User simone virili - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T04:35:51Z http://mathoverflow.net/feeds/user/24891 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120120/what-is-the-dual-of-a-pre-injective-map What is the dual of a pre-injective map? Simone Virili 2013-01-28T15:51:27Z 2013-06-08T09:22:00Z <p>In [M. Gromov, Endomorphisms of symbolic algebraic varieties, J. Eur. Math. Soc. (JEMS) 1 (1999), 109–197], Gromov introduces the notion of pre-injective map. Recasting this notion in the setting of Abelian groups we get the following:</p> <p>Definition. Given a finite discrete Abelian group $G$, an (infinite) index set $X$ and taking the product $G^X$ endowed with the product topology, we say that a continuous endomorphism $\phi:G^X\to G^X$ is <em>pre-injective</em> if the restriction of $\phi$ to $G^{(X)}$ (the subgroup of elements with finite support) is injective. </p> <p>The Pontryagin-Van Kampen dual of $\phi$ is just an endomorphism $\widehat\phi$ of the discrete group $G^{(X)}$. Can we say that $\phi$ is pre-injective just looking at $\widehat\phi$? In other words, what is the algebraic property of $\widehat\phi$ which corresponds to the pre-injectivity of $\phi$? </p> http://mathoverflow.net/questions/132398/surjectivity-of-the-natural-map-of-injective-module-to-its-localization/132399#132399 Answer by Simone Virili for Surjectivity of the natural map of injective module to its localization Simone Virili 2013-05-31T10:05:42Z 2013-05-31T13:05:52Z <p>I've not Hartshorne's book with me now so I cannot check the exact context of your question, anyway this is a very general fact about localization. Take a Grothendieck category $\frak C$ and a hereditary torsion subclass $\frak T$, then you have a localization of categories $Q:\frak C\to \frak C/\frak T$ (in the sense of Gabriel). There is always a fully faithful functor $S:\frak C/\frak T\to\frak C$ such that $(Q,S)$ is an adjoint pair. By the closure properties of $\frak T$, the quotient functor $Q$ is automatically exact. </p> <p>Now, identifying $\frak C/\frak T$ with a subcategory of $\frak C$ via the section functor $S$, you have a very explicit construction of the localized objects using the torsion-theoretic machinery. Indeed, denote by $T:\frak C\to \frak T$ the torsion functor (for an object $C\in \frak C$, $T(C)$ is the direct union of all the sub-objects belonging to $\frak T$). </p> <p>Given $X\in \frak C$, let $X'=X/T(X)$, then $Q(X)$ is isomorphic to $\pi^{-1}(T(E(X')/X'))$, where $\pi:E(X')\to E(X')/X'$ is the natural projection.</p> <p>Now, in some cases you can prove that the class $\frak T$ is closed under taking injective envelopes. This is the case for example the case of localization at prime ideals in commutative Noetherian rings. This is not always the case in non-commutative (even Noetherian) rings, something can be said in the case of FBN rings. If $X$ is injective from the beginning and the torsion class is closed under taking injective envelopes, then it is an exercise to prove that $T(X)=X_t$ is again injective and so you have $X=X_t\oplus X_f$ for some torsion-free injective object $X_f$. Thus $Q(X)\cong Q(X_f)$. Furthermore, $E(X_f)/X_f=0$ and so $Q(X_f)\cong X_f$ is injective.</p> <p>Anyway, not every "localization" is the localization with respect to a hereditary torsion theory and, even in that case, the torsion class is not always closed under taking injective envelopes so that the above argument does not always apply... In this moment it is not clear to me what happens for a general universal localization or for Cohn's localization.</p> http://mathoverflow.net/questions/131761/amenable-group-rings-embeddable-in-skew-fields Amenable group rings embeddable in skew fields Simone Virili 2013-05-24T19:12:28Z 2013-05-24T20:17:47Z <p>I've made this question on math.stackexchange.com (also offering a bounty) but I did not receive any answer:</p> <p>I'm looking for a reference of the following fact:</p> <p>given a (countable?) amenable group $G$ and a (skew) field $K$, the following are equivalent:</p> <p>(1) the group ring $K[G]$ is a domain;</p> <p>(2) $K[G]$ is a (left and right) Ore domain.