User anatoly preygel - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T09:24:16Z http://mathoverflow.net/feeds/user/1921 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/74943/constructible-sheaves-and-dg-modules/75115#75115 Answer by Anatoly Preygel for Constructible sheaves and dg-modules Anatoly Preygel 2011-09-10T20:22:31Z 2011-09-12T07:42:42Z <p>This is not a <em>helpful</em> answer to your main question, but merely a negative answer to your "In particular..what happens.." question. But, the general idea may be helpful in figuring out what more precise things (weaker than Keller's result) would be reasonable to ask for.</p> <p>$\newcommand{\RR}{\mathbb{R}}\newcommand{\RHom}{\mathrm{RHom}}\newcommand{\pt}{\mathrm{pt}}\DeclareMathOperator{\deg}{deg}\newcommand{\RGamma}{R\Gamma}\renewcommand{\mod}{\text{-mod}}\newcommand{\C}[1]{C^\bullet(#1)}$</p> <p>Since I'll lapse into this notation anyway, let me make it explicit: Identity $\gamma_M$ with the functor $$\RGamma(M, -)\colon D \to \C{M}\mod$$ where by $\mod$ I'm implicitly working in a dg-setting.</p> <p>At some point below, I'll assume that $N$ is <em>compact oriented</em> of dimensions $n$. (This is not strictly necessary, but allows me to avoid extra notation.)</p> <p><strong>Claim:</strong> Suppose that $M$ is simply connected, $\dim M \geq 2$, and that $i \colon N \hookrightarrow M$ is not the identity. Then, the functor $\gamma_M \colon D= \langle \RR_M, i_* \RR_N \rangle \to D_{A_M}$ is not fully faithful.</p> <h2><em>Fuzzy Remark:</em></h2> <p>Before sketching an argument, here's a "philsophical" remark about why things will go wrong:</p> <p>The category $D_c(M)$ feels the <strong>topology</strong> (or maybe even geometry) of $M$. In particular, it has a Proper Base-change Theorem saying something like $q^* p_! = (p')_! (q')^*$ where the maps take part in a fiber-square of (actual) <em>topological spaces</em>. The category $D_{A_M} = \C{M}\mod$ feels only the <strong>homotopy theory</strong> of $M$. You should expect a Base-Change Theorem in this context, but now with a fiber-square of homotopy types -- more correctly, a homotopy fiber-square. </p> <p>A simpler sort of 'no go' result that this heuristic implies: Suppose you had wanted to include two sub-manifolds $i_k\colon N_k \hookrightarrow M$, $k=1,2$. The $\C{M}\mod$ images would be unable to tell them apart if the $i_k$ were homotopic -- e.g., the inclusion of any two points. While the constructible theory would certain care whether the two points were the same or not.</p> <h2><em>Sketch of Claim</em>:</h2> <p>To see this, note that $$\RHom_{D_c(M)}(i_* \RR_N, i_* \RR_N) = \RHom_{D_c(N)}(i^* i_* \RR_N, \RR_N) = \RHom_{D_c(N)}(\RR_N, \RR_N) = \C{N}$$ while</p> <p><strong>Sub-Claim</strong>: Letting $\stackrel{h}\times_M$ denote the homotopy fiber product, $$\RHom_{A_M}(\gamma_M(i_* \RR_N), \gamma_M(i_* \RR_N)) =\RHom_{\C{M}}\left(\C{N}, \C{N}\right) \approx C_{\bullet}\left(N \stackrel{h}\times_M N\right)[-n]$$</p> <p>Assuming the sub-claim: to conclude it suffices to produce homology classes on $\Omega M$ in arbitrarily positive degrees, whose images under the composite $$H(\Omega M) \to H_*(N \stackrel{h}\times_M N) \to H_*(\Omega (M/N))$$ are non-zero. I think the following should do this upon filling in the details: Equip $M$ with a base-point in $N$, take some non-zero element of $\pi_i M$, with $i \geq 2$, that remains non-zero in $\pi_i (M/N)$. Use it to produce an $(i-1)$-homology class on $\Omega M$, and then take its Pontrjagin products.</p> <p><em>Sketch of sub-claim:</em> Underlying the Eilenberg-Moore spectral sequence is the statement that, letting $\boxtimes$ denote derived co-tensor of co-modules over a co-algebra, $$C_\bullet(N) \boxtimes_{C_\bullet(M)} C_\bullet(N) \approx C_\bullet(N \stackrel{h}\times_M N)$$ Poincare duality gives an equivalence $\C{N} \approx C_\bullet(N)[-n]$ of $\C{M}$-modules (or $C_\bullet(M)$-comodules). It remains to identify $$\RHom_{\C{M}}(\C{N}, \C{N}) \approx C_\bullet(N) \boxtimes_{C_\bullet(M)} \C{N}$$ by term-wise identifying the co-simplicial cobar constructions on both sides.</p> <p><strong>Example:</strong> Note that if $N = \pt$ the sub-claim is a familiar statement in Koszul duality: That for $M$ simply-connected $\RHom_{\C{M}}(\RR,\RR) \approx C_\bullet(\Omega M)$. In certain cases, e.g. $M = S^{2k+1}$, you can just see it. As an aside: $C_\bullet(\Omega M)\mod$ knows about all locally-constant things, not just local systems finitely-buildable from the trivial one. (But will run into the same issues if you try to include submanifolds without explicitly adding in extra generators for the strata.)</p> <p><strong>Remark:</strong> Though $\gamma_M$ is not fully-faithful here, it <em>does</em> get the maps into/out of $\RR_M$ right. Logic as above shows that $$\RHom_D(\RR_M, i_* \RR_N) = \C{N} = \RHom_{\C{M}}(\C{M}, \C{N})$$ and then Verdier Duality (for $D_c(M)$) + something like Grothendieck Duality (for $\C{M}$) give the other direction as well.</p> http://mathoverflow.net/questions/66170/derived-categories-of-coherent-sheaves-of-modules-exceptional-images-gluing/66209#66209 Answer by Anatoly Preygel for Derived categories of (coherent) sheaves of modules: exceptional images, gluing, and proper descent? Anatoly Preygel 2011-05-27T18:20:38Z 2011-05-27T20:19:09Z <p>Briefly re 2 and 4: the Ind-completion IndDCoh has shriek pullbacks and star pushforwards. Moreover, it has what one might call "derived h-" descent with respect to shriek pullback. This includes "derived proper" descent -- note that derived is modifying the topology, not just the functors/categories; more on this below.</p> <p>Just to clarify: by DCoh I mean the "bounded coherent" dg- or infinity-category, and by Ind the infinity-categorical Ind-completion (e.g., close up the dg-Yoneda image by filtered homotopy colimits). This is modelled by Krause's homotopy category of objectives, Positselski's coderived category, and variations. In particular, it will not coincide with the usual quasi-coherent derived category unless one imposes regularity: The usual version is IndPerf, so one needs Perf=DCoh. But the gain of this definition is shriek pullback without restricting to cohomologically bounded below complexes, and things like proper descent. (And both IndDCoh and IndPerf have flat descent..)