User theo johnson-freyd - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T07:22:27Z http://mathoverflow.net/feeds/user/78 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/131066/in-cell-decomposed-manifolds-how-easy-is-it-to-arrange-for-the-tubular-neighborh In cell-decomposed manifolds, how easy is it to arrange for the tubular neighborhood of a diagonal to contract onto the diagonal? Theo Johnson-Freyd 2013-05-18T17:37:00Z 2013-05-18T17:37:00Z <p>Suppose that you have decomposed a manifold $M$ into cells (I care most, if it matters, about compact oriented smooth manifolds; but if my question can be solved in the PL category, all the better). By this, I have in mind some sort of combinatorial description, so there probably should be various words like "regular" or whatnot. For various reasons, I cannot restrict my attention to simplicial decompositions &mdash; I must allow cubes, for example. In any case, if you have done so, then on the set of cells in $M$ there is a metric, generated by declaring that $\operatorname{distance}(a,b) = 1$ if $a$ is a facet of $b$ (i.e. if the cell $a$ lies in the closure of the cell $b$). Then given any collection $Y$ of cells and any $\ell \in \mathbb N = \lbrace 0,1,2,\dots\rbrace$, I can define the set $\mathrm{B}_\ell(Y)$ to consist of the closure of the union of all cells at distance at most $\ell$ from some element of $Y$. It is a sub-cell-complex of $M$. Suppose that the cell decomposition is very fine compared to the topologies of $Y$ and $M$: then one should expect that $\mathrm{B}_\ell(Y)$ contracts onto the closure of $Y$. Note that the closure of $Y$ is precisely $\mathrm{B}_0(Y)$.</p> <p>My specific situation is as follows. I have a manifold (compact, oriented, etc., if it matters) $X$. If I choose a cell decomposition of $X$, then I can induce a cell decomposition of $M = X^n$ by declaring that a cell of $M$ is an $n$-tuple of cells in $X$. (Certainly, if my cells were simplices with ordered vertices, then I could make other choices, but for my application this is most natural.) The diagonal map $X \hookrightarrow X^n$ is not a map of cell complexes, but still for each cell in $X$, there is a corresponding diagonal cell of $M$, and I will define $Y$ to be this "diagonal" copy of $X$.</p> <p>My question is the following:</p> <blockquote> <p>For fixed $\ell$, but letting $n$ vary, how can one find a cell decomposition of $X$ such that, with the notation above, $\mathrm B_\ell(Y)$ has the homotopy type of $X$? Or at least the same rational homology?</p> </blockquote> <p>Almost surely, the $\ell$-fold barycentric subdivision of any cell decomposition of $X$ will do the trick &mdash; probably the $\lceil \log_2\ell \rceil$-fold barycentric subdivision would work &mdash; but I find myself unable to prove this, even after talking to various friends who know more topology than I do. Or perhaps I'm supposed to find a Riemannian metric for which I would have the appropriate result, and then choose a cell decomposition in which all vertices are at distance roughly $1$ from each other. Or something. In any case, I know that my intuition for high-dimensional manifolds is poor.</p> <p>I do know how to prove that after one barycentric subdivision, $\mathrm{B}_0(Y)$ has the rational homology of $X$.</p> http://mathoverflow.net/questions/130890/can-distinct-open-knots-correspond-to-the-same-closed-knot/130902#130902 Answer by Theo Johnson-Freyd for Can distinct open knots correspond to the same closed knot? Theo Johnson-Freyd 2013-05-17T03:27:54Z 2013-05-17T03:27:54Z <p>I will ignore many important questions about wildness and the like, and instead suppose that you have enough regularity to make work the following argument (which I learned from John H. Conway).</p> <p>Take a closed knot. Place a small bead somewhere along it, and make the bead out of a very shiny material: silver, say. Clearly, moving the bead along the knot doesn't make a difference up to isotopy of knots. Now, the point of making the bead very shiny is that if you look at it, what you see is the mirror image of the knot, now drawn inside the sphere of the bead. (Implicitly I have compactified $\mathbb R^3$ to $S^3$ by adding a point at infinity, which the mirror maps to the center of the bead. Which is to say, what I'm really doing is geometric inversion through the surface of the bead.)</p> <p>Then inside the bead you see what you've called an "open" knot, and is also called a "long" knot. Any isotopy of the original knot can be performed so that the strings never have to pass through the very small bead, and so these reflect through to isotopies of the open knot. It follows that the isotopy class of the open knot is the same as the isotopy class of its closure.</p> <p>As an aside: it is a theorem of Dehn's that the left and right trefoils are not equivalent.</p> http://mathoverflow.net/questions/129758/when-does-an-even-dimensional-manifold-fiber-over-an-odd-dimensional-manifold When does an even-dimensional manifold fiber over an odd-dimensional manifold? Theo Johnson-Freyd 2013-05-05T19:01:08Z 2013-05-06T17:51:43Z <blockquote> <p>Are there simple necessary and sufficient conditions for an (oriented) even-dimensional compact smooth manifold to fiber over an (oriented) odd-dimensional manifold (with oriented fibers)?</p> </blockquote> <p>For example, if $M \to N$ is a fiber bundle of compact manifolds with fiber $F$, then their Euler characteristics satisfy $\chi(M) = \chi(N)\chi(F)$. But if $N$ and $F$ are odd-dimensional, $\chi(N) = \chi(F) = 0$, and so $\chi(M)$ must also vanish. Is the converse true? I.e. if $M$ has vanishing Euler class, does it fiber over an odd-dimensional manifold? Or perhaps there's a hint that $\chi(M) = 0^2$, not just $0$, and so maybe some combination of Massey products also must vanish?</p> http://mathoverflow.net/questions/129752/does-a-connected-manifold-with-vanishing-euler-characteristic-admit-a-nowhere-van Does a connected manifold with vanishing Euler characteristic admit a nowhere-vanishing vector field? Theo Johnson-Freyd 2013-05-05T18:32:40Z 2013-05-05T18:53:52Z <p>A version of the "hairy ball" theorem, due probably to Chern, says that the Euler-characteristic of a closed (i.e. compact without boundary) manifold $M$ can be computed as follows. Choose any vector field $\vec v \in \Gamma(\mathrm T M)$. If $p$ is a zero of $\vec v$, then the matrix of first derivatives at $p$ makes sense as a linear map $\partial\vec v|_p : \mathrm T_p M \to \mathrm T_p M$. By perturbing $\vec v$ slightly, assume that at every zero, $\partial\vec v|_p$ is invertible. Then $\chi(M) = \sum_{\vec v(p)=0} \operatorname{sign}\bigl( \det \bigl(\partial\vec v|_p\bigr)\bigr)$.</p> <p>I am curious about the following potential converse: "If $M$ is closed and connected and $\chi(M) = 0$, then $M$ admits a nowhere-vanishing vector field."</p> <p>Surely the above claim is false, or else I would have learned it by now, but I am not sufficiently creative to find a counterexample. Moreover, I can easily see an outline of a proof in the affirmative, which I will post as an "answer" below, in the hopes that an error can be pointed out. Thus my question:</p> <blockquote> <p>What is an example of a compact, connected, boundary-free manifold with vanishing Euler characterstic that does not admit a nowhere-vanishing vector field? (Or does no such example exist?)</p> </blockquote> http://mathoverflow.net/questions/129752/does-a-connected-manifold-with-vanishing-euler-characteristic-admit-a-nowhere-van/129756#129756 Answer by Theo Johnson-Freyd for Does a connected manifold with vanishing Euler characteristic admit a nowhere-vanishing vector field? Theo Johnson-Freyd 2013-05-05T18:52:54Z 2013-05-05T18:52:54Z <p>Here is an outline of a proof that a compact manifold $M$ with vanishing Euler characteristic has a nonvanishing vector field. I'll post it as an "answer" to provide a convenient place for comments where errors might be pointed out (or, you know, a reference given that "that's the proof that ABC used in their paper LMNOP").</p> <p>Call a vector field $\vec v$ with only regular zeros (i.e. at every $p$ with $\vec v|_p = 0$, the linear map $\partial \vec v|_p : \mathrm T_p M \to \mathrm T_p M$ is invertible) <i>divergence-free</i> if for every zero $p$, $\partial \vec v|_p$ has only real eigenvalues. A zero $p$ of a divergence-free vector field $\vec v$ has a <i>Morse index</i> $\mu(p) = \#\lbrace$negative eigenvalues of $\partial\vec v|_p\rbrace$. Of course, $\operatorname{sign}(\det(\partial \vec v|_p)) = (-1)^{\mu(p)}$. By choosing a Morse function and a metric, our manifold $M$ certainly has a divergence-free vector field.</p> <p>Pick some vector field $\vec v$ on $M$, and some little neighborhood without any zeros. I claim I can modify $\vec v$ by some vector field with compact support in that neighborhood to introduce two new zeros, with Morse indexes $\mu$ and $\mu+1$, for any $\mu = 0,\dots,\dim M - 1$. In one dimension, which is trivial: in a neighborhood, $\vec v = \nabla(x^3+x)$ for some coordinate $x$, and I can perturb this to $x^3 - x$. In higher dimensions, it is not much harder, and I could probably write out formulas is necessary.</p> <p>Conversely, and here's the crux of the argument, where I'm not sure it's correct: Suppose I have a divergence-free vector field $\vec v$, with nearby zeros at consecutive Morse index. Then I think I can cancel them, by undoing the insertion step in the previous paragraph.</p> <p>If so, then choose any divergence-free vector field $\vec v$ on $M$. Since $M$ by assumption has vanishing Euler characteristic, $\vec v$ has the same number of zeros with odd Morse index as with even Morse index. If $\vec v$ has no zeros, we're done; otherwise, choose two with Euler characteristics $\mu$ and $\mu+k$, for $k\geq 1$ odd. By assumption, $M$ is connected; choose a simply path between the two zeros, and a small neighborhood thereof. Now, along this path, insert in pairs zeros with Morse index $\mu +1$, $\mu + 2$, ... $\mu + (k-1)$. Now cancel the zeros in pairs but this time cancel the zero with Morse index $\mu$ with the new one with index $\mu + 1$, and so on. After all this, you end up with a vector field with two fewer zeros than you started with.</p> <p>Then the result follows by induction, provided the crux in the 4th paragraph is correct.</p> http://mathoverflow.net/questions/127779/higher-homotopy-groups-and-representation-theory/127911#127911 Answer by Theo Johnson-Freyd for HIgher Homotopy Groups and Representation Theory Theo Johnson-Freyd 2013-04-18T04:00:40Z 2013-04-18T04:00:40Z <p>When $n = 2,3$:</p> <p>For any simply-connected <em>finite-dimensional</em> Lie group $G$, $\pi_2(G) = 1$ is trivial. So it can have no applications.</p> <p>If $G$ is simply-connected, then $\pi_3(G) = \mathrm{H}_3(G,\mathbb Z)$ by Hurewicz's theorem. This is a copy of $\mathbb Z$ whenever $G$ is simple. Classes here can be used to construct central extensions of the loop group of $G$.</p> <p>I do not know of applications of higher $\pi_n$.</p> http://mathoverflow.net/questions/81775/whats-an-example-of-a-commutative-algebra-over-mathbb-q-that-fails-to-satisfy What's an example of a commutative algebra over $\mathbb Q$ that fails to satisfy this version of the "PBW theorem" Theo Johnson-Freyd 2011-11-24T03:32:36Z 2013-04-03T21:51:58Z <p>In a <a href="http://mathoverflow.net/questions/81446/for-which-algebras-does-differential-operators-satisfy-a-pbw-like-theorem" rel="nofollow">recent question</a>, I recalled the notion of <em>differential operator</em>, <em>polyderivation</em>, and <em>principal symbol</em> for a commutative algebra $A$ over some fixed commutative ring $k$. (I will not repeat those definitions here, because you can read them there.) In that question, I asked for sufficient conditions to assure that $A$ satisfied the following version of the "PBW theorem": whether the principal symbol map (which I called $s_n$) from ($n$th order) differential operators to ($n$th order) polyderivations is a surjection.