User hiro - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:17:10Z http://mathoverflow.net/feeds/user/16968 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/124166/are-loop-spaces-of-homotopically-equivalent-spaces-homotopically-equivalent Are loop spaces of homotopically equivalent spaces homotopically equivalent? Hiro 2013-03-10T17:39:15Z 2013-03-10T18:43:56Z <p>Let $f:X \to Y$ be a homotopy equivalence of pointed topological spaces.</p> <p>Then, is the induced map of pointed loop spaces $\Omega (f): \Omega X \to \Omega Y$ a homotopy equivalence?</p> <p>Here, loop spaces are equipped with the compact-open topologies.</p> <p>Is there any counterexample?</p> <p>I do not know even whether the induced map $\Omega (f)$ is continuous or not in general.</p> http://mathoverflow.net/questions/118827/examples-of-sheafification-via-hypercovers Examples of Sheafification via Hypercovers Hiro 2013-01-13T17:09:16Z 2013-01-13T17:31:40Z <p>For a presheaf $F$ on a category equipped with a pretopology, one has the sheafification $F^{\sharp}$ of $F$.</p> <p>I know well the plus-construction of sheafification, which is presented in Artin's paper "Grothendieck Topologies", for example.</p> <p>QUESTION</p> <p>I have heard that there is another construction of sheafification by using hypercovering, but I can not find good explanation of it.</p> <p>So, I want some examples to feel the essence of the construction.</p> <p>In particular, I am interested in the following situation:</p> <p>Let $C$ be a category with a pretopology $T$.</p> <p>For an object $X \in C$, denote by $F_{X}$ the presheaf on $C$ represented by $X$ i.e. $F_{X}=Hom_{C}(-,X)$.</p> <p>Now, let $X_{\bullet} \to X$ be a $T$-hypercover of $X \in C$.</p> <p>Then, how can the sheafification of $F_{X}$ be written by using $F_{X_{\bullet}}$ ?</p> http://mathoverflow.net/questions/117658/endomomorphisms-of-chain-complexes-of-vector-spaces-and-determinants Endomomorphisms of Chain Complexes of vector spaces and determinants Hiro 2012-12-30T17:51:38Z 2012-12-30T20:13:57Z <p>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 $K$.</p> <p>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 $n$.</p> <p>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 $g_{\ast}$).</p> <p>Moreover, assume that $\det (f_{n}) = \det (g_{n})$ for any $n$.</p> <p>My question is:</p> <hr> <p>QUESTION</p> <p>Under the above conditions, does the next equation hold up to sign?</p> <p>$\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}} = \prod_{n} \det (H_{n}(g_{\ast}))^{(-1)^{n}}$.</p> <hr> <p>Note that if the chain complex $C_{\ast}$ is bounded above, the statement can be proved as follows: </p> <p>$\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}} = \prod_{n} \det (f_{n})^{(-1)^{n}} = \prod_{n} \det (g_{n})^{(-1)^{n}} = \prod_{n} \det (H_{n}(g_{\ast}))^{(-1)^{n}}$</p> <p>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.</p> <p>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.</p> <h1>By Sawin's answer</h1> <p>There is a counterexample if one does not admit the difference of sign.</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/117472/morphisms-of-spectral-sequences-and-alternating-products Morphisms of Spectral Sequences and alternating products Hiro 2012-12-29T03:34:40Z 2012-12-30T16:16:27Z <p>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.</p> <p>Assume that morphisms $f_{a,b}^{1} : E_{a,b}^{1} \to F_{a,b}^{1}$ of the first page are all isomoprhisms.</p> <p>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:</p> <p>$\prod_{n}(\det (f_{n}))^{(-1)^{n}} = \prod_{a,b}(\det (f_{a,b}^{1}))^{(-1)^{a+b}}$.</p> <p>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 $E^{\infty}_{i,n-i}$).</p> <p>(In order to define $\det$, one has to fix bases, or assume $F=E$ and $f$ is an endomorphism of $E$.)</p> <p>My question is :</p> <p>QUESTION</p> <p>Assume that $E_{n}$ and $F_{n}$ are zero for any $n$ large enough.</p> <p>Then, the left hand side of the above equation is well-defined.</p> <p>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?</p> <p>If so, then how can it be done?</p> <h1>Maybe</h1> <p>this question can be essentially reduced to the following:</p> <p>Let $C_{\ast}$ be a chain complex of $K$-vector spaces of the form $\cdots \to A_{2} \to A_{1} \to A_{0} \to 0$.</p> <p>And let $f_{\ast} : C_{\ast} \to C_{\ast}$ be an endomorphism of the chain complex $C_{\ast}$.