User unknown - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T17:37:03Z http://mathoverflow.net/feeds/user/15015 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/130339/hodge-filtration-over-mathbb-z-p Hodge filtration over $\mathbb Z_p$ unknown 2013-05-11T14:20:21Z 2013-05-11T14:20:21Z <p>Let $p$ be a prime number. Let $X\to\operatorname{Spec}\mathbb Z_p$ be smooth and proper. Is it true that the map $H^i(X,\Omega^{\bullet\geq j}_{X/\mathbb Z_p})\to H^i(X,\Omega^\bullet_{X/\mathbb Z_p})$ induced by the inclusion of complexes is injective? This is easy to see if the groups $H^i(X,\Omega^j_{X/\mathbb Z_p})$ are $p$-torsion free, but I think it should be true in general.</p> <p>Thanks!</p> http://mathoverflow.net/questions/129942/how-to-prove-this-algebra-is-flat How to prove this algebra is flat? unknown 2013-05-07T10:35:30Z 2013-05-07T12:16:01Z <p>Hi,</p> <p>Let $S = R[T_1,\dots,T_n]/(f_1,\dots,f_r)$ where $\det(\partial f_i/\partial T_j)_{i,j=1,\dots,r}\in S^\times$. Then $S$ is flat over $R$. How to prove it? I am not looking for an answer like: "$Spec(S)$ is smooth over $Spec(R)$, hence flat.".</p> <p>Thanks!</p> http://mathoverflow.net/questions/128853/algebraic-de-rham-cohomology-of-singular-varieties algebraic de Rham cohomology of singular varieties unknown 2013-04-26T19:43:05Z 2013-04-26T20:22:26Z <p>Hi,</p> <p>Is there a simple example of an (affine) algebraic variety $X$ over $\mathbb C$ where the $H^*_{dR}(X/\mathbb C) = H^*(\Omega^\bullet_{A/\mathbb C})$ differs from the singular cohomology $H^*_{sing}(X(\mathbb C)^{an},\mathbb C)$?</p> <p>Such an example has to be singular (by a theorem of Grothendieck), but I am having a hard time finding one. The case $xy = 0$ doesn't work (both cohomology theories give the same answer).</p> <p>Thanks!</p> <p>EDIT: I would like a reduced example if possible...</p> http://mathoverflow.net/questions/128077/needless-axiom-for-grothendieck-topologies Needless axiom for Grothendieck topologies? unknown 2013-04-19T12:31:07Z 2013-04-19T12:31:07Z <p>Hi,</p> <p>The first axiom for a Grothendieck (pre)topology on a category $C$ says that for every object $X\in C$, the family consisting of just the identity $1_X : X\to X$ should be a covering family.</p> <p>Why is this axiom needed? Obviously a functor $F : C^{opp}\to Sets$ will satisfy the sheaf condition with respect to this family, so nothing seems to be gained by adding this covering family...</p> <p>Am I missing something?</p> <p>Thanks!</p> http://mathoverflow.net/questions/127807/exponential-map-for-finite-group-schemes exponential map for finite group schemes? unknown 2013-04-17T08:39:16Z 2013-04-17T08:39:16Z <p>Hi,</p> <p>I am trying to define an exponential map for finite abelian group schemes. The following looks like it should work, but doesn't (see below). I am putting up this question hoping that someone will know how to fix this.</p> <p>Let $R$ be a $\mathbb Q$-algebra and let $I\subset R$ be a nilpotent ideal. Let $G$ be a finite free group scheme over $R$. Let $A = \Gamma(G,\mathcal O_G)$. It is an $R$-algebra and $A^\vee = Hom_R(A,R)$ is likewise an $R$-algebra (using the comultiplication corresponding to the group operation of $G$). Let $\epsilon : A\to R$ correspond to the unit element $e\in G(R)$ of $G$. Let $Lie\ G$ be the Lie algebra of $G$: $$(Lie\ G)(R) = \ker(G(R[t]/t^2)\to G(R)) = [ x\in A^\vee : x(ab) = \epsilon(a)x(b) +\epsilon(b)x(a)]\subset A^\vee.$$ Next we want to define $\exp : I\cdot (Lie\ G)(R)\subset I\cdot A^\vee \to G(R)$. Let $x\in I\cdot (Lie\ G)(R)$. Then we put $$\exp (x) = \sum_{n=0}^\infty \frac{x^n}{n!}$$ Note that the sum is finite because $I$ is nilpotent. The element $x^n$ corresponds to the multiplication in $A^\vee$ and is defined by $x^n(a) = x(a)^n\in R$.