User dan petersen - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:57:07Z http://mathoverflow.net/feeds/user/1310 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130833/generalization-of-the-lefschetz-fixed-point-theorem/130837#130837 Answer by Dan Petersen for Generalization of the Lefschetz fixed point theorem Dan Petersen 2013-05-16T14:39:44Z 2013-05-16T14:45:20Z <p>One of the standard proofs of the Lefschetz formula proceeds by showing that the RHS of your equation computes the intersection number between the graph of $f$ and the diagonal inside $X \times X$. The usual case is when this intersection has expected dimension. The general case requires the excess intersection formula. Now, the normal bundle of the diagonal $\Delta$ in $X \times X$ is the tangent bundle of $\Delta = X$. After dividing by the normal bundle of $e(g)$ in $X$, the excess intersection formula shows that the intersection number is given by the top Chern class of the tangent bundle of $e(g)$. But this is just the topological Euler characteristic of $e(g)$. (This last fact is just the special case of your formula when $g = \mathrm{id}$!)</p> http://mathoverflow.net/questions/33282/can-we-have-aa/33283#33283 Answer by Dan Petersen for Can we have A={A} ? Dan Petersen 2010-07-25T15:20:59Z 2013-05-09T19:34:02Z <p>In standard set theory (ZF) this kind of set is forbidden because of the <a href="http://en.wikipedia.org/wiki/Axiom_of_foundation" rel="nofollow">axiom of foundation</a>.</p> <p>There are alternative axiomatisations of set theory, some of which do not have an equivalent of the axiom of foundation. This is called non-well-founded set theory. See e.g. <a href="http://en.wikipedia.org/wiki/Aczel%27s_anti-foundation_axiom" rel="nofollow">Aczel's anti-foundation_axiom</a>, where there is a unique set such that <code>$x = \{x\}$</code>.</p> http://mathoverflow.net/questions/129811/frobenius-weights-on-etale-cohomology-and-purity/129813#129813 Answer by Dan Petersen for Frobenius weights on etale cohomology and purity Dan Petersen 2013-05-06T09:25:08Z 2013-05-06T12:48:59Z <p>You remember correctly, that the effect of Tate twisting the sheaf $\mathcal F$ is just to multiply the Frobenius eigenvalues by a factor of $q$. This result is not surprising because the Tate twist $\mathcal F(m)$ has no relation to the Serre twist on a projective variety, they are just denoted the same way. </p> <p>The analogous operation to Tate twist in the classical cohomology of a complex variety is defined on the singular cohomology, not in coherent cohomology. There it has the effect of changing the natural $\mathbf Z$-structure on the complex cohomology by a factor $(2\pi i)$, shifting the weight filtration on the mixed Hodge structure by two steps and the Hodge filtration by one step.</p> <p>Purity in $\ell$-adic cohomology has no direct relation to cohomological purity. But if you interpret a Frobenius eigenvalue of absolute value $q^{k/2}$ as something "k-dimensional", then both are statements that one thing or another is equidimensional, that it has pure dimension.</p> http://mathoverflow.net/questions/128965/thom-gysin-sequences-and-stratifications/128978#128978 Answer by Dan Petersen for Thom-Gysin Sequences and Stratifications Dan Petersen 2013-04-28T05:59:40Z 2013-04-28T05:59:40Z <p>I haven't read Kirwan's thesis but I think this is what you are after. If $U \subset V$ is an open subvariety (no conditions on the variety $V$) then there is a long exact sequence (Thom-Gysin) $$H^\bullet_c(U) \to H^\bullet_c(V) \to H^\bullet_c(V \setminus U) \to H^{\bullet+1}_c(U).$$</p> <p>Let's now declare a stratification of $X$ to be a collection of disjoint locally closed subvarieties $X_\alpha$ ("strata") whose union is $X$, and such that the closure of a stratum is a union of strata. Then for every such stratification the above LES can in principle be used to inductively study the compactly supported cohomology of $X$, one stratum at a time: start with the smallest strata and in each step let $V \setminus U$ be what you have already computed and $V$ the union with one more stratum.</p> <p>If the stratification is actually a filtration by closed subvarieties, $$X = T_d \supset T_{d-1} \supset \cdots \supset T_0,$$ then there is a spectral sequence $$E_1^{pq} = H^{p+q}_c(T_p\setminus T_{p-1})\implies H^{p+q}_c(X)$$ which encodes this procedure.</p> <p>So one needs almost no conditions at all on the stratification. A slogan is that compactly supported cohomology always behaves better w.r.t. stratified spaces.</p> <p>But probably you are interested in ordinary cohomology; one should add hypotheses relating compactly supported and usual cohomology. For instance if $X$ is compact then $H^\bullet_c(X) = H^\bullet(X)$, same for all closures of strata, and if all strata are smooth varieties then their cohomology with compact support is determined from the usual cohomology by Poincare duality. This covers many cases of interest.</p> http://mathoverflow.net/questions/128712/is-the-moduli-space-of-ppavs-smooth/128718#128718 Answer by Dan Petersen for Is the moduli space of ppAVs smooth? Dan Petersen 2013-04-25T13:46:41Z 2013-04-25T13:46:41Z <p>The answer is no, it is not smooth for any $g \geq 2$. For $g \geq 3$ the singular locus is precisely the locus of PPAVs with automorphism group greater than $\pm \mathrm{id}$, as proven in Oort, Frans: "Singularities of coarse moduli schemes". For $g=2$ there is IIRC a unique singular point which is in $M_2$ (the open subvariety of $A_2$ of Jacobians), I think this is in Igusa's paper "Arithmetic variety of moduli for genus two".</p> http://mathoverflow.net/questions/127737/smooth-modular-compactification-of-moduli-of-curves/127742#127742 Answer by Dan Petersen for smooth modular compactification of moduli of curves Dan Petersen 2013-04-16T19:42:19Z 2013-04-16T19:49:28Z <p>Sure: see Eduard Looijenga, "Smooth Deligne-Mumford compactifications by means of Prym level structures", which completely answers your question.</p> <p>There is also later work by de Jong-Pikaart, Boggi-Pikaart and Abramovich-Corti-Vistoli where more general non-abelian level structures are considered, and over more general base schemes.