</p> <p>Assume that the homology $H_{n}(C_{\ast})$ is zero for any $n>>0$.</p> <p>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.</p> <p>Under this situation, can the alternating product $\prod_{n} \det (H_{n}(f_{\ast}))^{(-1)^{n}}$ be expressed as "$\prod_{n}\det (f_{n})^{(-1)^{n}}$"?</p> <p>The latter product is a priori an infinite product, so one will need to make some modifications.</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/116513/vanishing-of-motivic-cohomology-with-finite-coefficients-in-negative-degrees Vanishing of motivic cohomology with finite coefficients in negative degrees Hiro 2012-12-16T06:36:42Z 2012-12-16T14:19:06Z <p>I wonder whether the "finite coefficient version" of Beilinson-Soule conjecture i.e. the following statement holds or not.</p> <p>STATEMENT:</p> <p>Let $X$ be a smooth and projective scheme over a finite field $\mathbb{F}_{p}$.</p> <p>Then, Bloch's higher chow group $CH_{0}(X,i,\mathbb{Z}/n)$ vanishes for $n$ satisfying (n,p) =1 and $i>2dim(X)$.</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/116467/finite-uniquely-divisible-abelian-groups Finite / uniquely divisible abelian groups Hiro 2012-12-15T17:21:09Z 2012-12-15T23:04:05Z <p>Is there any counter example for the following statement?</p> <p>STATEMENT:</p> <p>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.</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/116440/special-values-of-zeta-functions-and-extensions-of-base-fields Special values of zeta functions and extensions of base fields. Hiro 2012-12-15T11:13:23Z 2012-12-15T11:59:41Z <p>Let $X$ be a scheme of finite type over a finite field $k=\mathbb{F}_{q}$ of $q$ elements.</p> <p>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 $X$.</p> <p>Assume that the structure morphism $X \to Spec(k)$ factors through $Spec(k^{\prime})$, the spectrum of a finite extension field $k^{\prime} \supset k$.</p> <p>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 $X$.</p> <p>My question is: Does the following equation holds?:</p> <p>(the special value of $Z_{X/k}$ at $T=1$) = $[k^{\prime}:k] \cdot ($the special value of $Z_{X/k^{\prime}}$ at $T=1$)</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/110812/what-kind-of-spectral-sequences-come-from-double-complexes What kind of spectral sequences come from double complexes? Hiro 2012-10-27T05:54:39Z 2012-10-29T18:27:44Z <p>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.</p> <p>My question is: When is a (homological or cohomological) spectral sequence coming from a double complex?</p> http://mathoverflow.net/questions/108819/is-there-a-mayer-vietoris-spectral-sequence-of-motivic-cohomology-for-closed-cove Is There a Mayer-Vietoris Spectral Sequence of Motivic Cohomology for Closed Coverings? Hiro 2012-10-04T13:59:10Z 2012-10-04T19:12:06Z <p>For etale cohomology, there is a spectral sequence of the following form ("Mayer-Vietories spectral sequence for closed covers"):</p> <p>$E_{1}^{p,q}=\oplus_{i_{0}&lt; \cdots &lt; i_{p}} H_{ Y_{i_{0} \cdots i_{p} }}^{q} (X, F) \Longrightarrow H_{Y}^{q-p}(X, F).$</p> <p>Here, $X$ is a scheme, $Y\to X$ is a closed subscheme, and $Y_{i}$'s are a closed covering of $Y$.</p> <p>$Y_{i_{0} \cdots i_{p}}$ denotes $Y_{i_{0}} \cap \cdots \cap Y_{i_{p}}.$</p> <p>$F$ is an etale sheaf on $X$, and $H_{Z}^{\ast}$ denotes the cohomology with supports in a closed subscheme $Z$ on $X$.</p> <p>My questions are: is there any spectral sequence of the similar form for motivic cohomology (or higher Chow group)?</p> <p>If yes, then how one can prove it?</p> <p>If no, then why is it?</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/99908/does-mapping-cylinder-category-have-enough-injectives Does mapping cylinder category have enough injectives? Hiro 2012-06-18T15:24:26Z 2012-06-18T15:30:12Z <p>Let $A, B$ be two abelian categories, and $\tau : A \to B$ a left exact functor.</p> <p>We define a category $C$ as follows:</p> <p>objects: triples $(M, N, \varphi)$ where $M\in A, N\in B$ and $\varphi: N\to \tau M$ is a morphism in $B$.</p> <p>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 $(\tau f) \circ \varphi = \varphi ^{\prime} \circ g$.</p> <p>Then, is the following statement true?</p> <p>If so, then how can one prove it?</p> <p>STATEMENT: If $A, B$ have enough injectives, then so does $C$.</p> <p>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.