</p> <p>I think to remember that this result is due to Beno Eckmann but, unfortunately, I cannot remember in which paper. I tried to look for this result and I'm not able to find it at the moment. Any reference would be strongly appreciated!</p> http://mathoverflow.net/questions/112574/cocomplete-but-not-complete-abelian-category Cocomplete but not complete abelian category Simone Virili 2012-11-16T13:36:09Z 2013-03-29T15:48:11Z <p>This is a duplicate of the following question to which I did not receive any answer: <a href="http://math.stackexchange.com/questions/238247/complete-but-not-cocomplete-category" rel="nofollow">http://math.stackexchange.com/questions/238247/complete-but-not-cocomplete-category</a></p> <p>Let $\mathfrak C$ be an abelian, cocomplete category. If $\mathfrak C$ has a generator and colimits are exact (i.e., $\mathfrak C$ is Grothendieck) then $\mathfrak C$ is the torsion-theoretic localization of a full category of modules (by the Gabriel-Popescu Theorem) and so it is also complete. Anyway I'm not aware of any counter-example showing that a cocomplete abelian category may not be complete. So my question is: could you provide such example or a reference to a proof of the bicompleteness of cocomplete abelian categories?</p> <p>My first idea was to look for counterexamples in non-Grothendieck subcategories of a Grothendieck category. After some attempt I realized the following </p> <p><strong>Lemma.</strong> <em>Let $\mathfrak C$ be a Grothendieck category and $\mathcal T$ a full hereditary torsion subcategory (i.e. $\mathcal T$ is closed under taking sub-objects, quotient objects, extensions and coproducts). Then $\mathcal T$ is bicomplete.</em></p> <p><em>Proof.</em> Let $T:\mathfrak C\to \mathcal T$ be the hereditary torsion functor associated to $\mathcal T$. Now, given a family {$C_i:i\in I$} of objects in $\mathcal T$ we can take the product $(P,\pi_i:P\to C_i)$ of this family in $\mathfrak C$. We claim that $(T(P), T(\pi_i))$ is a product in $\mathcal T$. Indeed, let $X\in \mathcal T$ and choose maps $\phi_i:X\to C_i$. By the universal property of products in $\mathfrak C$, there exists a unique morphism $\phi:X\to P$ such that $\pi_i\phi=\phi_i$ for all $i\in I$. Now, since $X\in\mathcal T$, there is an induced map $T(\phi):X\to T(P)$ which is clearly the unique possible map satisfying $T(\pi_i)T(\phi)=T(\phi_i)=\phi_i$. \\\</p> <p>Thus there are lots of non-Grothendieck bicomplete abelian categories.</p> <p>EDIT: notice that in the lemma we never use the hypothesis that the subcategory $\mathcal T$ is closed under taking extensions or subobjects. In fact, if $\mathcal T$ is just closed under taking coproducts and quotients, one defines the functor $T:\mathfrak C\to \mathcal T$ such that, for all object $X\in\mathfrak C$, $T(X)\in \mathcal T$ is the direct union of all the subobjects belonging to $\mathcal T$ (image (which is a quotient) of the coproduct of all the subobject of $X$ belonging to $\mathcal T$ under the universal map induced by the inclusions of the subobjects in $X$). Clearly $T(X)$ is fully invariant as a subobject of $X$ (by the closure of $\mathcal T$ under taking quotients and the construction of $T$) and so $T$ can be defined on morphisms by restriction. It is also clear that $T(X)=X$ if $X\in\mathcal T$ so the proof of the lemma can be easily adapted to this case.