</p> http://mathoverflow.net/questions/125735/prorepresentable-functors-repres-by-alg-spaces-covering-spaces-by-alg-spaces/125741#125741 Answer by Dan Petersen for Prorepresentable functors repres. by alg. spaces? Covering spaces by alg. spaces. Dan Petersen 2013-03-27T16:49:14Z 2013-03-27T16:49:14Z <p>One nontrivial example where the universal cover is actually a scheme (locally of finite type) is when $X$ is the nodal cubic curve over a field $k$, i.e. $\mathbf P^1$ with two points identified. Note that when $k = \mathbf C$, we have $\pi_1(X(\mathbf C)) \cong \mathbf Z$. The universal cover $\widetilde X$ is an infinite chain of $\mathbf P^1$, where the point $0$ on the $i$th curve has been glued to the point $\infty$ on the $(i+1)$st curve, for all $i \in \mathbf Z$; the action of $\mathbf Z$ by deck transformation is translating along this chain.</p> http://mathoverflow.net/questions/124950/f-f-fj-is-mathbb-c-linearly-independent-if-f-is-a-modular-form/124951#124951 Answer by Dan Petersen for $f,f',...,f^{(j)}$ is $\mathbb C$-linearly independent if $f$ is a modular form Dan Petersen 2013-03-19T12:16:36Z 2013-03-19T12:16:36Z <p>The only functions $f$ for which $f,f ',\ldots, f^{(j)}$ are linearly dependent are linear combinations of exponential functions, as one learns in a basic course on ordinary differential equations. </p> <p>But maybe you meant to ask instead about whether $f$ and its derivatives are <em>algebraically independent</em>. Then it turns out that there always exists such an algebraic dependence. The reason is that $f$ and all its derivatives are quasimodular forms, and the ring of quasimodular forms has transcendence degree 3 over $\mathbb C$. So $f$ always satisfies some <em>nonlinear</em> third order differential equation. See e.g. Zagier's chapter in "The 1-2-3 of Modular Forms", section 5.</p> http://mathoverflow.net/questions/124816/alexander-duality-theorem/124817#124817 Answer by Dan Petersen for Alexander duality theorem Dan Petersen 2013-03-17T20:28:35Z 2013-03-17T20:46:31Z <p>Yes: whenever $U \subset M$ is an open subspace with complement $Z$, then there is a long exact sequence $$\ldots\to H^\bullet_c(U) \to H^\bullet_c(M) \to H^\bullet_c(Z) \to H^{\bullet+1}_c(U)\to \ldots$$ which in this case gives $$H^{n-1}_c(\mathbf R^n) = 0 \to H^{n-1}_c(\Sigma) \to H^n_c(\mathbf R^n \setminus \Sigma) \to H^n_c(\mathbf R^n) \to 0.$$ By Poincaré duality we have $H^n_c(\mathbf R^n \setminus \Sigma) \cong H^0(\mathbf R^n \setminus \Sigma)^\vee$ and $H^{n-1}_c(\Sigma) = H^0(\Sigma)^\vee$, so the ranks of these cohomology groups are just the numbers of connected components of $\mathbf R^n \setminus \Sigma$ resp. $\Sigma$. Since $H^n_c(\mathbf R^n)$ is one-dimensional this shows that $\mathbf R^n \setminus \Sigma$ has exactly one more connected component than $\Sigma$. Compactness of $\Sigma$ was an unnecessary hypothesis.</p> http://mathoverflow.net/questions/124390/a-universal-elliptic-curve/124392#124392 Answer by Dan Petersen for A Universal Elliptic Curve Dan Petersen 2013-03-13T06:28:05Z 2013-03-13T07:17:03Z <p>Every point of $X$ gives you an embedding of $\mathbb Z^2$ as a lattice in $\mathbb C$ (via the canonical inclusion $\mathbb Z^2 \subset \mathbb R^2$), and hence by quotienting an elliptic curve. This produces a universal diagram $$X \times \mathbb Z^2 \to X \times \mathbb C \to E,$$ where $E$ is a family of elliptic curves over $X$. The analytic sections of this diagram over $X$ give you Deligne's short exact sequence, you can think of it just as a sequence of sheaves of abelian groups.</p> http://mathoverflow.net/questions/122740/cohomology-of-configuration-space-of-a-compact-manifold/122789#122789 Answer by Dan Petersen for Cohomology of configuration space of a compact manifold Dan Petersen 2013-02-24T08:06:55Z 2013-02-24T08:06:55Z <p>I would be extremely surprised if there was anything other than the Cohen-Taylor spectral sequence that you can do in this generality.</p> <p>As you know, the first nontrivial page of Cohen-Taylor spectral sequence depends only on the ring $H^\bullet(M)$. I think it's written up somewhere that the higher order differentials are defined by Massey products on $M$. In particular the result you refer to in a comment that $H^\bullet(F(M,n))$ (or its associated graded) can be computed explicitly from $H^\bullet(M)$ extends to arbitrary formal manifolds $M$, not just smooth projective varieties. And if you give yourself a minimal model of $M$ you should be able to compute $H^\bullet(F(M,n))$ for all $n$ as well (but not a minimal model of $F(M,n)$ -- see the paper of Longoni and Salvatore).</p> http://mathoverflow.net/questions/122703/what-do-generic-torelli-theorems-claim/122715#122715 Answer by Dan Petersen for What do generic Torelli theorems claim? Dan Petersen 2013-02-23T11:56:00Z 2013-02-23T11:56:00Z <p>I would formulate it as follows: global Torelli says that the period map is injective/an immersion (but not necessarily an isomorphism!!), local Torelli says that the period map has injective differential, and generic Torelli says that the period map is generically injective (injective on a Zariski open dense set). </p> http://mathoverflow.net/questions/122537/what-is-the-intuition-inertia-orbifold-or-stack/122543#122543 Answer by Dan Petersen for What is the intuition inertia orbifold (or stack)? Dan Petersen 2013-02-21T15:30:52Z 2013-02-21T18:22:16Z <p>I am not sure what kind of answer you are looking for. But if you have a stack $X$, then the inertia stack $IX$ is basically the gadget parametrizing pairs $(x,\sigma)$ where $x$ is a point of $X$ and $\sigma$ is in the isotropy group at $x$ (an automorphism of $x$). Informally, the locus where you have automorphism group $G$ becomes "doubled" $|G|$ times. Example: for the stack $[\mathbb A^1/\mu_2]$ the inertia stack is $[\mathbb A^1/\mu_2] \sqcup B\mu_2$, the extra $B\mu_2$ corresponding to the origin being doubled because it has an extra automorphism. </p> <p>One way to think about it is as a kind of "infinitesimal loop space", where instead of taking maps to $X$ from a circle we take maps from the homotopically equivalent object $B\mathbb Z$. This is pleasant because the inertia stack is the fibered product $X \times_{X\times X} X$, and for a topological space $X$ the homotopy fibered product $X \times_{X\times X}^h X$ is the space of free loops on $X$. </p> <p>You can motivate the inertia stack through Gromov--Witten theory. If $X$ is a variety, then there is an evaluation map from the stack of stable $n$-pointed maps to $X$ to $X^n$. If $X$ is a stack, then the correct notion is that of a twisted stable map, and in this case the evaluation maps do not land on $X$ but in its inertia stack $IX$! (In fact it lands on the rigidified inertia stack, where some automorphisms have been removed from the picture, but nevermind this). So quantum cohomology of a stack is not extra structure on the cohomology ring of $X$ itself, but on the cohomology ring of $IX$. </p> <hr> <p>Since it was discussed in some now deleted comments, let me say a few words about the example $X = BG$. Then $X$ has a single point (point = $\mathbb C$-point), corresponding to the trivial torsor over a point. What are the automorphisms of the trivial torsor? For every $g \in G$ we have an automorphism given by the action of $G$. But these will not correspond to distinct points of the inertia stack $IX$ in general, because these automorphisms may be isomorphic. So we should figure out what is a morphism between automorphisms of the trivial torsor. By thinking a bit one sees that morphisms are given by conjugation in the group, i.e. a morphism between the automorphisms "acting by $g$" and "acting by $g'$" is an element $h$ such that $g = hg'h^{-1}$. In other words, $IX$ is given by the stack quotient $[G/G]$, where the first $G$ is the underlying set of $G$, and the group $G$ acts on itself by conjugation. Equivalently, <code>$IX = \coprod_{[g]} BC_G(g)$</code> where the disjoint union is taken over conjugacy classes in $G$ and $C_G(g)$ is the centralizer of $g$. Using unnecessarily fancy words, $IX$ is the classifying stack of the "loop groupoid" of the finite group $G$. This illustrates the informal statement earlier, that the locus with automorphism group $G$ becomes "doubled" $|G|$ times: the stack $[G/G]$ is of course $|G|$ times larger than $[pt/G]$, in a natural sense. </p> http://mathoverflow.net/questions/122511/are-constructible-derived-categories-invariant-up-to-weak-homotopy-equivalence/122518#122518 Answer by Dan Petersen for Are constructible derived categories invariant up to weak homotopy equivalence? Dan Petersen 2013-02-21T08:32:39Z 2013-02-21T13:40:08Z <p>I don't know much about these things but I think 1. fails even if $X$ is a point and $Y$ is a line. Let's say also $R$ is a field. Then the category of constructible sheaves on $X$ is just $R$-Vect which is in particular semisimple, but the category of constructible sheaves on $Y$ is not semisimple: if we choose a sheaf on a point $p$ and on the complement $U$ of that point, then there are in general many non-isomorphic choices of a sheaf on $Y$ with given restrictions to $p$ and $U$ which fit in a short exact sequence. </p> <p>So their derived categories are not equivalent either.</p> <hr> <p>Addendum: Maybe I can say something about how the usual homotopy invariance of cohomology is visible in the 'six functor' formalism. For any space $X$ let $\pi_X$ be the projection to a point, then the cohomology of $X$ (with any coefficients) just corresponds to functor $R \pi_{X\ast} \circ \pi_X^\ast$. If $f \colon X \to Y$ is an arbitrary map, then note that $\pi_Y \circ f = \pi_X$ which implies $$R\pi_{Y\ast} \circ Rf_\ast \circ f^\ast \circ \pi_Y^\ast = R \pi_{X\ast} \circ \pi_X^\ast;$$ and now the morphism $\mathbf 1 \to Rf_\ast \circ f^\ast$ coming from the adjunction gives a map <code>$f^\# \colon R \pi_{Y\ast} \circ \pi_Y^\ast \to R \pi_{X\ast} \circ \pi_X^\ast$</code>. Of course evaluating this equation on a choice of coefficients we get the usual map $H^\bullet(Y) \to H^\bullet(X)$. The correct way to express homotopy invariance is now that if $f$ and $g$ are homotopic maps $X \to Y$, then <code>$f^\#$</code> and <code>$g^\#$</code> are equal. This can be generalized to the relative situation, when $X$ and $Y$ are spaces over some base space $S$ and we consider the derived pushforward to $S$ instead of to a point, and we take homotopies over $S$.</p> http://mathoverflow.net/questions/122289/flatness-of-projective-bundles/122291#122291 Answer by Dan Petersen for Flatness of projective bundles Dan Petersen 2013-02-19T12:28:58Z 2013-02-19T12:28:58Z <p>For any $X,Y$ the projection $X \times Y \to Y$ is an open map. Since being open is a local property you know this holds also for any fiber bundle and in particular any projective bundle.</p> http://mathoverflow.net/questions/120604/classification-of-first-order-deformations-of-n-pointed-non-singular-variety/122053#122053 Answer by Dan Petersen for Classification of first order deformations of n-pointed non-singular variety Dan Petersen 2013-02-17T10:55:44Z 2013-02-17T10:55:44Z <p>It shouldn't be hard to adapt the proof for unpointed curves. The idea is the following. If $\newcommand \G {\mathcal G} \G$ is a sheaf of groups on $X$, say a topological space, then $H^1(X,\G)$ always classifies isomorphism classes of "locally trivial things over $X$ such that their sheaf of automorphisms over $X$ coincides with $\G$". Example: if $G$ is a finite group, then $H^1(X,G)$ classifies $G$-torsors over $X$. The proof is purely formal and is easiest to carry out in Cech cohomology.</p> <p>Let's say deformations = 1st order deformations to avoid repeating myself. All deformations of smooth affine varieties are trivial. This means that deformations of any smooth variety are locally trivial, and so can be classified by their sheaf of automorphisms as above. An automorphism of a trivial deformation is the same thing as an infinitesimal automorphism which (on a smooth variety) is the same thing as a vector field (a section of $T_X$). So $H^1(X,T_X)$ classifies isomorphism classes of deformations of $X$.