</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/99394/are-these-connecting-homomorphisms-commutative Are these connecting homomorphisms commutative? Hiro 2012-06-13T00:43:58Z 2012-06-13T01:58:34Z <p>Are the connecting homomorphism induced by Kummer sequence and that of localization sequence commutative?</p> <p>In other words, is the following statement true?</p> <p>If it is true, then, how can one prove it?</p> <p>Please give me any advice.</p> <p>STATEMENT: Let $X$ be a scheme, $n$ an integer invertible on $X$, $Z$ a closed subscheme of $X$ and $U=X\setminus Z$.</p> <p>Let $G_{m}$ be an etale sheaf on $X$ defined by sections invertible.</p> <p>And let $\mu_{X}$ be an etale sheaf on $X$ defined by sections which are n-th roots of 1.</p> <p>Then, the following two homomorphisms are the same.(Sorry, I don't know how to draw a diagram...)</p> <p>$H^{i}(U, G_{m}) \stackrel{localization}{\to} H_{Z}^{i+1}(X, G_{m}) \stackrel{Kummer}{\to} H_{Z}^{i+2}(X, \mu_{X})$, </p> <p>$H^{i}(U, G_{m}) \stackrel{Kummer}{\to} H^{i+1}(U, \mu_{X}) \stackrel{localization} \to H_{Z}^{i+2}(X, \mu_{X})$.</p> <p>Here, morphisms are connectiong homomorphisms of long exact sequences.</p> http://mathoverflow.net/questions/85643/nonstandard-methods-or-model-theory-and-arithmetic-geometry Nonstandard Methods ( or Model Theory ) and Arithmetic Geometry Hiro 2012-01-14T09:02:16Z 2012-04-16T18:19:18Z <p>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.</p> <p>So, I have become interested in using nonstandard methods to my research areas, which are in and around arithmetic geometry.</p> <p>Questions:</p> <ol> <li><p>What kind of useful applications of nonstandard methods to arithmetic geometry exist?</p></li> <li><p>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.)</p></li> <li><p>Is there any relationship between the transfer principle and Hasse principle?</p></li> </ol> <p>Please give me any advice.</p> http://mathoverflow.net/questions/89444/do-disjoint-unions-and-fiber-products-commute Do Disjoint Unions and Fiber Products Commute? Hiro 2012-02-24T20:52:53Z 2012-02-26T21:32:28Z <p>Do disjoint unions and fiber products commute?</p> <p>In other words, is the following statement true?</p> <p>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, $U\times_{X}Y \cong \coprod_{i}(U_{i}\times_{X}Y)$.</p> <p>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.</p> <p>However, I could not prove this by using universality (i.e. in categorical settings).</p> <p>My questions are:</p> <ol> <li><p>Is the above statement true? If so, then how can one prove it?</p></li> <li><p>If the statement is false, what kind of counter example exists?</p></li> <li><p>If the statement is false, then, please change the statement replacing "coproducts" by "disjoint unions". Is the NEW statement true?</p></li> </ol> <p>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.)</p> <h1>Later</h1> <p>Counterexamples for the first statement exist (e.g. the category of pointed sets or the opposite of the category of sets).</p> <p>However, those are not for the refined statement in my question 3. Does anybody have ideas for it?</p> http://mathoverflow.net/questions/89568/is-sheafification-functor-exact Is Sheafification Functor Exact? Hiro 2012-02-26T10:06:27Z 2012-02-26T20:41:37Z <p>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.</p> <p>My question is:</p> <p>How about the sheafification functor from the category of presheaves of "sets" on $C$ to the category of sheaves of "sets" on $C$?</p> <p>Is this an exact functor? (i.e. preserving finite limits and finite colimits?)</p> <p>If so, how can one prove it?</p> <p>In fact, I want to know whether sheafification functor preserves cartesian products or not.</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/89508/do-categorical-quotients-preserve-covering-maps Do Categorical Quotients Preserve Covering Maps? Hiro 2012-02-25T19:11:20Z 2012-02-26T04:04:31Z <p>Before asking a question, please let me write down settings.</p> <p>SETTINGS:</p> <p>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 $B$).</p> <p>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 $B$).</p> <p>Moreover, assume that the topology $T$ satisfies the following two conditions:</p> <p>Condition I</p> <p>Let $f,g,h$ be arbitrary morphisms in $C$, and assume that $h=gf$ and $h$ is in $B$. </p> <ol> <li><p>If $g \in B$, then $f \in B$.</p></li> <li><p>If ${f} \in CovT$, then $g \in B$.