User hiro lee tanaka - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T10:38:51Z http://mathoverflow.net/feeds/user/3593 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/3237/japanese-chinese-for-mathematicians/130720#130720 Answer by Hiro Lee Tanaka for japanese/chinese for mathematicians? Hiro Lee Tanaka 2013-05-15T14:00:29Z 2013-05-15T14:00:29Z <p>This may not be the answer you're looking for, but I thought I'd share my experience as someone who was born in Japan but was transplanted quickly into the United States. My Japanese is not nearly as good as it should be, but is certainly good enough to read math.</p> <p>A beautiful part of reading Japanese/Chinese math is that you can grasp the meaning without knowing how to pronounce anything. I don't know any technical Chinese, but in Japanese,</p> <blockquote> <p>写像</p> </blockquote> <p>is the word for "mapping" or "function", and the literal meaning of its characters hints at this. Let me explain.</p> <p><a href="http://en.wiktionary.org/wiki/%25E5%2586%2599" rel="nofollow">The first character</a> means to transcribe, to picture, or to give a visual form--poetically, it can mean to simply give an abstract form to something, rather than a visual one. (For instance, the word 写真 means photograph, where <a href="http://en.wiktionary.org/wiki/%25E7%259C%259F" rel="nofollow">the second character</a> in this particular word means "truth". It might be silly to think the word for photograph is to "picture something truly/in its true form", but that's a beautiful translation to ponder on another occasion.) </p> <p><a href="http://en.wiktionary.org/wiki/%25E5%2583%258F" rel="nofollow">The final character</a> in 写像 means figure, or image, or an embodiment. For instance, the word 画像 means "image" in the computer sense of file type. In fact the character 像 alone can mean "image" in the sense of mathematics, as in the image of something under a map.</p> <p>In short, the word for "function" or "map" can be literally and clumsily translated back into English as "forming an image" or "creating a figure" or "realizing a form", most abstractly. I doubt any Japanese person ever thinks in these terms, no more than we think of the word "projection" deeply in terms of its Latin roots. But to harzard a guess at the meanings of these words can be a beautiful experience, and one unique to those weirdos who know the meanings of things without knowing how to say them.</p> <p>So it may be a really interesting experience to simply learn the meaning of each commonly occurring (math) character---I'll list a few below---and to get a feel for the mathematical meanings of their combinations via intuition. When I've read Japanese math books, the feeling of knowing the meaning on a page without knowing how to pronounce a word has been the most rewarding and beautiful part. If you choose to do this, the best tip I have is to simply write: Make sure you copy and write the characters over and over again, so you begin to distinguish subtle differences between them.</p> <p>For the enjoyment of some, here are examples of Japanese math words and the meanings of their constituent characters. I'll list some irrelevant meanings of some characters--though characters often only take on one of many meanings based on context, I still think it's fun to know their other possible meanings.</p> <blockquote> <p>空間 (space)</p> </blockquote> <p>空 = sky, emptiness, space, air</p> <p>間 = between, the space between, an interval of time</p> <blockquote> <p>位相 (topology)</p> </blockquote> <p>位 = rank (as in seniority or importance in an organization), a word for counting dead souls, decimal place, position. As a verb, it can mean to locate--i.e., to determine the position of.</p> <p>相 = form, shape, appearance, the relationship of one thing to another.</p> <p>Strangely enough, 位相　can also mean the phase of something, as in the angle or phase of a complex number or a wave. It also mean the phase of something as in "solid/liquid/gaseous". I would assume that the term first came to use to describe the states of matter, was tangentially used to describe the phase of wave-like phenomena since the English term "phase" was used in both instances. </p> <blockquote> <p>微分 (derivative, to take the derivative of)</p> </blockquote> <p>微 = infinitesimal, tiny, slight</p> <p>分 = to divide, an amount of something.</p> <p>In learning language so much emphasis is placed on the sounds of things, rather than on the abstract units of meaning. I suppose Chinese characters were developed exactly to avoid this aural emphasis, but it is always a joy to have zero verbal understanding with a Chinese or Korean person, but to be able to communicate by writing characters in the air.</p> <p>Well, perhaps this was not helpful in the least, but maybe it will at least entertain some non-Japanese-speakers. (By the way, I'd be very curious to hear if the Chinese technical terms are the same, as almost all technical terms in Japanese utilize kanji, or Chinese characters.)</p> http://mathoverflow.net/questions/4796/braided-monoidal-2-categories-with-duals/130647#130647 Answer by Hiro Lee Tanaka for Braided Monoidal 2-categories with duals Hiro Lee Tanaka 2013-05-14T23:23:38Z 2013-05-15T00:24:13Z <p>Let $A$ be an $E_3$-algebra, so that $A$ is an $E_2$-algebra in the category of $E_1$-algebras by Dunn additivity. The functor $$E_1-alg \to Cat$$ $$A \mapsto A-mod$$ is symmetric monoidal, so it will send a "banana" algebra in $E_1$-alg to a "banana" algebra in categories. In particular, the category of left $E_1$-modules over $A$ is an $E_2$-category; i.e., a braided monoidal (but not symmetric monoidal) category.</p> <p>If you want the target to be 2-cats, rather than Cat, you can enhance by considering an $E_4$-algebra $A$, forget it to an $E_2$ algebra $A'$ and looking at the Morita 2-category of algebras over $A'$, or at the category of $E_2$-algebras over $A'$.</p> http://mathoverflow.net/questions/116476/where-does-the-notion-of-pseudoholomorphic-curve-come-from/116480#116480 Answer by Hiro Lee Tanaka for Where does the notion of pseudoholomorphic curve come from? Hiro Lee Tanaka 2012-12-15T20:27:15Z 2012-12-16T16:02:28Z <p>As many have pointed out: Gromov <strike>introduced it</strike>* wrote a seminal paper utilizing it, and we continue to use it today because it's an incredibly useful tool. I've never spoken to Gromov about why he introduced it (who knows how great mathematicians come up with great ideas) but I can try to give some (probably historically false) motivations as to why someone might have come up with the notion. For instance, if Gromov hadn't discovered it, you might have come up with it as follows:</p> <p>(1) First, complex geometry--if you like, you can think of algebraic geometry--has a lot of rigidity. The fact that we can even give a discrete count to sub-objects (like how many curves pass through n fixed points) is special -- the question takes on a totally different nature in more flimsy geometries.</p> <p>Now, is there a way to relax the background of complex geometry, and still come up with a useful, fun theory? For instance, how necessary is the integrability condition on J (the complex structure) to still make sense of curve-counting?