</p> <p>Now do the same thing for infinitesimal automorphisms that fix the $p_i$. This is the same thing as a vector field which vanishes at each $p_i$, which is the same as a section of $T_X(-p_1-\ldots -p_n)$. Intuitively (differential-geometrically) this is clear: a vector field on a manifold is an infinitesimal automorphism because you can flow along it, and this flow fixes precisely those points where the vector field vanishes. This assertion is enough to finish the proof. So I would look at how Hartshorne identifies automorphisms of $X \times_k k[\varepsilon]$ over $X$ with sections of $T_X = \mathrm{Hom}(\Omega^1,\mathcal O_X) = \mathrm{Der}(\mathcal O_X,\mathcal O_X)$ and modify that proof to take into account that the automorphism should fix a number of points.</p> http://mathoverflow.net/questions/120875/ring-with-three-binary-operations/120877#120877 Answer by Dan Petersen for Ring with three binary operations Dan Petersen 2013-02-05T16:45:14Z 2013-02-05T16:45:14Z <p>An important example is the notion of a <a href="http://en.wikipedia.org/wiki/Gerstenhaber_algebra" rel="nofollow">Gerstenhaber algebra</a>. It is simultaneously a commutative ring and a Lie algebra, such that the product and bracket satisfy the Poisson identity, except all these things need to be understood in a differential graded sense.</p> http://mathoverflow.net/questions/120836/motive-of-unlabeled-fulton-macpherson-configuration-space/120876#120876 Answer by Dan Petersen for Motive of unlabeled fulton-macpherson configuration space? Dan Petersen 2013-02-05T16:36:12Z 2013-02-05T16:36:12Z <p>If we identify $T_x^2 / T_x$ with $\mathbb A^n$, then the action of $S_2$ is multiplication by $-1$. So $S_2$ acts trivially on the projectivization and you are just asking about the existence of a cell decomposition of projective space. </p> <hr> <p>A useful reference might be the last parts of Ezra Getzler's old preprint "Mixed Hodge structures of configuration spaces". He derives explicit formulas for the $S_n$-equivariant "motivic" Euler characteristic of $X[n]$ in terms of the motivic Euler characteristic of $X$. Here "motivic" means that the Euler characteristic is taken in a suitable ring of the form $K$ tensored with the representation ring of $S_n$, where $K$ is e.g. the Grothendieck group of $\ell$-adic Galois representations or the Grothendieck group of Hodge structures. So his formulas include in particular the equivariant Hodge-Deligne polynomial. Then taking $S_n$-invariants just corresponds to the substitution $p_n \mapsto x^n$, where $p_n$ is the $n$th power sum and $x$ is an indeterminate.</p> http://mathoverflow.net/questions/120567/on-the-coarse-moduli-space-of-a-stack/120603#120603 Answer by Dan Petersen for On the coarse moduli space of a stack Dan Petersen 2013-02-02T16:49:26Z 2013-02-04T08:04:14Z <p>Yes, this would imply that $\newcommand{\X}{\mathcal X}\X^s$ is the coarse moduli space, but I don't think this is the "right" question to ask -- I believe that $\X^s$ will not even form a sheaf unless $\X$ happens to be a scheme/algebraic space to begin with. </p> <p>Anyway, any morphism from a groupoid to a set factors through $\pi_0$ of the groupoid. This implies in particular that any morphism from $\X$ to an algebraic space factors through the presheaf $\X^s$. And the map $\X \to \X^s$ is a bijection on geometric points because it's in fact a bijection on $S$-points for any scheme $S$. So if $\X^s$ is a scheme/algebraic space then it is the coarse moduli space.</p> <hr> <p>Addendum. I think you are confused about some basic issues. Let us see why $BG^s$ is not the coarse moduli space of $BG$. Let $G$ be a nontrivial finite group, say.</p> <p>Consider for simplicity the topological setting, so we have a topological space $X$ and an open cover <code>$\{U_i\}$</code>. If we have a $G$-torsor on $X$ then we can restrict to a $G$-torsor on each $U_i$, and on each overlap $U_i \cap U_j$ we have isomorphisms between the restrictions from $U_i$ and from $U_j$. These isomorphisms satisfy cocycle relation. Conversely, if we have $G$-torsors on each $U_i$ and isomorphisms satisfying the cocycle relation, we can reconstruct a $G$-torsor on the whole of $X$, unique up to canonical isomorphism. What this paragraph says is exactly that the functor $BG$ which sends a space to the groupoid of $G$-torsors over it is a <em>sheaf of groupoids</em>, that is, a <em>stack</em>. (In the usual Grothendieck topology on the category of topological spaces, where open covers are, well, open covers. And when I call $BG$ a "functor" I should say "pseudofunctor" or "fibered category".)</p> <p>On the other hand we can consider $BG^s$, which is now a priori just a presheaf of sets, mapping a space to the <em>set</em> of <em>isomorphism classes</em> of $G$-torsors over it. If we have an isomorphism class of $G$-torsor on $X$ then we get well defined isomorphism classes of $G$-torsors on each $U_i$ with compatible restrictions to each $U_i \cap U_j$. But it is NOT true that if we have an isomorphism class of $G$-torsor on each $U_i$ which agree on double overlaps, then we can reconstruct a unique isomorphism class on all of $X$: consider the case when $G$ is nontrivial on $X$ and <code>$\{U_i\}$</code> is a trivializing cover! What this says is that $BG^s$ is in fact only a presheaf - it is NEVER a sheaf of sets. Put simply, <em>one can not glue together isomorphism classes</em>.</p> <p>What this shows is in fact that if we sheafify $BG^s$, then we get a point. If we only remember isomorphism classes of torsors then every $G$-torsor becomes equivalent to the trivial torsor on some open covering of your space, which means that these torsors are identified under sheafification.</p> <p>The same arguments work verbatim in algebraic geometry, since every $G$-torsor is locally trivial in the étale topology.</p> <p>In any case, this is why I said above that this is not the "right" question to ask: it is not natural to expect $\X^s$ to be a sheaf in the first place. I would suggest reading Heinloth or Fantechi's notes on stacks (they are somewhere online) and thinking over just what question it is you want to ask.