</p> <p>This all seems like it should work, but unfortunately you will see that $\exp(x):A\to R$ is not a ring homomorphism so it doesn't define an element in $G(R)$.</p> <p>How to fix this?</p> http://mathoverflow.net/questions/127351/simple-proof-of-relation-between-h1-crystalline-and-dieudonne-module simple proof of relation between H^1 crystalline and Dieudonne module? unknown 2013-04-12T11:56:18Z 2013-04-12T11:56:18Z <p>Hi,</p> <p>Let $k$ be a perfect field of characteristic $p > 0$. Let $A/k$ be an abelian variety. Then the first crystalline cohomology group of $A$ with respect to $W(k)$ (= Witt vectors) is canonically isomorphic to the (contravariant) Dieudonne functor of the $p$-divisible group of $A$.</p> <p>This is well-known result and I am wondering if there is a simple proof of it. Specifically, I am wondering if there is a proof that doesn't require defining the Dieudonne <em>crystal</em> for a $p$-divisible group as an intermediary.</p> <p>Thanks!</p> http://mathoverflow.net/questions/126973/de-rham-complex-of-closed-immersion-between-smooth-schemes de Rham complex of closed immersion between smooth schemes unknown 2013-04-09T12:33:53Z 2013-04-09T15:05:41Z <p>Hi,</p> <p>Let $R$ be a $\mathbb Q$-algebra and let $P$ and $Q$ be (EDIT: smooth) $R$-algebras such that there is a surjective map of $R$-algebras $Q\to P$. The following proof cannot possibly be correct, but I can't find the mistake:</p> <p>''Theorem'': The natural map $\Omega^\bullet_{Q/R}\to \Omega^\bullet_{P/R}$ is a quasi-isomorphism of $Q$-modules.</p> <p>''Proof''. Since this assertion is local on $Spec(Q)$, we can assume there is a cartesian diagram of rings</p> <pre><code>Q ---&gt; P ^ ^ | | F ---&gt; G </code></pre> <p>where $F = R[T_1,\dots,T_{r+n}]$ and $G = F/(T_{r+1},\dots,T_n)$ and both vertical maps are etale. Since etale maps are flat, it is enough to prove the assertion when $Q = F$ and $P = G$. This case is well-known to be true (because $R\supset\mathbb Q$). QED?</p> <p>Do you see my mistake??</p> <p>Thanks!</p> http://mathoverflow.net/questions/54593/diagonal-map-and-infinitesimal-points/126956#126956 Answer by unknown for Diagonal map and "infinitesimal points" unknown 2013-04-09T10:26:19Z 2013-04-09T10:26:19Z <p>I would like to add another answer to this old question. Consider the case $X = Spec(A)$, $Y = Spec(R)$. Just to fix ideas, suppose that $A = R[T]$. If $f\in A$ and $a_0\in R$, one can consider the Taylor expansion of $f$ around $a_0$:</p> <p>$$f(T) = \sum_i \frac{f^{(i)}(a_0)}{i!}\cdot (T-a_0)^i\in R[T].$$</p> <p>Now there is no reason why we should take a rational point $a_0 : R[T] \to R$ and in fact we can consider the Taylor expansion around an arbitrary $S$-valued point $a_0 : R[T]\to S$. The Taylor expansion will then be naturally an element of $S\otimes_R R[T]$. Taking the <em>universal</em> point $S = R[T_0]$, $a_0 = T_0$, we see that the universal Taylor expansion'' of $f$ is $$f(T) = \sum_i\frac{f^{(i)}(T_0)}{i!}\cdot (T-T_0)^i\in R[T_0,T].$$ If we write $R[T_0,T] = R[T]\otimes_R R[T]$, then we rewrite the above as $$1\otimes f(T) = \sum_i\left(\frac{f^{(i)}(T)}{i!}\otimes 1\right)\cdot(1\otimes T-T\otimes 1)^i$$ Looking mod $(1\otimes T-T\otimes 1)^2$ we get: $$1\otimes f(T) \equiv f(T)\otimes 1 + (f'(T)\otimes 1)\cdot (1\otimes T-T\otimes 1)\pmod{(1\otimes T-T\otimes 1)^2}$$</p> <p>Now, in this particular case, $I =\ker(A\otimes_R A\to A)$ is generated by $1\otimes T-T\otimes 1$. Hence we see that $I/I^2$ is simply the space of linear terms of Taylor expansions and the canonical map $d : A\to I/I^2$ is simply sending a function $f\in A$ to the linear term in its Taylor series. Note that $1\otimes T-T\otimes 1$ is usually denoted by $dT$.