</p> <p>An important point is that one has to pass to derived schemes (not just complexes or dg-categories) to have any hope in the proper case. For non-flat proper maps, the descent condition will thus look very different from the classical versions: one needs to take derived fiber products in forming the Cech nerve.</p> <p>An example giving a feeling for the previous paragraph: Consider the map from a point to a non-reduced point. It is a proper cover, and its classical and derived Cech nerves look very different! Also considering this example one sees that any sort of proper *hyper*descent will always be too much to ask for as it forces insensitivity to nilpotents.</p> <p>I sketch some things like the above in an appendix to <a href="http://arxiv.org/abs/1101.5834" rel="nofollow">this preprint</a> though a much better reference seems to now be available: <a href="http://www.math.harvard.edu/~gaitsgde/GL/IndCohtext.pdf" rel="nofollow">here</a> eg Section 7.2</p> http://mathoverflow.net/questions/47952/proving-interesting-theorems-about-s-n-using-its-character-table/48005#48005 Answer by Anatoly Preygel for Proving interesting theorems about S_n using its character table. Anatoly Preygel 2010-12-02T06:43:54Z 2010-12-02T06:43:54Z <p>The following is not strictly speaking something that can be read off from the character table. However, it is an elementary combinatorial identity about partitions which one can deduce from understanding the character theory of symmetric groups well enough, and looking at the character table does play a central role:</p> <p>For $\lambda \vdash n$ a partition of $n$ (i.e., $n = 1 \lambda_1 + 2 \lambda_2 + \cdots + n \lambda_n$) define $$A(\lambda) = \prod_{i=1}^{n} n^{\lambda_n}, \qquad B(\lambda) = \prod_{i=1}^{n} (\lambda_n)!$$ <strong>Claim:</strong> $$\prod_{\lambda \vdash n} A(\lambda) = \prod_{\lambda \vdash n} B(\lambda)$$</p> <p>The character-theoretic proof proceeds as follows:</p> <ol> <li>For an element in the conjugacy class of $S_n$ indexed by the partition $\lambda$, it's centralizer has cardinality $A(\lambda) B(\lambda)$, i.e., the number of elements in the conjugacy class is $$\frac{n!}{A(\lambda) B(\lambda)}$$</li> <li>Take the character matrix $M$. The orthogonality relations tells us that a suitable rescaling of the character matrix is orthogonal, so has $\det = \pm 1$. From this, together with 1 to find the scaling factors for the columns, we obtain $$(\det M)^2 = \prod_{\lambda} A(\lambda) B(\lambda)$$</li> <li>$M$ relates two bases for the spaces of class-functions: The characters of irreps (indexed in the Schur ordering by partitions), and the delta functions on conjugacy classes (indexed obviously by partitions). For symmetric groups, there is a third nice basis: For $\lambda \vdash n$, let $$S_\lambda = \prod_{i=1}^{n} S_i^{\lambda_i}$$ and consider the characters of the induced reps $Ind_{S_\lambda}^{S_n} \mathbb{C}$.</li> <li>Consider the change of basis matrix relating characters of induced reps and the delta functions on conjugacy classes: Easy character theory shows that it is triangular with diagonal entries equal to $B(\lambda)$.</li> <li>Consider the change of basis matrix relating characters of induces reps and characters of irreps: Knowing how character theory for symmetric groups works over $\mathbb{Z}$ (i.e., that both span integrally), we can show that it has determinant $\pm 1$. More precisely, knowing the character theory well enough we can show that the change of basis matrix between them is upper triangular with ones on the diagonal.</li> </ol> <p>Putting together 2, 4, 5 we obtain $$\det(M)^2 = \prod_{\lambda \vdash n} A(\lambda) B(\lambda) = \left(\prod_{\lambda \vdash n} B(\lambda)\right)^2$$ and so the claimed identity.</p> http://mathoverflow.net/questions/41428/can-deligne-mumford-stacks-be-characterized-by-their-restriction-to-a-small-subca/41443#41443 Answer by Anatoly Preygel for Can Deligne-Mumford stacks be characterized by their restriction to a small subcategory? Anatoly Preygel 2010-10-07T18:59:31Z 2010-10-07T18:59:31Z <p>Restricting to Aff is certainly enough, but Aff isn't small (there are e.g., polynomial algebras on arbitrary sets). If your DM stack is <em>finitely presented</em> over $k$ (which is probably good to include in the definition, to avoid these issues), then it is determined by it's restriction to finitely-presented affines (which is essentially small).</p> <p>Without some finiteness hypothesis, no set of finitely-presented algebras can suffice (even for affine schemes, nevermind DM stacks). (And I suppose no small category of test objects can suffice: Take Spec of a field generated by a set of cardinality larger than that of global sections of any of your test schemes.)</p> <p>The set you give is insufficient even for smooth varieties over an alg. closed field: you will have a morphism whenever you have an (arbitrary) map on $k$-points.</p> http://mathoverflow.net/questions/36659/interdependence-between-a1-homotopy-theory-and-algebraic-cobordism/36698#36698 Answer by Anatoly Preygel for Interdependence between A^1 homotopy theory and algebraic cobordism Anatoly Preygel 2010-08-25T22:00:30Z 2010-08-25T22:00:30Z <p>The two topics are logically, if not morally, independent of one another. $\mathbb{A}^1$-homotopy encodes objects like motivic cohomology &amp; it's relatives which are of interest regardless of the framework. There's no way for algebraic cobordism to supersede that -- algebraic cobordism is more directly comparable to Chow theory and $K^0$ than to motivic cohomology.</p> <p>Conversely, algebraic cobordism provides a more geometric viewpoint on (a piece of) $MGL$ -- surely a valuable thing to have around as well as being of independent geometric interest. That said, if your interests are more motivic than geometric, you could get by without knowing the details of algebraic cobordism provided that you know all the classical statements in complex cobordism that inspired it.</p> <h2>Motivic vs algebraic cobordism.</h2> <p>The $\mathbb{P}^1$-spectrum $MGL$, or "motivic cobordism," enjoys a privileged role in the world of $\mathbb{A}^1$-homotopy similar to that of $MU$ in classical homotopy. There is a relationship between this "motivic cobordism" and "algebraic cobordism." The former is a bigraded theory, and Levine showed that $MGL^{2n,n}(X) = \omega^n(X)$. (This bigrading issue is analogous to how Chow theory occurs as the $(2n,n)$-graded piece of motivic cohomology, and explains why you don't get a long exact sequence in algebraic cobordism, etc...)