</p> http://mathoverflow.net/questions/119861/commutativity-of-the-chow-ring-in-positive-characteristic Commutativity of the Chow ring in positive characteristic Dan Petersen 2013-01-25T17:08:01Z 2013-01-26T14:39:41Z <p>I was looking in <a href="http://math.stanford.edu/~vakil/245/" rel="nofollow">Ravi Vakil's notes on Intersection Theory</a>, Class 20, where he introduces the bivariant intersection theory, in particular the Chow ring $A^\ast (X)$.</p> <p>On p. 2, he writes the following which I had no idea about:</p> <blockquote> <p><strong>Alarming fact:</strong> This ring is apparently not known to be commutative in general, because the argument requires resolution of singularities. (It is known to be commutative in characteristic 0, and for smooth things in positive characteristic, and a few more things.) I think it should be possible to show that the ring is commutative in general using technology not available when this theory was first developed, using Johan de Jong’s “alteration theorem” in positive characteristic. If you would like to patch this hole, then come talk to me.</p> </blockquote> <p>Vakil's notes for this class are from 2004, so plenty could have happened since then. Is this still an open problem?</p> http://mathoverflow.net/questions/119726/discovering-and-selecting-conferences/119746#119746 Answer by Dan Petersen for Discovering and selecting conferences Dan Petersen 2013-01-24T11:25:11Z 2013-01-24T11:25:11Z <p>Kiran Kedlaya's list of conferences in arithmetic geometry. <a href="http://scripts.mit.edu/~kedlaya/wiki/index.php?title=Conferences_in_Arithmetic_Geometry" rel="nofollow">http://scripts.mit.edu/~kedlaya/wiki/index.php?title=Conferences_in_Arithmetic_Geometry</a></p> http://mathoverflow.net/questions/61732/semidirect-products-and-props Semidirect products and PROPs Dan Petersen 2011-04-14T18:29:19Z 2013-01-23T15:44:16Z <p>$\newcommand{\p}{\mathcal{P}}$Let <em>G</em> be a group, and let $G_n$ denote the wreath product $G^n \rtimes \Sigma_n$. </p> <p>There seems to be a notion of a PROP $\p$ where the role of the symmetric groups $\Sigma_n$ is instead played by the groups $G_n$. An example would be $G=SO(N)$ and $\p$ the PROP associated to the framed little <em>N</em>-disc operad. Here $\p(1,n)$ has an action of $G_1^{op} \times G_n$ -- the copy of $G_1^{op}$ rotates the entire disc "counterclockwise" (i.e. an element $g \in SO(N)$ acts via $g^{-1}$), and the action of $G_n$ on $\p(1,n)$ is the evident one. The gluing maps are then suitably equivariant under this simultaneous action of $G$ on the input/output legs, so to speak.</p> <p>Is there a standard name for this kind of PROP/operad? Cf. how one calls an operad where $\Sigma_n$ has been replaced by $B_n$ a braided operad. I would also be happy to hear of any paper where this kind of gadget has been defined and/or studied.</p> <hr> <p>Addendum, Jan 23 2013. Sorry if it is poor form to bump an inactive question only to advertise your own work, but I just ran across this old question. When I had thought some more about this I realized eventually that what I described above is really a special case of the notion of a <em>colored</em> PROP/operad, except the collection of colors do not form a set but a category. In this case there is only one color but its automorphism group is $G$, that is, the collection of colors is exactly the one-object category corresponding to the group $G$. </p> <p>The notion of a PROP/operad which is colored by a category is defined in my paper <a href="http://arxiv.org/abs/1205.0420" rel="nofollow">http://arxiv.org/abs/1205.0420</a> . This is actually somewhat more general than what Salvatore and Wahl define, even when the category in question is a group.</p> http://mathoverflow.net/questions/118865/spaces-parametrizing-ramified-covers-of-surfaces Spaces parametrizing ramified covers of surfaces Dan Petersen 2013-01-14T10:36:54Z 2013-01-15T10:22:23Z <p>Let $\Sigma$ be a surface (let's say oriented and of finite type). We can consider the configuration space $F(\Sigma,n)$ of $n$ ordered distinct points on $\Sigma$, i.e. $\Sigma^n\setminus \Delta$ where $\Delta$ is the "big diagonal". The cohomology of $F(\Sigma,n)$ can be computed in a very explicit way, as explained in Burt Totaro's paper "Configuration spaces of algebraic varieties": the Leray spectral sequence for $F(\Sigma,n) \hookrightarrow \Sigma^n$ degenerates after the first differential and can be written down in a concrete way, which gives us a completely explicit differential graded algebra whose cohomology is $H^\bullet(F(\Sigma,n))$. </p> <p>Now fix a finite group $G$ and consider the space $F(\Sigma,G,n)$ which parametrizes $n$ points on $\Sigma$ and a principal $G$-bundle over the complement of the $n$ points on $\Sigma$. So we have a finite sheeted covering $F(\Sigma,G,n) \to F(\Sigma,n)$ such that the fiber over the point $(s_1,\ldots,s_n) \in \Sigma^n$ is the set <code>$\mathrm{Hom}(\Pi_1(\Sigma \setminus \{s_1,\ldots,s_n\}),G)/G$</code>, where $G$ acts on the set of maps by conjugation. </p> <p>Q1. Is there a good way to describe or compute the cohomology of $F(\Sigma,G,n)$?</p> <p>Q2. (A vaguer question.) Let's say $\Sigma$ is compact for simplicity and let $FM(\Sigma,n)$ be the Fulton-MacPherson compactification of $F(\Sigma,n)$. I think there is a natural compactification $$F(\Sigma,G,n) \hookrightarrow FM(\Sigma,G,n)$$ covering $F(\Sigma,n) \hookrightarrow FM(\Sigma,n)$ and such that the covering $FM(\Sigma,G,n) \to FM(\Sigma,n)$ ramifies along the boundary; $FM(\Sigma,G,n)$ should parametrize principal $G$-bundles which are allowed to ramify over the nodes. Is there a natural smaller compactification of $F(\Sigma,G,n)$ which is still smooth, analogous to the compactification $$F(\Sigma,n) \hookrightarrow \Sigma^n$$ (which of course is smaller than the Fulton-MacPherson)?</p> <hr> <p>Craig Westerland suggests an alternative description of the cohomology of $F(\Sigma,G,n)$. The covering $p \colon F(\Sigma,G,n) \to F(\Sigma,n)$ satisfies $R^ip_\ast\mathbf Z =0$ for $i>0$, so $$H^\bullet(F(\Sigma,G,n),\mathbf Z) = H^\bullet(F(\Sigma,n),p_\ast\mathbf Z).