</p> <p>I thought I had an example to show that it is not always a surjection, but in the comments <a href="http://mathoverflow.net/users/1306/vladimir-dotsenko" rel="nofollow">Vladimir Dotsenko</a> pointed out that I was wrong. In fact, I have since been unable to come up with any counterexample to this "PBW theorem".</p> <p>Because for some other part of my project I must restrict to the situation when $k\supseteq \mathbb Q$ (essentially because I only know how to define the commutator of polyderivations in that case), I will ask my question there:</p> <blockquote> <p><strong>Question:</strong> What's an explicit example of a commutative algebra $A$ over a commutative ring $k\supseteq \mathbb Q$ such that there exists a polyderivation $A^{\otimes n} \to A$ which is not the principal symbol of any differential operator.</p> </blockquote> <p>If you can show that no such algebra exists, then I will gladly accept your answer both here and as an answer to my previous question.</p> http://mathoverflow.net/questions/126060/on-mentioning-recommenders-names-in-cover-letter-for-postdoctoral-applications/126065#126065 Answer by Theo Johnson-Freyd for On mentioning recommenders' names in cover letter for postdoctoral applications Theo Johnson-Freyd 2013-03-31T02:36:59Z 2013-03-31T02:36:59Z <p>I have made my answer CW, because I believe the question should be.</p> <p>My answer should carry very little weight, because I have never served on a postdoc committee: I am in my final months as a graduate student, and my experience is only as a (successful, thankfully) postdoc applicant in the US.</p> <p>What I did, and have seen my friends do, is to include in my cover letter: </p> <ol> <li>my name, current position, and contact information; </li> <li>explicit mention of the position I am applying to in that application; </li> <li>a very short (two sentences) description of my research; </li> <li>names of two or three faculty at the institution I'm applying to who might be interested in my research</li> <li>a list of the materials included in my application.</li> </ol> <p>In item 5. above, I would certainly include a sentence of the form "I have arranged for letters of recommendation from ...".</p> <p>Some examples of successful (and a few non-successful) applications have been collated by the Secret Blogging Seminar, e.g. <a href="http://sbseminar.wordpress.com/2011/04/19/a-plea-for-putting-grant-applications-online/" rel="nofollow">http://sbseminar.wordpress.com/2011/04/19/a-plea-for-putting-grant-applications-online/</a>, although a few of those links are no longer good.</p> http://mathoverflow.net/questions/124753/are-there-results-from-gauge-theory-known-or-conjectured-to-distinguish-smooth-fr Are there results from gauge theory known or conjectured to distinguish smooth from PL manifolds? Theo Johnson-Freyd 2013-03-17T05:38:57Z 2013-03-18T05:21:18Z <p>My question begins with a caveat: I sometimes spend time with topologists, but do not consider myself to be one. In particular, my apologies for any errors in what I say below — corrections are encouraged.</p> <p>My impression is that manifold topologists like to consider three main categories of (finite-dimensional, paracompact, Hausdorff) manifolds, which I will call $\mathcal C^0$, $\mathrm{PL}$, and $\mathcal C^\infty$, corresponding to manifolds whose atlases have transition functions that are, respectively, homeomorphisms, piecewise-affine transformations, and diffeomorphisms. The latter two categories obvious map faithfully into the first, and a theorem of Whitehead says that every $\mathcal C^\infty$ manifold admits a unique PL structure.</p> <p>These categories are not equivalent in any reasonable sense. The generalized Poincare conjecture is true in $\mathcal C^0$, true (except possibly in dimension $4$) in $\mathrm{PL}$, and false in many dimensions including $7$ in $\mathcal C^\infty$. $\mathcal C^0$ is the realm of surgery and h-cobordism. In $\mathcal C^\infty$, and in particular in $4$ dimensions, there is a powerful tool called "gauge theory", which provides the main technology used to prove examples of homeomorphic but not diffeomorphic manifolds.</p> <p>By definition, <em>gauge theory</em> is that part of PDE that studies connections on principal $G$-bundles for Lie groups $G$. The most important gauge theories for distinguishing between the $\mathcal C^\infty$ and $\mathcal C^0$ worlds are Donaldson Theory (which studies the moduli space of $\mathrm{SU}(2)$ connections with self-dual curvature) and the conjecturally equivalent Seiberg–Witten Theory (which studies an abelian gauge field along with a matter field, and which I understand less well). Another important gauge theory that I understand much better is (three-dimensional) Chern–Simons Theory, whose PDE picks out the moduli space of flat $G$ connections; for example, counting with sign the flat $\mathrm{SU}(2)$ connections on a 3-manifold is supposed to correspond to the Casson invariant.</p> <p>My impression, furthermore, has been that the categories $\mathcal C^\infty$ and $\mathrm{PL}$ are in fact quite close. There are more objects in the latter, certainly, but in fact many of the results separating $\mathcal C^0$ from $\mathcal C^\infty$ in fact separate $\mathcal C^0$ from $\mathrm{PL}$. A side version of my question is to understand in better detail the distance between $\mathcal C^\infty$ and $\mathrm{PL}$. But my main question is whether the technology of <em>gauge theory</em> (possibly broadly defined) can be used to separate them. A priori, the whole theory of PDE is based on smooth structures, so it would not be unreasonable, but I am not aware of examples.</p> <blockquote> <p>Are there gauge-theoretic invariants of smooth manifolds that distinguish nondiffeomorphic but $\mathrm{PL}$-isomorphic manifolds?</p> </blockquote> <p>Of course, a simple answer would be something like "For any $X,Y \in \mathcal C^\infty$, the inclusion $\mathcal C^\infty \hookrightarrow \mathrm{PL}$ induces a homotopy equivalence of mapping spaces $\mathcal C^\infty(X,Y) \to \mathrm{PL}(X,Y)$." This would explain the impression I have that $\mathcal C^\infty$ and $\mathrm{PL}$ are close — if it is true, it is not something I recall having been told.</p> http://mathoverflow.net/questions/123942/how-many-flat-connections-has-a-line-bundle-in-algebraic-geometry How many flat connections has a line bundle in algebraic geometry? Theo Johnson-Freyd 2013-03-08T06:02:12Z 2013-03-09T03:06:01Z <p>Suppose $X$ is a projective variety over $\mathbb C$. I am happy to entertain more or different adjectives — I'm not looking for the most general statement, but rather to understand when and how smooth-manifold intuition leads me astray. I know very little algebraic geometry, and so please forgive and correct me if a statement below is mistaken.</p> <p>It is rare for a line bundle $\mathcal L \to X$ to have a nowhere-vanishing section, and when it does, there are usually very few (only $\mathbb C^\times$ many). Suppose instead that I ask for a weaker structure than for $\mathcal L$ to have a section, but rather let me ask only that it has a flat connection. My question is:</p> <blockquote> <p>In algebraic geometry, how often does a line bundle have a flat connection? When it has a flat connection, how many flat connections can it have?</p> </blockquote> http://mathoverflow.net/questions/21415/what-algebraic-group-does-tannaka-krein-reconstruct-when-fed-the-category-of-modu What algebraic group does Tannaka-Krein reconstruct when fed the category of modules of a non-algebraic Lie algebra? Theo Johnson-Freyd 2010-04-15T04:06:49Z 2013-03-08T16:55:08Z <p>Let <code>$\mathfrak g$</code> be a finite-dimensional Lie algebra over <code>$\mathbb C$</code>, and let <code>$\mathfrak g \text{-rep}$</code> be its category of finite-dimensional modules. Then <code>$\mathfrak g\text{-rep}$</code> comes equipped with a faithful exact functor "forget" to the category of finite-dimensional vector spaces over <code>$\mathbb C$</code>. Moreover, <code>$\mathfrak g\text{-rep}$</code> is symmetric monoidal with duals, and the forgetful functor preserves all this structure. By Tannaka-Krein duality (see in particular the excellent paper <a href="http://www.maths.mq.edu.au/~street/CT90Como.pdf" rel="nofollow">André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, 1991</a>), from this data we can reconstruct an affine algebraic group <code>$\mathcal G$</code> such that <code>$\mathfrak g \text{-rep}$</code> is equivalent (as a symmetric monoidal category with a faithful exact functor to vector spaces) to the category of finite-dimensional representations of <code>$\mathcal G$</code>.</p> <p>However, <a href="http://mathoverflow.net/questions/124/is-every-finite-dimensional-lie-algebra-the-lie-algebra-of-an-algebraic-group" rel="nofollow">it is not true that every finite-dimensional Lie algebra is the Lie algebra of an algebraic group</a>. So it is not true that <code>$\mathcal G$</code> is, say, necessarily the simply-connected connected Lie group with Lie algebra <code>$\mathfrak g$</code>, or some quotient thereof. So my question is:</p> <blockquote> <p>Given <code>$\mathfrak g$</code>, what is an elementary description of <code>$\mathcal G$</code> (that avoids the machinery of Tannaka-Krein)?</p> </blockquote> <p>For example, perhaps <code>$\mathcal G$</code> is some Zariski closure of something...?</p> http://mathoverflow.net/questions/123738/contraction-of-graded-vector-fields-on-de-rham-complex/123949#123949 Answer by Theo Johnson-Freyd for Contraction of graded vector fields on de Rham complex Theo Johnson-Freyd 2013-03-08T07:34:04Z 2013-03-08T07:34:04Z <p>I don't know whether to take your last line as an invitation or a slight, but I'll do my best with an answer. As you probably already know, I am (often uselessly) verbose.</p> <p>I like to think of $\Omega^\bullet$ as a bigraded commutative algebra. The monoidal category of bigraded vector spaces can be equipped with a symmetric structure which is determined up to unique equivalence of symmetric monoidal categories by the request that for the generating lines $L_{(0,1)}$ and $L_{(1,0)}$ in bidegrees $(0,1)$ and $(1,0)$ respectively, the braiding $L \otimes L \to L\otimes L$ is minus the identity. This forces the line $L_{(0,1)} \otimes L_{(1,0)}$, along with any isomorph of it, to braid with itself by $+1$. More generally, a line in bidegree $(m,n)$ is forced to braid with itself for a factor of $(-1)^{m+n}$.</p> <p>Note that I never said how different lines are supposed to braid with each other. That's because the question is not well-posed. If you skeletalize your category and fix a monoidal structure, then you do get an isomorphism between, say, $L_{(0,1)} \otimes L_{(1,0)} \cong L_{(1,1)} \cong L_{(1,0)} \otimes L_{(0,1)}$, and you can ask by what factor the braiding acts relative to this isomorphism. But from the point of view of category theory (i.e. without choosing extra data, like a skeletalization), there is no distinguished isomorphism between $L_{(0,1)} \otimes L_{(1,0)}$ and $L_{(1,0)} \otimes L_{(0,1)}$ except for the braiding. I will come back to this point in a moment.