</p> <p>What Gromov showed is that if the complex structure is tame' in the sense that one has a compatible symplectic form, questions about curve-counting can still have nice answers. Really, the difference between a pseudoholomorphic curve and a holomorphic curve isn't in their definitions, it's in the nature of J in the target. Relaxing the J from "integrable complex structure" to "complex structure tamed by a symplectic form" is the generalization that's happening.</p> <p>(1') Put another way, we already had a famous 2-out-of-3 principle recognizing the relationship between Riemannian, complex, and symplectic structures on a vector space. Studying curves on complex projective varieties take on rigidity, in some sense, because we study maps between manifolds with Kahler structure: manifolds both symplectic and complex, and further, each structure is integrable--in that the Nijenhaus tensor vanishes, and omega is closed. It's natural to ask whether we can still find interesting structure in the 2-out-of-3 world by studying manifolds whose tangential structures are compatibly Riemannian, complex, and symplectic, but which do not satisfy a global condition like integrability of J or closedness of $\omega$. And when you get rid of the integrability of J, it turns out that you can find such a structure on any symplectic manifold. (In fact, once you fix $\omega$, there's a <em>contractible</em> space of compatible $J$. That's why pseudoholomorphic curves can be applied widely in the symplectic world.)</p> <p>(2) There might be another motivation from physics. In mirror symmetry, one predicts the existence of mirror Calabi-Yau manifolds. A field theory that relies on the symplectic structure of one manifold should correspond to a field theory that relies on the complex structure of the mirror. And the correlation functions count J-holomorphic curves in the symplectic manifold. Historically though, I'm not sure if physics alone would be able to motivate the study of these field theories on just symplectic manifolds with almost-complex structure, as opposed to Calabi-Yaus. Somebody with more background could probably comment on this.</p> <p>*As I learned from Antoine and Dmitri, there were previous works utilizing pseudo-holomorphic curves. For instance: </p> <p>A.Nijenhuis, W.Wolf. Some integration problems in almost-complex and complex manifolds, Ann. Math. 77 (1963), </p> <p>J. Eells and S. Salamon. Twistorial construction of harmonic maps of surfaces into four-manifolds. (1985).</p> http://mathoverflow.net/questions/115567/covering-maps-in-real-life-that-can-be-demonstrated-to-students/115611#115611 Answer by Hiro Lee Tanaka for Covering maps in real life that can be demonstrated to students Hiro Lee Tanaka 2012-12-06T14:17:36Z 2012-12-08T20:52:23Z <p>Here's an example I learned from John Franks. This is a nice example because it's used to produce an example of Smale's sphere eversion problem. It also generalizes to include (for example) Will Sawin's comment above.</p> <p>Consider the Boy's surface. This is an immersion of $\mathbb{R}P^2$ into $\mathbb{R}^3$. If you look at its normal bundle, there's no sense of +1 or -1 (because it's non-orientable) but you can look at its associated unit sphere bundle. This is the orientation double cover--a.k.a. the sphere--immersed into $\mathbb{R}^3$. By scaling the fibers of the normal bundle from 1 to 0, you see the "covering homotopy" of $S^2$ onto the Boy's surface, as you ask for.</p> <p>So that's the thing you're looking for, but let's go further--instead of just scaling from 1 to 0, scale from 1 to -1. This is a homotopy, through immersions, of $S^2$ to itself, and it leads to one way in which you can evert the sphere (i.e., turn it inside out). I think this strategy originally came from Shapiro, though any historical corrections are welcome!</p> <p>More generally, if you immerse any 2-dimensional object (i.e., an un/orientable surface with or without boundary) you can perform the same trick, examining the unit-length elements of its normal bundle. In Will Sawin's example, take an embedded Mobius band and examine its unit normal bundle--this gives the non-standard embedding of the cylinder you're asking for, and scaling the normal bundle to the zero section gives you the "covering homotopy" you seek.</p> http://mathoverflow.net/questions/106955/internal-homs-in-infinity-categories/107050#107050 Answer by Hiro Lee Tanaka for Internal Homs in Infinity Categories Hiro Lee Tanaka 2012-09-13T00:36:58Z 2012-09-13T00:36:58Z <p>I'm sure someone might give a better answer, but here are my two cents. I'd like to start out by pointing out that your question is difficult already for n=0:</p> <p>First, it would help to know what one means by an enriched oo-category. Other than special, incredibly natural cases like Spaces, Chain complexes, and Spectra, it's hard to know what we mean by this. You should send an e-mail to Rune Haugseng and David Gepner--they're working on the theory of enriched oo-categories. </p> <p>Second, though I think your question is a cool theoretical question, I don't know of any example of an $\infty$-category where the higher morphisms ($n \geq 2$) form anything but spaces in a reasonable sense. For instance, though spectra and chain complexes are enriched over themselves, 2-morphisms seem most naturally a space, and nothing more. I may just be unaware of some higher enrichment, though. (If so, someone please comment--I'd love to learn.)</p> <p>So as a result, it seems natural to ask about $(\infty,n)$-categories with various forms of enrichment, rather than just $\infty$-categories. If people are still figuring out enriching $\infty$-categories, I might suspect that enriching $(\infty,n)$-categories is also difficult. A suggestion is to think of examples of higher categories that seem to have a symmetric monoidal structure. If the tensor product seems to want a right adjoint, you might have a chance at understanding internal Homs via some tensor-hom adjunction.</p> http://mathoverflow.net/questions/106705/2d-problems-which-are-easier-to-solve-in-3d/106758#106758 Answer by Hiro Lee Tanaka for 2D Problems Which are Easier to Solve in 3D Hiro Lee Tanaka 2012-09-09T22:56:49Z 2012-09-10T15:59:22Z <p>There's a famous problem posed by Erdos that has an easy 3-D solution, but a very difficult 2-D solution. The problem is to prove the following: Given a decomposition of an n-cube into finitely many n-cubes $Q_1, ... Q_k$ ($k>1$), prove that there exist two distinct cubes $Q_i, Q_{i'}$, of equal size.</p> <p>The above statement is certainly true for $n=3$ (this is a simple exercise), but it is in fact untrue for $n=2$. I think this is known as the "Squared square" problem, and you can <a href="http://squaring.net/history_theory/sprague.html" rel="nofollow">read more about it here</a>. Below is the first counter-example, due to Sprague, to the problem.