</p> <p>This also explains nicely what happens in higher degree. We can introduce the algebras $P^n = (A\otimes_R A)/I^{n+1}=R[T_0,T]/(T-T_0)^{n+1}$, the <em>ring of Taylor expansions of degree $\leq n$</em> where the terms of degree at most $n$ of the Taylor expansion live. There is a natural map $d^n : A\to P^n$, sending $a$ to $1\otimes a$ which is simply sending $a$ to its Taylor expansion.</p> <p>This explanation works exactly the same if $A/R$ is smooth (instead of $A = R[T]$), because locally on $A$ there is an etale map $F\to A$ where $F$ is a polynomial $R$-algebra and this map induces an isomorphism on $I/I^2$ and $P^n$ more generally.</p> http://mathoverflow.net/questions/126949/smooth-algebras-and-triviality-of-de-rham-complex smooth algebras and triviality of de Rham complex unknown 2013-04-09T09:46:23Z 2013-04-09T09:46:23Z <p>Hi,</p> <p>Let $R$ be a $\mathbb Q$-algebra and let $A$ be a smooth $R$-algebra. If $A$ is a polynomial algebra $A = R[T_1,\dots,T_n]$, then it is easy to see that the natural map $R \to \Omega^\bullet_{A/R}$ is a quasi-isomorphism of $R$-modules.</p> <p>Question: is the same true for $A$ a smooth $R$-algebra if there is an etale map $F\to A$ with $F = R[T_1,\dots,T_n]$ (this is always the case locally on $A$)?</p> <p>Thanks!</p> http://mathoverflow.net/questions/123684/arthur-clozel-prop-3-1-for-function-fields Arthur-Clozel Prop 3.1 for Function Fields? unknown 2013-03-05T23:29:27Z 2013-03-06T08:43:06Z <p>The subject says it all. I would like to know if Proposition 3.1 in Arthur-Clozel's book on the trace formula holds for local fields of positive characteristic.</p> <p>Thanks!</p> <p>EDIT: Here is Prop 3.1 of Arthur-Clozel: (Notation will be explained after the statement)</p> <h3>Proposition 3.1</h3> <p>Assume $\phi\in C_c^\infty(GL_n(E))$. Then there exists $f\in C_c^\infty(GL_n(F))$ such that, for regular $\gamma\in GL_n(F)$, $O_\gamma(f) = 0$ if $\gamma$ is not a norm and $O_\gamma(f) = TO_{\sigma\delta}(\phi)$ if $\gamma = N\delta$.</p> <h3>Notation</h3> <ul> <li><p>$E/F$ is an finite unramified extension of local fields (hence cyclic) of degree $r$. Let $\sigma\in Gal(E/F)$ be a generator.</p></li> <li><p>Then there is a <em>norm</em> map $N : \sigma\text{-conjugacy classes in } G(E)\to \text{conjugacy classes in } G(F)$, where we say that $\delta, \delta'\in G(E)$ are <em>$\sigma$-conjugate</em> if there is $h\in G(E)$ such that $\delta' = h^{-1}\delta\sigma(h)$. The map $N$ is defined by sending the class of $\delta$ to the class of $\delta\sigma(\delta)\cdots\sigma^{r-1}\delta$.</p></li> <li><p>$O_\gamma(f)$ is an orbital integral: $O_\gamma(f) = \int_{GL_n(F)_\gamma\backslash GL_n(F)}f(g^{-1}\gamma g)\ dg$, where $GL_n(F)_\gamma\subset GL_n(F)$ is the centralizer of $\gamma$.</p></li> <li><p>$TO_{\delta\sigma}(\phi)$ is the <em>twisted orbital integral</em>: $TO_{\delta\sigma}(\phi) = \int_{GL_n(E)_{\delta\sigma}\backslash GL_n(E)}\phi(h^{-1}\delta \sigma(h))\ dh$. Here $GL_n(E)_{\delta\sigma}\subset GL_n(E)$ is the <em>twisted centralizer</em> of $\delta$: $GL_n(E)_{\delta\sigma} = {h\in GL_n(E) : h^{-1}\delta\sigma(h) = \delta}$.</p></li> </ul> http://mathoverflow.net/questions/123236/does-this-follow-from-the-fundamental-lemma-of-ngo-laumon-waldspurger does this follow from the Fundamental Lemma of Ngo, Laumon, Waldspurger, ...? unknown 2013-02-28T16:36:18Z 2013-02-28T16:36:18Z <p>Hi,</p> <p>Does the following follow from FL?</p> <p>Recall some definitions:</p> <p>Let $E/F$ be an unramified extension of degree $r$ of local fields <em>of positive characteristic.</em> Let $\theta\in Gal(E/F)$ be a generator. Let $G = GL_{n}/F$. Let $N : G(E)\to G(F)$ be the <em>norm:</em> $N\delta = \delta\theta(\delta) \cdots \theta^{r-1}(\delta)$. It is known that $N\delta$ is conjugate to en element of $G(F)$ and this defines a map: $$N : {\text{\theta-conjugacy classes in G(E)}}\to{\text{conjugacy classes in G(F)}}.