User hiro - MathOverflow

Are loop spaces of homotopically equivalent spaces homotopically equivalent?

2013-03-10T17:39:15Z

Let $f:X \to Y$ be a homotopy equivalence of pointed topological spaces.

Then, is the induced map of pointed loop spaces $\Omega (f): \Omega X \to \Omega Y$ a homotopy equivalence?

Here, loop spaces are equipped with the compact-open topologies.

Is there any counterexample?

I do not know even whether the induced map $\Omega (f)$ is continuous or not in general.

Examples of Sheafification via Hypercovers

2013-01-13T17:09:16Z

For a presheaf $F$ on a category equipped with a pretopology, one has the sheafification $F^{\sharp}$ of $F$.

I know well the plus-construction of sheafification, which is presented in Artin's paper "Grothendieck Topologies", for example.

QUESTION

I have heard that there is another construction of sheafification by using hypercovering, but I can not find good explanation of it.

So, I want some examples to feel the essence of the construction.

In particular, I am interested in the following situation:

Let $C$ be a category with a pretopology $T$.

For an object $X \in C$, denote by $F_{X}$ the presheaf on $C$ represented by $X$ i.e. $F_{X}=Hom_{C}(-,X)$.

Now, let $X_{\bullet} \to X$ be a $T$-hypercover of $X \in C$.

Then, how can the sheafification of $F_{X}$ be written by using $F_{X_{\bullet}}$ ?

Endomomorphisms of Chain Complexes of vector spaces and determinants

2012-12-30T17:51:38Z

Let $C_{\ast} : \cdots \to A_{2} \to A_{1} \to A_{0} \to 0$ be a chain complex of finite dimensional vector spaces over a field $K$.

And let $f_{\ast} : C_{\ast} \to C_{\ast}$ and $g_{\ast} : C_{\ast} \to C_{\ast}$ be two chain endomorphisms of $C_{\ast}$ satisfying $\det (f_{n}) \neq 0$ and $\det (g_{n}) \neq 0$ for any $n$.

Assume that the homology group $H_{n}(C_{\ast})$ is zero for all but finitely many $n$. Then, the induced homomorphism $H_{n}(f_{\ast}) : H_{n}(C_{\ast}) \to H_{n}(C_{\ast})$ is zero for all but finitely many $n$, and so the alternating product $\prod_{n}\det (H_{n}(f_{\ast}))^{(-1)^{n}}$ is well-defined (of course, the same holds for $g_{\ast}$).

Moreover, assume that $\det (f_{n}) = \det (g_{n})$ for any $n$.

My question is:

QUESTION

Under the above conditions, does the next equation hold up to sign?

$\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}} = \prod_{n} \det (H_{n}(g_{\ast}))^{(-1)^{n}}$.

Note that if the chain complex $C_{\ast}$ is bounded above, the statement can be proved as follows:

$\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}} = \prod_{n} \det (f_{n})^{(-1)^{n}} = \prod_{n} \det (g_{n})^{(-1)^{n}} = \prod_{n} \det (H_{n}(g_{\ast}))^{(-1)^{n}}$

Here, the first and the last equation can be shown by using induction on the length of $C_{\ast}$, snake lemma and the multiplicativity of $\det$ for short exact sequences. The middle equation is the result of the assumption.

The problem is that, in general, $\prod_{n} \det (f_{n})^{(-1)^{n}}$ and $\prod_{n} \det (g_{n})^{(-1)^{n}}$ are not well-defined.

By Sawin's answer

There is a counterexample if one does not admit the difference of sign.

Please give me any advice.

Morphisms of Spectral Sequences and alternating products

2012-12-29T03:34:40Z

Let $E_{a,b}^{r}, F_{a,b}^{r}$ be two (co)homologica first quadrant spectral sequences of vector spaces over a field $K$, and $f : E \to F$ be a morphism of spectral sequences.

Assume that morphisms $f_{a,b}^{1} : E_{a,b}^{1} \to F_{a,b}^{1}$ of the first page are all isomoprhisms.

If the numbers of non-zero terms appearing in the first pages of $E$ and $F$ are finite, then one can show that the following equation holds:

$\prod_{n}(\det (f_{n}))^{(-1)^{n}} = \prod_{a,b}(\det (f_{a,b}^{1}))^{(-1)^{a+b}}$.