</p> <p>So one can view Morel-Levine's (or Levine-Pandharipande's) algebraic cobordism as giving an axiomatic (or geometric) viewpoint on the motivic theory, like we had for $MU$. Unlike the case of complex cobordism, where one can directly compare it to $MU$-cohomology using transversality results, here the comparison is much more difficult and computational. The proof of this comparison relies on a (currently unpublished) spectral sequence due to Hopkins-Morel. It should be noted that constructing this spectral sequence is hard, and by the time you've constructed it you've had to independently check lots of things that you might've wanted to deduce from the comparison with algebraic cobordism (for instance you pretty much end up computing $MGL^{2*,*}(Spec k)$, you can see the comparison of cobordism to Chow theory, etc.).</p> <h2>Degree formula (or, application to B-K)</h2> <p>The reference to open problems likely refers to the use of cobordism and Rost's degree formula in the final steps of proving Bloch-Kato for $\ell \neq 2$. Cobordism is a tool in the proof, but introducing algebraic cobordism is not strictly necessary. (One can get by with explicit computations with the characteristic numbers of interest. It'd certainly be fair to like the Levine-Morel proof of the degree formula, though.)</p> <p>The Bloch-Kato conjecture is concerned with the "Galois symbol" map $$K^n_M(k)/\ell = H^n(k, \mathbb{Z}/\ell(n)) \to H_{et}^n(k, \mathbb{Z}/\ell(n)) = H_{et}^n(k, \mu_\ell^{\otimes n})$$ No cobordism in sight yet. Suslin-Merkurjev's proof for $n=2$ and Voevodsky's proof for $\ell=2$ made use of "splitting varieties" that one could write down pretty much explicitly and then proceed to study: Brauer-Severi varieties and Pfister quadrics, respectively. This doesn't seem to work for the general case, and instead one writes down a minimalist wishlist for splitting varieties and then has to show that they exist --- it is in this step where cobordism (or really, characteristic numbers) play a role.</p> <p>A "splitting variety" for a non-zero symbol $0 \neq u = u_1 \otimes \cdots \otimes u_n \in K^n_M(k)/\ell$ should be a smooth variety $X/k$ such that $u$ pulls back to zero in $H^n(k(X), \mathbb{Z}/\ell(n))$ ("$X$ splits $u$"), with $\dim X = \ell^{n-1}-1$, and some more technical conditions, including a partial "universality" for this property: $X'$ splits $u$ iff there is a rational map from (a degree prime-to-$\ell$ cover of) $X'$ to $X$; it follows that $X$ must have no degree prime-to-$\ell$ zero-cycles (or else $Spec k$ splits $u$, i.e., $u = 0$). </p> <p>Cobordism (of whatever flavor you like: complex cobordism suffices) enters when relating this to characteristic numbers: namely, to the property of being a $v_n$-variety (=representing a $v_n$ class in complex or algebraic cobordism, up to decomposables). Here, one needs something like "Rost's degree formula", which implies for instance that the property of being a $v_n$-variety with no prime-to-$\ell$ zero-cycles is invariant under prime-to-$\ell$ degree covers.</p> http://mathoverflow.net/questions/32248/extraordinary-cohomology-as-a-derived-functor/32252#32252 Answer by Anatoly Preygel for Extraordinary cohomology as a derived functor? Anatoly Preygel 2010-07-17T02:46:59Z 2010-07-17T02:46:59Z <p>I'd like to propose that the answer is "No, but..". The viewpoint I'd like to suggest is that thinking of "derived functors" is probably insufficient here (because we're secretly interested in homotopical categories that are not derived categories of abelian categories), but that we are relying essentially on the observation that <code>$R\Gamma(X, -) = R\Gamma(Y, Rf_* -)$</code>. So what follows is a sketch-construction that gives a positive answer to a related question you could've been asking: "We can get the Serre spectral sequence from the Leray spectral sequence. Can we get the Atiyah-Hirzebruch spectral sequence in a similarly sheaf-theoretic way?"</p> <hr> <p>We get the Leray spectral sequence by studying sheaves of complexes of $\mathbb{Z}$-modules. The "derived" category (of sheaves of complexes of $\mathbb{Z}$-modules) is the derived category of its heart (sheaves of $\mathbb{Z}$-modules) w.r.t. the usual $t$-structure, so it's reasonable that we get lots of mileage from looking at derived functors, composites of derived functors, etc.</p> <p>Analogously, we can think of more general Atiyah-Hirzebruch type spectral sequences as arising from studying sheaves of <em>spectra</em>. This is not the derived category of its heart w.r.t. the usual $t$-structure (this heart is again sheaves of $\mathbb{Z}$-modules). But, this does give a way of thinking about Atiyah-Hirzebruch type spectral sequences:</p> <p>We start with (say) the "constant sheaf of spectra" $\mathbf{E}$ on $X$, and we're interested in computing the (homotopy groups of the spectrum) $$R\Gamma(X, \mathbf{E}) = R\Gamma(Y, Rf_* \mathbf{E})$$ The spectral sequence of interests arises by filtering $Rf_* \mathbf{E}$ using the $t$-structure (Postnikov sections "on values"), $$\cdots \to \tau_{\geq k} Rf_* \mathbf{E} \to \tau_{\geq (k-1)} Rf_* \mathbf{E}\to \cdots \to \mathbf{E}$$ The $k$-th "associated graded piece" of this filtration is $\pi_k Rf_* \mathbf{E}$ (which, recall, is an object in the heart --- i.e., a sheaf of $\mathbb{Z}$-modules on $Y$). This gives rise to a spectral sequence (excuse the funny indexing!, and no comment on convergence) $$E^2_{p,q} = \pi_{-p} R\Gamma\left(Y, \pi_{-q} Rf_* \mathbf{E}\right) \Rightarrow \pi_{-p-q} R\Gamma\left(Y, Rf_*\mathbf{E}\right) = \pi_{-p-q} R\Gamma(X,\mathbf{E})$$ The funny indexing was picked so that I can rewrite it as $$H^p(Y, \pi_{-q} Rf_* \mathbf{E}) \Rightarrow E^{p+q}(X)$$</p> <p>To recover the usual form of AH-SS we need the following observation (analogous to what we'd need to get the Serre spectral sequence via sheaf theory): If $f$ is nice (i.e., a fibration between reasonable spaces), then $\pi_{-q} Rf_* E$ will be the locally constant sheaf associated to $E^{q}(F)$ with its monodromy action.