$$ (here $\delta$ and $\delta'$ are <em>$\theta$-conjugate</em> in $G(E)$ if there exists $h\in G(E)$ such that $\delta = h^{-1}\delta'\theta(h)$).</p> <p>We say that $\gamma\in G(F)$ <em>is a norm</em> if it is conjugate to to $N\delta$ for some $\delta\in G(E)$.</p> <p>Next, functions $f\in C^\infty_c(G(F))$ and $\phi\in C^\infty_c(G(E))$ are said to be <em>associated</em> if the following condition holds: for every semi-simple $\gamma\in G(F)$ the orbital integral $O_\gamma(f)$ if $\gamma$ is not a norm, and if there exists $\delta\in G(E)$ such that $N\delta = \gamma$ then $$O_\gamma^{G(F)}(f) = TO_{\delta\theta}^{G(E)}(\phi).$$ Here $TO_{\delta\theta}^{G(E)}(\phi)$ is the twisted orbital integral of $\phi$.</p> <p>Theorem: Suppose that $\phi\in C^\infty_c(G(E))$. Then there exists $f\in C^\infty_c(G(F))$ such that $f$ and $\phi$ are associated.</p> <p>Thanks!</p> http://mathoverflow.net/questions/122795/unramified-base-change-in-characteristic-p-0 unramified base change in characteristic p > 0? unknown 2013-02-24T13:31:35Z 2013-02-25T14:59:54Z <p>Hi,</p> <p>Suppose that $E/F$ is a unramified extension of local fields of characteristic zero. Let $G = GL_n$. Then it is well-known (due to Clozel?) that base change of tempered representations from $G(F)$ to $G(E)$ holds.</p> <p>Question: does the same result hold in the case of characteristic $p > 0$?</p> <p>Thanks!</p> <p>EDIT: As Olivier says, this actually seems to follow immediately from LLC for function fields. Thanks!</p> http://mathoverflow.net/questions/122424/taking-invariants-under-pro-p-group-is-exact Taking invariants under pro-p-group is exact? unknown 2013-02-20T16:58:14Z 2013-02-21T00:00:49Z <p>Let $l$, $p$ be primes. Is it true that the functor of taking invariants under pro-$p$-group $P$ of finite-dimensional $\mathbb Q_l$-vector spaces ($l\neq p$) is an exact functor?</p> <p>Thanks!</p> <p>NOTE 1: I am not assuming that the action is discrete.</p> <p>NOTE 2: I am assuming the action of the group on the vector space is continuous.</p> <p>ANSWER: The answer is yes. The proof is as follows. First, note that if $V$ is a continuous $\mathbb Q_l[P]$-module, then there exists a $\mathbb Z_l[P]$-lattice $L\subset V$ (use the compactness of $P$). Then $L = \varprojlim L/l^nL$ and $H^1_{cont}(P,L) = \varprojlim_n H^1_{cont}(P,L/l^nL) = 0$ [N, Prop. 2.3.5]. Since $H^1(P,V) = H^1(P,L)\otimes_{\mathbb Z_l}\mathbb Q_l$ [N, Prop. 2.3.10], we conclude.</p> <p>[N] Neukirch, Schmidt, Wingberg, Cohomology of Number Fields.</p> http://mathoverflow.net/questions/122351/grothendieck-monodromy-theorem-for-l-adic-sheaves Grothendieck monodromy theorem for l-adic sheaves unknown 2013-02-19T21:37:08Z 2013-02-20T19:44:46Z <p>Hi,</p> <p>Suppose that $F$ is a local field, $G_F$ its Galois group, $I$ the inertia subgroup, $k$ its residue field. Let $X$ be a finite type scheme over $k$. Let $C$ be a constructible $l$-adic sheaf on $X$ equipped with an action of $G_F$ compatibly with the action on $X_{\overline k}$ via $G_F\to G_k$.</p> <p>Is it true that there exists a filtration $0 = C_0 \subset \dots \subset C_n = C$ such that $I$ acts through a finite quotient on the associated graded?</p> <p>The statement when $X = Spec k$ is well-known (Grothendieck monodromy theorem).</p> <p>Thanks!</p> <p>Answer to Olivier's comment: As you mention, the object of interest is $H^*_{et}(X\otimes_k\overline k,C)^I$. I would like to have the result on the level of sheaves for some intermediate manipulations in my argument.</p> http://mathoverflow.net/questions/121593/weight-monodromy-conjecture-for-curves weight monodromy conjecture for curves? unknown 2013-02-12T13:15:31Z 2013-02-13T18:01:04Z <p>Hi,</p> <p>Is there a simple proof of the weight monodromy conjecture in the case of a curve over a mixed characteristic local field?