</p> <p>This category-theoretic language is the cleanest way I know to set up the basic definitions. For example, given any bigraded commutative algebra $\Omega^\bullet$ with multiplication $m$, there is a bigraded vector Lie algebra $\operatorname{Der}(\Omega^\bullet)$ of derivations. It is the universal object with a map $\triangleright : \operatorname{Der}(\Omega^\bullet) \otimes \Omega^\bullet \to \Omega^\bullet$ such that for any $\phi : \Phi \to \operatorname{Der}(\Omega^\bullet)$ and $\alpha : A\to \Omega^\bullet$ and $\beta : B \to \Omega^\bullet$, the following is an equality of maps $\Phi \otimes A \otimes B \to \Omega^\bullet$: $$\triangleright \circ (\phi \otimes m) \circ (\alpha \otimes \beta) = m \circ ((\triangleright \circ(\phi \otimes \alpha))\otimes \beta) + m\circ(\alpha \otimes (\triangleright \circ(\phi \otimes \beta)))$$ You can clean up the notation a bit by inventing "generalized elements". What's important to mention is that the second summand on the RHS as written is not a map $\Phi \otimes A \otimes B \to \Omega^\bullet$, but rather a map $A \otimes \Phi \otimes B \to \Omega^\bullet$; to add them, I need them to have the same codomain, and so I insist that you identify $\Phi \otimes A \otimes B \cong A \otimes \Phi \otimes B$ by using the symmetry, and not any other way. Finally, you've distinguished a differential $d_{dR}$ on $\Omega^\bullet$ of bidegree $(0,1)$ say, by which I actually mean that you've distinguished a particular line $L_{(0,1)}$ and a map $d_{dR} : L_{(0,1)} \to \Omega^\bullet$; such a choice gives you a map $\iota: (L_{(0,1)})^{\otimes (-1)} \otimes \operatorname{Der}(\mathcal A) \to \operatorname{Der}(\Omega^\bullet)$, where $(L_{(0,1)})^{\otimes (-1)}$ comes equipped as the tensor inverse of $L_{(0,1)}$, and $\mathcal A$ is your original algebra.</p> <p>Ok, so I said this was the cleanest way to set up the definitions, but by no means is it always the smoothest way to perform computations. Sometimes, generalized elements (which I have not defined, but I have used implicitly) are the best way to do some computations. But sometimes you do need to sit down and use homogeneous elements.</p> <p>Talking about homogeneous elements is basically the same as skeletalizing the category of bigraded vector spaces, and thus requires you to make more choices than you would otherwise have to. There are two reasonable choices (and many unreasonable ones), and as I asserted above, the two choices give you equivalent categories, and hence equivalent theorems. But since there are two reasonable choices, you need to be careful that you make one and use it consistently, and don't trust formulas in other papers, because they probably used different choices.</p> <p>The two choices are to declare, after skeletalizing, that the generators $L_{(0,1)}$ and $L_{(1,0)}$ braid past each other as $+1$ relative to the skeletalization, or as $-1$. The former of these implies that if $\alpha$ is of bidegree $(p,q)$ and $\beta$ is of bidegree $(m,n)$ then $\alpha$ and $\beta$ commute for a factor of $(-1)^{pm + qn}$. The latter choice would have them commute for a factor of $(-1)^{(p+q)(m+n)}$. This latter choice makes it easier to present the symmetric monoidal functor that collapses the gradings. The former choice makes it easier to see that a bigraded vector space is the same as a graded (graded vector space).</p> <p>See, you never said your conventions even for how to think of $\Omega^\bullet$ as a commutative algebra. I recommend that you use the same convention there as for derivations.</p> <p>Anyway, if your vector field $X$ on $\mathcal A$ was of degree $m$, then $\iota_X$ will be of bidegree $(m,-1)$. If $\alpha$ is of bidegree $(p,q)$, then you should use one of: $$\iota_X(\alpha \wedge\beta) = \iota_X(\alpha)\wedge\beta + (-1)^{mp + q}\alpha \wedge \iota_X(\beta)$$ or: $$\iota_X(\alpha \wedge\beta) = \iota_X(\alpha)\wedge\beta + (-1)^{(m+1)(p + q)}\alpha \wedge \iota_X(\beta)$$ Either rule will work if you apply it consistently. Probably other rules will work as well.</p> http://mathoverflow.net/questions/86852/when-can-i-assure-that-the-representation-theory-of-a-prop-is-faithful When can I assure that the representation theory of a PROP is faithful? Theo Johnson-Freyd 2012-01-27T18:50:24Z 2013-03-01T17:45:40Z <p>Recall that a <a href="http://ncatlab.org/nlab/show/PROP" rel="nofollow">PROP</a> is a symmetric monoidal category whose object-set is freely $\otimes$-generated by a single object $x\in P$. A <em>representation</em> or <em>algebra</em> for a prop $P$ in a symmetric monoidal category $C$ is a symmetric monoidal functor $f: P \to C$ — i.e. it is an object $f(x)\in C$ and some morphisms between tensor powers of $f(x)$ that satisfy all the relations in $P$. I am interested in the $\mathbb Q$-linear versions of all of these: a <em>$\mathbb Q$-linear prop $P$</em> is a $\mathbb Q$-linear symmetric monoidal category whose object-set is freely $\otimes$-generated by a single object $x\in P$, and a <em>representation</em> thereof in a $\mathbb Q$-linear category $C$ is a $\mathbb Q$-linear symmetric monoidal functor. I will henceforth leave implicit the word "$\mathbb Q$-linear". By $\mathrm{Vect}$ I mean the category of $\mathbb Q$-vector spaces.</p> <p>A general question you can ask is the following. Suppose you are given a prop $P$, and some presentation of it in generators and relations. Suppose you write some expression in the generators — i.e. you pick some morphism in the prop. Suppose furthermore that for every representation $P \to \mathrm{Vect}$, this morphism evaluates to $0$. Does it follow that the morphism is $0$ in $P$? Put another way: <em>Does every prop have a faithful representation in $\mathrm{Vect}$?</em></p> <p>The answer, of course, is "NO!". An example: let $P$ be the prop generated by the relation that the braiding $x\otimes x \to x\otimes x$ is minus the identity. Then the only representation of $P$ in $\mathrm{Vect}$ is $x = 0$, and in particular the identity map $x \to x$ evaluates to $0$ in this representation. On the other hand, $P$ has a non-zero representation in the category of super vector spaces.</p> <blockquote> <p>My question, then, is for (checkable) conditions on a prop $P$ to assure that it does, in fact, have faithful representations in $\mathrm{Vect}$. </p> </blockquote> <p>For example, the prop that I happen to care about has a presentation in which it is generated by (at most) one morphism between any two objects, and the relations are all (homogeneous) linear and quadratic in the generators. I could imagine this to be the type of condition that might assure faithfulness of representations. I would like to know that my prop has a faithful $\mathrm{Vect}$-representation, because I can prove that in any $\mathrm{Vect}$-representation a certain morphism evaluates to $0$, by choosing a basis for the underlying vector space of the representation. Of course, this proof does not universalize, but maybe some other results assure me that the morphism is universally $0$.</p> http://mathoverflow.net/questions/23026/how-can-i-understand-the-groupoid-quotient-of-a-group-action-as-some-sort-of-p How can I understand the "groupoid" quotient of a group action as some sort of "product"? Theo Johnson-Freyd 2010-04-29T19:36:23Z 2013-02-14T21:37:03Z <p>Recall the notion of <strong>groupoid</strong> (<a href="http://en.wikipedia.org/wiki/Groupoid" rel="nofollow">Wikipedia</a>, <a href="http://ncatlab.org/nlab/show/groupoid" rel="nofollow">nLab</a>). An important construction of groupoids is as "action groupoids" for group actions. Namely, let $X$ be a groupoid and $G$ a group, and suppose that $G$ acts on $X$ by groupoid automorphisms. Then we can form a new groupoid $X//G$, which has as objects the objects of $X$, but the morphisms include, in addition to the original morphisms of $X$, a morphism $x \overset g \to gx$ for each $g\in G$ and $x\in X$. The composition of morphisms is well-defined if the action is by groupoid automorphisms. (When $X$ is a set, then $X//G$ is equivalent to the skeletal groupoid whose objects are the elements of the "coarse" quotient $X/G$, and with ${\rm Aut}(\bar x) = {\rm Stab}_G(x)$.)</p> <p>(Probably there is a fancier construction, in which the conditions on the word "group action" be relaxed to an "action" up to specified natural isomorphism, and then $G$ could act on $X$ by autoequivalences, rather than autoisomorphisms, but this generalization won't concern me.)</p> <p>Let $1$ denote the one-point set, thought of as a groupoid with only identity morphisms. Then any group $G$ acts uniquely on $1$, and so we have the groupoid $1//G$. In general, although $X\times 1 \cong X$, we do not have $X \times (1//G) \cong X//G$ for arbitrary $G$-actions on $X$ unless the action is trivial. (Here $\times$ denotes the <strong>groupoid product</strong>, which is just what you think it is.) However, the construction provides natural bijections between the objects of $X//G$ and the objects of $X \times (1//G)$, and between the morphisms of $X//G$ and the morphisms of $X \times (1//G)$.</p> <blockquote> <p><strong>Question:</strong> Is there some sort of "semidirect" or "crossed" product of groupoids, which presumably depends on extra data, so that we do have $X//G \cong X \rtimes (1//G)$? By which I mean, what is the correct notion of "action" of a groupoid $Y$ on a groupoid $X$ and what is the corresponding correct notion of $X \rtimes Y$?</p> </blockquote> <p>I see that the page <a href="http://ncatlab.org/nlab/show/semidirect+product" rel="nofollow">semidirect product in nLab</a> defines $X \rtimes G$ as something closely related to $X//G$. But clearly this ought to be called $X\rtimes (1//G)$, but then I do not know what the right definition for $X\rtimes Y$ is, hence the question. And really I'd like to know about a "double crossed product" <code>$X\bowtie Y$</code>.</p> <p>My motivation for this question is from my answer to <a href="http://mathoverflow.net/questions/22860/do-rational-numbers-admit-a-categorification-which-respects-the-following-dualit/22879#22879" rel="nofollow">Do rational numbers admit a categorification which respects the following “duality”?</a>.</p> http://mathoverflow.net/questions/121375/exactness-is-often-an-open-condition-how-often Exactness is often an open condition. How often? Theo Johnson-Freyd 2013-02-10T05:54:09Z 2013-02-10T05:54:09Z <p>Let $\mathcal C$ denote an abelian category. Consider a formal $\mathbb Z$-graded object $V_\bullet$ in $\mathcal C$, by which I simply mean a $\mathbb Z$-indexed list $n\mapsto V_n$ of objects. A <em>differential</em> on $V_\bullet$ is a sequence of maps $\partial : V_n \to V_{n-1}$ such that $\partial^2 = 0$; for example, the sequence of all $0$s works. A differential is <em>exact</em> if for every $n$, $\ker \partial|_{V_n} = \partial(V_{n+1})$. Clearly, for fixed $V_\bullet \in \mathcal C$, the collection of differentials is the intersection of some quadrics in some abelian group. I have less of an idea of what type of space the exact differentials constitute, which is one of the motivations for my question.</p> <p>Thus, fix a $\mathbb Z$-graded object $V_\bullet \in \mathcal C$. Suppose that you fix also an exact differential $\partial_0$. There are many situations where one is interested in "varying" $\partial_0$ to some new differential $\partial$. One situation that arises is the "formal deformation": we can formally extend $\mathcal C$ to a new category $\mathcal C[\![\hbar]\!]$ by tensoring every hom set with $\mathbb Z[\![\hbar]\!]$, and then consider differentials $\partial$ on $V_\bullet[\![\hbar]\!]$ such that $\partial = \partial_0 \mod\hbar$. I can also imagine more algebrogeometric contexts: I expect that there is a natural way to give the hom spaces in an abelian category, and thus the space of differentials, the structure of affine algebraic varieties; then I could probe this space by, say, local Artinian rings. I could also, of course, probe non-infinitesimal neighborhoods.