Here, $f_{n}$ denotes the morphism $f_{n} : E_{n} \to F_{n}$, where $E_{n}$ and $F_{n}$ are the $n$th terms to which $E$ and $F$ converge(i.e. $E_{n}$ is a vector space equipped with filtration whose $i$th congruent is isomorphic to $E^{\infty}_{i,n-i}$).

(In order to define $\det$, one has to fix bases, or assume $F=E$ and $f$ is an endomorphism of $E$.)

My question is :

QUESTION

Assume that $E_{n}$ and $F_{n}$ are zero for any $n$ large enough.

Then, the left hand side of the above equation is well-defined.

If the numbers of non-zero terms appearing in the first pages of $E$ and $F$ are not necessarily finite, is there any way to calculate the left hand side by using $f_{a,b}^{1}$s?

If so, then how can it be done?

Maybe

this question can be essentially reduced to the following:

Let $C_{\ast}$ be a chain complex of $K$-vector spaces of the form $\cdots \to A_{2} \to A_{1} \to A_{0} \to 0$.

And let $f_{\ast} : C_{\ast} \to C_{\ast}$ be an endomorphism of the chain complex $C_{\ast}$.

Assume that the homology $H_{n}(C_{\ast})$ is zero for any $n>>0$.

Then, enoting by $H_{n}(f_{\ast})$ the homomorphism $H_{n}(C_{\ast}) \to H_{n}(C_{\ast})$ induced by $f$, the alternating product $\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}}$ is well-defined if each $\det (H_{n}(f_{\ast}))$ is non-zero.

Under this situation, can the alternating product $\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}}$ be expressed as "$\prod_{n}\det (f_{n})^{(-1)^{n}}$"?

The latter product is a priori an infinite product, so one will need to make some modifications.

Please give me any advice.

Vanishing of motivic cohomology with finite coefficients in negative degrees

2012-12-16T06:36:42Z

I wonder whether the "finite coefficient version" of Beilinson-Soule conjecture i.e. the following statement holds or not.

STATEMENT:

Let $X$ be a smooth and projective scheme over a finite field $\mathbb{F}_{p}$.

Then, Bloch's higher chow group $CH_{0}(X,i,\mathbb{Z}/n)$ vanishes for $n$ satisfying (n,p) =1 and $i>2dim(X)$.

Please give me any advice.

Finite / uniquely divisible abelian groups

2012-12-15T17:21:09Z

Is there any counter example for the following statement?

STATEMENT:

Let $0 \to F \to A \to Q \to 0$ be a short exact sequence of abelian groups. Assume that $F$ is a finite group, and $Q$ is a uniquely divisible abelian group. Then, this exact sequence splits.

Please give me any advice.

Special values of zeta functions and extensions of base fields.

2012-12-15T11:13:23Z

Let $X$ be a scheme of finite type over a finite field $k=\mathbb{F}_{q}$ of $q$ elements.

Then, one can define the zeta function $Z_{X/k}(T)$ of $X$ ovet $k$ as $\prod_{x\in |X|}\frac{1}{1-T^{deg_{k}(x)}}$, where $deg_{k}(x)=[k(x):k]$ is the degree of extension $k(x)/k$ for any closed point $x$ of $X$.

Assume that the structure morphism $X \to Spec(k)$ factors through $Spec(k^{\prime})$, the spectrum of a finite extension field $k^{\prime} \supset k$.

Then, one can define the zeta function $Z_{X/k^{\prime}}(T)$ of $X$ ovet $k^{\prime}$ as $\prod_{x\in |X|}\frac{1}{1-T^{deg_{k^{\prime}}(x)}}$, where $deg_{k^{\prime}}(x)=[k(x):k^{\prime}]$ is the degree of extension $k(x)/k^{\prime}$ for any closed point $x$ of $X$.

My question is: Does the following equation holds?:

(the special value of $Z_{X/k}$ at $T=1$) = $[k^{\prime}:k] \cdot ($the special value of $Z_{X/k^{\prime}}$ at $T=1$)

Please give me any advice.

What kind of spectral sequences come from double complexes?

2012-10-27T05:54:39Z

Given a double complex in the first quadrant, one can derive from it a (homological or cohomological) spectral sequence converging to the (co)homology of the total complex of the double complex.

My question is: When is a (homological or cohomological) spectral sequence coming from a double complex?

Is There a Mayer-Vietoris Spectral Sequence of Motivic Cohomology for Closed Coverings?

