User unknown - MathOverflow

Hodge filtration over $\mathbb Z_p$

2013-05-11T14:20:21Z

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.

Thanks!

How to prove this algebra is flat?

2013-05-07T10:35:30Z

Hi,

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.".

Thanks!

algebraic de Rham cohomology of singular varieties

2013-04-26T19:43:05Z

Hi,

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 $H^*_{sing}(X(\mathbb C)^{an},\mathbb C)$?

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).

Thanks!

EDIT: I would like a reduced example if possible...

Needless axiom for Grothendieck topologies?

2013-04-19T12:31:07Z

Hi,

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.

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...

Am I missing something?

Thanks!

exponential map for finite group schemes?

2013-04-17T08:39:16Z

Hi,

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.

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 $x^n(a) = x(a)^n\in R$.

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 $G(R)$.

How to fix this?

simple proof of relation between H^1 crystalline and Dieudonne module?

2013-04-12T11:56:18Z

Hi,

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 $A$.

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 crystal for a $p$-divisible group as an intermediary.

Thanks!

de Rham complex of closed immersion between smooth schemes

2013-04-09T12:33:53Z

Hi,

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:

''Theorem'': The natural map $\Omega^\bullet_{Q/R}\to \Omega^\bullet_{P/R}$ is a quasi-isomorphism of $Q$-modules.

''Proof''. Since this assertion is local on $Spec(Q)$, we can assume there is a cartesian diagram of rings

Q ---> P
^      ^
|      |
F ---> G

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?

Do you see my mistake??

Thanks!

Answer by unknown for Diagonal map and "infinitesimal points"

2013-04-09T10:26:19Z

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 $a_0$:

$$f(T) = \sum_i \frac{f^{(i)}(a_0)}{i!}\cdot (T-a_0)^i\in R[T].$$

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 universal 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: $$ 1\otimes f(T) \equiv f(T)\otimes 1 + (f'(T)\otimes 1)\cdot (1\otimes T-T\otimes 1)\pmod{(1\otimes T-T\otimes 1)^2} $$

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 $dT$.

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 ring of Taylor expansions of degree $\leq n$ 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.

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.

smooth algebras and triviality of de Rham complex

2013-04-09T09:46:23Z

Hi,

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.

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 $A$)?

Thanks!

Arthur-Clozel Prop 3.1 for Function Fields?

2013-03-05T23:29:27Z

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.

Thanks!

EDIT: Here is Prop 3.1 of Arthur-Clozel: (Notation will be explained after the statement)

Proposition 3.1

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 $\gamma = N\delta$.

Notation

$E/F$ is an finite unramified extension of local fields (hence cyclic) of degree $r$. Let $\sigma\in Gal(E/F)$ be a generator.
Then there is a norm 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 $\sigma$-conjugate 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 $\delta\sigma(\delta)\cdots\sigma^{r-1}\delta$.
$O_\gamma(f)$ 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 $\gamma$.
$TO_{\delta\sigma}(\phi)$ is the twisted orbital integral: $TO_{\delta\sigma}(\phi) = \int_{GL_n(E)_{\delta\sigma}\backslash GL_n(E)}\phi(h^{-1}\delta \sigma(h))\ dh$. Here $GL_n(E)_{\delta\sigma}\subset GL_n(E)$ is the twisted centralizer of $\delta$: $GL_n(E)_{\delta\sigma} = {h\in GL_n(E) : h^{-1}\delta\sigma(h) = \delta}$.

does this follow from the Fundamental Lemma of Ngo, Laumon, Waldspurger, ...?

2013-02-28T16:36:18Z

Hi,

Does the following follow from FL?

Recall some definitions:

Let $E/F$ be an unramified extension of degree $r$ of local fields of positive characteristic. Let $\theta\in Gal(E/F)$ be a generator. Let $G = GL_{n}/F$. Let $N : G(E)\to G(F)$ be the norm: $N\delta = \delta\theta(\delta) \cdots \theta^{r-1}(\delta)$. 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 $\theta$-conjugate in $G(E)$ if there exists $h\in G(E)$ such that $\delta = h^{-1}\delta'\theta(h)$).

We say that $\gamma\in G(F)$ is a norm if it is conjugate to to $N\delta$ for some $\delta\in G(E)$.

Next, functions $f\in C^\infty_c(G(F))$ and $\phi\in C^\infty_c(G(E))$ are said to be associated 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 $\phi$.

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.

Thanks!

unramified base change in characteristic p > 0?

2013-02-24T13:31:35Z

Hi,

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.

Question: does the same result hold in the case of characteristic $p > 0$?

Thanks!

EDIT: As Olivier says, this actually seems to follow immediately from LLC for function fields. Thanks!

Taking invariants under pro-p-group is exact?