</p> <p>Thanks!</p> http://mathoverflow.net/questions/121434/vanishing-of-spectral-term-in-arthur-selberg-trace-formula-for-gl2 vanishing of spectral term in Arthur-Selberg trace formula for GL(2)? unknown 2013-02-11T00:09:34Z 2013-02-11T14:45:00Z <p>Hi,</p> <p>In the Arthur-Selberg trace formula for $G = GL(2)/\mathbf Q$ (as seen for example in Gelbart's "Lectures on the Trace Formula"), the spectral side includes terms like: $$\int_{-\infty}^\infty tr (\rho(\mu,it)(f))dt$$ where $f$ is the test function, $\mu$ is a Hecke character, and $\rho(\mu,s)$ is the induced representation $$Ind_{B(\mathbf A)}^{G(\mathbf A)}\mu(a_1/a_2)|a_1/a_2|^s_{\mathbf A}$$ where we write elements of $B(\mathbf A)$ (= the standard Borel of $G$) as having $a_1$, $a_2$ in the diagonal. This term appears both in the hyperbolic and unipotent spectral terms.</p> <p>I would like to understand under which condition this term vanishes. In particular, is there a condition on $f_{\mathbf R}$ (the infinite component of my test function $f$) that would imply the vanishing of this term?</p> <p>Thanks!</p> http://mathoverflow.net/questions/120951/intersection-cohomology-and-etale-cohomology intersection cohomology and etale cohomology unknown 2013-02-06T08:10:48Z 2013-02-06T08:10:48Z <p>Hello,</p> <p>Can someone explain or give a reference on the comparison between intersection cohomology and l-adic etale cohomology of a variety over a field of characteristic zero?</p> <p>Thanks!</p> http://mathoverflow.net/questions/120251/cech-cohomology-in-topos cech cohomology in topos unknown 2013-01-29T19:50:51Z 2013-01-29T21:39:56Z <p>Hi,</p> <p>The following result seems to be well known, but I can't come up with a proof.</p> <p>Suppose that $C$ is a topos and that $F\to G$ is an effective epimorphism in $C$. If $P$ is any abelian sheaf on $C$, then the object $RHom(G,P)$ is computed by the bicomplex $$RHom(F,P)\to RHom(F\times_G F,P)\to RHom(F\times_G F\times_G F, P)\to \dots$$</p> <p>Any ideas? Thanks!</p> http://mathoverflow.net/questions/107602/difference-between-automorphic-forms-for-sl2-and-gl2 Difference between automorphic forms for SL(2) and GL(2)? unknown 2012-09-19T18:31:35Z 2012-11-23T16:08:36Z <p>Hi,</p> <p>Let $A$ denote the adeles of $Q$. I know how to decompose $L^2(SL(2,A)/SL(2,Q))$ into irreducible $SL(2,A)$-representations. What is the difference between this decomposition and the corresponding decomposition for $GL(2)$? Can I deduce the $GL(2)$-case from the $SL(2)$-case?</p> <p>Thanks for answering this basic question.</p> http://mathoverflow.net/questions/110462/poincare-duality-in-crystalline-cohomology-over-general-base-rings poincare duality in crystalline cohomology over general base rings unknown 2012-10-23T19:39:42Z 2012-10-23T19:39:42Z <p>Hi,</p> <p>Is there a reference for poincare duality for crystalline cohomology over rings more general than $W(k)$ (Witt vectors over a perfect field $k$)? In Berthelot's thesis, he only treats this case.</p> <p>Thanks!</p> http://mathoverflow.net/questions/94857/moduli-problem-for-flag-varieties moduli problem for flag varieties? unknown 2012-04-22T16:26:14Z 2012-04-23T19:03:55Z <p>Hi,</p> <p>Suppose $G$ is a reductive group over an algebraiclly closed field $k$ (suppose $k$ of char zero if you want at first). Let $X$ be its flag variety.</p> <p>Question: What is the moduli problem that $X$ represents?</p> <p>EDIT (to clarify): What is the functor of points of $X$?</p> <p>Thanks!