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<img src="http://squaring.net/history_theory/gfx/sprague.jpeg" alt="A squared square with no equal sub-squares"></p> http://mathoverflow.net/questions/97377/codimension-two-embeddings-in-goodwillie-weiss-manifold-calculus-and-the-difficu Codimension Two Embeddings in Goodwillie-Weiss Manifold Calculus, and the Difficulty of Fundamental Groups Hiro Lee Tanaka 2012-05-19T06:53:48Z 2012-06-20T16:08:25Z <p>In manifold calculus, there are various analyticity estimates which run into trouble for codimension two embeddings. For instance, the functor $\operatorname{Emb}(M,N)$ is analytic in $M$ if $\dim M \leq \dim N - 3$. Another example is given by the usual multiple disjunction lemma, which gives estimates on connectivity so long as we study the disjunction of manifolds $L_i$ which have dimension $\leq \dim N-3$. </p> <p>At the same time, when I think of codimension 2 embeddings, I think of introducing $\pi_1$ complications. (For instance, think of a 3-manifold, and removing a link. More simply: Remove a point from $\mathbb{R}^2$.) And as a general philosophy of topology, spaces with $\pi_1 \neq 0$ are more difficult to study.</p> <p>This is a somewhat vague question: Are these two complications related in an obvious or philosphical way, deeper than what I've said here? That is, is there a specific sense in which the codimension-two complications of manifold calculus are a manifestation of a general viewpoint, that non-simply-connected spaces are complicated?</p> http://mathoverflow.net/questions/93969/how-to-make-the-category-of-chain-complexes-into-an-infty-category/100143#100143 Answer by Hiro Lee Tanaka for how to make the category of chain complexes into an $\infty$-category Hiro Lee Tanaka 2012-06-20T15:56:46Z 2012-06-20T15:56:46Z <p>As everybody's said, there's an obvious thing to do. As Yosemite Sam cites, it's done in Section 13 of the ArXiv version of DAG I -- you think of chain complexes as enriched over simplicial sets via Dold-Kan, and then apply the nerve construction. </p> <p>But there's an explicit thing you can do for any dg category, and I find it useful because it's given in terms of formulas. Moreover there's an obvious (if tedious) way to generalize this formula for any $A_\infty$-category so it's a cool thing to know. It's in the latest (February 2012) version of Higher Algebra. Since chain complexes obviously form a dg-category, this explicit method might be what you're looking for in case you want to produce some simplices in your quasi-category.</p> <p>Specifically, Construction 1.3.1.6 tells you how to get a quasi-category from any dg category. Then Construction 1.3.13 and Remark 1.3.1.12 should convince you that it's equivalent to the "Dold-Kan + Simplicial nerve" construction cited by everybody else. (Lurie summarizes this equivalence in Proposition 1.3.1.17.) I would write out the formulas here but I don't want to re-TeX the long discussions. So here's at least <a href="http://www.math.harvard.edu/~lurie/papers/HigherAlgebra.pdf" rel="nofollow">a link</a> to the latest Higher Algebra.</p> http://mathoverflow.net/questions/100001/how-to-construct-maps-between-cofibre-sequences-in-a-stable-infty-category/100076#100076 Answer by Hiro Lee Tanaka for How to construct maps between (co)fibre sequences in a stable $\infty$-category? Hiro Lee Tanaka 2012-06-20T05:58:30Z 2012-06-20T05:58:30Z <p>Hi dhagbert,</p> <p>Everything you say seems right. In particular the map $c \to z$ is determined up to contractible choice by a choice of homotopy from the composite $$c[-1] \to a \to x \to y$$ to the zero map. (Depending on how you diagram-chase, you can rephrase this as: "the map $b \to y$ is determined up to contractible choice by a choice of homotopy..." and you can also replace the above composite by another one.) As you seem to intuit, one uses the universal property of both push-outs and pull-backs to see this (hence the necessity of being stable).</p> <p>And yes, I think you point out a richness. In the Verdier axiom (TR3), you only remember that a certain homotopy commutative square exists (without specifying the homotopy) so you know that the vertical map $c \to z$ exists, but nothing more. In the stable oo-category setting, however, we know from the above paragraph that the space of certain homotopies is equivalent to the space of vertical maps. So once you specify the homotopy commutative square, then you specify the vertical map (up to contractible choice).</p> <p>You asked what choices are involved in the extension you're after, and that choice is precisely the homotopy of the above composite to the 0 map. As you and I have both pointed out, there are various equivalent ways to phrase this choice (such as filling in a homotopy from a different composite to the 0 map) but it seems you're familiar enough -- and I'm sleepy enough -- that I won't need to spell out all these equivalent phrasings.</p> http://mathoverflow.net/questions/91634/simplicial-spaces-without-degeneracies/91640#91640 Answer by Hiro Lee Tanaka for simplicial spaces without degeneracies Hiro Lee Tanaka 2012-03-19T17:08:17Z 2012-03-20T18:05:59Z <p>[ I now realize that the OP is asking about functors in the image of the forgetful functor from simplicial spaces to semisimplicial spaces, though I focus my answer mainly on a general semisimplicial space. I'll keep the answer up anyway in case it's helpful.</p> <p>To clarify, by a simplicial blah, I'm as usual talking about a functor $\Delta^{op} \to Blahs$. A semisimplicial blah is given by a functor $\Delta_0^{op} \to Blahs$, where $\Delta_0$ is the non-full subcategory where we discard the non-identity surjections.]</p> <p>The degeneracies are critical when considering products--it is in fact false that the geometric realization commutes with products (up to homotopy equivalence) without the degenerate simplices. A simple example is given by taking the product of $S^1$ with itself---if you take the one-vertex, one-edge semisimplicial structure, you'll never recover the torus via geometric realization. (An obvious problem arises when now your higher simplices $X_n$ can equal the empty set.) Note this is an example in semisimplicial sets, not even spaces. In short, the functor $||\bullet||: semiSSpace \to Spaces$ is not well-behaved with respect to products.</p> <p>However, these "incomplete" simpicial sets/spaces arise really naturally. For instance, a lot of categories may not have a natural unit/identity, so there are no degeneracy maps in the nerve of the category---these are modeled most easily by "semi"simplicial sets/spaces. But so long as you're not taking products, the theory of such things (I think) work out just fine. You can still talk about the classifying space of a non-unital category by taking geometric realization of the corresponding semisimplicial set (i.e., its nerve).</p> <p>As for your second question, I'm fairly certain that what you say is correct--if you have two semisimplicial spaces $Y_\bullet, X_\bullet$ with a natural transformation that induces level-wise equivalences, then the geometric realizations will be (weakly) homotopy equivalent. I think you can see this by taking a filtration by simplicial index and seeing that the associated gradeds are equivalent.