2013-02-20T16:58:14Z

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?

Thanks!

NOTE 1: I am not assuming that the action is discrete.

NOTE 2: I am assuming the action of the group on the vector space is continuous.

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.

[N] Neukirch, Schmidt, Wingberg, Cohomology of Number Fields.

Grothendieck monodromy theorem for l-adic sheaves

2013-02-19T21:37:08Z

Hi,

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 $G_F\to G_k$.

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?

The statement when $X = Spec k$ is well-known (Grothendieck monodromy theorem).

Thanks!

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.

weight monodromy conjecture for curves?

2013-02-12T13:15:31Z

Hi,

Is there a simple proof of the weight monodromy conjecture in the case of a curve over a mixed characteristic local field?

Thanks!

vanishing of spectral term in Arthur-Selberg trace formula for GL(2)?

2013-02-11T00:09:34Z

Hi,

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.

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?

Thanks!

intersection cohomology and etale cohomology

2013-02-06T08:10:48Z

Hello,

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?

Thanks!

cech cohomology in topos

2013-01-29T19:50:51Z

Hi,

The following result seems to be well known, but I can't come up with a proof.

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 $$ RHom(F,P)\to RHom(F\times_G F,P)\to RHom(F\times_G F\times_G F, P)\to \dots $$

Any ideas? Thanks!

Difference between automorphic forms for SL(2) and GL(2)?

2012-09-19T18:31:35Z

Hi,

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?

Thanks for answering this basic question.

poincare duality in crystalline cohomology over general base rings

2012-10-23T19:39:42Z

Hi,

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.

Thanks!

moduli problem for flag varieties?

2012-04-22T16:26:14Z

Hi,

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.

Question: What is the moduli problem that $X$ represents?

EDIT (to clarify): What is the functor of points of $X$?

Thanks!

semisimplicity of automorphic Galois representations

2012-04-14T00:48:29Z

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?

Sorry for the simple question.

Thanks

false elliptic curves and principal polarizations

2012-04-09T19:34:55Z

Hi,

Let $\Delta$ be a quaternion algebra over $\mathbf Q$ and let $\mathcal O_\Delta$ be a maximal order in $\Delta$. Recall that a false elliptic curve 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.

Question 1: is it possible to find a (necessarily unique) principal polarization with this property?

Question 2: is it possible to find such a (principal) polarization over a field $K$ of non-zero characteristic?

Thanks!

CM abelian variety from an algebraic Hecke character?

2012-03-10T02:07:56Z

Hi,

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 $\chi$...

Thanks!

algebraic de Rham cohomology functoriality

2012-02-02T18:52:44Z

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 maps: 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...

Possible Borel subgroups of GL_n?

2011-11-20T21:30:51Z

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?

algebraic proof of Atiyah-Bott fixed point formula?

2011-11-15T03:58:05Z

Hi,

Atiyah and Bott apparently proved the following theorem:

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 $x\in X$.

Where can one find an algebraic proof of this result?

Thanks!

Coherent cohomology of G/U, G = reductive group, B = TU Borel subgroup

2011-11-14T02:24:18Z

Hi,

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....

Any suggestions?

Thanks!

Answer by unknown for P-adic representations

2011-11-11T00:41:25Z

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 (thanks Arjit) $G$-module (the limit runs over open compact subgroups $H$ of $G$). Look in p. 38 for the definition.

Jacobian criterion for smoothness of schemes

2011-11-03T18:09:13Z

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.

My question is:

Will the Jacobian of ANY set of $n-r$ generators of $I$ be of maximal rank?

Comment by unknown

2013-05-07T12:26:03Z

Thanks for the link, the proof there is nice.

Comment by unknown

2013-04-26T19:50:30Z

Right, I was hoping for a reduced one...

Comment by unknown

2013-04-09T15:05:26Z

Sorry, I forgot: $P$ and $Q$ are smooth over $R$.

Comment by unknown

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.

Comment by unknown

2013-02-20T21:46:54Z

Yes, it is easy to see that there always is a $G$-invariant $\mathbb Z_l$-lattice.

Comment by unknown

2013-02-20T21:46:18Z

Note that I am assuming that the action is continuous (I have amended the statement in my question).

Comment by unknown

2013-02-20T18:38:21Z

Note that I am not assuming that the modules are discrete.

Comment by unknown

2012-04-18T21:21:42Z

Thanks for your answer. I meant the global Galois representation.

Comment by unknown

2011-11-15T11:48:49Z

Thanks for the helpful answer.

Comment by unknown

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).

Comment by unknown

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...

Comment by unknown

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...

Comment by unknown

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.

Comment by unknown

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.

Comment by unknown

2011-10-24T12:40:37Z

Thanks for your helpful answer, Olivier.