, take pairs of pants, and their associated moduli spaces. If you glue a bunch of pants together then you get a surface and the dual graph to the pants decomposition is a trivalent graph. Pants aren't closed under composition because if you glue two you get a sphere with 4 holes instead of 3. So consider moduli spaces of genus zero surfaces with boundaries - these now have an operad structure. If you start gluing these genus zero guys together you get a surface, and the dual graph is an arbitrary graph. That's the sort of picture I meant. http://mathoverflow.net/questions/66803/minimal-models-with-local-coefficients Comment by Jeffrey Giansiracusa Jeffrey Giansiracusa 2011-06-03T09:42:46Z 2011-06-03T09:42:46Z I think you are asking for the impossible here because the cohomology with local coefficients does not generally have a cup product ring structure. However, $H^*(X;\rho)$ is of course a module over $H^*(X)$ and so you can ask for a dg-module over a cdga model for $X$.