</p> http://mathoverflow.net/questions/29919/how-ugly-is-the-isomorphism-rgxh-rg-x-rh-for-groups-g-h/29974#29974 Answer by Anatoly Preygel for How ugly is the isomorphism R[GxH] = R[G] (X) R[H] for groups G, H? Anatoly Preygel 2010-06-29T21:52:57Z 2010-06-29T21:52:57Z <ol> <li>You can somewhat lift the algebraic closedness assumption: You have to allow an auxillary ring (actually, division algebra) to act equivariantly on both representation and tensor over it.</li> <li>Such a decomposition should hold whenever one of the groups has semi-simple representation category (the division rings in 1 are endomorphisms of simples). Then, the decomposition can be made canonical precisely up to choosing representative simple objects. If $V$ is a $G \times H$-rep, and $\rho$ are representative simples for $G$, then the natural map <code>$$\bigoplus_{\rho} \rho \otimes_{D_\rho} Hom_G(\rho, V) \to V$$</code> with $D_\rho = End_G(\rho)$ will be an isomorphism of $G \times H$-modules. (Conversely, given applying such a decomposition to $k[G]$ viewed as $G \times G$-module one would have to recover a representative set of simples.)</li> <li>For symmetric groups (in char. $0$), the endomorphism rings of simples are just the field (i.e., the simples remain irreducible over the alg. closure), so in particular you get such decompositions. Moreover, there are explicit representative simples that one can write down (the Specht modules). I don't know of the combinatorial theory to say if this gives any sort of satisfactory answer to your question 3.</li> </ol> http://mathoverflow.net/questions/21370/where-does-the-splitting-principle-come-from-and-does-it-generalize/21381#21381 Answer by Anatoly Preygel for Where does the splitting principle come from and does it generalize Anatoly Preygel 2010-04-14T19:50:09Z 2010-04-14T19:50:09Z <p>We can think of the splitting principle as a <em>condition</em> on a "cohomology theory" (of some sort) <code>$E^*$</code>, coming about when working with Chern classes for instance, and then ask: When does $E^*$ satisfy this condition? First, let's make the condition more precise and reformulate it:</p> <p><strong>Condition 1:</strong> Given $X$ and a vector bundle $V$ on $X$, there exists $f: X' \to X$ such that $f^* V'$ has a filtration with subquotients line bundles, and <code>$f^*: E^*(X) \to E^*(X')$</code> is injective.</p> <p>But there is a <em>universal</em> choice for $X'$, namely the <em>flag variety of $V$</em>: $p: Fl(V) \to X$. Any $f: X' \to X$ with $f^* V'$ filtered with line bundle subquotients will factor through $p$, and so we're really just asking if <code>$p^*: E^*(X) \to E^*(Fl(V))$</code> is injective.</p> <p><strong>Condition 1':</strong> For all $X$ and $V$, <code>$p^*: E^*(X) \to E^*(Fl(V))$</code> is injective.</p> <p>At this point there are two ways this answer can go, depending on ones tastes:</p> <ol> <li>$Fl(V)$ is a very geometric object over $X$, so we might as well ask that we actually have a formula for <code>$E^*(Fl(V))$</code> in terms of <code>$E^*(X)$</code>. If <code>$E^*$</code> is "reasonable" (i.e., has Chern classes giving rise to a "projective bundle formula") then iteratively applying the projective bundle formula will give such a thing, and in fact show that <code>$E^*(X)$</code> is a direct summand of <code>$E^*(Fl(V))$</code>.</li> <li>(My favorite:) There's a nice way of strengthening Condition 1' that also holds in all reasonable cases, and that looks rather natural. You can ask that $Fl(V) \to X$ behave like a "covering", i.e. that (<strong>Condition 2:</strong>) <code>$$E^*(X) \to E^*(Fl(V)) \to E^*\left(Fl(V) \times_X Fl(V)\right)$$</code> is an equalizer diagram. (So not only is pullback injective, but you can identify its image...) (In fact, in reasonable cases it'll be a split equalizer diagram, related to the direct summand thing above.)</li> </ol> <p>If your question is one of proof + generalization (which I think it is), rather than vague motivation, then I haven't addressed it yet:</p> <p>In topology. one can show that any complex-oriented cohomology theory (i.e., one with Chern classes for line bundles) $E^*$ has a projective bundle formula, satisfies all the conditions, etc. </p> <p>In more-algebro-geometric contexts, you could deduce the Chow + K-theory (I don't know anything about the $\gamma$-filtration) statements by either </p> <ol> <li>Constructing $c_1$ + proving a projective bundle formula, and then feeding this into a general argument using these to prove the rest.</li> <li>Going to the universal example of <em>algebraic cobordism</em> and then deducing the results for Chow + K-theory from the known relationships between them and algebraic cobordism. (Though this second approach is not so great, since those relationships hold under much more stringent hypotheses than are necessary to run the argument.)</li> </ol> <p>One could also ask to generalize this in another direction, replacing vector bundles and $Fl(V)$ by more general $G$-bundles and their associated $G/B$-bundles. In general, that's a more complicated story...</p> http://mathoverflow.net/questions/18331/flatness-of-relative-canonical-bundle/18399#18399 Answer by Anatoly Preygel for Flatness of relative canonical bundle Anatoly Preygel 2010-03-16T17:10:52Z 2010-03-16T17:10:52Z <p>It sounds like you may want Exercise 9.7 in Hartshorne's "Residues and Duality". I paraphrase the statement:</p> <hr> <p><strong>Exercise 9.7 (RD):</strong> Let $f: X \to B$ be a flat morphism of finite type of locally Noetherian preschemes. Then, $f^!(\mathcal{O}_B)$ has a unique non-zero cohomology sheaf, which is flat over $B$, iff all the fibers of $f$ are Cohen-Macaulay schemes of the right dimension. Moreover $f^!(\mathcal{O}_B)$ is isomorphic to (a shift of) an invertible sheaf (on $X$) iff all the fibers of $f$ are Gorenstein schemes of the right dimension.</p> <hr> <p>In particular, this addresses the case you mention in your comment ($f$ Gorenstein), since then $f^!(\mathcal{O}_Y)$ is locally free on $X$ and, since $f$ is flat, certainly flat over $B$.</p> <p>[Aside: I believe I learned this reference from Brian's book "Grothendieck Duality and Base Change", which I think also contains a proof of this.]</p> http://mathoverflow.net/questions/17937/algebraic-de-rham-cohomology-vs-analytic-de-rham-cohomology/17940#17940 Answer by Anatoly Preygel for Algebraic de Rham cohomology vs. analytic de Rham cohomology Anatoly Preygel 2010-03-12T03:48:12Z 2010-03-12T03:48:12Z <p>If $X$ is smooth and proper, GAGA does in fact suffice (despite the observation that $d$ is not <code>$\mathcal{O}_X$</code>-linear: One obtains a comparison map of hypercohomology spectral sequences; it is an isomorphism on the $E_2$ page by GAGA, and thus on the $E_\infty$ page.