2012-10-04T13:59:10Z

For etale cohomology, there is a spectral sequence of the following form ("Mayer-Vietories spectral sequence for closed covers"):

$E_{1}^{p,q}=\oplus_{i_{0}< \cdots < i_{p}} H_{ Y_{i_{0} \cdots i_{p} }}^{q} (X, F) \Longrightarrow H_{Y}^{q-p}(X, F).$

Here, $X$ is a scheme, $Y\to X$ is a closed subscheme, and $Y_{i}$'s are a closed covering of $Y$.

$Y_{i_{0} \cdots i_{p}}$ denotes $Y_{i_{0}} \cap \cdots \cap Y_{i_{p}}.$

$F$ is an etale sheaf on $X$, and $H_{Z}^{\ast}$ denotes the cohomology with supports in a closed subscheme $Z$ on $X$.

My questions are: is there any spectral sequence of the similar form for motivic cohomology (or higher Chow group)?

If yes, then how one can prove it?

If no, then why is it?

Please give me any advice.

Does mapping cylinder category have enough injectives?

2012-06-18T15:24:26Z

Let $A, B$ be two abelian categories, and $\tau : A \to B$ a left exact functor.

We define a category $C$ as follows:

objects: triples $(M, N, \varphi)$ where $M\in A, N\in B$ and $\varphi: N\to \tau M$ is a morphism in $B$.

morphisms: pairs $(f, g): (M, N, \varphi) \to (M^{\prime}, N^{\prime}, \varphi^{\prime})$ where $f:M\to M^{\prime}$ and $g:N\to N^{\prime}$ satisfying $(\tau f) \circ \varphi = \varphi ^{\prime} \circ g$.

Then, is the following statement true?

If so, then how can one prove it?

STATEMENT: If $A, B$ have enough injectives, then so does $C$.

For example, let $X$ be a scheme, $Y$ be a closed subscheme of $X$, and $U=X\setminus Y$. If $A$=(etale sheaves on $X$), $B$=(etale sheaves on $U$), then the cagegory $C$ is equivalent to the category of etale sheaves on $Y$. So, $C$ has enough injectives , of course. I wonder whether this kind of situation happens in the general setting above.

Please give me any advice.

Are these connecting homomorphisms commutative?

2012-06-13T00:43:58Z

Are the connecting homomorphism induced by Kummer sequence and that of localization sequence commutative?

In other words, is the following statement true?

If it is true, then, how can one prove it?

Please give me any advice.

STATEMENT: Let $X$ be a scheme, $n$ an integer invertible on $X$, $Z$ a closed subscheme of $X$ and $U=X\setminus Z$.

Let $G_{m}$ be an etale sheaf on $X$ defined by sections invertible.

And let $\mu_{X}$ be an etale sheaf on $X$ defined by sections which are n-th roots of 1.

Then, the following two homomorphisms are the same.(Sorry, I don't know how to draw a diagram...)

$H^{i}(U, G_{m}) \stackrel{localization}{\to} H_{Z}^{i+1}(X, G_{m}) \stackrel{Kummer}{\to} H_{Z}^{i+2}(X, \mu_{X})$,

$H^{i}(U, G_{m}) \stackrel{Kummer}{\to} H^{i+1}(U, \mu_{X}) \stackrel{localization} \to H_{Z}^{i+2}(X, \mu_{X})$.

Here, morphisms are connectiong homomorphisms of long exact sequences.

Nonstandard Methods ( or Model Theory ) and Arithmetic Geometry

2012-01-14T09:02:16Z

I hear that the nonstandard methods are applied to many problems in various fields of mathematics such as functional analysis, topology, probability theory and so on.

So, I have become interested in using nonstandard methods to my research areas, which are in and around arithmetic geometry.

Questions:

What kind of useful applications of nonstandard methods to arithmetic geometry exist?
Is there any recommendation of introductory textbook or PDF file to study nonstandard methods in arithmetic geometry? (I heve studied the "nonstandard analysis" to a certain extent: construction of ultraproducts, the transfer principle etc. But I have few knowledge of nonstandard methods for algebra or algebraic geometry.)
Is there any relationship between the transfer principle and Hasse principle?

Please give me any advice.

Do Disjoint Unions and Fiber Products Commute?

2012-02-24T20:52:53Z

Do disjoint unions and fiber products commute?

In other words, is the following statement true?