</p> http://mathoverflow.net/questions/94001/semisimplicity-of-automorphic-galois-representations semisimplicity of automorphic Galois representations unknown 2012-04-14T00:48:29Z 2012-04-16T03:36:52Z <p>Is it known that the Galois representation constructed by Harris and Taylor in their book is semisimple? I can't see this proven in the book, but on the other hand, everywhere else the representation is taken to be semisimple... Are they considering its semisimplification?</p> <p>Sorry for the simple question.</p> <p>Thanks</p> http://mathoverflow.net/questions/93591/false-elliptic-curves-and-principal-polarizations false elliptic curves and principal polarizations unknown 2012-04-09T19:34:55Z 2012-04-09T20:12:44Z <p>Hi,</p> <p>Let $\Delta$ be a quaternion algebra over $\mathbf Q$ and let $\mathcal O_\Delta$ be a maximal order in $\Delta$. Recall that a <em>false elliptic curve</em> over a field $K$ is a pair $(A/K,i)$ consisting of an abelian surface $A/K$ and a ring homomorphism $i : \mathcal O_\Delta\to End_K(A)$. Suppose that $\Delta$ is indefinite, i.e., $\Delta\otimes\mathbf R \simeq M_2(\mathbf R)$. There is an involution $*:\Delta\to\Delta$ that coincides with taking transpose under the previous isomorphism. It is well-known (easy?) that if $K$ is of characteristic zero there is a polarization of $A/K$ such that the corresponding Rosatti involution in $End_K(A)\otimes\mathbf Q$ corresponds to $* : \Delta\to\Delta$ by $i$. Moreover, this polarization is unique up to a rational number.</p> <p>Question 1: is it possible to find a (necessarily unique) principal polarization with this property?</p> <p>Question 2: is it possible to find such a (principal) polarization over a field $K$ of non-zero characteristic?</p> <p>Thanks!</p> http://mathoverflow.net/questions/90784/cm-abelian-variety-from-an-algebraic-hecke-character CM abelian variety from an algebraic Hecke character? unknown 2012-03-10T02:07:56Z 2012-03-10T02:07:56Z <p>Hi,</p> <p>Given an algebraic Hecke character $\chi$ of a number field $k$ there should be a "rank 1 CM-motive" $M$ with $\overline Q$-coefficients such that $L(s,M) = L(s,\chi)$. This follows from the general theory of the Taniyama group and its quotient, the Serre group. My question is if there is a simpler way of constructing the motive $M$ starting from $\chi$...</p> <p>Thanks!</p> http://mathoverflow.net/questions/87358/algebraic-de-rham-cohomology-functoriality algebraic de Rham cohomology functoriality unknown 2012-02-02T18:52:44Z 2012-02-02T20:54:38Z <p>Suppose that $Y/k$ is a an algebraic variety over a field $k$ of characteristic zero and that $Y\subseteq X$ is a closed embedding into a smooth variety over $k$. Then the completion of the de Rham complex of $X/k$ along $Y$ is independent (up to quasi-isomorphism) of $X$. This is proven in Hartshorne's paper on de Rham cohomology. I want to understand the analogous statement for <em>maps</em>: suppose that $f:Y'\to Y$ is a morphism and that we can find smooth embeddings $Y'\subseteq X'$, $Y\subseteq X$ and maps $g_1,g_2 : X'\to X$ lifting $f$. Then they induce two maps of abelian sheaves on $Y'$: $$g_1^*, g_2^* : f^{-1}\hat{\Omega^*_X} \to \hat{\Omega^{*}_{X'}}$$ (hats mean completion along $Y'$, resp. $Y$). These two maps should be homotopic. I can't quite see this. Any ideas? I would like to write down the homotopy explicitly if possible as well...</p> http://mathoverflow.net/questions/81456/possible-borel-subgroups-of-gl-n Possible Borel subgroups of GL_n? unknown 2011-11-20T21:30:51Z 2011-11-21T00:55:14Z <p>I am trying to understand the interaction between Borel subgroups of $GL_n$ and its roots. Is it correct to say that for any choice of roots among each pair of reciprocal roots there is a Borel subgroup containing those root subgroups? (Meaning that exactly those roots appear on the decompsition of its Lie algebra) If not, what is the precise condition on the choice of roots that is needed for such a Borel to exist?