</p> <p>In any case, one should expect that if $\partial_0$ is exact that $\partial$ is "close enough" to $\partial_0$, then $\partial$ is also exact. I know how to prove a special case of this. Say that $\mathcal C$ <em>has the Axiom of Choice</em> if every epimorphism in $\mathcal C$ splits. Then, since $\mathcal C$ is abelian, it must also have the Axiom of "co-Choice", and one can always choose a deformation retraction of any complex onto its homology, for example, and also one can always choose an isomorphism between a (Hausdorff exhaustive) filtered object and its associated graded. Under these conditions, I can prove that formal deformations of exact differentials remain exact. Another somewhat trivial example is when $V_\bullet = (V_0,V_1)$ is two terms long. Then an exact differential is an isomorphism $\partial_0 : V_1 \overset\sim\to V_0$. Consider the situation when $\mathcal C$ is, say, the category of finite-dimensional vector spaces. Then the space of isomorphisms is naturally a Zariski-open (and hence analytically open, if we are over $\mathbb R$) subset of the space of all maps.</p> <blockquote> <p><strong>Question:</strong> Is there a way to give the the space of differentials on $V_\bullet$ a natural (Zariski, say, but I'm agnostic) topology, perhaps under some conditions on $\mathcal C$, and if so, how? Under what conditions does the slogan "exactness is an open condition" hold? As a special case, does it hold for "infinitesimal" (either in the local-Artinian or $\hbar$ sense) deformations in the absence of Choice?</p> </blockquote> <p>A further version of my question recognizes the fact that many categories are naturally enriched in topological abelian groups; categories of Banach spaces, for example. Then one can also ask: when is exactness an open condition for "analytic" topologies?</p> <p>Lest you say that exactness is <em>always</em> an open condition, let me end with a trivial parting comment. I could, if I so chose, give every space the indiscrete topology. For this topology, exactness certainly is not open, as deforming all the way to $\partial = 0$ would be a "small" deformation.</p> http://mathoverflow.net/questions/119849/quantization-of-a-classical-system-e-g-the-case-of-a-billard/119912#119912 Answer by Theo Johnson-Freyd for Quantization of a classical system (e.g. the case of a billard) Theo Johnson-Freyd 2013-01-26T04:13:39Z 2013-01-26T04:13:39Z <p>As to the bulk of your question, which I take to be a reference request for mathematical accounts of quantum mechanics, I am partial to the book <a href="http://www.amazon.com/Quantum-Mechanics-Mathematicians-Graduate-Mathematics/dp/0821846302" rel="nofollow">Quantum Mechanics for Mathematicians</a> by L. Takhtajan.</p> <p>Be sure to look over the MathOverflow thread <a href="http://mathoverflow.net/questions/2917/where-does-a-math-person-go-to-learn-quantum-mechanics" rel="nofollow">Where does a math person go to learn quantum mechanics?</a>, as it contains many good references.</p> http://mathoverflow.net/questions/119849/quantization-of-a-classical-system-e-g-the-case-of-a-billard/119911#119911 Answer by Theo Johnson-Freyd for Quantization of a classical system (e.g. the case of a billard) Theo Johnson-Freyd 2013-01-26T04:07:43Z 2013-01-26T04:07:43Z <p><a href="http://mathoverflow.net/questions/8606/what-does-quantization-is-not-a-functor-really-mean" rel="nofollow">Quantization is not a functor.</a></p> http://mathoverflow.net/questions/119813/when-is-a-solution-to-an-ode-determined-by-its-average-value When is a solution to an ODE determined by its average value? Theo Johnson-Freyd 2013-01-25T05:37:18Z 2013-01-25T17:51:49Z <p>Fix a smooth vector field $\vec v$ on $\mathbb R^n$. It is well-known that any trajectory $\vec x : [0,1]\to \mathbb R^n$ solving the ODE $\dot x(t) = \vec v(\vec x(t))$ is determined by its evaluation $\vec x(t)$ for any $t\in [0,1]$. My question is about when a solution is determined by its average value.</p> <p>More precisely, let $\phi$ be a smooth function on $\mathbb R$ vanishing identically outside $[0,1]$ and satisfying $1 = \int \phi(t)\ \mathrm{d}t$. For any $f : [0,1] \to \mathbb R$, let $\langle f\rangle = \int f(t)\ \phi(t)\ \mathrm{d}t$. My question is:</p> <blockquote> <p>Under what conditions does the data of the vector field $\vec v$, the averaging function $\phi$, and the value $\langle \vec x\rangle \in \mathbb R^n$ determine the classical trajectory $\vec x$?</p> </blockquote> <p>Here are two examples. Fix $\vec a \in \mathbb R^n$ and consider the constant vector field $\vec v(x) = \vec a$. Then $$\vec x(t) = \vec a t + \langle \vec x \rangle - \vec a \int t\ \phi(t)\ \mathrm{d}t$$ and so the answer is that any $\phi$ works.</p> <p> On the other hand, on $\mathbb R^2$ consider vector field $\vec v \bigl( \begin{smallmatrix} x_1 \\ x_2 \end{smallmatrix} \bigr) = \bigl( \begin{smallmatrix} 4\pi\ x_2 \\ -4\pi\ x_1 \end{smallmatrix} \bigr)$. Then classical trajectories are of the form $$x_1(t) = r\cos (4\pi(t - \theta)), \quad x_2(t) = r\sin(4\pi(t-\theta))$$ for fixed $r,\theta$. Choose a smooth function $\varphi$ which is identically $0$ on $(-\infty,0]$ and identically $1$ on $[1,\infty)$, and set $\phi(t) = 2\varphi(2t) - 2\varphi(2t - 1)$. If I haven't made an arithmetic error, then for any $r,\theta$ we have $\langle \vec x \rangle = \bigl( \begin{smallmatrix} 0 \\ 0 \end{smallmatrix} \bigr)$. </p> <p>I could imagine answers of the following forms:</p> <ul> <li>For any vector field $\vec v$, there exists an averaging function $\phi$ that works: take a sufficiently good approximation of a delta-function.</li> <li>If $\vec v$ is bounded in some appropriate norm (or if the trajectory $\vec x$ is known a priori to stay in a region in which $\vec v$ is bounded), then there is some $\epsilon$ depending on the bound such that any $\phi$ with domain $(0,\epsilon)$ works. Or perhaps what works is any $\phi$ which is "within $\epsilon$ of a delta-function" in the appropriate sense.</li> <li>For any $\phi$, the only vector fields that fail to have their solutions determined in the above way have such-and-such property, and hence are few and far between.</li> <li>...?</li> </ul> http://mathoverflow.net/questions/119726/discovering-and-selecting-conferences/119729#119729 Answer by Theo Johnson-Freyd for Discovering and selecting conferences Theo Johnson-Freyd 2013-01-24T06:04:19Z 2013-01-24T06:04:19Z <p>Preamble: I am a graduate student at a large program, about to start a postdoc, so I can speak about what has worked for me at the (hopefully) beginning of my career.</p> <p>One thing I do occasionally is to check the upcoming conferences at the <a href="http://www.mathinstitutes.org/" rel="nofollow">NSF Mathematical Science Institutes</a>. These institutes tend to run many strong conferences. Since I'm in mathematical physics, I also try to monitor the upcoming conferences at the <a href="http://www.scgp.stonybrook.edu/" rel="nofollow">Simons Center for Geometry and Physics</a> and the <a href="http://qgm.au.dk/" rel="nofollow">Centre for the Quantum Geometry of Moduli Spaces</a>, and I'm sure there are centers that I'm not aware of.</p> <p>The second way I find out about workshops and conferences is via announcements forwarded to our department's mailing lists. My adviser occasionally forwards announcements to me. But the most important way I find out about workshops and conferences is by talking about upcoming opportunities with my peers. Socializing at department teas is a good way to do this, and you can certainly directly ask people whose interests overlap with yours about their upcoming plans. Moreover, as you attend conferences, you should talk to the other participants about related upcoming activities.</p> http://mathoverflow.net/questions/119479/different-hopf-algebra-structures-on-same-graded-algebra/119492#119492 Answer by Theo Johnson-Freyd for Different Hopf algebra structures on same graded algebra Theo Johnson-Freyd 2013-01-21T17:12:49Z 2013-01-21T17:12:49Z <p>There are generalizations of the following construction, but here is a simple one. Let $\mathfrak n$ denote the Lie algebra of strictly-upper-triangular $n\times n$ matrices. It is $\mathbb Z_{>0}$-graded, by declaring that for $1\leq k \leq n$, the degree-$k$ part of $\mathfrak n$ consists of those matrices supported on the diagonal that is $k$ steps above the main diagonal. (So $\mathfrak n_k$ has a basis consisting of those matrices with a $1$ in the $(i,i+k)$th spot and $0$s elsewhere.) This is a bosonic grading, so to match Hatcher's conventions, you perhaps should double the grading.</p> <p>In any case, consider the polynomial algebra $\operatorname{Sym}(\mathfrak n^\ast)$ generated by the dual vector space — so this is the algebra $\mathcal{O}(\mathfrak n)$ of polynomial functions on $\mathfrak n$. It is connected and $\mathbb Z_{\geq 0}$-graded. It has many compatible coproducts. One is the canonical coproduct generated by $x \mapsto x\otimes 1 + 1\otimes x$ for $x\in \mathfrak n^\ast$. Another is the Baker–Campbell–Hausdorff formula $\operatorname{BCH}$. In detail, let $N$ denote the group of matrices which have $1$s on the diagonal and $0$s below it, and let $\operatorname{m} : N\times N \to N$ denote the map of matrix multiplication. Then the matrix $\exp : \mathfrak{n} \to N$ is a degree-$n$ polynomial, as is its inverse $\log: N \to \mathfrak{n}$; we can therefore pull back polynomial functions along these, so for example we have $\operatorname{m}^\ast : \mathcal{O}(N) \to \mathcal{O}(N)\otimes \mathcal{O}(N)$. Then: $$\operatorname{BCH} = (\exp^\ast \otimes \exp^\ast) \circ \operatorname{m}^\ast \circ \log^\ast.$$ It is a standard fact that $\operatorname{BCH}$ preserves the grading on $\operatorname{Sym}(\mathfrak n^\ast)$, and makes $\operatorname{Sym}(\mathfrak n^\ast)$ into a Hopf algebra.</p> http://mathoverflow.net/questions/119037/is-the-metaplectic-group-not-a-matrix-group-counterexample/119040#119040 Answer by Theo Johnson-Freyd for Is the metaplectic group not a matrix group - counterexample Theo Johnson-Freyd 2013-01-16T05:46:58Z 2013-01-16T05:46:58Z <p>Keep in mind that any finite-dimensional representation of a Lie group determines a finite-dimensional representation of its Lie algebra, and for a connected Lie group the induced Lie algebra representation determines the Lie group representation.</p> <p>However, every finite-dimensional representation of $\operatorname{Lie}(\mathrm{Mp}(2,\mathbb R)) = \mathfrak{sl}(2,\mathbb R)$ comes from a representation of $\mathrm{SL}(2,\mathbb R)$, and so does not come from a faithful rep of $\mathrm{Mp}(2,\mathbb R)$. One way to see this by directly classifying all finite-dimensional $\mathfrak{sl}(2,\mathbb R)$ representations, which is not too difficult. A better way is to observe that any $\mathfrak{sl}(2,\mathbb R)$-representation $V$ embeds in an $\mathfrak{sl}(2,\mathbb C)$-representation $V \otimes \mathbb C$, but $\mathrm{SL}(2,\mathbb C)$ is simply connected, so $V \otimes \mathbb C$ is a representation of $\mathrm{SL}(2,\mathbb C)$, and so the $\mathrm{Mp}(2,\mathbb R)$-representation that gave rise to $V$ factors through $\mathrm{SL}(2,\mathbb C)$, and on the other hand the map $\mathrm{Mp}(2,\mathbb R) \to \mathrm{SL}(2,\mathbb C)$ factors through $\mathrm{SL}(2,\mathbb R)$ and is not faithful.</p> http://mathoverflow.net/questions/116502/integer-triangle/116510#116510 Answer by Theo Johnson-Freyd for Integer triangle Theo Johnson-Freyd 2012-12-16T05:03:29Z 2012-12-16T05:03:29Z <p>Clearly it is enough to find a triangle in which all seven points are rational, as then you can make them integral by rescaling. But given <em>any</em> triangle with rational coordinates, aren't the centroid, orthocenter, and circumcenter all automatically at rational coordinates?</p> <ol> <li>The centroid is the arithmetic average of the coordinates, and hence rational.