Statement: Let $C$ be a category with (infinite) coproducts and fiber products. Let {$U_{i}$} be a family of objects in $C$, and denote the coproduct of them by $U = \coprod_{i}U_{i}$. Moreover, let $U_{i} \to X$ and $Y\to X$ be morphisms in $C$, and $U\to X$ be the morphism induced by the universality of coproduct. Then, $U\times_{X}Y \cong \coprod_{i}(U_{i}\times_{X}Y)$.

If $C$ is the category of schemes, this statement will be true. This is because fiber products of schemes are constructed locally at first, and glued together.

However, I could not prove this by using universality (i.e. in categorical settings).

My questions are:

Is the above statement true? If so, then how can one prove it?
If the statement is false, what kind of counter example exists?
If the statement is false, then, please change the statement replacing "coproducts" by "disjoint unions". Is the NEW statement true?

Here, disjoint union of $U_{i}$'s means coproduct $U=\coprod_{i}U_{i}$ satisfying that the fiber products $U_{i}\times_{U}U_{j}$ are the strict initial object if $i \neq j$. Here, strict initial object means initial object $\phi$ such that for any object $X$, the set of morphisms $Hom(X, \phi)$ is the empty set if $X$ is not isomorphic to $\phi$. (This is the generalization of empty set in the category of sets or schemes.)

Later

Counterexamples for the first statement exist (e.g. the category of pointed sets or the opposite of the category of sets).

However, those are not for the refined statement in my question 3. Does anybody have ideas for it?

Is Sheafification Functor Exact?

2012-02-26T10:06:27Z

I know that sheafification functor from the category of abelian presheaves on $C$ to the category of abelian sheaves on $C$. Here, $C$ is a category with Grothendieck pretopology.

My question is:

How about the sheafification functor from the category of presheaves of "sets" on $C$ to the category of sheaves of "sets" on $C$?

Is this an exact functor? (i.e. preserving finite limits and finite colimits?)

If so, how can one prove it?

In fact, I want to know whether sheafification functor preserves cartesian products or not.

Please give me any advice.

Do Categorical Quotients Preserve Covering Maps?

2012-02-25T19:11:20Z

Before asking a question, please let me write down settings.

SETTINGS:

Let $C$ be a category with fiber products and $B$ be a closed subcategory of $C$ (i.e. $B$ contains any isomorphism of $C$, and any base change of any morphism in $B$ is also a morphism in $B$).

Let $T$ be the topology on $C$ associated to $B$ (i.e. $Cov T$ consists of universal effective epimorphic families {$f_{i}:U_{i}\to U$} in $C$ such that each $f_{i}$ is a morphism in $B$).

Moreover, assume that the topology $T$ satisfies the following two conditions:

Condition I

Let $f,g,h$ be arbitrary morphisms in $C$, and assume that $h=gf$ and $h$ is in $B$.

If $g \in B$, then $f \in B$.
If ${f} \in CovT$, then $g \in B$.

Condition II

For any family of maps {$f_{i}:U_{i}\to X$} in $C$ for which there exists "disjoint union" $\coprod_{i} U_{i}$, then the induced map $\coprod_{i}U_{i}\to X$ is in $B$ if and only if $f_{i} \in B$ for all $i$.

Under the situation above,

Let $R$ be a categorical equivalence relation on an object $U \in C$ such that the two canonical projections $\pi_{i}: R\to U$ ($i=1,2$) are both covering maps of $U$ in the topology $T$.

And assume that this categorical equivalence has $T$-quotient $X$. This means that there exists a categorical quotient $p:U\to X$ of $\pi_{i}:R\to U$ such that the induced morphism of associated sheaves (on $T$) $p_{\ast}:h_{U} \to h_{X}$ is a categorical quotient of $\pi_{i \ast}:h_{R}\to h_{U}$ in the category of sheaves of sets on $T$.

(Then, one can prove that $R \cong U\times_{X}U$.)

Now, this is my QUESTION:

Is the map $p:U\to X$ a covering map in $T$ ?

In other words, is $p$ a universal effective epimorphism and satisfying $p\in B$ ?

I could prove a "converse" statement i.e. if $p:U\to X$ is a covering map of the topology $T$, then $\pi_{i}: U\times_{X}U \to U$ ($i=1,2$) is a categorical equivalence relation such that each $\pi_{i} \in CovT$ and has $p$ as a $T$-quotient.

But I could not prove the statement in my question.

Please give me any advice.