</p> <p>It is to prove the general case (i.e., $X$ smooth but not necessarily proper) that one needs to do additional work.</p> http://mathoverflow.net/questions/15904/whitehead-products-without-base-points/15934#15934 Answer by Anatoly Preygel for Whitehead Products without Base Points? Anatoly Preygel 2010-02-20T21:52:35Z 2010-02-20T21:52:35Z <p>As I posted in my comment, I think Paul's suggestion does work. Here's a (sloppy) description of how I think things will work:</p> <p>The local systems you describe can be obtained, by passing to homotopy groups, from a "local system of loop spaces" $$\Omega: \Pi_{\leq \infty} X \to \Omega\mathbf{Spaces}$$ One can imagine that this corresponds under the Grothendieck construction to the free loop-space fibration $\Omega X \to LX \to X$. Alternatively, if we fix a basepoint and identify $X = BG$ for a simplicial group $G$, then this is just encoding the simplicial conjugation action of $G$ on itself.</p> <p>Rather than think about (strangely-graded) Whitehead products, I prefer to think about (reasonably graded) Samelson products: We think of the structure (Whitehead product) on <code>$\pi_{*+1} X$</code> as really being a structure (Samelson product) on <code>$\pi_{*} \Omega X$</code>. I claim that Samelson products give a functor $$\pi_*: \Omega\mathbf{Spaces} \to \mathbf{grqLie}$$ so that composing with the above gives our desired "local system of graded (quasi-)Lie algebras".</p> <p>For convenience, I'll replace loop spaces with (strict) simplicial groups. Then, the Samelson product comes from noticing that the commutator map $[,]: G^2 \to G$ is trivial if one of the factors is the identity, and so factors through a pointed map $[,]: G \wedge G \to G$. This pointed map goes on to induce the (quasi-)Lie structure on homotopy. A group homomorphism $H \to G$ preserves commutators and identities, and so induces a map $H \wedge H \to G \wedge G$ compatible with the brackets, so that this construction is indeed functorial.</p> http://mathoverflow.net/questions/15897/in-what-topology-dm-stacks-are-stacks/15910#15910 Answer by Anatoly Preygel for In what topology DM stacks are stacks Anatoly Preygel 2010-02-20T17:36:47Z 2010-02-20T17:36:47Z <p>The rule of thumb is this: Your DM (or Artin) stack will be a sheaf in the fppf/fpqc topology if the condition imposed on its diagonal is fppf/fpqc local on the target ("satisfies descent"). </p> <p>In other words, in condition 2 you asked that the diagonal be a relative scheme/relative algebraic space perhaps with some extra properties. If there if fppf descent for morphisms of this type (e.g., "relative algebraic space", "relative monomorphism of schemes"), you'll have something satisfying fppf descent. If there is fpqc descent for morphisms of this type (e.g., "relative quasi-affine scheme"), then you'll have something satisfying fpqc descent.</p> <p>See for instance LMB (=Laumon, Moret-Bailly. Champs algebriques), Corollary 10.7. Alternatively: earlier this year I wrote up some <a href="http://www.math.harvard.edu/~gaitsgde/grad%5F2009/SeminarNotes/Sept15-17%28stacks%29.pdf" rel="nofollow">notes (PDF link)</a> that included an Appendix collecting in one place the equivalences of some standard definitions of stacks, including statements of the type above.</p> http://mathoverflow.net/questions/15850/characterization-of-cofibrations-in-cw-complexes-with-g-action/15880#15880 Answer by Anatoly Preygel for characterization of cofibrations in CW-complexes with G-action Anatoly Preygel 2010-02-20T07:30:36Z 2010-02-20T07:30:36Z <p>In the model structure you describe, the cofibrations should be the retracts of the free relative G-cell maps: i.e., retracts of maps obtained by attaching cells of the form $G \times S^{n} \to G \times D^{n+1}$.</p> <p>One way to see this is via the following general machine: There is an adjoint pair $$G \times -: \mathbf{Top} \leftrightarrow \mathbf{GTop}: forget$$ $\mathbf{Top}$ is a cofibrantly generated model category and one can check that this adjoint pair satisfies the conditions of the standard Lemma for transporting cofibrantly generated model structures across adjoint pairs (see e.g., Hirschorn's "Model categories and their localizations" Theorem 11.3.2). Thus, it gives rise to a model structure on $\mathbf{GTop}$ such that a map in $\mathbf{GTop}$ is an equivalence(resp. fibration) iff its image under the right adjoint (forget) is so. Moreover, the generating (acyclic/)cofibrations are precisely the images under the left adjoint ($G \times -$) of the generating (acyclic/)cofibrations in $\mathbf{Top}$. This yields the description of the cofibrations as retracts of (free G-)"cellular" maps.</p> <p>Also, some context:</p> <p>The model structure you describe (which I'd like to call "Spaces over BG") is a localization of a model structure "G-Spaces" (where the weak equivalences are maps inducing weak equivalences on all fixed point sets). An argument along the lines of the above constructs this other model structure and identifies its cofibrations with retracts of (arbitrary) relative G-cell maps: i.e., retracts of maps obtained by attaching cells of the form $G/H \times S^n \to G/H \times D^{n+1}$ for $H$ a closed subgroup of $G$.</p> http://mathoverflow.net/questions/11117/h-space-structure-on-infinite-projective-spaces/11122#11122 Answer by Anatoly Preygel for H-space structure on infinite projective spaces Anatoly Preygel 2010-01-08T08:01:29Z 2010-01-08T08:01:29Z <p>There's also a different way of writing down the $H$-space structure, that I like for its algebro-geometric flavor. (I'll talk about $\mathbb{C}P^\infty$ here, and $\mathbb{R}P^\infty$ should be analogous.)