</li> <li>The slopes of all sides are rational. Hence the altitude of through any vertex has rational slope (-1/the slope of the opposite side) and passes through a rational point, and so has a rational equation. The orthocenter is the intersection of any two altitudes, and the intersection of two lines with rational equations is necessarily rational.</li> <li>By a similar argument, the perpendicular bisectors of each side are rational, and so the circumcenter is rational.</li> </ol> <p>So the only non-automatic point is the incenter. If $A$, $B$, and $C$ are the coordinates of the corners, and $a$, $b$, and $c$ are the side lengths of the respective opposite sides (so that $a = \|B-C\|$, for example) then the incenter is at $(aA + bB + cC) / (a + b + c)$. This is not necessarily rational — the triangle with coordinates $(0,0)$, $(1,0)$, and $(0,1)$ is a counterexample — but is rational as soon as the three sidelengths are.</p> <p>So Noam D. Elkies's answer in the comments has many generalizations, including all Pythagorean triples, and also including combinations of them.</p> http://mathoverflow.net/questions/116313/homotopy-transfer-theorem-for-differential-graded-associative-algebras/116321#116321 Answer by Theo Johnson-Freyd for Homotopy Transfer Theorem for Differential Graded Associative Algebras Theo Johnson-Freyd 2012-12-13T20:43:58Z 2012-12-13T20:43:58Z <p>There is a systematic graphical notation that allows the tracking of signs, which I will mention at the end of this answer. But before doing so, let me outline the situation when $2 = 0$.</p> <p>Note first that since $(A,\nu)$ is strictly associative, $\mu_0 = 0$ and $\mu_1 = d_H = d : H \to H$. By convention, if I have multilinear maps $f: H^{\otimes k}\to H$ and $g: H^{\otimes l}\to H$, then I will write $f\circ g : H^{\otimes(k+l -1)} \to H$ for: $$(f\circ g) (x_1\otimes \cdots \otimes x_{k+l-1}) = \sum_{i=1}^k f\bigl(x_1\otimes \dots \otimes x_{i-1} \otimes g(x_i\otimes \dots \otimes x_{i+l-1}) \otimes x_{i+l} \otimes \cdots \otimes x_{k+l-1}\bigr)$$ This has a useful graphical notation, wherein the composition is the sum over all rooted planar trees with an $f$ at the bottom node and precisely one $g$ at one of the upper nodes. </p> <p>Then axiom to be an $A_\infty$-algebra in characteristic $2$ is $0 = \sum_{j=0}^{n+1} \mu_j\circ \mu_{n+1-j}$, or, since $\mu_0 = 0$ and $\mu_1 = d$: $$[d,\mu_n] = \sum_{j=2}^{n-1} \mu_j \circ \mu_{n+1-j}$$ The right-hand side is a sum over all rooted planar trees with $n$ leaves and precisely two nodes, each of which has at least two branches from it. </p> <p>To check this, the first thing to convince yourself is that the operator $[d,-] : f \mapsto d\circ f + f\circ d$ is a derivation of composition and tensor, so that to apply $[d,-]$ to some large diagram, you sum all diagrams you get by replacing one component of your original diagram by $[d,-]$ of it. Note also that $[d,-]$ comutes with (i.e. annihilates) $i$, $p$, and $\nu$. So when you work out $[d,\mu_n]$, you get a sum over diagrams that look like $\mu_n$ (i.e. planar rooted trees with $n$ leaves, each node has two branches, and interior edges labeled by $h$), except one of the interior edges has been replaced by $[d,h] = \mathrm{id}_A + ip$.</p> <p>Now, it should be completely clear that the diagrams where the $h$ is replaced by an $ip$ are precisely the diagrams appearing in $\sum_{j=2}^{n-1} \mu_j \circ \mu_{n+1-j}$. (If this is not clear, let me know, and I will try to make it clearer.)</p> <p>Finally, we must dispense with the diagrams in which an $h$ is replaced by an $\mathrm{id}$. For any such diagram, consider contracting it along the offending $\mathrm{id}$ vertex, to produce a node with three branches. Except the resulting diagram with the trivalent vertex can be produced in two ways, corresponding to the two planar ways of blowing up a rooted node with three branches into two two-branch nodes. So, after sorting all of your offending diagrams into such pairs, you get a sum of diagrams that looks like a $\mu_n$-type sum, except one vertex is has three branches. What is this vertex labeled by? Why, $\nu \circ \nu$, of course, which is a sum of two terms. On the other hand, $\nu \circ \nu = 0$ by the associativity law for $(A,\nu)$.</p> <hr> <p>In characteristic not equal to $2$, the exact same argument works, but you must find a good convention / notation for signs. The best notation that I know is as follows. It should, of course, already by understood that the solid "$H$" or "$A$" edges extend to "infinity" at the top and bottom of the page. You should additionally draw diagrams with some other color of edge (I usually used "dashed") that records the degrees of operators — so this "dashed" edge should carry an arrow denoting its direction. A vertex that raises homological degree by $n$ is required to receive $n$ dashed edges, and a vertex that lowers homological degree by $n$ is required to emit $n$ dashed edges. Free dashed edges are sent off to "infinity" at (say) the left-hand side of the page, and the order from top to bottom that the free dashed edges arrive is important. Just as you cannot add diagrams whose numbers of input and output "$H$"$strands mismatch, you similarly must have the same sequence of dashed edges. (In categorical language, what I'm saying is that you only work with "global" elements of endomorphism spaces, which is to say actual morphisms in the category of homologically-graded abelian groups, but that you give yourself access to the <em>objects</em> of this category which are lines in degree$\pm 1$.)</p> <p>Now whenever two edges cross, something happens with signs. The notation basically takes care of this, but if you ever insist on working with "homogeneous elements" (which is a bad habit — it's better to work more categorically) the convention is that as an element runs down the "wire" of a solid edge, when it passes through a dashed edge it remains unchanged if it is of even degree and changes sign if it is of odd degree. The notations that do matter are: </p> <ol> <li>A closed dashed circle can be removed for a factor of$-1$.</li> <li>A dashed crossing can be resolved for a factor of$-1$. (Since dashed edges are directed, any dashed crossing has a unique resolution.)</li> </ol> <p><img src="http://math.berkeley.edu/~theojf/conventions1.gif" alt="Resolving a crossing for a sign"> <img src="http://math.berkeley.edu/~theojf/conventions2.gif" alt="Loops, etc., cost signs"></p> <p>For example, the operator$d$emits a dashed edge, and the operator$h$receives a dashed edge. Thus an equation like "$\mathrm{id} - ip = dh + hd$" is nonsense: the left-hand side has no dashed edges running to infinity, whereas on the right-hand side the first summand emits an edge and then receives one, and the second summand does those in the opposite order. To sum the two terms on the right-hand side you have to at least get them into their edges-at-infinity into the same order, which you can do by adding a crossing (but remember that resolving that crossing changes a sign). To make the two sides agree, you get connect up the two dashed edges, and again you should think a moment about signs. The correct right-hand side to "$\mathrm{id} - ip = dh + hd$" is: </p> <p><img src="http://math.berkeley.edu/~theojf/conventions3.gif" alt="the commutator"></p> <p>Yes, the sign is correct. (Incidentally, I'm lifting these images from my thesis, which works out a slightly different question, and so the colors and numbers are not for this post.)</p> <p>Anyway, I'll leave it as an exercise to write out diagrams for$A_\infty$algebras in this notation, and to get all the signs right. (Hint: there should be no "weird" signs.) Part of the reason that I'll leave it as an exercise is that there's not really a unique correct answer: you make a sign convention, and work with it.</p> http://mathoverflow.net/questions/116215/is-there-an-algebraic-derived-mapping-space-construction-that-encompasses-both Is there an algebraic "derived mapping space" construction that encompasses both Hochschild homology and loop spaces of non-simply-connected spaces? Theo Johnson-Freyd 2012-12-12T23:47:00Z 2012-12-13T00:36:03Z <p>I'm looking for directions to the literature that might contain fairly explicit constructions that might be called (the algebra of functions on) the "derived mapping space" from a simplicial set to a simplicial (affine) scheme. To make the question reasonably self-contained and to give a sense of my background and current understanding, I will begin with some general abstract nonsense, and then point to a construction that I have found in the literature and that does not work to my satisfaction.</p> <h2>A little abstract nonsense</h2> <p>Let$C$and$D$be categories. In a bit, I will give them specific values, but for now I will ask only that$D$be small, and that$C$have any necessary limits and colimits. A <em>(generalized)$D$-object in$C$</em> is a presheaf on$D$valued in$C$, i.e. a functor$X : D^{\mathrm{op}}\to C$. Each$d\in D$determines (and is determined by) a$D$-object in$\mathrm{SET}$, by the usual Yoneda embedding$d \mapsto \operatorname{hom}_D(-,d)$. It will be convenient for me to denote the presheaf$\operatorname{hom}_D(-,d)$by$[d]$, and given$k\in D$and$X : D^{\mathrm{op}}\to C$, I will write$X_k$for$X(k)$.</p> <p>For$x\in C$and$s\in \mathrm{SET}$, there is an object$\operatorname{maps}(s,x) = x^s = \prod_s x \in C$, which is the$s$-fold cartesian product of$x$with itself. Now, let$X : D^{\mathrm{op}} \to C$be a$D$-object in$C$, and$S: D^{\mathrm{op}} \to \mathrm{SET}$a$D$-set. Then there is an object$\operatorname{hom}_D(S,X) \in C$, which is built as a certain limit ranging over the objects$\operatorname{maps}(S_k,X_k)$for$k\in D$. Even better, the categories of$D$-sets and$D$-objects in$C$have products — the ("categorical") cartesian product of functors is constructed by taking the product for each — and so we can define an enriched hom by: $$\underline{\operatorname{hom}}_D(S,X) : D^{\mathrm{op}} \to C, \quad d \mapsto \operatorname{hom}_D(S \times [d],X).$$ Finally, there is one more, much more naive "mapping space" between$D$-objects, which I will denote by$\operatorname{maps}(S,X) : D \times D^{\mathrm{op}} \to C$, sending$(d,k) \mapsto \operatorname{maps}(S_d,X_k)$.</p> <h2>A little concrete nonsense</h2> <p>I will be interested in the situation where$D = \Delta$is the category of finite nonempty totally-ordered sets (and monotonic maps). It has a skeletalization with objects indexed by the natural numbers, given by$[n] = \lbrace 0 &lt; \dots &lt; n \rbrace$. Note that the$\Delta$-set$[0]$is terminal, so$\underline{\operatorname{hom}}_D(S,X)_0 = \operatorname{hom}_D(S,X)$. An object$X : D^{\mathrm{op}} \to C$determines, among other data, two maps$X_1 \rightrightarrows X_0$, corresponding to the two inclusions$[0] \rightrightarrows [1]$. By definition,$\pi_0(X) \in C$is the coequalizer of the two arrows$X_1 \rightrightarrows X_0$.</p> <p>Fix a commutative ring$\mathbb K$. I will not be upset if you would like to make further assumptions on$\mathbb K$, e.g. that$\mathbb K \supseteq \mathbb Q$, or that$\mathbb K$is an algebraically closed field. I believe that I am primarily interested in the following two values for$C$, but I am open to being convinced otherwise:</p> <ol> <li>$C = \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$is the category of affine schemes over$\mathbb K$.</li> <li>$C = \mathrm{Mod}_{\mathbb K}$is the category of$\mathbb K$-modules.</li> </ol> <p>There is a well-known contravariant forgetful functor$\mathcal O$from$\mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$to$\mathrm{Mod}_{\mathbb K}$.</p> <p>I will also take inspiration from the case$C = \mathrm{Top}$of nice enough topological spaces.