Two Definitions of "Character" of topological groups

2012-01-19T10:46:58Z

When I first met the concept of "characters" of topological groups in Pontryagin's book "Topological groups", it was defined as follows:

Let $G$ be a topological group. A character of $G$ is a homomorphism of topological groups from $G$ to the torus $T=\mathbb{R}/\mathbb{Z}$. Here, the torus $T$ has the induced topology from the usual topology of the real line $\mathbb{R}$.

I found the same definition on Wikipedia. So, I think this is a standard definition of character.

But, in a lot of modern articles, it seems to me that characters are defined as follows:

Let $G$ be a topological group. A character of $G$ is a homomorphism of topological groups from $G$ to the discrete additive group $\mathbb{Q}/\mathbb{Z}$.

Clearly, these two definitions cannot be the same for all topological groups.

However, if $G$ is a (discrete) finite group, then the two definitions agree.

Question:

What kind of conditions on a topological group $G$ one needs in order to identify the two definitions of character?
If $G$ is a "profinite group", then, do the two definitions agree? If the answer is yes, then how can one prove it?

Please give me any advice.

Later

I found a way to answer the second question. One only needs to show the following: for any profinite group $G$ and any continuous homomorphism $f:G \to T$, the image of $f$ is finite. This statement can be shown as follows. The torus $T$ has an open neighborhood $U$ of $0$ which contains no non-tirivial subgroup of $T$. Since $G$ is profinite, there exists an open normal subgroup $H$ of $G$ satisfying $H \subset f^{-1}(U)$. This implies $H \subset Ker(f)$. So, the map $f$ factors through the finite group $G/H$. This concludes that the image of $f$ is finite.

Hence the two definitions of character agree for any profinite group.

What does Gal(Q_p/Q) mean?

2011-11-13T11:39:55Z

What does $\mathrm{Gal}(\overline{\mathbb{Q}}_{p}/\mathbb{Q})$ mean? ($p$ is a prime number.)

If it is defined as $\mathrm{Aut}(\overline{\mathbb{Q}}_{p}/\mathbb{Q})$, then does it have any property like usual galois groups?

For example, is the following statement true? :

Let $\alpha \in \overline{\mathbb{Q}}_{p}$, and assume that for any $\sigma \in \mathrm{Aut} (\overline{ \mathbb{Q} } _{p} / \mathbb{Q})$, $\sigma (\alpha) = \alpha$. Then, $\alpha \in \mathbb{Q}$.

If this is true, then how can I prove it?

Please give me any advice.

Thanks!

What kind of conditions we need to make morphisms of schemes quasi-projective?

2011-09-22T00:46:40Z

What kind of conditions we need to make morphisms of schemes quasi-projective?

I am really interested in the following case:

If $f : X \to Y$ is an etale, of finite type and separated morphism of schemes, then is it quasi-projective?

If so, which conditions we use?

If necessary, please assume that the scheme $Y$ is locally noetherian.

Does n-multiplication maps of cohomology groups vanish if it vanishes at the 0th cohomology?

2011-08-07T14:33:45Z

In general, we know that a morphism $f=(f ^ {q})$ between universal (cohomological) $\delta$ functors $S=(S ^ {q}),T=(T ^ {q})\ $vanishes if and only if $f ^ {0} \ \colon \ S^{0} \to T^{0}$ vanishes.

However, for a fixed object $F$, we cannot say that $f^{q} (F) = 0$ for every $q$ even if $f^{0} (F) = 0$.

Now, in order to make this statement true, what kind of condition we need for the morphism $f \ $?

In particular, I am interested in the following situation:

Let $n$ be an integer, $X$ be a scheme, $F\ \colon \ X _ {et} \to Ab \ $ be a $n$-torsion etale sheaf on $X$, and $S=T=H^{\ast}(X,-)$, then can we say that the morphism $H^{q}(X,F)\to H^{q}(X,F)$ induced by the $n$-multiplication map vanishes? Thanks!

Can the zero-degree part of Mf ⊗Sf Nf be identified with M(f) ⊗S(f) N(f) ?

2011-08-05T10:16:03Z

The isomorphism ${(M \otimes _ {S} N)} _ {f} = M _ {f} \otimes _ {S _ {f}} N _ {f}$ is well-known. Here, $S$ is a graded ring, and $M,N$ are graded $S$ modules.

Now, let $f$ be any homogeneous element of $S$ of degree 1. I want to have ${(M \otimes _ {S} N)} _ {(f)} = M _ {(f)} \otimes _ {S _ {(f)}} N _ {(f)}$. Here, $L _ {(f)}$ means the zero-degree part of $L _ {f}$ for any graded $S$-module $L$.