</p> <p>Regarding $\mathbb{C}P^\infty$ as a classifying space for complex line bundles, we know that this $H$-space structure is supposed to implement "tensor product of line bundles". In a (not very explicit) sense this tells us the homotopy class of $\mathbb{C}P^\infty \times \mathbb{C}P^\infty \to \mathbb{C}P^\infty$: It represents the line bundle $\mathcal{O}(1,1) = p_1^* \mathcal{O}(1) \otimes p_2^* \mathcal{O}(1)$. We can use this description to write down a much more explicit (and classical) explicit representative. </p> <p>First, let's recall what the analogous picture looks like for finite projective spaces. The line bundle $\mathcal{O}(1,1)$ determines (upon picking generating sections) the <strong>Segre map</strong> $\mathbb{C}P^n \times \mathbb{C}P^m \to \mathbb{C}P^{nm+n+m}$ which takes (in homogeneous coordinates) </p> <p>$([X_0:\ldots:X_n] , [Y_0:\ldots:Y_m]) \mapsto[X_0 Y_0: \ldots : X_i Y_j: \ldots: X_n Y_m]$ </p> <p>where I'm choosing to be vague on the precise ordering of the coordinates. (In the end this won't matter up to homotopy, as the maps will become homotopic upon composing with $\mathbb{C}P^{nm+n+m} \hookrightarrow \mathbb{C}P^\infty$.)</p> <p>The analogous formula with infinitely many homogeneous coordinate makes just as much sense, one just has to a good ordering of pairs of non-negative integers. Such an infinite Segre map gives another realization of the $H$-space structure.</p> http://mathoverflow.net/questions/7363/cyclic-spaces-and-s1-equivariant-homotopy-theory/7394#7394 Answer by Anatoly Preygel for Cyclic spaces and S^1-equivariant homotopy theory Anatoly Preygel 2009-12-01T08:19:37Z 2009-12-01T08:19:37Z <p>I don't know if this is exactly what you're looking for (and there's a good chance you already know what I'm going to write) but let me give it a try:</p> <p>The realization functor of cyclic sets (not spaces!) to $S^1$-spaces can be made part of a Quillen equivalence for two of the three commonly desired model structures on $S^1$-spaces: The model structure that gives you "Spaces over $BS^1$" is given in a 1985 paper of Dwyer-Hopkins-Kan, while a model structure that gives you the equivalences that you want (i.e., checked on fixed sets for finite subgroups) is given in a 1995 paper "Strong homotopy theory of cyclic sets" by Jan Spalinksi.</p> <p>(Irrelevant to your question, but along the same lines: A recent paper of Andrew Blumberg describes how one can throw in some extra--still combinatorial--data and obtain a combinatorial model of the third desirable model structure on $S^1$-spaces, namely where equivalences are those that induce equivalences on fixed sets for all closed subgroups.)</p> <p>Spalinksi's model structure depends on the following construction of <code>$|X_{.}|^{C_n}$</code> (as a space-over-$BS^1$) in terms of the subdivision construction: The simplicial set <code>$(sd_r X)_n = X_{r(n+1)-1}$</code> has an action of <code>$C_r$</code>--since <code>$C_r$</code> is a subgroup of the copy of <code>$C_{r(n+1)}$</code> acting on <code>$X_{r(n+1)-1}$</code>; taking fixed points (in sSet) and then realizing gives <code>$|X_{.}|^{C_n}$</code>.</p> <p>This suggests (though I haven't checked too carefully) that remembering each <code>$X_n$</code> as a <code>$C_{n+1}$</code>-space (in the sense you suggest, with subgroups) is enough, as you expected. </p> <p><hr /></p> <p>Now begins the speculative (and probably wrong) part of this answer: I have nothing too certain to say about writing this as a functor category, but it doesn't seem too unreasonable (to me, right now, at least) based on the above simplicial subdivision construction that we might be able to construct a reasonable candidate: some sort of mix of the cyclic category and the orbit categories for the cyclic groups. Purely combinatorially, this seems to get tricky.</p> <p>But, I think we can realize this geometrically: Let $(S^1)_r$ be the circle equipped with a $\mathbb{Z}$ action given by the rotation by $2\pi/r$. We could try to define <code>$Hom'([m-1]_r, [m'-1]_{r'})$</code> along the lines of "(htpy classes of) degree $r'/r$, increasing $\mathbb{Z}$-equivariant maps $S^1 \to S^1$ sending the $mr$-torsion points to the $m' r'$-torsion points". This should correspond to taking all the $r$-cyclic categories and sticking them together, and in particular is bigger than what we want. But, the $\mathbb{Z}$-action on the circles should induce one on the $Hom'$-sets and the composition should respect it. Taking the quotient, we seem to get something that looks like a reasonable candidate. For each fixed $r$, we should be getting a copy of the cyclic category. And, e.g. <code>$Hom([m-1]_r, [mr-1]_1)$</code> should contain <code>$Hom_{orbit}(Z/mr, Z/r)$</code>. (Disclaimer: It's late and I haven't checked any of this too carefully!)</p> http://mathoverflow.net/questions/6891/gerbes-for-a-cyclic-group-or-maybe-g-m-too/6910#6910 Answer by Anatoly Preygel for Gerbes for a cyclic group. (or maybe G_m too) Anatoly Preygel 2009-11-26T19:44:34Z 2009-11-26T19:44:34Z <p>A bit of a response to your "Commentary": </p> <p>As you point out, the failure of your construction to hit all <code>$\mu_n$</code>-gerbes is governed by the exact sequence <code>$H^1(X, \mathbb{G}_m) \to H^2(X, \mu_n) \to H^2(X,\mathbb{G}_m)[n] \to 0$</code> so answering your question is related to producing torsion <code>$\mathbb{G}_m$</code>-torsors.</p> <p>The question of doing so has been studied as part of the theory of Brauer groups:</p> <p>Let $Br(X)$ ("Brauer group") denote the group of <em>Azumaya algebras</em>, which are a generalization of the central simple algebras over a field (that is the classical Brauer group).</p> <p>Let $Br'(X)$ ("Cohomological Brauer group") denote the <em>torsion part</em> of <code>$H^2(X, \mathbb{G}_m)$</code>.</p> <p>In the case of $X = Spec k$, $k$ a field, the equivalence of these two groups is classical and they can be computed in various cases of number-theoretic interest (e.g., number fields/local fields/finite fields). In this case, <code>$H^1(X, \mathbb{G}_m)=0$</code> by Hilbert's Theorem 90, and yet there are plenty of examples where $Br(X)$ is very much non-trivial. Grothendieck studied the relation between $Br(X)$ and $Br'(X)$ in general in Dix Exposes. The upshot is that there is an injective map <code>$Br(X) \to Br'(X)$</code> and it is an isomorphism in reasonable cases (e.g., I think $X$ quasi-projective over a field). (See Dix Exp, or Ch. IV of Milne's "Etale Cohomology".)