</p> <p>There is a further functor$\operatorname{ch}: \Delta\mathrm{Mod}_{\mathbb K} \to \mathrm{DGMod}_{\mathbb K}$(the category of homologically-graded chain complexes of$\mathbb K$-modules) which sets$\operatorname{ch}(X)_k = X_k$with differential a certain well-known alternating sum. There is a standard symmetric monoidal structure on$\mathrm{DGMod}_{\mathbb K}$which sums the homological degrees, and for this structure$\operatorname{ch}$is not strongly monoidal, but there is a canonical <i>Eilenberg–Zilber</i> map$\operatorname{ch}(X) \otimes \operatorname{ch}(Y) \to \operatorname{ch}(X \otimes Y)$, which sums over all$(k+\ell)$-simplices in a product of a$k$-simplex with an$\ell$-simplex (closely related is the fact that for simplicial sets, the geometric realization of a product is <i>homeomorphic</i> to the product (in the category of compactly-generated spaces) of geometric realizations), making$\operatorname{ch}$into a "lax symmetric monoidal functor". The Eilenberg–Zilber map is a quasi-isomorphism, and one choice of quasi-inverse is the (non-symmetric) <i>Alexander–Whitney</i> map; if$\mathbb K \supseteq \mathbb Q$, there are other more symmetrical choices. In any case, the Eilenberg–Zilber map means that any simplicial commutative algebra determines canonically a dg commutative algebra. Of course, when$C = \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$, a simplicial affine scheme$X : \Delta^{\mathrm{op}} \to \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$determines a <i>co</i>simplicial commutative algebra$\mathcal{O}(X)$, and so$\operatorname{ch}(\mathcal{O}(X))$is not quite a dgca (the Alexander–Whitney map makes it into a dga). Anyway, this all won't matter much for me. </p> <p>What I wanted to mention about all this is that if$X : \Delta^{\mathrm{op}} \to \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$is a simplicial affine scheme, then$\operatorname{H}_\bullet(\operatorname{ch}(\mathcal{O}(X)))$is canonically a graded commutative algebra (supported in nonpositive homological degrees; and there is more algebraic data in the form of Massey products) and $$\operatorname{H}_0(\operatorname{ch}(\mathcal{O}(X))) = \mathcal{O}(\pi_0(X)).$$ </p> <h2>Examples</h2> <p>Let$A$be a commutative$\mathbb K$-algebra, with corresponding affine scheme$X = \operatorname{spec}(A) \in \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$. If you want, you can extend$X$to a constant functor$X : \Delta^{\mathrm{op}} \to \mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$. Let$S^1$denote the simplicial set generated by one nondegenerate$0$-simplex and one nondegenerate$1$-simplex. Then$\operatorname{maps}(S^1, X)$is a cosimplicial affine scheme (or simplicial cosimplicial, but constant in the simplicial direction), and so$\mathcal{O}(\operatorname{maps}(S^1, X))$is a simplicial commutative algebra. By definition, $$\operatorname{HH}_\bullet(A) = \operatorname{H}_\bullet(\operatorname{ch}(\mathcal{O}(\operatorname{maps}(S^1, X))))$$ is the <i>Hochschild homology</i> of$A$. The complex$\operatorname{ch}(\mathcal{O}(\operatorname{maps}(S^1, X)))$can be alternately defined by making a certain choice of resolution of$A$as an$(A\otimes A)$-module, and using this resolution to construct the derived tensor product$A \otimes_{A\otimes A} A$. </p> <p>Let$G$be an affine algebraic group over$\mathbb K$(e.g. a finite group). There is a well-known simplicial affine scheme$X = \mathrm{B}G$whose space of$k$-simplices is$G^k$, with boundary maps that encode the multiplication. Let$M$be a simplicial set, and I am primarily interested in the case that$M$is a simplicial finite set describing the homotopy type of a finite-dimensional compact manifold. The simplicial affine scheme$\underline{\operatorname{hom}}_\Delta(M,\mathrm{B}G)$is the space if <i>$G$-local systems on$M$</i>. In particular,$\pi_0(\underline{\operatorname{hom}}_\Delta(M,\mathrm{B}G))$is the <i>character variety</i> of$M$. </p> <h2>My question</h2> <p>I am looking for a general construction, of the flavor above, that incorporates both examples. More specifically, the construction should:</p> <ul> <li>input a simplicial (finite) set$M$and a a simplicial affine scheme$X$over$\mathbb K$</li> <li>output a chain complex$V(M,X)$over$\mathbb K$, supported in both directions, that deserves to be thought of as a "derived space of global functions on the space of maps from$M$to$X$"</li> <li>have good functoriality and monoidality properties in both variables (implying for instance that$V(M,X)$has a strongly-homotopy commutative dg algebra structure, coming from various diagonal and Eilenberg–Zilber-like maps)</li> <li>if$X = \operatorname{spec}(A)$is a constant simplicial scheme, then$V(M,\operatorname{spec}(A))$is the generalized Hochschild homology of$A$determined by$M$</li> <li>$\operatorname{H}_0(V(M,X)) = \mathcal{O}(\pi_0(\underline{\operatorname{hom}}_\Delta(M,X)))$</li> </ul> <h2>Some near misses</h2> <p>The problem seems to be when$X$is not "simply connected". In particular, I have not come across a construction that works even when$X = \mathrm{B}G$for$G$a finite simple group.</p> <p>Greg Ginot and collaborators (see e.g. <a href="http://www.math.jussieu.fr/~ginot/papers/Higher-Order-Hochschild-Long.pdf" rel="nofollow">Higher order Hochschild cohomology</a>, <a href="http://arxiv.org/abs/1011.6483" rel="nofollow">Derived Higher Hochschild Homology, Topological Chiral Homology and Factorization algebras</a>, and <a href="http://arxiv.org/abs/0905.2231" rel="nofollow">A Chen model for mapping spaces and the surface product</a>) have extended <a href="http://archive.numdam.org/ARCHIVE/ASENS/ASENS_2000_4_33_2/ASENS_2000_4_33_2_151_0/ASENS_2000_4_33_2_151_0.pdf" rel="nofollow">work by Pirashvili</a> defining the generalized Hochschild homology. Let$A$be a cdga over$\mathbb K \supseteq \mathbb Q$and let$M$be a simplicial set. Then there is a simplicial cdga$\int_M A = \mathcal{O}(\operatorname{maps}(M,\operatorname{spec}(A)))$with good functoriality and monoidality properties, which agrees up to quasi-isomorphism with Lurie's "topological chiral homology." </p> <p>By definition, a <i>quasi-isomorphism</i> of cdgas is a morphism that induces isomorphisms on homology. One of the things that Ginot et al prove is that a quasi-isomorphism$A \to B$induces a quasi-isomorphism$\int_M A \to \int_M B$. Thus in particular when$A = \mathcal{O}(\mathrm{B}G) = \operatorname{Ext}_G(\mathbb K,\mathbb K)$, for any meaning of this, and$G$is a finite simple group, then the canonical map$\mathbb K \to A$is a quasi-isomorphism, and so the chain complex$\int_M A$will never contain data. So this construction fails my last condition, e.g.:$\pi_0(\underline{\operatorname{hom}}_\Delta(S^1,\mathrm{B}G)) = G/G^{\mathrm{conj}}$and$\mathcal{O}(\pi_0(\underline{\operatorname{hom}}_\Delta(S^1,\mathrm{B}G))) = \mathcal{O}(G)^G$is the algebra of class functions on$G$, whereas$\operatorname{H}_0(\int_{S^1}\mathcal{O}(\mathrm{B}G)) = \mathbb K$.</p> <p>Ben-Zvi and Nadler have discussed <a href="http://arxiv.org/abs/1002.3636" rel="nofollow">loop spaces and connections</a> their relationships to Hochschild homology and <a href="http://arxiv.org/abs/1004.5120" rel="nofollow">representations</a>. They run into what I believe are related issues, but work primarily with not the space of loops in a derived scheme, but rather the infinitesimal neighborhood of the constant loops within that space. I should also mention that for my particular application, I really am looking for an explicit one-categorical construction (akin to the Pirashvili-style work), rather than quickly moving to model or$\infty$categories.</p> <p>Finally, perhaps the result I should have started with is one I learned from <a href="http://arxiv.org/abs/1110.0405" rel="nofollow">a review by Loday</a> (original references are included there). Suppose that$M$is a simplicial approximation of an$n$-dimensional manifold, and that$X$is a simplicial set which is <i>$n$-connected</i>, in$\pi_{\leq n}(X)$is trivial. (So <i>$1$-connected</i> means connected simply-connected.) I can build a cosimplicial simplicial set$\operatorname{maps}(M,X)$, and a simplicial set$\underline{\operatorname{hom}}_\Delta(M,X)$, as discussed above. There is a _free$\mathbb K$-module_ functor$\mathbb K : \mathrm{SET} \to \mathrm{Mod}_{\mathbb K}$, and with is I get a cosimplicial simplicial$\mathbb K$-module$\mathbb K\operatorname{maps}(M,X)$and a simplicial$\mathbb K$-module$\mathbb K\underline{\operatorname{hom}}_\Delta(M,X)$. Of course, given a cosimplicial simplicial$\mathbb K$-module, I can apply the "alternating sum of boundaries" functor$\operatorname{ch}$to get a bicomplex, which I can then totalize. Unless I have made a mistake, I believe the statement is that under the conditions on$M$and$X$, the canonical map of chain complexes between$\operatorname{ch}(\mathbb K\operatorname{maps}(M,X))$and$\operatorname{ch}(\mathbb K\underline{\operatorname{hom}}_\Delta(M,X))$is a quasi-isomorphism. This example specifically does not include classifying spaces of finite groups.</p> <h2> Final comments and examples</h2> <p>Truth be told, I am most interested in the case$X = \mathrm B G$for$G = \mathrm{SL}(2)$and$M$a simplicial approximation of a three-manifold. Note that the topological space$\mathrm{B}(\mathrm{SL}(2,\mathbb C))$is$3$-connected (it is homotopy equivalent to$\mathrm{B}(\mathrm{Spin}(3,\mathbb R))$), but I don't have a good sense about notions like "3-connected" for algebraic stacks. And, besides, I would like a robust construction.</p> <p>Let me end with an example that does work. A much easier category than$\mathrm{CAlg}^{\mathrm{op}}_{\mathbb K}$is the category$\mathrm{CCog}_{\mathbb K}$of cocommutative coalgebras (or "cogebres" in French, hence the name). A group object in$\mathrm{CCog}_{\mathbb K}$is a <i>cocommutative</i> Hopf algebra, and a good example is the universal enveloping algebra$U\mathfrak g$of a Lie algebra$\mathfrak g$. Then$\mathrm B U\mathfrak g = \mathrm B \mathfrak g$is a simplicial cocommutative coalgebra, which is quasi-isomorphic to the dg cocomutative coalgebra$\mathrm{CE}(\mathfrak g)$of Chevalley–Eilenberg cochains with trivial coefficients. I believe that it _is_ true that Hochschild homology of$\mathrm{CE}(\mathfrak g)$is the Chevalley–Eilenberg cochain complex with coefficients in$U\mathfrak g$, as it should be if you think about loop spaces. </p> http://mathoverflow.net/questions/114631/are-annihilation-modules-in-the-quantum-torus-necessarily-principal Are annihilation modules in the quantum torus necessarily principal? Theo Johnson-Freyd 2012-11-27T05:05:58Z 2012-11-27T18:01:38Z <p>I hope that my question yields some standard fact from (noncommutative) ring theory. In discussions with other graduate students, we have outlined some approaches to tackling the question, but haven't come up with an answer.</p> <p><strong>Notational caveats:</strong> I will work over some commutative ring that I will call$\mathbb Z$, but if you like, you may let$\mathbb Z$denote your favorite algebraically closed field, or whatever — I am agnostic about that type of thing. I will denote by$\mathbb Z_q$the Laurent polynomial ring$\mathbb Z_q = \mathbb Z[q^{\pm 1}]$in a commuting parameter$q$, but again I am reasonably agnostic: if you need to use the field of rational functions, say, then so be it.</p> <p><strong>The quantum torus:</strong> I am interested in the noncommutative (but "$q$-mutative") Laurent polynomial ring of "functions on the quantum torus", defined by: $$R = \mathbb Z_q\langle x^{\pm 1},y^{\pm 1}\rangle / (yx = qxy)$$ It is an "exponential" version of the Heisenberg ring$\mathbb Z[\hbar]\langle \xi,\upsilon\rangle / ([\upsilon,\xi] = \hbar)$, with$q = e^\hbar$,$x = e^\xi$, and$y = e^\upsilon$.