</p></li> </ol> <p>Condition II</p> <p>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 $i$.</p> <p>Under the situation above, </p> <p>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 $T$.</p> <p>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 $T$.</p> <p>(Then, one can prove that $R \cong U\times_{X}U$.)</p> <p>Now, this is my QUESTION:</p> <p>Is the map $p:U\to X$ a covering map in $T$ ?</p> <p>In other words, is $p$ a universal effective epimorphism and satisfying $p\in B$ ?</p> <p>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.</p> <p>But I could not prove the statement in my question.</p> <p>Please give me any advice.</p> http://mathoverflow.net/questions/86089/two-definitions-of-character-of-topological-groups Two Definitions of "Character" of topological groups Hiro 2012-01-19T10:46:58Z 2012-01-19T23:52:00Z <p>When I first met the concept of "characters" of topological groups in Pontryagin's book "Topological groups", it was defined as follows:</p> <p>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 $\mathbb{R}$.</p> <p>I found the same definition on Wikipedia. So, I think this is a standard definition of character.</p> <p>But, in a lot of modern articles, it seems to me that characters are defined as follows:</p> <p>Let $G$ be a topological group. A character of $G$ is a homomorphism of topological groups from $G$ to the discrete additive group $\mathbb{Q}/\mathbb{Z}$.</p> <p>Clearly, these two definitions cannot be the same for all topological groups.</p> <p>However, if $G$ is a (discrete) finite group, then the two definitions agree.</p> <p>Question:</p> <ol> <li><p>What kind of conditions on a topological group $G$ one needs in order to identify the two definitions of character?</p></li> <li><p>If $G$ is a "profinite group", then, do the two definitions agree? If the answer is yes, then how can one prove it?</p></li> </ol> <p>Please give me any advice.</p> <h1>Later</h1> <p>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.</p> <p>Hence the two definitions of character agree for any profinite group.</p> http://mathoverflow.net/questions/80811/what-does-galq-p-q-mean What does Gal(Q_p/Q) mean? Hiro 2011-11-13T11:39:55Z 2011-11-13T13:08:11Z <p>What does $\mathrm{Gal}(\overline{\mathbb{Q}}_{p}/\mathbb{Q})$ mean? ($p$ is a prime number.)</p> <p>If it is defined as $\mathrm{Aut}(\overline{\mathbb{Q}}_{p}/\mathbb{Q})$, then does it have any property like usual galois groups?</p> <p>For example, is the following statement true? :</p> <p>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, $\alpha \in \mathbb{Q}$.</p> <p>If this is true, then how can I prove it?</p> <p>Please give me any advice.</p> <h1>Thanks!</h1> http://mathoverflow.net/questions/76097/what-kind-of-conditions-we-need-to-make-morphisms-of-schemes-quasi-projective What kind of conditions we need to make morphisms of schemes quasi-projective? Hiro 2011-09-22T00:46:40Z 2011-09-22T01:37:28Z <p>What kind of conditions we need to make morphisms of schemes quasi-projective?</p> <p>I am really interested in the following case:</p> <p>If $f : X \to Y$ is an etale, of finite type and separated morphism of schemes, then is it quasi-projective?</p> <p>If so, which conditions we use?</p> <p>If necessary, please assume that the scheme $Y$ is locally noetherian.</p> http://mathoverflow.net/questions/72287/does-n-multiplication-maps-of-cohomology-groups-vanish-if-it-vanishes-at-the-0th Does n-multiplication maps of cohomology groups vanish if it vanishes at the 0th cohomology? Hiro 2011-08-07T14:33:45Z 2011-08-07T15:08:12Z <p>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.</p> <p>However, for a fixed object $F$, we cannot say that $f^{q} (F) = 0$ for every $q$ even if $f^{0} (F) = 0$.</p> <p>Now, in order to make this statement true, what kind of condition we need for the morphism $f \$? </p> <p>In particular, I am interested in the following situation:</p> <p>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!</p> http://mathoverflow.net/questions/72164/can-the-zero-degree-part-of-mf-sf-nf-be-identified-with-mf-sf-nf Can the zero-degree part of Mf ⊗Sf Nf be identified with M(f) ⊗S(f) N(f) ? Hiro 2011-08-05T10:16:03Z 2011-08-05T10:24:00Z <p>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.</p> <p>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 $L$.</p> <p>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 $M _ {(f)} \otimes _ {S _ {(f)}} N _ {(f)}$ ?