$$ Now $F(\Sigma,n)$ is a $K(\pi,1)$ space where $\pi$ is by definition the pure braid group on $n$ strands of the surface $\Sigma$, $P_n(\Sigma)$. Hence this cohomology is given by <code>$$H^\bullet(P_n(\Sigma), \mathbf Z [ \mathrm{hom}(\pi_1(\Sigma \setminus \{s_1,\ldots,s_n\},G)/G]).$$</code> Also the action of $P_n(\Sigma)$ is the restriction of an action of the $n$-strand surface braid group $B_n(\Sigma)$ (i.e. where the points are unordered). </p> <p>The simplest example should be $\Sigma = \mathbf R^2$, where we get the usual pure braid group. Since the fundamental group of the punctured plane is free we find that <code>$\mathrm{hom}(\pi_1(\Sigma \setminus \{s_1,\ldots,s_n\},G)/G = G^n/G$</code>, where $G$ acts by elementwise conjugation on $G^n$. The action of the braid group $B_n$ can be written down in terms of Artin's generators $\sigma_i$: the element $\sigma_i$ acts by $$(g_1,\ldots,g_i,g_{i+1},\ldots,g_n) \mapsto (g_1,\ldots,g_ig_{i+1}g_i^{-1},g_i,\ldots,g_n)$$ (which is well defined on equivalence classes modulo conjugation). Then I guess even these cohomology groups are hard to compute in general?</p> http://mathoverflow.net/questions/118708/primitive-cohomology-useful/118712#118712 Answer by Dan Petersen for Primitive Cohomology Useful? Dan Petersen 2013-01-12T06:45:26Z 2013-01-12T06:45:26Z <p>One application is Deligne's theorem on the degeneration of the Leray spectral sequence. Strictly speaking this is not an interesting fact about smooth projective varieties but about <em>families</em> of smooth projective varieties, and their <em>relative</em> cohomology $\mathrm R^if_\ast \mathbf Q$.</p> <p>More precisely, the result is that for $f \colon X \to S$ a smooth and projective morphism of $\mathbf C$-varieties, there is an isomorphism $$\mathrm Rf_\ast\mathbf Q \cong \bigoplus_i \mathrm R^i f_\ast\mathbf Q[-i]$$ in the derived category of $S$. In particular, the Leray spectral sequence $E_2^{pq} = H^p(S,\mathrm R^q f_\ast \mathbf Q) \implies H^{p+q}(X)$ degenerates, but the derived category result is "universal". </p> <p>The proof is very formal and the key idea is the existence of a Lefschetz operator and a decomposition into primitive parts on the relative cohomology. Deligne's paper is very readable: <a href="http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1968__35_/PMIHES_1968__35__107_0/PMIHES_1968__35__107_0.pdf" rel="nofollow">Théorème de Lefschetz et critères de dégénérescence de suites spectrales</a></p> http://mathoverflow.net/questions/118458/homology-of-symmetric-product/118462#118462 Answer by Dan Petersen for Homology of symmetric product Dan Petersen 2013-01-09T16:20:45Z 2013-01-09T16:20:45Z <p>Yes, this is true. See Albrecht Dold, "Homology of symmetric products and other functors of complexes", Ann. of Math. (1958).</p> http://mathoverflow.net/questions/117432/splitting-of-the-weight-filtration/117509#117509 Answer by Dan Petersen for Splitting of the weight filtration Dan Petersen 2012-12-29T13:07:22Z 2012-12-29T13:07:22Z <p>I don't have a general answer, but let me add some more examples.</p> <ol> <li><p>For your second question, examples of smooth varieties with $H^i$ pure of the 'wrong' weight, a good example is the complement of an affine arrangement of hyperplanes in $\mathbf C^n$. In this case, the Hodge structure on $H^i$ is pure of type $(i,i)$. One way to see this is that the cohomology is generated as an algebra by $H^1$, and $H^1$ is spanned by logarithmic forms $$\omega_H = \frac 1 {2\pi i} \mathrm d \log H,$$ where $H=0$ is the defining equation of one of the hyperplanes; the class of $\omega_H$ is of type $(1,1)$. A reference for this is Brieskorn's "Sur les groupes des tresses [d'après V.I. Arnol'd]". See also the simple and "motivic" proof in "Weights in cohomology groups arising from hyperplane arrangements" by Minhyong Kim, as well as near-simultaneous papers by Boris Shapiro and Gus Lehrer which present basically the same result.</p></li> <li><p>Another example is when $X$ is an abelian variety, and $F(X,n)$ is the configuration space of $n$ points in $X$, i.e. the complement of the "big diagonal" in $X^n$. Then the mixed Hodge structure on $H^i(F(X,n))$ is always a direct sum of pure Hodge structures. A reference for this is Gorinov's preprint "Rational cohomology of the moduli spaces of pointed genus 1 curves" (on his webpage), Section 3. </p></li> <li><p>Let $Y(N)$ be the open modular curve parametrizing full level $N$ structures on elliptic curves. Then $H^1(Y(N))$ is a sum of pure Hodge structures, by the criterion in Donu Arapura's answer and the theorem of Drinfel'd and Manin.</p></li> </ol> http://mathoverflow.net/questions/113924/line-bundle-on-s2/113926#113926 Answer by Dan Petersen for Line bundle on $S^2$ Dan Petersen 2012-11-20T11:08:21Z 2012-11-20T11:08:21Z <p>Since $GL(1,\mathbf R)$ and the symmetric group $S_2$ are homotopy equivalent as topological groups, there is a bijective correspondence between isomorphism classes of real line bundles and double covers. It follows that a simply connected space has no nontrivial real line bundles.</p> http://mathoverflow.net/questions/113812/eilenberg-steenrod-axioms-of-sheaf-cohomology Eilenberg-Steenrod axioms of sheaf cohomology Dan Petersen 2012-11-19T10:52:17Z 2012-11-19T15:09:33Z <p>Cohomology of a space is often defined axiomatically: a cohomology theory is a functor from pairs of spaces to abelian groups satisfying the Eilenberg-Steenrod axioms. Is there a similar characterization of sheaf cohomology, where the domain of the functor is now a category of pairs $(A,X,\mathcal F)$ with $A \subset X$ a pair and $\mathcal F$ an abelian sheaf on $X$ (with the obvious morphisms)?</p> <p>Are there extraordinary sheaf cohomology theories?!</p> http://mathoverflow.net/questions/112271/noncontractible-domain-with-trivial-cohomology/112274#112274 Answer by Dan Petersen for Noncontractible domain with trivial cohomology Dan Petersen 2012-11-13T12:16:18Z 2012-11-13T12:16:18Z <p>There are a million examples. You should google "acyclic space". Here is one: if you remove a point from a homology sphere you get a manifold whose cohomology is trivial in positive degrees. Take a tubular neighbourhood of it in some $\mathbf R^n$ and you get an open domain.