</p> <p>I am also interested in a distinguished left$R$-module: $$M = \mathbb Z_q\langle x^{\pm 1}\rangle$$ $$x \triangleright f(x) = x\ f(x), \quad y \triangleright f(x) = f(qx)$$ Again, this is an exponential version of the action of differential operators on polynomials, in which$\upsilon$acts by$\frac{\partial}{\partial \xi}$.</p> <p>Fix$f\in M$. It defines a left ideal$\operatorname{Ann}(f) \subseteq R$consisting of those$r\in R$such that$r\triangleright f = 0$.</p> <p><strong>My question:</strong> (When) is$\operatorname{Ann}(f)$principal? I.e. (for which$f$) does there exist an element$a_f \in R$such that$\operatorname{Ann}(f) = Ra_f$?</p> <p>If it exists, then$a_f$is determined up to units in$R$, which are precisely the monomials (unless I am mistaken). Is there an algorithm that computes$a_f$from$f$?</p> <p>For example, I believe that$\operatorname{Ann}(x^n)$is generated by$y - q^n$. On the other hand, included within$\operatorname{Ann}(x-1)$are the elements$(y-1)(y-q)$and$x(y-q) - (y-1)$, and I don't see immediately a principal generator.</p> <p><strong>Not the main question, but an interesting topic for discussion:</strong> The classical limit$q \to 1$gives the classical torus$R|_{q=1} = \mathbb Z[x^{\pm 1},y^{\pm 1}]$, with the symplectic structure$\omega = \frac{\mathrm dy}{y}\wedge \frac{\mathrm dx}{x}$as the 1-jet of the noncommutativity. If I coarsely take the classical limit before computing the annihilation ideal, then for all non-zero$f$I get the ideal generated by$(y-1)$, because$M$is a domain. The more interesting thing to do is to compute$\operatorname{Ann}(f)$in the quantum ring$R$, and then specialize$q \to 1$. This gives the ideal generated by$a_f(q=1)$, provided$\operatorname{Ann}(f)$is principal in$R$. Can the vanishing locus of$\operatorname{Ann}(f)|_{q=1}$be described in terms of the (symplectic) geometry of the function$f$? </p> http://mathoverflow.net/questions/114363/impact-of-lhc-on-math/114366#114366 Answer by Theo Johnson-Freyd for Impact of LHC on math ? Theo Johnson-Freyd 2012-11-24T19:33:01Z 2012-11-24T19:33:01Z <p>I am not an expert on the following aspect of LHC, so I'll bring it up but hope others will elaborate (feel free to use the community wiki features!):</p> <p>One of the major contributions of LHC has been to develop better data-management technology. LHC generates way too much data to <em>store</em>, let alone to transmit, and so quite a lot of statistics, mathematics, and computer research went into designing hardware that makes instantaneous decisions about what data to save. Once the robot has culled the data, transmitting it to humans for further processing is quite a task, and it's reasonable to think that the hardware developed to do so will provide one of the models for a next-generation internet.</p> http://mathoverflow.net/questions/112013/morita-semi-equivalences/112040#112040 Answer by Theo Johnson-Freyd for Morita semi-equivalences Theo Johnson-Freyd 2012-11-10T22:49:16Z 2012-11-10T22:49:16Z <p>This was slightly too long to be a comment, but it certainly is not a complete answer.</p> <p>Since module categories behave very much like abelian groups, I expect you can say a lot by thinking in analogy with the lower-categorical-level situation: you have two abelian groups$A$and$B$, and homomorphisms$n : A \to B$and$m: B \to A$, such that$nm = \mathrm{id}_B$. Then you certainly can say that$mn : A\to A$is a projection onto a direct summand isomorphic to$B$, and$\mathrm{id}_A - mn$is a projection onto the other direct summand.</p> <p>Back to module categories, you may not always be able to make sense of the difference${_A A_A} \ominus ({_A {M\otimes_B N}_A})$. Two best cases are: perhaps you have a natural inclusion${_A {M\otimes_B N}_A}\hookrightarrow {_A A_A}$, in which case the difference is the quotient; or perhaps you have a natural projection${_A A_A} \twoheadrightarrow {_A {M\otimes_B N}_A}$, in which case the difference is the kernel. More generally, you may be in the situation where the functors$M\otimes_B$and$N\otimes_A$are one-sided adjoints, in which case you can use whichever of the unit or counit of the adjunction is appropriate. Recall also that the cleanest way to take differences is to work in the derived category, in which case the difference <i>is</i> the cone of the (co)unit. (In the derived category, there are many available differences, one of which is the cone of the zero map, but that almost certainly will not be useful to you.) There probably will be some conditions on the map between$A$and$M\otimes_B N$coming from the request that${_A A_A} \ominus ({_A {M\otimes_B N}_A})$is a projection. </p> http://mathoverflow.net/questions/111497/can-i-bound-the-degree-of-a-contracting-homotopy-in-an-exact-filtered-complex Can I bound the degree of a contracting homotopy in an exact filtered complex? Theo Johnson-Freyd 2012-11-05T00:17:45Z 2012-11-05T03:36:51Z <p>Suppose that I have given you a bigraded vector space$V = \bigoplus_{i,j} V_{i,j}$. The first grading is a "homological"$\mathbb Z$-grading, and the second is an independent$\mathbb Z$-grading. Furthermore, I will supply a differential$\partial : V \to V$(i.e.$\partial^2 = 0$). With respect to the homological grading, the differential does what you expect, lowering homological degree by$1$:<br> $$\forall i,\quad \partial : \bigoplus_j V_{i,j} \to \bigoplus_j V_{i-1,j}$$ You probably would expect that the differential preserves (say) the independent grading, but alas, with respect to this independent grading the differential is only filtered, which is to say "nonincreasing". Actually, I will guarantee to you a little bit more. Namely, there is some bound$b \in \lbrace 0,1,2,\dots\rbrace$such that the differential$\partial$does not decrease the "$j$"-degree by more than$b$: $$\forall i\forall m, \quad \partial : V_{i,m} \to \bigoplus_{m-b \leq j \leq m} V_{i-1,j}$$</p> <p>Finally, suppose that I guarantee to you that$\partial$is <em>exact</em>, meaning that$\operatorname{im}(\partial) = \operatorname{ker}(\partial)$, so that all homology groups vanish.</p> <p>My question is: is it possible to bound the degrees of a homotopy witnessing this exactness?</p> <p>More specifically, recall that the best way to prove that a differential$\partial$is exact is to demonstrate a <em>homotopy</em>$h : V \to V$such that$\partial h + h\partial = 0$. As always, I will demand that$h$raises the homological "$i$" degree by$1$.</p> <blockquote> <p>Can such an$h$be found which does not increase the "$j$" degree by more than, say,$b$?</p> </blockquote> <p>If$b = 0$, then since I am working in vector spaces the answer is YES. Specifically, if$b=0$, then I can break my complex into a direct sum indexed by$j$, and for each$j$the subcomplex$\partial : V_{\bullet,j} \to V_{\bullet-1,j}$is exact, and therefore has a contracting homotopy.</p> <p>Note that I do need to allow$h$to increase degree in general. For example, if the only non-zero terms in$V$are $$V_{1,1} = V_{0,0} = \mathbb k = \text{ground field}$$ and$\partial$is an isomorphism$V_{1,1} \to V_{0,0}$, then the only contracting homotopy is the inverse isomorphism$h : V_{0,0} \to V_{1,1}$, which increases "$j$"-degree by$1$.</p> <p>On the other hand,$h$cannot be bounded below. For example, consider the complex in which $$V_{1,j} = V_{0,j} = \mathbb k \text{ for } 0 \leq j \leq n \text{ and } 0 \text{ otherwise}$$ and $$\partial = \begin{pmatrix} 1 & -1 & \\ & 1 & -1 & \\ && 1 & \ddots & \\ &&& \ddots & -1 \\ &&&& 1 \end{pmatrix}$$ then $$h = \partial^{-1} = \begin{pmatrix} 1 & 1 & 1 & \cdots & 1 \\ & 1 & 1 & & 1\\ && 1 & \ddots & \vdots \\ &&& \ddots & 1 \\ &&&& 1 \end{pmatrix}$$ </p> <p>Similarly, if$\partial$is bounded above, but not by$0$, then then answer is NO.</p> <p>Finally, note that since$\partial$is non-increasing, its$j$-degree-preserving piece is also a differential on$V$; specifically, it is the <em>associated graded</em> differential$\operatorname{gr}(\partial)$. If$\operatorname{gr}(\partial)$is exact, then so is$\partial$, and moreover the answer to my question is YES, and in fact you can find an$h$which is non-increasing. The example above of the invertible upper-triangular matrix is typical.</p> http://mathoverflow.net/questions/111100/a-geometric-characterization-of-rees-algebras-in-categories-without-choice A geometric characterization of Rees algebras in categories without Choice Theo Johnson-Freyd 2012-11-01T03:14:58Z 2012-11-01T03:14:58Z <p>Before asking my question, a caveat: The category theorist in me would like me to ask this question in more generality, but I will restrict my scope since what I'm really after is some geometric intuition.</p> <p>Let$R$be a commutative ring. A <em>$\mathbb Z$-filtered$R$-module</em> is a diagram$\cdots \hookrightarrow X_{\leq -1} \hookrightarrow X_{\leq 0} \hookrightarrow X_{\leq 1} \hookrightarrow \dots$, where each$X_{\leq i}$is an$R$-module and each map is an inclusion. I will abuse notation and denote by$X$both the diagram and its direct limit (the union of the$X_{\leq i}$s). I am happy to assume that the inverse limit (the intersection of the$X_{\leq i}$s) is trivial. The <em>associated graded</em> of$X$is the$\mathbb Z$-graded$R$-module$\operatorname{gr}(X) = \bigoplus (X_{\leq i}/X_{\leq i-1})$, where the piece$(X_{\leq i}/X_{\leq i-1})$is put in grading$i$.</p> <p>The <em>Rees algebra</em> of$X$is the$R[\epsilon]$-module $$\operatorname{Rees}(X) = \sum_i \epsilon^i X_{\leq i}[\epsilon] \subseteq X[\epsilon^{\pm 1}].$$ I.e. you look at the$X[\epsilon^{\pm 1}] = X \otimes_R R[\epsilon,\epsilon^{-1}]$, and inside it you take the$R[\epsilon]$-module generated by all elements of the form$\epsilon^i x$for$x\in X_{\leq i}$.</p> <p>Then the Rees algebra interpolates between$X$and$\operatorname{gr}(X)$in the following sense: $$\operatorname{Rees}(X)/(\epsilon=1) = \operatorname{Rees}(X) \otimes_{R[\epsilon]} R[\epsilon]/(\epsilon-1) \cong X$$ $$\operatorname{Rees}(X)/(\epsilon=0) = \operatorname{Rees}(X) \otimes_{R[\epsilon]} R[\epsilon]/(\epsilon) \cong \operatorname{gr}(X)$$ Here the "$=$" signs are definitions and the "$\cong$" signs are canonical isomorphisms. The first of these should be completely obvious; the second is an important calculation.</p> <p>The Rees algebra has the following geometric interpretation. The <em>affine line</em> is$\mathbb A^1 = \operatorname{spec}(R[\epsilon])$, and so any$R[\epsilon]$-module is a sheaf over$\mathbb A^1$. Consider the sheaf defined by$\operatorname{Rees}(X)$. The above isomorphisms say that$\operatorname{gr}(X)$is the fiber of$\operatorname{Rees}(X)$over$0 \in \mathbb A^1$, and (the union of)$X$is the fiber over$1\in \mathbb A^1$.</p> <p>Moreover,$\mathbb A^1$has an action by the group$\mathbb G_m = \mathrm{GL}(1) = \operatorname{spec}(R[\lambda^{\pm 1}])$given by$\epsilon \mapsto \lambda \epsilon$. This action extends to$\operatorname{Rees}(X)$, by restricting the action on$X[\epsilon^{\pm 1}]$. Thus:</p> <blockquote> <p>$\operatorname{Rees}(X)$is a$\mathbb G_m$-equivariant sheaf over$\mathbb A^1$.</p> </blockquote> <p>Since$\mathbb G_m$fixes$0 \in \mathbb A^1$, the fiber$\operatorname{gr}(X)$of$\operatorname{Rees}(X)$over$0$inherits a$\mathbb G_m$-action. This is precisely the grading, wherein the$i$th graded piece is the weight space on which$\lambda \in \mathbb G_m$acts by$\lambda^i$. I have been told that one should think of$X$-as-a-filtered-thing not as the fiber over$1$but rather as the fiber over the generic point in$\mathbb A^1$.</p> <p>Now, not every$\mathbb G_m$-equivariant sheaf over$\mathbb A^1$arises as the Rees algebra of a filtered$R$-module. When$R$is a field, I believe the correct statement is that:</p> <blockquote> <p>Over a field, the$\mathbb G_m$-equivariant sheaves over$\mathbb A^1$that arise as Rees algebras of filtered modules are precisely the$\mathbb G_m$-equivariant vector bundles.