</p> <p>Given a semisimplicial space, there are some ways to get an actual simplicial space (this is akin to formally adding units in a category, or to an algebra) but I'm not sure how well-behaved this functor is. </p> <p>However, the composite functor $$SSpace \to semiSSpace \to Spaces$$ (forget, and then realize) will satisfy all the properties you asked for, mainly because the fat geometric realization doesn't differ in homotopy type from the usual geometric realization (provided the simplicial space is good, in the sense of Segal!).</p> <p>As always if anybody has questions or more enlightening comments, please share.</p> <p>Hiro</p> http://mathoverflow.net/questions/86120/reshetikhin-turaev-as-a-3-2-1-theory/86127#86127 Answer by Hiro Lee Tanaka for Reshetikhin-Turaev as a 3-2-1-theory Hiro Lee Tanaka 2012-01-19T18:41:48Z 2012-01-20T21:31:15Z <p>Hi Ulrich,</p> <p>I think that's a good way to think about it. There's also a good reason why we have the associations</p> <blockquote> <p><em>1-manifold &lt;--> Linear Category</em></p> <p><em>2-manifolds F &lt;--> Functors &lt;--> Vector Spaces (when F is closed)</em></p> <p><em>3-manifolds X &lt;--> Natural Transformations &lt;--> Numbers (when X is closed),</em></p> </blockquote> <p>which I can try to answer now. In what follows, I'll let $Z$ be the topological field theory. It's a functor</p> <p>$Z: Cob_1^3 \to LinearCat$</p> <p>between 2-categories, in your language. What the Reshetikhin-Turaev TFT assigns to the circle should be the linear category of fixed-level, positive energy representations of a loop group, or the linear category of (certain) representations of a quantum group. But those details won't matter for these general comments:</p> <p><strong>(Surfaces)</strong> First you should pin down what you mean by a morphism in the category of linear categories. (i.e., What kinds of functors you want to allow.) If you impose, for instance, the condition that all morphisms must preserve finite colimits, then you see that a functor</p> <p>$Z(F): Vect --> Vect$</p> <p>is determined completely by what $Z(F)$ does to the one-dimensional vector space $k$. So $Z(F)(k) \cong V$ for some vector space $V$, and one can simply think of $Z(F)$ as $V$ itself. Vect is the unit in linear categories, so a closed surface (a cobordism from the empty manifold to the empty manifold) is hence assigned a vector space. </p> <p><strong>(3-manifolds)</strong> Now a natural transformation from $Z(F)$ to $Z(F')$ is given by a map of vector spaces. When the three manifold is closed, its boundaries F and F' are both empty manifolds. These are assigned the 1-dimensional vector space $k \cong Z(\emptyset)$, which is the unit in the category of functors from Vect to Vect. (Tensor product of vector spaces gives rise to a monoidal structure on the category of functors from Vect to itself.) Hence a natural transformation corresponds to a a linear transformation </p> <p>$Z(X): k \to k$,</p> <p>and this is just an element of $k$, since it's determined by what it does on the unit $1 \in k$. (This is the analogue of the statement I made above about how a functor $Vect \to Vect$ is given by what it does to the unit $k \in Vect$.)</p> <p><strong>(On 3-2-1-0 and 4-3-2-1 and 4-3-2-1-0)</strong> Chern-Simons is interesting because it might extend up and it might extend down. First, to extend to zero-manifolds, there are some notes by Dan Freed from <a href="http://www.ma.utexas.edu/users/dafr/stringsmath_np.pdf" rel="nofollow">a lecture he gave at UPenn Strings this year</a>. He says that, since Chern-Simons has an anomaly, one can think of Chern-Simons as a 3-2-1-0 TFT with a twist, and he gives a concrete mathematical definition of this notion.</p> <p>Another fun direction is seeing Chern-Simons as coming from a field theory in higher dimensions. The tip of this iceberg is visible through Khovanov homology--in Khovanov, we recover a link invariant whose Euler characteristic recovers the Jones poynomial. Why is this a suggestion of extending "up?" On the one hand, the Jones polynomial comes out of $SU(2)$ Chern-Simons theory, and on the other hand, Khovanov homology assigns morphisms to cobordisms between links, where a 2-manifold inside a 4-manifold is a cobordism between embedded links. I think Witten has written about how this comes from a 5- or 6-dimensional field theory in <a href="http://arxiv.org/abs/1101.3216" rel="nofollow">this pre-print</a>, and also in a lecture at the Simons center: <a href="http://media.scgp.stonybrook.edu/video/video.php?f=20101103_4-Edward_Witten.mp4" rel="nofollow">Video</a>, <a href="http://media.scgp.stonybrook.edu/presentations/20101103_Witten_-_A_New_Look_At_Khovanov_Homology.pdf" rel="nofollow">Slides</a>. Charlie above commented about the Crane-Yetter invariant, but I don't know enough to talk about connections to that.</p> <p>PS I am using Chern-Simons and Resthetikhin-Turaev interchangeably because in my mind they are philosophically the same, but some schools may complain about this.</p> http://mathoverflow.net/questions/84485/on-lifts-in-kan-simplicial-sets/84490#84490 Answer by Hiro Lee Tanaka for On Lifts in Kan Simplicial Sets Hiro Lee Tanaka 2011-12-29T03:42:42Z 2011-12-29T03:42:42Z <p>The answer is "No, it is not unique." As a simple example consider the triangulation of a circle with two vertices and two non-degenerate edges. You can write down the associated Kan complex. If you fix the two vertices you find more than one edge that fills them.</p> <p>A more flimsy example: Just take the singular complex of any topological space X that's not a point. If you take a null-homotopic map of S^n into X, there are in general a whole ton of ways to fill in S^n by a disc. It's an easy exercise to translate this argument into the language of filling in a simplex with specified boundary.</p> http://mathoverflow.net/questions/84381/computations-in-infty-categories/84418#84418 Answer by Hiro Lee Tanaka for Computations in $\infty$-categories Hiro Lee Tanaka 2011-12-27T22:52:30Z 2011-12-27T22:52:30Z <p>What a fun question!</p> <p>I'd like to first mention a speculation on my part: When most people think of $\infty$-categories, the categories they think of probably exist in another formulation. For instance, the category of spaces (or Kan complexes), of chain complexes, of commutative DGAs, of spectra, et cetera, all have well-developed theory. And we're used to doing computations in these categories based on approaches that predate quasi-categories or Lurie's HTT or Higher Algebra. For such well-studied categories, I think you'll almost always find some pre-quasi-category (e.g., model categorical) "computation" that'll get the job done. There's no need to think of them as weak Kan complexes first.</p> <p>(To compute a homotopy fiber of a map, for instance, we'll probably just replace the map by a standard fibration and compute its fiber thereafter. This is a silly example, but one in which I think working with a model-categorical framework is easier. This is much faster than proving that some over-category has a terminal object.)