</p> <p>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 $M _ {f} \otimes _ {S _ {f}} N _ {f}$.</p> <p>However, I cannot prove that this homomorphism is injective.</p> http://mathoverflow.net/questions/72155/pullback-of-sheaves-associated-to-graded-modules Pullback of sheaves associated to graded modules Hiro 2011-08-05T07:33:00Z 2011-08-05T07:33:00Z <p>Please let me ask some questions about a proposition in R.Hartshorne's textbook "Algebraic Geometry."</p> <p>In the proof of Proposition 5.12(c), Chapter II, Hartshorne says,</p> <p>"For any graded $S$-module $M$, $f^*(\tilde M )=\widetilde{(M\otimes_S T)} |_U$."</p> <p>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 $\phi (S_{+})$.</p> <p>Now, I have two questions:</p> <ol> <li><p>How can this assertion be proved?</p></li> <li><p>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.</p></li> </ol> <p>I have tried to prove these, but I couldn't. I have little knowledge about localizing graded modules, or tensor products of those.</p> http://mathoverflow.net/questions/124166/are-loop-spaces-of-homotopically-equivalent-spaces-homotopically-equivalent/124167#124167 Comment by Hiro Hiro 2013-03-10T18:56:45Z 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? http://mathoverflow.net/questions/124166/are-loop-spaces-of-homotopically-equivalent-spaces-homotopically-equivalent/124171#124171 Comment by Hiro Hiro 2013-03-10T18:53:51Z 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? http://mathoverflow.net/questions/117658/endomomorphisms-of-chain-complexes-of-vector-spaces-and-determinants/117663#117663 Comment by Hiro Hiro 2012-12-30T20:55:09Z 2012-12-30T20:55:09Z Thank you for your kind! I did not think one can make any difference! http://mathoverflow.net/questions/117658/endomomorphisms-of-chain-complexes-of-vector-spaces-and-determinants/117663#117663 Comment by Hiro Hiro 2012-12-30T20:10:17Z 2012-12-30T20:10:17Z Tanks for your answer! How about if one admits the difference of sign? http://mathoverflow.net/questions/117472/morphisms-of-spectral-sequences-and-alternating-products Comment by Hiro Hiro 2012-12-30T15:34:23Z 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 &quot;Maybe&quot;. http://mathoverflow.net/questions/117472/morphisms-of-spectral-sequences-and-alternating-products Comment by Hiro Hiro 2012-12-30T12:15:27Z 2012-12-30T12:15:27Z Tanks, but I could not get the point... Will you tell me the important part? http://mathoverflow.net/questions/117472/morphisms-of-spectral-sequences-and-alternating-products Comment by Hiro Hiro 2012-12-29T21:18:38Z 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 &quot;sign problem&quot; ? Is there any reference to learn the method? http://mathoverflow.net/questions/117472/morphisms-of-spectral-sequences-and-alternating-products Comment by Hiro Hiro 2012-12-29T18:32:06Z 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. $E_{a,b}^{1} \Longrightarrow E_{n}$. http://mathoverflow.net/questions/116513/vanishing-of-motivic-cohomology-with-finite-coefficients-in-negative-degrees Comment by Hiro Hiro 2012-12-16T08:02:02Z 2012-12-16T08:02:02Z Sorry, I meant n is an integer prime to p. http://mathoverflow.net/questions/116440/special-values-of-zeta-functions-and-extensions-of-base-fields Comment by Hiro Hiro 2012-12-15T11:36:22Z 2012-12-15T11:36:22Z Thanks, do you know anything about special values? http://mathoverflow.net/questions/110812/what-kind-of-spectral-sequences-come-from-double-complexes/110820#110820 Comment by Hiro Hiro 2012-11-02T16:59:59Z 2012-11-02T16:59:59Z Thanks a lot! The couterexamples are interesting. http://mathoverflow.net/questions/107446/combinatorial-spectral-sequence-in-arithmetic-geometry/107456#107456 Comment by Hiro Hiro 2012-09-19T14:48:19Z 2012-09-19T14:48:19Z I do not understand how to induce the last equation... Can anyone explain it? http://mathoverflow.net/questions/99394/are-these-connecting-homomorphisms-commutative/99396#99396 Comment by Hiro Hiro 2012-06-13T15:57:43Z 2012-06-13T15:57:43Z Thank you so much! http://mathoverflow.net/questions/89503/when-is-homg-h-cyclic/89506#89506 Comment by Hiro Hiro 2012-02-25T20:04:01Z 2012-02-25T20:04:01Z @Qiaochu Yuan: I understood clearly, thanks! http://mathoverflow.net/questions/89503/when-is-homg-h-cyclic/89506#89506 Comment by Hiro Hiro 2012-02-25T19:54:20Z 2012-02-25T19:54:20Z Why is it that H=Z if ab=1?