</p> http://mathoverflow.net/questions/81851/what-is-a-twisted-modular-operad/111656#111656 Answer by Dan Petersen for What is a twisted modular operad? Dan Petersen 2012-11-06T15:43:58Z 2012-11-06T15:43:58Z <p>Here is a long and belated answer, mostly written so that I can work out some details I should have worked out long ago. The short version of it is the following: when $g > 0$, one needs the notion of a twisted modular operad in order to describe the precise sense in which the "gravity operad" is an operad. I do hope all signs below are correct.</p> <p>The "standard" motivation for introducing twisting is that it is needed for the higher genus generalization of the bar transform (the "Feynman transform"): twisted modular operads appear naturally as "Feynman duals" of ordinary operads. The following is a different perspective.</p> <p>Remark: Since the gravity operad was introduced by Getzler, who also proved it dual to the hypercommutative operad, and since the gravity operad is the only twisted modular operad discussed in any detail in Getzler-Kapranov's paper, it seems plausible that this was one of their main motivating example for introducing the notion of twisting. </p> <p>The starting point is the situation in genus zero, and Getzler's paper "Operads and moduli of genus 0 Riemann surfaces". We consider the collection of spaces $\newcommand{\M}{\overline{M}} \M_{0,n}$, which form a cyclic operad in the category of algebraic varieties. Taking cohomology we get a cyclic co-operad $H^\bullet(\M_{0,n})$. So far nothing surprising. The surprise is that even though the spaces $M_{0,n}$ do not form an operad in any natural sense, their cohomologies still do - this is the "gravity" operad. Moreover, the two cohomology operads are dual in a precise sense (which is purely a genus zero phenomenon), but I will not talk about that. </p> <p>The magic words that make the cohomology of $M_{0,n}$ an operad are "Poincaré residue". For this read Deligne, "Hodge II", pp 31-32, or Peters-Steenbrink, pp 92-93. The short version is that the Poincaré residue is defined whenever you have an open subvariety $U \subset X$ where $X$ is smooth and the complement is a simple normal crossing divisor. Let the divisor be $D_1 \cap \cdots \cap D_N$, let $D_I = \bigcap_{i\in I} D_i$, and let $D_I^\circ$ denote the interior of the intersection (the complement in $D_I$ of all other components $D_j$). Then it is a map $H^\bullet(U) \to H^{\bullet-|I|}(D_I^\circ)$. Apply this to $X = \M_{0,n}$ and $U = M_{0,n}$: the set of possible intersections $D_I$ is exactly the set of all stable trees with $n$ legs, and $D_I^\circ$ is the corresponding product of open moduli spaces $M_0(\Gamma) = \prod_{v \in \mathrm{Vert}(\Gamma)} M_{0,n(v)}$ (where $n(v)$ = valence of the vertex).</p> <p>Hence we have co-composition maps $H^{\bullet+|\mathrm{Edge}(\Gamma)|}(M_{0,n}) \to H^\bullet(M_{0}(\Gamma))$, in particular $$H^{\bullet+1}(M_{0,n+n'}) \to H^\bullet(M_{0,n+1}) \otimes H^\bullet(M_{0,n'+1}),$$ and we would like to say that this makes the cohomology of $M_{0,n}$ a co-operad. There is one obvious problem here, which is that there is a degree shift in the definition. This is not such a big deal: we can get rid of it by declaring instead that the collection <code>$$\{ H^{\bullet-1}(M_{0,n})\}$$</code> should form a co-operad of graded vector spaces. </p> <p>But even this does not give us a co-operad. The issue is in the very definition of Poincaré residue: it is only defined up to an ordering of the boundary divisors, which potentially introduces a sign ambiguity. This is dealt with explicitly in Deligne who twists everything by a kind of orientation sheaf $\varepsilon^n$ to get things defined canonically. However, there is a simple operadic solution also to this problem: the correct statement is instead that the collection <code>$$\{ H^{\bullet-1}(M_{0,n}) \otimes \mathrm{sgn}_n\}$$</code> <em>does</em> form a cyclic co-operad of graded vector spaces. Here $\mathrm{sgn}_n$ is the sign representation of the symmetric group.</p> <p>So far so good. Now we wish to generalize this to higher genera. Again the spaces $\M_{g,n}$ form a modular operad, and their cohomologies $H^\bullet(\M_{g,n})$ a co-operad. We would like to play the same game again to get a co-operad structure on the cohomology of $M_{g,n}$. The self-glung gives us maps $$H^{\bullet+1}(M_{g+1,n-2}) \to H^{\bullet}(M_{g,n})$$ that we would like to use to define the modular co-operad structure. Hence we are led to introducing a second degree shift, now depending on the genus: one might hope that the spaces <code>$$\{ H^{\bullet+g-1}(M_{g,n}) \otimes \mathrm{sgn}_n\}$$</code> will form a suitable modular co-operad. This is unfortunately not true! The self-gluing does not work with this definition, as the $\mathbb{S}_2$-action on the two points that get identified is not the right one. In fact no strategy like the one used in genus zero will work in this situation.</p> <p>What one should then do is to define one's way out of the situation. Instead of an "ordinary" operad $\newcommand{\P}{\mathcal{P}}\P$ with structure maps $$\P(\Gamma) \to \P(n)$$ where $\Gamma$ is a rooted tree with $n$ inputs, or an "ordinary" modular operad with structure maps $\P(\Gamma) \to \P(g,n)$ where $\Gamma$ now has genus $g$ and $n$ legs, one introduces a "twisting" $\newcommand{\D}{\mathfrak{D}}\D$ such that there are maps $$\D(\Gamma) \otimes \P(\Gamma) \to \P(g,n).