</p> <p>But it would be a lie to say that most "computations" one needs to perform in an $\infty$-category can already be done in a non-quasi-category world. Here are two examples:</p> <p>(1) <strong>If you're working with a new $\infty$-category for which someone hasn't done the model-categorical prep work for you.</strong> Depending completely on your math-path, you might come across (or define) a category which is most naturally defined as a weak Kan complex. One degree of separation away, you might define a category which is most naturally enriched over Kan complexes, and apply the simplicial nerve construction to obtain a quasi-category. But if you're just a tramp like me, suddenly face-to-face with a new category with some homotopical flavor, you may not have a natural candidate for a model structure, nor have any intuition for how to prove that something really is a "homotopy fiber" for some map, or more generally a homotopy (co)limit for a homotopy coherent diagram. (Whatever a "homotopy (co)limit" means for your category.) </p> <p>With such a new category, all you a priori have is whatever led you to define this combinatorial gadget (a simplicially enriched category, or a quasi-category) and maybe some interpretations of your morphisms depending on what motivated your definition. So when trying to prove that some object is the homotopy fiber for some morphism in your quasi-category, it might be easiest for you to simply prove that an over-category has a terminal vertex. </p> <p>I should admit that, while I imagine that examples like this will come up more and more, the only example I have in mind comes from joint work with David Nadler, where we compute kernels in a category we define, which happens to be most naturally a quasi-category.</p> <p>(By the way, if someone has techniques that make it very easy to prove something is a limit of a diagram in a fibrant simplicial category, please post it as a comment to this post! I'd love to know more techniques.)</p> <p>(2) <strong>If the diagrams you're working with are homotopy coherent but hard to make sense of at a level of strict commutativity.</strong> Another example I've come across is to prove that two homotopy coherent diagrams have the same homotopy colimit. Again because there is a concrete model for over/under categories in the $\infty$-categorical model, I could write down an equivalence between the over-categories associated to the two diagrams.</p> <p>Summary:</p> <p>If the non-$\infty$-categorical computations are made possible by model-categorical ideas (like knowing how to replace morphisms), then $\infty$-categorical computations are possible because you reduce your computations to simplicial-set ideas, often proving that a simplicial set is contractible or that it has a terminal vertex. In some mathematical universe both paths may amount to the same thing, but it's my impression that the former approach is hard to follow if you're working in a new category without model structure, and is also not as easy to work with when the higher homotopy coherences of your diagrams are subtle. (Please feel free to let me know if I'm mistaken on this point, I'd love to hear more views.) The quasi-category framework allows you to avoid much of that difficulty by passing the buck to the geometry of simplicial sets.</p> <p>By the way, you also mentioned something about computing spaces of $k$-morphisms. I have no idea how to do such a thing in the $\infty$-category world, or in either world, really. (Please educate me if someone does know how.)</p> http://mathoverflow.net/questions/79542/limits-in-an-infty-1-category/79571#79571 Answer by Hiro Lee Tanaka for Limits in an $(\infty,1)$-category Hiro Lee Tanaka 2011-10-31T01:18:20Z 2011-10-31T10:24:53Z <p><strong>Definitions</strong></p> <p>I think there is a definition that should fit into most models of $(\infty,1)$-categories. If you want an "elevator speech" answer, it's:</p> <p><strong>Definition.</strong> <em>A limit of a diagram $\mathcal D \to \mathcal C$ is a terminal object in the $(\infty,1)$-category of objects living over $\mathcal D$.</em></p> <p>This is the (almost naive) generalization of one definition for limits in usual category theory. But let me elaborate on this definition.</p> <p>(Full disclosure, I almost always work with quasicategories, so I anticipate that a better answer can be given by someone who's worked with all models. I hope this answer will be helpful regardless. Also, when you say "limit," I assume you mean homotopy limit. I sometimes distinguish between the two since not everybody is happy when I use classical terminology with an implicit "$\infty$" or "homotopy" before every word.)</p> <p>Morally speaking, a good model for "$(\infty,1)$-categories" should have definitions for the following ideas:</p> <ol> <li><strong>Mapping spaces.</strong> That is, given an $(\infty,1)$-category $\mathcal C$, between any two objects $X,Y$ of $\mathcal C$, a topological space of morphisms ${\mathcal C}(X,Y)$. This is a fairly obvious pre-requisite because $(\infty,1)$-categories are supposed to be like categories enriched in spaces.</li> <li><strong>Terminal objects.</strong> Morally, these are objects $\ast$ such that for any other object $Y$ in your $(\infty,1)$-category $\mathcal C$, the mapping space ${\mathcal C}(Y,\ast)$ is contractible. There may be more subtle issues involved in defining terminal objects properly, depending on your model, but at least in the case of quasi-categories, it turns out this moral definition is perfectly fine as an actual one. (See Corollary 1.2.12.5 of HTT.)</li> <li><strong>Under/Over-Categories, aka Cone Categories.</strong> Given two $(\infty,1)$-categories $\mathcal C$ and $\mathcal D$, an $(\infty,1)$-category of ( $(\infty,1)$-) functors between them. And in our discussion, we specifically want the following: Given a diagram ${\mathcal D} \to \mathcal C$, a good notion of an ( $(\infty,1)$-)category whose objects are functors from $\ast \star {\mathcal D}$ to $\mathcal C$, where $\ast \star {\mathcal D}$ is the category obtained by affixing an initial object to $\mathcal D$. This is the same thing as the category of objects of $\mathcal C$ equipped with a map to the diagram $\mathcal D \to \mathcal C$.</li> </ol> <p>You can see why this third point, about cone categories, is so simple in the quasi-category model. It is as simple as defining the join of simplicial sets, and knowing what the mapping space is between simplicial sets.</p> <p>Anyhow, if you believe that your model (whatever it is) has definitions for the above three things, you can define a limit to be a terminal object in a cone category. You can dually define colimits as initial objects in an undercategory.</p> <p><strong>Actually proving that (co)limits are preserved.