</p> <p><hr /></p> <p>I won't say more about the general picture, but I'll work out in detail the simple case of $X = Spec \mathbb{R}$. In this case, <code>$Br(Spec \mathbb{R}) = \mathbb{Z}/2\mathbb{Z}$</code> generated by the class of the usual quaternions, viewed as a central simple algebra over $\mathbb{R}$. We can give a geometric description of the resulting $\mathbb{G}_m$-torsor: </p> <p>Start with the smooth plane conic <code>$C = Proj \mathbb{R}[x,y,z]/(x^2+y^2+z^2)$</code>. It's a smooth genus $0$ curve, but has no $\mathbb{R}$-points and so is not isomorphic to <code>$\mathbb{P}^1_{\mathbb{R}}$</code>. However, after base-change to $\mathbb{C}$ it attains a point and so becomes isomorphic to <code>$\mathbb{P}^1_{\mathbb{C}}$</code>; such a Galois-twisted form of <code>$\mathbb{P}^n_{\mathbb{C}}$</code> is known as a Brauer-Severi variety and the elements of the Brauer-group (of a field) can also be thought of as corresponding to them (the group structure is then a bit strange). Since <code>$Aut(\mathbb{P}^n) = PGL_{n+1}$</code>, these correspond to <code>$PGL_{n+1}$</code>-torsors and the relation to <code>$\mathbb{G}_m$</code>-torsors is via the exact sequence for <code>$PGL_{n+1}=GL_{n+1}/\mathbb{G}_m$</code>. So, a $T$-point of the corresponding <code>$\mathbb{G}_m$</code>-torsor for a $\mathbb{R}$-scheme $T$ consists of the following data:</p> <p>It is a rank $2$ vector bundle $V$ over $T$, together with an isomorphism of $T$-schemes <code>$C_T \simeq \mathbb{P}(V)$</code> where <code>$C_T = C \times_{Spec \mathbb{R}} T$</code> is the pullback of our genus $0$ curve to $T$, and $\mathbb{P}(V)$ is the associated projective space (here $\mathbb{P}^1$) bundle of our vector bundle $V$. </p> <p>Why is this a <code>$\mathbb{G}_m$</code>-gerbe? Well, $\mathbb{P}(V) \simeq \mathbb{P}(V')$ iff $V$ and $V'$ differ by tensoring by a line bundle. The gerbe is non-trivial since it has no $\mathbb{R}$-points, since $C$ itself is not isomorphic to projective space. It has $\mathbb{C}$-points because the base-change is isomorphic to projective space.</p> http://mathoverflow.net/questions/6762/why-is-an-injective-quasi-coherent-sheafs-restriction-to-an-open-subset-still-an/6778#6778 Answer by Anatoly Preygel for Why is an injective quasi-coherent sheaf's restriction to an open subset still an injective object? Anatoly Preygel 2009-11-25T09:57:58Z 2009-11-25T09:57:58Z <p>The restriction-by-zero type arguments can actually be made to work, with some effort and an extra hypothesis. Suppose <code>$X$</code> is <em>locally Noetherian</em>, <code>$j: U \to X$</code> the inclusion of an open subscheme.</p> <p>Let $Mod(X)$ and $QCoh(X)$ be the categories of $O_X$-modules, and quasi-coherent $O_X$-modules, respectively.</p> <p>The "some effort" is the following Lemma</p> <p><strong>Lemma</strong> If $X$ is locally Noetherian, then the injective objects in $QCoh(X)$ are precisely the injective objects of $Mod(X)$ which are quasi-coherent as sheaves of modules.</p> <p><strong>Pf</strong>: Any injective object of $Mod(X)$ which is quasi-coherent must certainly be injective in the smaller category $QCoh(X)$. For the converse, it suffices to show that any injective object <code>$I$</code> of $QCoh(X)$ injects into some $I'$ which is a quasi-coherent injective object of $Mod(X)$, for then $I$ will be a retract of $I'$ and so injective in $Mod(X)$. This seems tricky, but is proved in Theorem 7.18 of Hartshorne's "Residues and duality".</p> <p><hr /></p> <p>Now, let's prove the result using the Lemma: If <code>$J$</code> is an injective object in $QCoh(X)$, then the hard direction of the Lemma implies that it is injective in $Mod(X)$. The restriction-by-zero argument applies in this category, allowing us to conclude that <code>$j^* J$</code> is injective in $Mod(U)$. It's clearly quasi-coherent, so applying the easy direction of the Lemma we see that it is injective in $QCoh(U)$ as desired.</p> <p>[Aside: On a <em>Noetherian</em> scheme, any quasi-coherent sheaf is a union of its coherent subsheaves and one can "extend" coherent sheaves on U to coherent sheaves on X (see e.g., Hartshorne Ex. II.5.15). Using these facts, one should be able to give a more direct argument in the Noetherian case.]</p> http://mathoverflow.net/questions/105753/universal-property-of-the-smash-product-of-pointed-spaces/105783#105783 Comment by Anatoly Preygel Anatoly Preygel 2012-08-29T06:50:05Z 2012-08-29T06:50:05Z Akhil's answer describes a particular formalization of the idea that the smash product is characterized by: do what you expect on (stable) cells, and then glue up. If no one in a millions years would have come up with that, we are all in trouble. http://mathoverflow.net/questions/77870/k-theory-and-k-theory-pushforward-in-topology-vs-in-algebraic-geometry Comment by Anatoly Preygel Anatoly Preygel 2011-10-12T00:55:57Z 2011-10-12T00:55:57Z And Remark 4.17 of op.cit., and the text around it, discusses the compatibility of that relationship with pushforwards. http://mathoverflow.net/questions/77870/k-theory-and-k-theory-pushforward-in-topology-vs-in-algebraic-geometry Comment by Anatoly Preygel Anatoly Preygel 2011-10-12T00:50:15Z 2011-10-12T00:50:15Z Re 1, elaborating on Daniel's comment: If you rationalize on both sides, this is asking to compare rational Chow theory with rational even-dimensional cohomology. In general, that won't go well (e.g., for $X$ a positive genus curve). It will be true only under some very restrictive hypotheses on $X$ (e.g., if $X$ is &quot;cellular&quot; in the sense of a locally closed decomposition by affine spaces). The result of Thomason is saying that things in fact get <i>better</i> if you look at higher and higher K-theory and look at torsion(/complete) instead of rationalizing. http://mathoverflow.net/questions/74405/poisson-algebras-as-deformations-vs-poisson-algebras-in-algebraic-topology Comment by Anatoly Preygel Anatoly Preygel 2011-09-03T01:57:11Z 2011-09-03T01:57:11Z I'm not entirely sure what you're looking for in the body of the question, but your &quot;For example&quot; admits a silly affirmative answer as follows. Going from chains to homology <i>is</i> a passage to an associated graded (for the &quot;Postnikov filtration&quot; $F_k V= \tau_{\geq k} V$). The $d$-Poisson algebra $H_*(\Omega^d X)$ is the associated graded of the $E_d$-algebra $C_*(\Omega^d X)$. (Here by $E_d$ I mean the operad of chains on the little $d$-disks operad. It's a filtered operad in the manner above, $C_*(\Omega^d X)$ is then a filtered algebra over this operad $E_d$, etc. etc.) http://mathoverflow.net/questions/60761/zero-dimensional-algebras-of-infinite-vector-space-dimension Comment by Anatoly Preygel Anatoly Preygel 2011-04-06T02:46:10Z 2011-04-06T02:46:10Z Take $A$ to be a field of infinite degree over the base field $k$. http://mathoverflow.net/questions/42623/are-the-stiefel-whitney-classes-of-a-vector-bundle-the-only-obstructions-to-its-b/42626#42626 Comment by Anatoly Preygel Anatoly Preygel 2010-10-18T18:23:19Z 2010-10-18T18:23:19Z Err, ha-ha? <a href="http://mathoverflow.net/questions/41833/is-every-artin-dm-algebraic-stack-fibered-in-sets-an-algebraic-space" rel="nofollow" title="is every artin dm algebraic stack fibered in sets an algebraic space">mathoverflow.net/questions/41833/&hellip;</a> http://mathoverflow.net/questions/39323/can-the-euler-product-and-weierstrass-factorization-of-a-function-yield-two-diffe Comment by Anatoly Preygel Anatoly Preygel 2010-09-19T20:22:33Z 2010-09-19T20:22:33Z It's unclear to me what's being asked, but maybe the following helps suggest how amazing it is that e.g. $L$-functions do admit two such product expansions: In the case of $\zeta$, the existence of the two substantially different products (Euler vs. Hadamard) underlies Riemann's relation of zeta zeroes and prime counting (mod non-trivial analysis). Take the two product expansions and take logarithmic derivatives; this gives you an equality of sums; taking a suitable integral transform (against $x^s ds/s$ along a suitable contour), one (i.e., Riemann) obtains Riemann's explicit formula! http://mathoverflow.net/questions/39323/can-the-euler-product-and-weierstrass-factorization-of-a-function-yield-two-diffe Comment by Anatoly Preygel Anatoly Preygel 2010-09-19T19:18:53Z 2010-09-19T19:18:53Z <a href="http://mathworld.wolfram.com/HadamardProduct.html" rel="nofollow">mathworld.wolfram.com/HadamardProduct.html</a> http://mathoverflow.net/questions/34848/are-topological-manifolds-homotopy-equivalent-to-smooth-manifolds Comment by Anatoly Preygel Anatoly Preygel 2010-08-07T18:44:21Z 2010-08-07T18:44:21Z @Ryan's Comment: It seems to me that the wiki article doesn't explicitly address (at least the stronger form of) this question. The triangulation obstruction is for lifting a given $BTop$ structure, and this structure (and thus the obstruction) is preserved by homeomorphism of topological manifolds but not homotopy equivalence. (There <i>are</i> non-triangulable topological manifolds homotopy equivalent to triangulable ones.) [It indirectly addresses it, by mentioning $E8$. But it doesn't mention the fact that it's ruled out by its signature form.] http://mathoverflow.net/questions/33345/if-a-colimit-of-distinguished-triangles-exists-is-it-also-a-distinguished-triang/33403#33403 Comment by Anatoly Preygel Anatoly Preygel 2010-07-27T05:11:13Z 2010-07-27T05:11:13Z Doesn't the Milnor $lim^1$-sequence measure the failure of sequential hocolims to be colims in the homotopy category? That is suppose $X = hocolim X_i$, so taking $\Sigma^\infty_+$ the same is true stably.The Milnor $lim^1$-sequence is identifying the, generally non-zero, difference between $E^0(X) = [\Sigma^\infty_+ X, E]$ and $\varprojlim E^0(X_i) = \varprojlim [\Sigma^\infty_+ X_i, E]$ (which would be $[colim \Sigma^\infty_+ X_i, E]$ if the colim existed). http://mathoverflow.net/questions/33345/if-a-colimit-of-distinguished-triangles-exists-is-it-also-a-distinguished-triang/33346#33346 Comment by Anatoly Preygel Anatoly Preygel 2010-07-26T01:49:53Z 2010-07-26T01:49:53Z Hi Ryan: The colimit of the third terms is a directed colimit along zero maps, so shouldn't it be just zero? I don't think a counterexample can exist in a category as nice as $\mathbb{Z}$-mod. The original statement wants to be true since &quot;(homotopy) colimits commute,&quot; and in $\mathbb{Z}$-mod that is more than a heuristic. http://mathoverflow.net/questions/32248/extraordinary-cohomology-as-a-derived-functor/32252#32252 Comment by Anatoly Preygel Anatoly Preygel 2010-07-20T00:32:20Z 2010-07-20T00:32:20Z The proof of the Theorem demonstrates how to get somewhat more &quot;hands on&quot; with sheaves of spaces (esp. 7.1.3). Another thing to keep in mind: The first step from sheaves of sets to sheaves of spaces, is sheaves of groupoids. This demonstrates the choices one can make in defining &quot;presheaves&quot; (fibered categories vs not), and what the higher bits of the sheaf condition encode. I don't know of a specific reference for sheaves of spectra, but the definitions are analogous to the space versions. (Also, I should repeat that the $R$ in $Rf_*$ and $R\Gamma$ was just analogy; it'd be $f_*$ in HTT.) http://mathoverflow.net/questions/32248/extraordinary-cohomology-as-a-derived-functor/32252#32252 Comment by Anatoly Preygel Anatoly Preygel 2010-07-20T00:16:28Z 2010-07-20T00:16:28Z I don't know a to-the-point reference, and the definitions <i>are</i> as in the classical case once the background is laid. For sheaves of spaces (for a Grothendieck topology, like say a usual topology) a definition appears in HTT 6.2.2. A better place to start looking may be HTT Section 7.1. Theorem 7.1.0.1 encodes, upon unraveling it, the relationship between the sheafy and representable perspectives (for sheaves of spaces). (Pullback from a point gets you the &quot;constant sheaf&quot;, and pushforward to a point is what I was calling $R\Gamma$. The $R$ was just by analogy, though.) .. (cont.) .. http://mathoverflow.net/questions/32248/extraordinary-cohomology-as-a-derived-functor/32252#32252 Comment by Anatoly Preygel Anatoly Preygel 2010-07-17T03:49:43Z 2010-07-17T03:49:43Z Finally, I should mention that in this example we only ever needed to work with &quot;locally constant&quot; sheaves of spectra. Just as locally constant sheaves of e.g., vector spaces admit a simpler description on nice spaces (as &quot;local systems&quot;), so it is for spectra. Details are in above refs, but here's the intuitive sketch: At each point, you have a spectrum. For each path, you have an equivalence between the spectra over them; for any two composable paths, you have a way of &quot;filling in the triangle&quot; (i.e., a homotopy between the two resulting maps); etc. http://mathoverflow.net/questions/32248/extraordinary-cohomology-as-a-derived-functor/32252#32252 Comment by Anatoly Preygel Anatoly Preygel 2010-07-17T03:43:41Z 2010-07-17T03:43:41Z .. some combination of Lurie's &quot;Higher Topos Theory&quot; (with more good stuff in Appendix A to DAG VI), or for a more classical perspective something like May-Sigurdsson &quot;Parameterized homotopy theory.&quot; What I call &quot;sheaves of spectra&quot; would go under &quot;ex-spectra&quot; in those terms.