By using the first formula, the problem reduces to the following assertion: Can the zero-degree part of $M _ {f} \otimes _ {S _ {f}} N _ {f}$ be identified with $M _ {(f)} \otimes _ {S _ {(f)}} N _ {(f)}$ ?

I see that there is a canonical homomorphism $M _ {(f)} \otimes _ {S _ {(f)}} N _ {(f)} \to M _ {f} \otimes _ {S _ {f}} N _ {f}$, and I also see that the image of this homomorphism equals to the zero-degree part of $M _ {f} \otimes _ {S _ {f}} N _ {f}$.

However, I cannot prove that this homomorphism is injective.

Pullback of sheaves associated to graded modules

2011-08-05T07:33:00Z

Please let me ask some questions about a proposition in R.Hartshorne's textbook "Algebraic Geometry."

In the proof of Proposition 5.12(c), Chapter II, Hartshorne says,

"For any graded $S$-module $M$, $f^*(\tilde M )=\widetilde{(M\otimes_S T)} |_U$."

Here, $S$ (resp $T$) is a graded ring generated by $S_{1}$ (resp $T_{1}$) as $S_{0}$ (resp $T_{0}$)-algebra. $T$ is regarded as $S$-algebra through a graded homomorphism $\phi \colon S \to T$, and $f \colon U\to Proj S$ is the morphism of schemes induced by the homomorphism $\phi$. Here, $U \subset Proj T$ is the open subset consisting of homogeneous prime ideals of T not containing $\phi (S_{+})$.

Now, I have two questions:

How can this assertion be proved?
Is it true that ${(M \otimes_S N)}_{(f)}$ $=$ $M _ {(f)} \otimes _ {S _ {(f)}} N _ {(f)}$, where $M$ and $N$ are graded $S$ -modules, and $f$ a homogeneous element of $S$ of degree 1? If it is, then how can it be proved? I think this kind of facts may be used to prove the assertion.

I have tried to prove these, but I couldn't. I have little knowledge about localizing graded modules, or tensor products of those.

Comment by Hiro

2013-03-10T18:56:45Z

Thanks for your comment. I do not know anything about homotopy limits, but how about if the spaces are CW complexes?

Comment by Hiro

2013-03-10T18:53:51Z

Thanks for your comment. I wonder how one can prove that $\Omega (F_{t})$ is actually a homotopy. Is it always continuous without any assumption such as locally compactness?

Comment by Hiro

2012-12-30T20:55:09Z

Thank you for your kind! I did not think one can make any difference!

Comment by Hiro

2012-12-30T20:10:17Z

Tanks for your answer! How about if one admits the difference of sign?

Comment by Hiro

2012-12-30T15:34:23Z

I mean, $E_{n}$ is a vector space equipped with filtration whose $i$th congruent is isomorphic to $E^{\infty}_{i, n-i}$. But if the notations are not clear, please think of the situation I added in the question below the "Maybe".

Comment by Hiro

2012-12-30T12:15:27Z

Tanks, but I could not get the point... Will you tell me the important part?

Comment by Hiro

2012-12-29T21:18:38Z

Tanks for your comment. Yes, I should heve fixed the bases. Or, if necessary, let $F=E$ and $f$ be a endomorphism of $E$. Then det is well-defined. What do you exactly mean by "sign problem" ? Is there any reference to learn the method?

Comment by Hiro

2012-12-29T18:32:06Z

Thanks for your question. $E_{n}$ denotes the $n$th term to which the spectral sequence $E$ converges i.e. $E_{a,b}^{1} \Longrightarrow E_{n}$.

Comment by Hiro

2012-12-16T08:02:02Z

Sorry, I meant n is an integer prime to p.

Comment by Hiro

2012-12-15T11:36:22Z

Thanks, do you know anything about special values?

Comment by Hiro

2012-11-02T16:59:59Z

Thanks a lot! The couterexamples are interesting.

Comment by Hiro

2012-09-19T14:48:19Z

I do not understand how to induce the last equation... Can anyone explain it?

Comment by Hiro

2012-06-13T15:57:43Z

Thank you so much!

Comment by Hiro

2012-02-25T20:04:01Z

@Qiaochu Yuan: I understood clearly, thanks!

Comment by Hiro

2012-02-25T19:54:20Z

Why is it that H=Z if ab=1?