</p> http://mathoverflow.net/questions/80951/algebraic-proof-of-atiyah-bott-fixed-point-formula algebraic proof of Atiyah-Bott fixed point formula? unknown 2011-11-15T03:58:05Z 2011-11-15T13:27:10Z <p>Hi,</p> <p>Atiyah and Bott apparently proved the following theorem:</p> <ul> <li>Let $X$ be a smooth projective complex variety and $L$ a line bundle on $X$. Let $f:X\to X$ be an automorphism of $(X,L)$ with finitely many fixed points $X^f$. Then $$\sum_{i=0}^{\dim X}(-1)^itr(f, H^i(X,L)) = \sum_{x\in X^f}\frac{tr(f,L_x)}{\det(1-T_xf)}$$ where $T_xf : T_xX\to T_xX$ is the derivative of $f$ at $x\in X$.</li> </ul> <p>Where can one find an algebraic proof of this result?</p> <p>Thanks!</p> http://mathoverflow.net/questions/80861/coherent-cohomology-of-g-u-g-reductive-group-b-tu-borel-subgroup Coherent cohomology of G/U, G = reductive group, B = TU Borel subgroup unknown 2011-11-14T02:24:18Z 2011-11-14T17:49:03Z <p>Hi,</p> <p>Let $G$ be an algebraic reductive group over an algebraically closed field $k$, $T$ a maximal torus and $B = TU$ a Borel subgroup containing it. I'm interested in computing $H^*(G/U,\mathcal O_{G/U})$ [corrected typo; I had written $B/U$] (coherent cohomology) (in terms of the representation theory of $G$?). I suppose this is well known, but I can't find it anywhere....</p> <p>Any suggestions?</p> <p>Thanks!</p> http://mathoverflow.net/questions/80637/p-adic-representations/80647#80647 Answer by unknown for P-adic representations unknown 2011-11-11T00:41:25Z 2011-11-11T01:11:08Z <p>They are using continuous cohomology, so that $$H^n(G,M) = \varinjlim H^n(G/H,M^H)$$ if $G$ is topological and $M$ is discrete (<em>thanks Arjit</em>) $G$-module (the limit runs over open compact subgroups $H$ of $G$). Look in p. 38 for the definition.</p> http://mathoverflow.net/questions/79956/jacobian-criterion-for-smoothness-of-schemes Jacobian criterion for smoothness of schemes unknown 2011-11-03T18:09:13Z 2011-11-04T08:21:10Z <p>An affine scheme $X = Spec(A)$ is said to be smooth if for any closed embedding $X\subset\mathbf A^n$, of ideal $I$, it is true that, locally on $x\in X$, the ideal $I$ can be generated by a sequence $f_{r+1},\dots,f_n$ such that their Jacobian has maximal rank.</p> <p>My question is:</p> <ul> <li>Will the Jacobian of ANY set of $n-r$ generators of $I$ be of maximal rank?</li> </ul> http://mathoverflow.net/questions/129942/how-to-prove-this-algebra-is-flat/129952#129952 Comment by unknown unknown 2013-05-07T12:26:03Z 2013-05-07T12:26:03Z Thanks for the link, the proof there is nice. http://mathoverflow.net/questions/128853/algebraic-de-rham-cohomology-of-singular-varieties Comment by unknown unknown 2013-04-26T19:50:30Z 2013-04-26T19:50:30Z Right, I was hoping for a reduced one... http://mathoverflow.net/questions/126973/de-rham-complex-of-closed-immersion-between-smooth-schemes Comment by unknown unknown 2013-04-09T15:05:26Z 2013-04-09T15:05:26Z Sorry, I forgot: $P$ and $Q$ are smooth over $R$. http://mathoverflow.net/questions/122424/taking-invariants-under-pro-p-group-is-exact Comment by unknown unknown 2013-02-21T13:14:48Z 2013-02-21T13:14:48Z You are right. I said it was not discrete because I was thinking of a general profinite group. It is true that for $P$ a pro-$p$-group acting on a pro-$l$-group, discrete iff continuous. Your observation makes it even simpler to prove that taking invariants under $P$ is an exact functor because after choosing a $\mathbb Z_l$-lattice, $P$ is acting thru a finite $p$-group quotient and then the claim is clear. http://mathoverflow.net/questions/122424/taking-invariants-under-pro-p-group-is-exact Comment by unknown unknown 2013-02-20T21:46:54Z 2013-02-20T21:46:54Z Yes, it is easy to see that there always is a $G$-invariant $\mathbb Z_l$-lattice. http://mathoverflow.net/questions/122424/taking-invariants-under-pro-p-group-is-exact/122460#122460 