$$ Then $\D(\Gamma)$ is required to depend functorially on $\Gamma$ in certain ways making it meaningful to talk about associativity and equivariance conditions. One has defined the notion of a twisted modular operad. </p> <p>The game we played in genus zero can now be re-interpreted: instead of saying that <code>$\{ H^{\bullet-1}(M_{0,n}) \otimes \mathrm{sgn}_n\}$</code> is a cyclic co-operad, we can say that the usual cohomology groups <code>$\{ H^{\bullet}(M_{0,n})\}$</code> form a <em>cyclic $\D$-co-operad</em>, where $\D(\Gamma)$ is given by a suspension for each edge on the graph, and tensoring with the sign representation of the symmetric group acting on the set of edges: in other words, $\D(\Gamma) = \mathrm{Det}(\mathrm{Edge}(\Gamma))^{-1}$, using the terminology introduced in Getzler-Kapranov's paper. This more abstract (but in a sense much more natural) definition now works without any changes also in higher genus!! </p> <p>In this framework one can also give a nice explanation of why these suspensions and sign changes worked in genus zero. The correct cocycle to twist with was $\D(\Gamma)$ defined in the preceding paragraph. But $\D$ is cohomologous, in an appropriate sense, to the cocycle $D(\Gamma) = \mathrm{Det}(H_1(\Gamma))^{-1}$, which is trivial when restricted to trees. In fact $\D$ and $D$ differ by a coboundary, and this coboundary is precisely given by putting a suspended copy of the sign representation in each spot $(g,n)$. So we have recovered exactly the recipe for making $H^\bullet(M_{0,n})$ into a co-operad that we wrote down in an ad hoc way above.</p> http://mathoverflow.net/questions/128269/internal-day-convolution/130014#130014 Comment by Dan Petersen Dan Petersen 2013-05-14T09:15:18Z 2013-05-14T09:15:18Z I don't think that's a reason not to accept it. For instance, I think a lot of people will happily upvote a question that they only partially understand, but will hesitate to do the same with an answer (because in this case the upvote could be seen as a &quot;certificate&quot; of correctness). http://mathoverflow.net/questions/129889/what-is-this-called-reverse-modulus-inverse-modulus Comment by Dan Petersen Dan Petersen 2013-05-06T20:38:47Z 2013-05-06T20:38:47Z This is $-n \pmod x$... http://mathoverflow.net/questions/129621/analogue-of-knudsen-clutching Comment by Dan Petersen Dan Petersen 2013-05-04T09:25:06Z 2013-05-04T09:25:06Z This question makes no sense as stated. Precisely what moduli stacks of pointed varieties are you considering? http://mathoverflow.net/questions/129344/dense-affine-subvarieties-of-algebraic-varieties/129346#129346 Comment by Dan Petersen Dan Petersen 2013-05-01T20:33:10Z 2013-05-01T20:33:10Z You can trivially reduce to the case when the connected components of $X$ are irreducible: just shrink $X$ to a smaller variety $X'$ by throwing out all points lying on more than one irreducible component. Then $X'$ is open and dense in $X$ and it suffices to find a dense affine in $X'$. http://mathoverflow.net/questions/129259/does-every-simplicial-polytope-have-a-topology-preserving-contractible-edge Comment by Dan Petersen Dan Petersen 2013-04-30T23:03:30Z 2013-04-30T23:03:30Z &quot;On a $d$-simplex, no edge is contractible. [...] Does every simplicial polytope have a contractible edge?&quot;. I think you just answered your own question. http://mathoverflow.net/questions/128712/is-the-moduli-space-of-ppavs-smooth/128718#128718 Comment by Dan Petersen Dan Petersen 2013-04-26T10:29:04Z 2013-04-26T10:29:04Z J. Martel, either I am misunderstanding you or you are confused. The locus of Jacobians is not closed in $A_g$ for any $g \geq 2$, its closure is the locus of products of Jacobians. When $g=2$ every ppav is a Jacobian or a product of two elliptic curves. http://mathoverflow.net/questions/128712/is-the-moduli-space-of-ppavs-smooth/128723#128723 Comment by Dan Petersen Dan Petersen 2013-04-25T14:42:09Z 2013-04-25T14:42:09Z I agree with all of this but it seems tough going to refer to Faltings--Chai for this result. The OP seems content to work over the complex numbers and then smoothness of the moduli stack amounts to saying that there exists a finite index subgroup of $\mathrm{Sp}(2g,\mathbf Z)$ which acts freely on Siegel space. http://mathoverflow.net/questions/128615/a-infinity-structure-on-cohomology-and-the-weight-filtration/128660#128660 Comment by Dan Petersen Dan Petersen 2013-04-24T21:18:14Z 2013-04-24T21:18:14Z The last paragraph doesn't sound right to me - shouldn't the $m_n$ be compatible with weights on the nose? http://mathoverflow.net/questions/128615/a-infinity-structure-on-cohomology-and-the-weight-filtration Comment by Dan Petersen Dan Petersen 2013-04-24T14:07:56Z 2013-04-24T14:07:56Z A relevant question is: <a href="http://mathoverflow.net/questions/22064/mixed-hodge-structure-on-the-rational-homotopy-type" rel="nofollow" title="mixed hodge structure on the rational homotopy type">mathoverflow.net/questions/22064/&hellip;</a> http://mathoverflow.net/questions/127843/contractibility-of-a-configuration-space Comment by Dan Petersen Dan Petersen 2013-04-17T13:38:47Z 2013-04-17T13:38:47Z In what sense is it a manifold? http://mathoverflow.net/questions/127495/cohomology-of-configuration-spaces Comment by Dan Petersen Dan Petersen 2013-04-15T08:03:02Z 2013-04-15T08:03:02Z @Nicholas Proudfoot: On the other hand, all Christin asked for was a procedure for computing the Betti numbers, not a closed formula. I know Orsola Tommasi wrote a computer program computing the Betti numbers of Totaro's DGA for an elliptic curve, you could e-mail her and ask for the code. http://mathoverflow.net/questions/127495/cohomology-of-configuration-spaces Comment by Dan Petersen Dan Petersen 2013-04-14T02:58:25Z 2013-04-14T02:58:25Z @Geoffroy: the paper you reference deals with configuration spaces of unordered points, not ordered... http://mathoverflow.net/questions/127495/cohomology-of-configuration-spaces Comment by Dan Petersen Dan Petersen 2013-04-14T02:57:00Z 2013-04-14T02:57:00Z Sorry, but why doesn't Totaro's paper answer your question? http://mathoverflow.net/questions/127223/are-period-domains-ever-contractible Comment by Dan Petersen Dan Petersen 2013-04-11T12:31:16Z 2013-04-11T12:31:16Z dhagbert: see <a href="http://mathoverflow.net/questions/67699/" rel="nofollow">mathoverflow.net/questions/67699</a> ... http://mathoverflow.net/questions/126947/minimal-compactification/126968#126968 Comment by Dan Petersen Dan Petersen 2013-04-10T18:37:03Z 2013-04-10T18:37:03Z Actually $A_g$ <i>is</i> a hermitian symmetric space, so the minimal compactification doesn't just mimic the construction of Satake-Baily-Borel: it is a special case of it.