</strong></p> <p>I assume you wanted an answer that was more specific about actually computing (homotopy) limits using different models (complete Segal spaces, quasi-categories, Kan simplicial categories, et cetera) but I'm afraid I don't know much about comparing homotopy limit computations in different models. Lurie does, however, prove in HTT (Theorem 4.2.4.1) that the usual homotopy (co)limits you'd compute in a category enriched over Kan complexes will agree with the homotopy (co)limits you'd compute in the quasi-category model. So that's a good start! And if you believe in the equivalences between different models of $(\infty,1)$-categories (see for instance Julie Bergner's <a href="http://arxiv.org/abs/math/0610239" rel="nofollow">"A Survey of $(\infty,1)$-Categories"</a>) then the equivalences should preserve initial objects of cone categories, so this would be an argument that all models preserve (homotopy) (co)limits.</p> <p><strong>Why "everybody" takes simplicial sets.</strong></p> <p>Actually, a lot of people prefer to use other models like the Segal space model. But you can see that with the combinatorics of quasi-categories, a lot of things can be defined and proved fairly cleanly, as I pointed out in some of my commentary above. So that's one advantage of Joyal's quasi-category model. But there are many situations in which the <em>space</em> of objects is so naturally a space that you might prefer a model which isn't based on weak Kan complexes. For instance, in <a href="http://arxiv.org/abs/math/0605249" rel="nofollow">Galatius-Madsen-Tillman-Weiss</a>, they think of the category of cobordisms as a category with a space of objects and a space of morphisms. This model might make it easier, for instance, to compute the classifying space of an $(\infty,1)$-category. And if you were interested in computing a (co)limit of a functor mapping such an $(\infty,1)$-category into another, you wouldn't want to say that your diagram comes from a simplicial set.</p> <p><strong>Simplicial Sets as the Diagram</strong></p> <p>Also, it seems you're interested in why Lurie takes as the diagram a map $\mathcal{D} \to \mathcal C$ in which $\mathcal D$ is a <em>simplicial set</em>. I don't think I would take "simplicial set" as the important idea here; I think it's more that $\mathcal D$ is an $(\infty,1)$-category, so for Lurie, it's a weak Kan complex, and in particular a simplicial set. I'm not sure (as Moosbrugger alludes to above) why the generality of an arbitrary simplicial set is so important, but in general I think we would take a diagram $\mathcal D \to \mathcal C$ to be <em>a map of $(\infty,1)$-categories</em>. That is, $\mathcal D$ being a simplicial set isn't so important for this definition. It just needs to be an $(\infty,1)$-category in whatever model you're using, and in the Lurie example, you should probably think of it as a quasi-category, rather than just an arbitrary simplicial set. And you can always replace an arbitrary simplicial set with a weak Kan complex (these are the fibrant-cofibrant objects in the Joyal model category.)</p> http://mathoverflow.net/questions/78939/two-questions-on-rational-homotopy-theory/79296#79296 Answer by Hiro Lee Tanaka for Two questions on rational homotopy theory Hiro Lee Tanaka 2011-10-27T18:32:06Z 2011-10-27T22:11:04Z <p>I'm not sure if this will still be helpful, but here is my understanding of the Quillen model. I'm a little more comfortable with the Sullivan approach, which replaces a space $X$ with a commutative DGA over $\mathbb{Q}$. So my understanding of the Quillen model might be a bit off (if so, someone please correct me!). Also, everything correct that I write below, I learned from John Francis. (Probably in the same lecture that Theo mentioned in his comment above.) Oh, but any mistakes are probably not his fault---more likely an error in my understanding.</p> <p><strong>Before we begin: Quillen v Sullivan.</strong></p> <p>As others have mentioned, Quillen gets you a DG Lie algebra, where as the Sullivan model will get you a commutative DG algebra. As you write, the passage from one to the other is (almost) Koszul duality. Really, a Lie algebra will get you a co-commutative coalgebra by Koszul duality, and a commutative algebra will get you a coLie algebra. You can bridge the world of coalgebras and algebras when you have some finiteness conditions--for instance, if the rational homotopy groups are finite-dimensional in each degree. Then I think you can simply take linear duals to get from coalgebras to algebras.</p> <p><strong>A way to find Lie algebras.</strong></p> <p>So where do (DG) Lie algebras come from? First let me point out that there is a natural place that one finds Lie algebras, before knowing about the Quillen model: Lie algebras arise as the tangent space (at the identity) of a Lie group $G$. </p> <p>Now, if you're an algebraist, you might claim another origin of Lie algebras: If you have any kind of Hopf algebra, you can look at the primitives of the Hopf algebra. These always form a Lie algebra. </p> <p>(Recall that a Hopf algebra has a coproduct $\Delta: H \to H \otimes H$, and a primitive of $H$ is defined to be an element $x$ such that $\Delta(x) = 1 \otimes x + x \otimes 1.$)</p> <p>One link between the algebraist's fountain of Lie algebras, and the geometer's, is that many Hopf algebras arise as functions on finite groups. If you are well-versed in algebra, one natural place to find Lie algebras, then, would be to take a finite group, take functions on that group, then take primitives. </p> <p>A cooler link arises when a geometer looks at <em>distributions</em> near the identity of $G$ (which are dual to 'functions on $G$') rather than functions themselves. This isn't so obviously the right thing to look at in the finite groups example, but if you believe that functions on a Lie group $G$ are like de Rham forms on $G$, then you'd believe that something like 'the duals to functions on $G$' (which are closer to vector fields) would somehow safeguard the Lie algebra structure. The point being, you should expect to find Lie structures to arise from things that look like 'duals to functions on a group'. So one should take 'distributions' to be the Hopf algebra in question, and look at its primitives to find the Lie algebra of 'vector fields.'</p> <p><strong>A (fantastical) summary of the Quillen model.</strong></p> <p>Let us assume for a moment that your space $X$ happens to equal $BG$ for some Lie group, and you want to make a Lie algebra out of it. Then, by the above, what you could do is take $\Omega X = \Omega B G = G$, then look at the primitives of the Hopf algebra known as distributions on $\Omega X$'.</p> <p>Now, instead of considering just Lie groups, let's believe in a fantasy world (later made reality) in which all the heuristics I outlined for a Lie group $G$ will also work for a based loop space $\Omega Y$. A loop space is like a group' because it has a space of multiplications, all invertible (up to homotopy). Moreover, any space $X$ is the $B$ (classifying space) of a loop space--namely, $X \cong B \Omega X$. So this will give us a way to associate a Lie algebra to any space, if you believe in the fantasy.</p> <p>Blindly following the analogy, functions on $\Omega X$' is like cochains on $\Omega X$, and the dual to this (i.e., distributions) is now <em>chains</em> on $\Omega X$. That is, $C_\bullet \Omega X$ should have the structure of what looks like a Hopf algebra. And its primitives should be the Lie algebra you're looking for.