Comment by unknown unknown 2013-02-20T21:46:18Z 2013-02-20T21:46:18Z Note that I am assuming that the action is continuous (I have amended the statement in my question). http://mathoverflow.net/questions/122424/taking-invariants-under-pro-p-group-is-exact/122433#122433 Comment by unknown unknown 2013-02-20T18:38:21Z 2013-02-20T18:38:21Z Note that I am not assuming that the modules are discrete. http://mathoverflow.net/questions/94001/semisimplicity-of-automorphic-galois-representations/94176#94176 Comment by unknown unknown 2012-04-18T21:21:42Z 2012-04-18T21:21:42Z Thanks for your answer. I meant the global Galois representation. http://mathoverflow.net/questions/80951/algebraic-proof-of-atiyah-bott-fixed-point-formula/80963#80963 Comment by unknown unknown 2011-11-15T11:48:49Z 2011-11-15T11:48:49Z Thanks for the helpful answer. http://mathoverflow.net/questions/80861/coherent-cohomology-of-g-u-g-reductive-group-b-tu-borel-subgroup/80912#80912 Comment by unknown unknown 2011-11-14T19:55:34Z 2011-11-14T19:55:34Z Thanks for the answer, Chuck. My original intention was to somehow invert the order and try to deduce the cohomology of $H^*(G/B,L)$ from the knowledge of the cohomology of $H^*(G/U,O)$. That is, I wanted to compute the latter cohomology group independently of Borel-Weil-Bott (I don't have any reason to think this should work). http://mathoverflow.net/questions/79967/derived-functors-and-triangulated-categories/79968#79968 Comment by unknown unknown 2011-11-03T23:39:34Z 2011-11-03T23:39:34Z ... composition, irrespective if the functors appearing are left exact or right exact or one and one: you just use that they are triangulated. From this perspective using the letters $R$ and $L$ seem to only serve the purpose of remembering that the negative cohomologies are zero (in the case of $R$) or its positive cohomologies are zero (in the case of $L$). Rereading what I wrote, I'm not sure anyone will understand anything... http://mathoverflow.net/questions/79967/derived-functors-and-triangulated-categories/79968#79968 Comment by unknown unknown 2011-11-03T23:37:30Z 2011-11-03T23:37:30Z Dear Sandor, thanks for your answer. My question had to do with the fact that if one stays in the derived/triangulated world there is no difference between left and right exact functors: they are both triangulated. For example, a left exact functor will give you a long exact sequence that goes to the left in cohomology. A right exact functor will give you a long exact sequence going to the right in cohomology. But in the derived category they both 'just' preserve triangles. There is also a spectral sequence that relates the cohomology of any two derived functors to the cohomology of their... http://mathoverflow.net/questions/79956/jacobian-criterion-for-smoothness-of-schemes/79958#79958 Comment by unknown unknown 2011-11-03T19:54:53Z 2011-11-03T19:54:53Z Thanks for your answer. I am interested in the case where the base could be a general ring, not necessarily a field. http://mathoverflow.net/questions/79693/why-does-a-group-action-on-a-scheme-induce-a-group-action-on-cohomology/79697#79697 Comment by unknown unknown 2011-11-01T12:14:55Z 2011-11-01T12:14:55Z Sheaf cohomology is only functorial on the pairs $(X, F)$ consisting of a scheme $X$ and a sheaf $F$, where a map $(X, F)\to (Y, G)$ is a pair of maps $(f,\phi)$, $f : X \to Y$ and $\phi : f^*G \to F$. Any such map of pairs induces a map on the cohomology $H^i(Y,G)\to H^i(X,F)$ as explained by Niels. http://mathoverflow.net/questions/78954/modular-form-fourier-coefficients-and-associated-automorphic-representation/78959#78959 Comment by unknown unknown 2011-10-24T12:40:37Z 2011-10-24T12:40:37Z Thanks for your helpful answer, Olivier.