</p> </blockquote> <p>Put another way, over a field, there always exist isomorphisms between$X$and$\operatorname{gr}(X)$(i.e. trivializations of the sheaf near$0$). The existence of such isomorphisms requires the <em>axiom of choice</em>, which says that every epimorphism splits. But this description doesn't make sense in generality. Hence:</p> <blockquote> <p><strong>Question:</strong> When$R$is not a field, so that I do not necessarily have isomorphisms of$R$-modules$X \cong \operatorname{gr}(X)$, what is the geometric characterization that determines when a$\mathbb G_m$-equivariant sheaf on$\mathbb A^1$is the Rees algebra of a filtered module?</p> </blockquote> http://mathoverflow.net/questions/110859/graphical-calculus-in-braided-g-crossed-fusion-categories-explanation-request-a/110927#110927 Answer by Theo Johnson-Freyd for Graphical calculus in braided G crossed fusion categories: Explanation request and a question Theo Johnson-Freyd 2012-10-28T18:17:46Z 2012-10-28T18:17:46Z <p>I recommend that you become familiar with graphical notation. As you will see from my answer, other ("formula") approaches are sub-par.</p> <p>Formula (4.1) on page 14 of the linked paper by Kirillov defines a morphism$T_X : A \to A$as follows. First,$A$is a rigid algebra object in a modular category$\mathcal C$(with all associators suppressed, and braiding denoted$\beta_{M,N} : M\otimes N \to N\otimes M$), and$X$is a rigid$A$-module. I will write the multiplication on$A$by$m_A$and the left-action of$A$on$X$by$m_X$. Not having read the paper carefully, I believe that "rigid" means that$A$and$X$are each isomorphic to their own duals, and that this isomorphism is chosen to have good compatibility properties. In particular,$A$should in fact be a Frobenius algebra for this isomorphism, and perhaps a symmetric one at that. I will write the unit as$u_X : 1 \to X\otimes X$and the counit as$\epsilon_X : X\otimes X \to 1$, and similarly for$A$. Then we can consider the following composition: $$\begin{eqnarray} A & \to & X \otimes X \otimes A & \quad\quad & (u_X \otimes \mathrm{id}_A) \\ & \to & X \otimes A \otimes X && (\mathrm{id}_X \otimes \beta_{X,A}) \\ & \to & X \otimes A \otimes A \otimes A \otimes X && (\mathrm{id}_A \otimes \mathrm{id}_X \otimes u_A \otimes \mathrm{id}_{X}) \\ & \to & X \otimes A \otimes X && (\mathrm{id}_X \otimes m_A \otimes m_X) \\ & \to & X \otimes X \otimes A && (\mathrm{id}_X \otimes \beta_{A,X}) \\ & \to & A && (\epsilon_X \otimes \mathrm{id}_A) \end{eqnarray}$$ Or, to put it another way, $$T_X = (\epsilon_X \otimes \mathrm{id}_A) \circ (\mathrm{id}_X \otimes \beta_{A,X}) \circ (\mathrm{id}_X \otimes m_A \otimes m_X) \circ (\mathrm{id}_A \otimes \mathrm{id}_X \otimes u_A \otimes \mathrm{id}_{X}) \circ (\mathrm{id}_X \otimes \beta_{X,A})$$ </p> <p>The point is, this is a mess, and is entirely unenlightening what's really happening. Moreover, the coherency axioms assure that there are many equivalent ways to write the above map. For example, I could have replaced the last two steps$(\epsilon_X \otimes \mathrm{id}_A) \circ (\mathrm{id}_X \otimes \beta_{A,X})$with$(\mathrm{id}_A \otimes \epsilon_X) \circ (\beta_{X,A}^{-1} \otimes \mathrm{id}_X)$. </p> <p>As for your second question, skimming did not for me reveal any place in the paper where that notation is used. I could make guesses, but perhaps someone else will have studied this paper more closely.</p> <p>A final, non-math remark: The phrase "punctured curve" has a technical meaning in various areas, including areas close to this paper, to mean a compact Riemann surface with finitely many points removed (or variations on this notion). The standard term for "font" in which Kirillov draws his$A$strands is "dashed", as opposed to "solid" for$X$. And a good term for the edges in such diagrams is "strands" — I have also seen "edges" and "strings", but the latter in particular is problematic because to a physicist a "string" is something that through time traces out a surface ("worldsheet"), whereas one meaning of these graphical calculi is some "particles"$A$and$X$traveling through time and thereby tracing out "worldlines".</p> http://mathoverflow.net/questions/131154/differential-form-on-a-compact-manifold-whose-exterior-derivative-is-nowhere-zero Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-05-20T04:29:02Z 2013-05-20T04:29:02Z It might be helpful for you to draw a picture. A 2-form on$S^2$can be thought of as a map$\mathrm T S^2 \to \mathrm T^* S^2$from the tangent to the cotangent bundle, and as such it has, at each point, a kernel which is either$0,1,$or$2$-dimensional. Since your 2-form is a restriction from$\mathbb R^3$, the kernel is the intersection of$\mathrm T S^2$with the kernel of the map$\mathrm T\mathbb R^3 \to \mathrm T^*\mathbb R^3$, and you can draw this. The 2-form vanishes exactly when the kernel is everything, so work out at which points$\ker \mathrm d\omega$is parallel to$S^2$. http://mathoverflow.net/questions/131110/what-are-the-main-structure-theorems-on-finitely-generated-commutative-monoids Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-05-20T04:22:36Z 2013-05-20T04:22:36Z @Noah S: I imagine that there are certain versions of the problem &quot;classify numerical monoids&quot; that really mean &quot;understand the primes&quot;. Certainly there are &quot;classification&quot; problems among prime numbers that are not known. http://mathoverflow.net/questions/131175/closed-geodesic-loops-around-points-in-compact-manifolds Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-05-20T00:32:53Z 2013-05-20T00:32:53Z (I take it you don't count the constant loop.) Of course, if$M$has nontrivial$\pi_1$, then you can find a geodesic with fixed endpoints representing any nontrivial homotopy class by minimizing the energy. In the simply-connected case I don't see an immediate proof. http://mathoverflow.net/questions/131066/in-cell-decomposed-manifolds-how-easy-is-it-to-arrange-for-the-tubular-neighborh Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-05-18T21:57:57Z 2013-05-18T21:57:57Z @Ryan: Thanks for the comment! I was unaware of their work, but I'll look it up. http://mathoverflow.net/questions/130783/reasons-to-prefer-one-large-prime-over-another-to-approximate-characteristic-zero Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-05-16T02:17:59Z 2013-05-16T02:17:59Z ... go into developing precisely this type of heuristic. http://mathoverflow.net/questions/130783/reasons-to-prefer-one-large-prime-over-another-to-approximate-characteristic-zero Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-05-16T02:17:11Z 2013-05-16T02:17:11Z I think this is an interesting question, to which I do not have an answer. I will point out that in some sense no prime is better than any other: for any particular finite set of primes, certainly there are sentences that fail exactly on that set. So your question presupposes something about &quot;interesting&quot; questions that can be answered by an algorithm, or about questions that are &quot;likely&quot; to come up in &quot;research&quot;. I doubt that pure model theory and pure number theory can give an absolute answer to things about &quot;interesting&quot; questions and &quot;likely research&quot;, but conversely much work does ... http://mathoverflow.net/questions/130537/reflection-lie-groups Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-05-14T04:16:37Z 2013-05-14T04:16:37Z I don't have a useful answer to your question, but I vaguely recall at least one conference talk in which$O(n)$was compared, with quite some success, to a finite reflection group, so this question is not entirely out of left field. http://mathoverflow.net/questions/130516/the-category-of-right-comodule-of-coalgebra-is-a-monoidal-category-why Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-05-14T04:13:20Z 2013-05-14T04:13:20Z Please look over <a href="http://mathoverflow.net/howtoask" rel="nofollow">mathoverflow.net/howtoask</a> and revise this question: it is missing definitions, should have better capitalization and punctuation, and might as well have the mathematics typeset as actual TeX (see the box &quot;How to write math&quot; on the right-hand column). But probably no change will fix your question, as in general there is no good monoidal structure for the category of comodules of a general coalgebra. Two situations in which comodule categories are monoidal are when the coalgebra in question is cocommutative, or when the coalgebra is given a Hopf algebra structure. http://mathoverflow.net/questions/128145/cocompleteness-of-the-category-of-small-a-infty-categories Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-04-20T05:43:11Z 2013-04-20T05:43:11Z The 1-category of small$A_\infty$categories is certainly cocomplete, as it is the category of representations of a Lawvere algebraic theory and hence presentable. But this is, of course, not the category that you care about. Presumably, you care instead about some version of an$(\infty,2)$-category, since$A_\infty$categories are some version of$(\infty,1)$-categories. http://mathoverflow.net/questions/128130/what-does-a-mathematician-expect-from-mathematics-education Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-04-20T05:26:41Z 2013-04-20T05:26:41Z I have voted to close this question as subjective and argumentative. If it is to be saved, at a minimum the first three sentences must go. I suppose that, being CW, I could just go in and change the question. Perhaps I will do so if the question is not closed, but even with the changes I see to make, I'm not sure I can make the question into an appropriate one for MO. http://mathoverflow.net/questions/128003/how-to-understand-chern-simons-action Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-04-19T01:30:20Z 2013-04-19T01:30:20Z I have no complaints about Urs's answer below, but if you want a less high-brow discussion, I recommend Freed's articles &quot;Classical Chern-Simons Theory Part I&quot; <a href="http://arxiv.org/abs/hep-th/9206021" rel="nofollow">arxiv.org/abs/hep-th/9206021</a> and &quot;Part II&quot; <a href="http://ace1.ma.utexas.edu/users/dafr/cs2.pdf" rel="nofollow">ace1.ma.utexas.edu/users/dafr/cs2.pdf</a> . http://mathoverflow.net/questions/127779/higher-homotopy-groups-and-representation-theory/127911#127911 Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-04-19T01:22:07Z 2013-04-19T01:22:07Z Well, I mean, there are many applications of the general statement that$\pi_2$is trivial. But for no$G$can elements of$\pi_2(G)$be applied to some particular construction, the way that$\pi_1$or$\pi_3$can. http://mathoverflow.net/questions/127779/higher-homotopy-groups-and-representation-theory Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-04-18T03:50:56Z 2013-04-18T03:50:56Z &quot;if G˜ is the simply connected universal cover of G then all representations of g can be integrated to representations of G&quot;. You should include &quot;finite-dimensional&quot; somewhere in that sentence. The Lie algebra$\mathbb R$acts on$\mathcal C^\infty(I)$, where$I$denotes the open unit interval$I = (0,1)$, by sending the basis vector to$\frac{\partial}{\partial x}$, but this representation is not integrable to a representation of$\mathbb R$on$\mathcal C^\infty(I)\$. http://mathoverflow.net/questions/127905/max-eigenvalue-property-of-a-matrix-of-a-sum-of-matrix-like-x-ix-jt Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-04-18T03:34:11Z 2013-04-18T03:34:11Z Hi unknown: As written, I think the correct question is &quot;no, no one has some clue&quot;. Please read <a href="http://mathoverflow.net/howtoask" rel="nofollow">mathoverflow.net/howtoask</a> and revise this question. http://mathoverflow.net/questions/127631/felder-kazhdan-classical-me Comment by Theo Johnson-Freyd Theo Johnson-Freyd 2013-04-16T03:01:16Z 2013-04-16T03:01:16Z Well, it has only been a few months...