</p> <p><strong>What Quillen Does.</strong></p> <p>So if that's the story, what else is there? Of course, there is the fantasy, which I have to explain. Loop spaces are most definitely not Lie groups. Their products have $A_\infty$ structure, and correspondigly, we should be talking about things like <em>homotopy</em> Hopf algebras, not Hopf algebras on the nose. What Quillen does is not to take care of all the coherence issues, but to change the models of the objects he's working with.</p> <p>For instance, one can get an actual simplicial <em>group</em> out of a space $X$ by Kan's construction $G$. This is a model for the loop space $\Omega X$, and I think this is what Quillen looks at instead of looking only at $\Omega X$, which is too flimsy. From this, taking group algebras over $\mathbb{Q}$ and completing (these are the simplicial chains, i.e., distributions), he obtains completed simplicial Hopf algebras. Again, instead of trying to make my fantasy precise in a world where one has to deal with higher algebraic structures (homotopy up to homotopy, et cetera) he uses this nice simplicial model. To complete the story, he takes level-wise primitives, obtaining DG Lie algebras.</p> <p><strong>Edit:</strong> This is from Tom's comment below. To recover a $k$-connected group or a $k$-connected Lie algebra from the associated $k$-connected complete Hopf algebra, you need $k \geq 0$. And $k$-connected groups correspond to $k+1$-connected spaces. This is why you need simply connected spaces in the equivalence.</p> <p>I'm not sure I gave any 'high concept' as to 'why Quillen's construction works', but this is at least a road map I can remember. </p> http://mathoverflow.net/questions/79004/homotopy-groups-of-spheres-in-a-infty-1-topos/79091#79091 Answer by Hiro Lee Tanaka for Homotopy groups of spheres in a $(\infty, 1)$-topos Hiro Lee Tanaka 2011-10-25T15:54:59Z 2011-10-25T16:11:28Z <p>Because Charles has already answered (incredibly nicely!) your other questions, I'll just answer your question about a natural group structure. My answer is really just an elaboration on Sam's comment.</p> <p>As Sam pointed out, there is in fact an $E_k$ structure on the space of all maps ${ * \to \Omega^k S^n}$; I'll illustrate this in a moment. But then by the usual Eckman-Hilton argument (or drawing pictures), $\pi_0$ of this space will have the structure of a group for $k\geq 1$, and of a commutative group for $k \geq 2$. The fact that we're taking $S^n$ is not so important here, it's true for any object $X$. </p> <p>In what follows, I'll let $H(A,B)$ denote the space of morphisms from $A$ to $B$ in the $\infty$-category $H$. In particular, when $A=\ast$ we get the space of global elements of $B$.</p> <p>By the universal property of pullbacks, a map $\ast \to \Omega X$ is the same as a homotopy coherent map from $\ast$ to the diagram $D := \ast \to X \leftarrow \ast$. Without loss of generality we assume that the two maps $\ast \to X$ in $D$ are the same map. We choose this to be the base point in the space of maps $H(\ast,X)$.</p> <p>Then a map from $\ast$ to the diagram $D$ is precisely a loop in the Hom-space $H(\ast,X)$. (This is the only key observation--it follows easily from the definition of the Hom Kan complex in an $\infty$-category, if you like.) In other words, $$H(\ast,\Omega X) \cong \Omega H(\ast,X)$$ where $\Omega$ in the right hand side actually means based loop space, in the usual sense of topology. By induction, the space of global elements of $\Omega^k X$ has the structure of a $k$-fold loop space. And we're finished.</p> http://mathoverflow.net/questions/42109/semicosimplicial-totalization/76130#76130 Answer by Hiro Lee Tanaka for Semicosimplicial totalization Hiro Lee Tanaka 2011-09-22T15:10:51Z 2011-09-22T15:10:51Z <p>I agree this should be standard, but I've only seen the proof in one place. See Lemma 6.5.3.7 of Lurie's Higher Topos Theory.</p> http://mathoverflow.net/questions/41285/how-does-one-interpret-the-naive-t-structure-on-constructible-sheaves-as-a-t-stru How does one interpret the naive t-structure on constructible sheaves as a t-structure on D-modules? Hiro Lee Tanaka 2010-10-06T15:54:19Z 2010-10-06T21:02:14Z <p>By the Riemann-Hilbert correspondence, there is an equivalence between </p> <p>(1) $\mathcal{D}\operatorname{-mod}(X)$ , the (derived) category of holonomic D-modules on a complex variety X, and </p> <p>(2) $D^b_c(X)$ , the (derived) category of constructible sheaves on X. </p> <p>There is a "naive" t-structure we can put on both categories. In $\mathcal{D}\operatorname{-mod}(X)$ , we can look at a t-structure whose heart $\mathcal{D}\operatorname{-mod}^\heartsuit$ is a complex (of D-modules) concentrated in degree 0. In $D^b_c(X)$ , we can look at the naive t-structure whose heart $D^{b \heartsuit}_c$ is a complex (of constructible sheaves) concentrated in degree 0. </p> <p>It's known that if we transfer the naive t-structure on $\mathcal{D}\operatorname{-mod}(X)$ to $D^b_c(X)$ (using the equivalence above), $\mathcal{D}\operatorname{-mod}^\heartsuit$ is identified with "perverse sheaves" on X.</p> <p>My question is: </p> <blockquote> <p>If we map $D^{b\heartsuit}_c$ to the category of D-modules using the Riemann-Hilbert correspondence, what subcategory of $\mathcal{D}\operatorname{-mod}$ do we get? Does this have a well-known name?</p> <p>More generally, is there some geometric/nice description of what the naive t-structure on $D^b_c$ becomes on $\mathcal{D}{\operatorname{-mod}}$ ?</p> </blockquote> http://mathoverflow.net/questions/4796/braided-monoidal-2-categories-with-duals/130580#130580 Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2013-05-14T23:09:48Z 2013-05-14T23:09:48Z I think you mean it generalizes Hochschild homology, not cohomology. Also, the statement about $E_n$-algebras is true if you want to construct a TFT in the sense <i>not</i> of the tangle hypothesis. That is, $E_n$ algebras give n-categories with duals, but not braided monoidal n-categories with duals. To get the structure the OP wants, you'd rather consider a pair $(A,M)$ where $A$ is an $E_4$-algebra and $M$ is an $E_2$-algebra with an action of $A$. The associated TFT constructed from top. chi. homology (aka factorization homology) will then evaluate on 2,3,4-mflds with codim 2 embedded objects. http://mathoverflow.net/questions/130585/homology-of-the-dg-nerve-vs-hochschild-homology-of-the-dg-category Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2013-05-14T22:29:10Z 2013-05-14T22:29:10Z @Chris, I think Yasha asks about $HH_\ast(C)$, not $HH^\ast$. Yasha, I see more naturally a relationship between $HH_\ast$ and the homotopy groups of the (free loop space of the) geometric realization; if you want to relate it to homology, you probably want some Hurewicz type assumption on the nerve of the category. Dually, the result Chris cites is fairly intuitive--$HH^\ast$ is endomorphisms of the identity of $C$, so you can see it at the $\pi_2$ level of the (largest) oo-groupoid contained in the oo-category of dg-categories, with basepoint $C$. http://mathoverflow.net/questions/130411/question-on-hartogss-extension-theorem Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2013-05-12T14:21:55Z 2013-05-12T14:21:55Z The argument that holomorphic = analytic is standard--you use Cauchy's Integral formula one variable at a time. The only difference between the one- and multi-variable cases is that you have more complicated polynomials showing up. (Products of the polynomials you see in the 1-D case.) A standard reference is Griffiths and Harris. http://mathoverflow.net/questions/70879/category-with-a-metric-for-arrow-composition Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2013-05-05T23:13:27Z 2013-05-05T23:13:27Z In Goodwillie calculus there's a heuristic notion of distance one uses fairly often: you can define it as something (more or less) inversely proportional to the connectnedness of a map $f: X \to Y$. So if $f$ is a weak homotopy equivalence, it's distance zero, for instance. It can also be used to talk about the radius of convergence of functors, but perhaps the metric in this example is too discrete for your interests. (And by the way, you can do the same thing for non-negatively graded chain complexes, or for connective spectra, more generally.) http://mathoverflow.net/questions/124816/alexander-duality-theorem Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2013-03-20T01:13:10Z 2013-03-20T01:13:10Z The definition of orientability is another matter--in cobordism theory for orientable manifolds, the two possible conventions are both strange: Either the empty manifold has a unique orientation, (any orientable manifold should have two!) or the empty manifold can be formally given two different orientations whilst the empty cobordism between them realizes an isomorphism. (When is an orientable manifold invertibly cobordant to its opposite?) http://mathoverflow.net/questions/117886/recovering-torsion-in-singular-homology-from-cplx-of-singular-chains Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2013-01-03T06:02:21Z 2013-01-03T06:02:21Z LMN, do you mean, for instance, that the cup product on cochains contains more information than its cohomology groups? Otherwise it's not true that the cochain complex has more information than the cohomology groups (as others have said). http://mathoverflow.net/questions/116476/where-does-the-notion-of-pseudoholomorphic-curve-come-from/116480#116480 Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2012-12-20T15:15:11Z 2012-12-20T15:15:11Z What a wonderful interview. Tangential to our discussion: &quot;Question: Is there one particular theorem or result you are the most proud of? Answer: Yes. It is my introduction of pseudo-holomorphic curves, unquestionably. Everything else was just understanding what was already known and to make it look like a new kind of discovery.&quot; http://mathoverflow.net/questions/116490/why-do-we-use-the-diagonal-for-diagonal-approximations Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2012-12-16T01:03:50Z 2012-12-16T01:03:50Z Nice observations, TJ. (1) The diagonal map is a gift. It makes any set (any space) into a coalgebra. So any symm. monoidal contravariant functor from sets (spaces) gives you an algebra for every set (space). This is why, morally, cohomology of a space has the structure of an algebra. So when you have a functor like &quot;cochains on a group,&quot; it's natural to examine the diagonal, motivated (for me) by this standard fact from topology, to yield an algebraic structure like the cup product. http://mathoverflow.net/questions/115508/a-infty-categories-and-their-equivalent-dg-categories-the-case-of-mathcalr Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2012-12-06T19:31:22Z 2012-12-06T19:31:22Z This doesn't stay solely in the dg world, but the proof I know of &quot;every Aoo category is equivalent to a dg category&quot; is more conceptual--it's the Yoneda Lemma. Consider the category of all contravariant Aoo functors from $\mathcal{A}$ to $Ch$, the category of chain complexes. The category of Aoo functors would normally be an Aoo category, but because $Ch$ is a dg category, this functor category is actually a dg category. A version of the usual Yoneda argument shows that $\mathcal{A}$ is in fact equivalent as an Aoo category to its image inside this dg category. http://mathoverflow.net/questions/115567/covering-maps-in-real-life-that-can-be-demonstrated-to-students/115576#115576 Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2012-12-06T16:00:37Z 2012-12-06T16:00:37Z Mark, I think Guntram's covering map wasn't to the folded sheet of paper, but to a single doll holding its own hand. But I agree--otherwise this is not at all an example of a covering map: It wouldn't even be a local homeomorphism. (Showing such an example to students could really hurt their understanding.) http://mathoverflow.net/questions/107857/is-there-a-universal-property-that-characterises-the-join-of-two-categories Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2012-09-22T19:51:13Z 2012-09-22T19:51:13Z Could you give a citation as to where this is called the join? I'd like to know because this is not what I've been calling the join of two categories--I hope I'm not in trouble! http://mathoverflow.net/questions/100990/homology-and-homotopy-type-for-knot-complements Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2012-06-30T12:32:54Z 2012-06-30T12:32:54Z The statement as written is clearly false -- the complement of the trefoil is not homotopy equivalent to S^1, as its fundamental group is not Z. If one knows (in addition to what you've written) that $\pi_1(E) = \mathbb Z$, we can conclude $E \simeq S^1$; otherwise it seems there's something missing. http://mathoverflow.net/questions/5353/how-to-respond-to-i-was-never-much-good-at-maths-at-school/5664#5664 Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2012-06-20T16:55:53Z 2012-06-20T16:55:53Z Why do you assume it's a she? http://mathoverflow.net/questions/100001/how-to-construct-maps-between-cofibre-sequences-in-a-stable-infty-category/100076#100076 Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2012-06-20T14:07:13Z 2012-06-20T14:07:13Z A great place to start is probably David Ben-Zvi's answer to this post: <a href="http://mathoverflow.net/questions/815/triangulated-vs-dg-a-infinity" rel="nofollow" title="triangulated vs dg a infinity">mathoverflow.net/questions/815/&hellip;</a>. It doesn't go into the details of (TR3) and its deficiencies, but it definitely highlights why you want to remember higher coherence data (like which homotopy commutative square you take in your map of fiber sequences), which triangulated categories don't do. http://mathoverflow.net/questions/100056/galois-groups-and-braid-groups Comment by Hiro Lee Tanaka Hiro Lee Tanaka 2012-06-20T00:24:18Z 2012-06-20T00:24:18Z It's not clear to me what you're asking exactly, but there are well-known analogies between the knot group (the fundamental group of a knot complement) and certain Galois groups. See for instance <a href="http://arxiv.org/abs/0904.3399" rel="nofollow">arxiv.org/abs/0904.3399</a>. I'm not aware of Galois-theoretic analogues of the braid group though.