</p> <p>REMARK: the new relaxed hypotheses of the lemma allow us to exclude other "exotic" examples... in particular, if you want to take the abelian subcategory of all the semisimple objects in a given Grothendieck category, this is closed under coproducts and quotients. </p> http://mathoverflow.net/questions/125213/minimal-prime-ideals-of-a-group-ring Minimal prime ideals of a group ring Simone Virili 2013-03-22T00:00:51Z 2013-03-22T00:00:51Z <p>Let $R$ be a left Noetherian ring (if you prefer you can just think to $R$ as a skew field, I'll be happy with an answer under that hypothesis) and $G$ be a polycyclic-by-finite (or, if you prefer Abelian-by-finite) group. Take a crossed product $R*G$, a left ideal $I$ and a subgroup $H$ of $G$. $H$ is said to control $I$ if $R*G(I\cap R*H)=I$. </p> <p>My questions are the following:</p> <p>(1) is it true that the minimal prime ideals of R*G are controlled by the finite subgroups of $G$?</p> <p>(2) if the answer to (1) is positive, is it true that for a minimal prime ideal $\mathfrak p\leq R*G$, there exists a finite subgroup $H$ of $G$ and a minimal prime ideal $\mathfrak q\leq R*H$ such that $(R*G)\mathfrak q=\mathfrak p$?</p> <p>A positive answer to (1) in the case of $R[G]$ with $G$ polycyclic is given in Corollary 22 of [J.E.Roseblade, Prime ideals in group rings of polycyclic groups, 1977]. There seems to be a wide literature on prime ideals of crossed products of group rings, I would really appreciate some precise references. Furthermore, in case the answers to my questions are positive, it would be very helpful for me to have some direct argument, i.e., not following from a much deeper result requiring heavy machinery.</p> http://mathoverflow.net/questions/122752/existence-of-nice-folner-sequences Existence of nice Folner sequences Simone Virili 2013-02-23T21:31:54Z 2013-02-23T22:15:14Z <p>I'm attempting a proof by induction and, for the inductive step, it would be very useful for me to have some control on a Folner sequence. Indeed, let $G$ be a finitely generated amenable group, fix a finite symmetric set of generators for $G$ and denote by $B_n$ the ball of radius $n$ in the Cayley graph of $G$ with respect of the fixed generators. </p> <p>Is it possible to find a Folner sequence $(F_n:n\in\mathbb N)$ for $G$ satisfying the following properties?</p> <p>(1) $F_1\subseteq F_2\subseteq F_3\subseteq \dots$ and $\bigcup_{n\in\mathbb N}F_n=G$;</p> <p>(2) for all $n\in \mathbb N$ there exists $k$ such that $B_k\subseteq F_n\subseteq B_{k+1}$.</p> http://mathoverflow.net/questions/122687/dual-concept-for-the-p-primary-component/122712#122712 Answer by Simone Virili for Dual concept for the p-primary component Simone Virili 2013-02-23T11:36:30Z 2013-02-23T11:43:35Z <p>EDIT: Note that in what follows I interpret "dual" as "dual in the sense of Pontryagin-Van Kampen's duality". If you had something different in mind I think you should have specified...</p> <p>Well, first of all you should start finding a dual notion for the torsion part. Indeed, a given Abelian discrete Abelian group $G$ is the central element in a short exact sequence: $$0\to t(G)\to G\to G/t(G)\to 0 ,$$ where $t(G)$ is the torsion part and $G/t(G)$ is torsion free. As the duality functor (in the Pontryagin-Van Kampen duality) is exact you get a short exact sequence of compact Abelian groups: $$0\to \widehat{G/t(G)}\to \widehat G\to \widehat{t(G)}\to 0 .$$ Recall also that the torsion part $t(G)$ can be definite to be the direct union of all the finite subgroups of $G$. Using that finite Abelian groups are self-dual and that the duality functor send direct to inverse limits, you obtain that $\widehat{t(G)}$ is an inverse limit of finite Abelian groups, that is, it is pro-finite. In fact, it is the maximal pro-finite quotient of $\widehat G$.</p> <p>Now, coming to your question, you have to recall that $t(G)$ is the direct sum of its $p$-primary components $t_p(G)$ (each one defined as the direct union of all the finite $p$-subgroups of $G$) and that the duality functor sends direct sums (i.e., coproducts) to products. Thus $\widehat {t(G)}$ is the (topological) product of the $\widehat{t_p(G)}$ and each $\widehat{t_p(G)}$ is the inverse limit of $p$-torsion finite groups, that is, pro $p$-finite. </p> <p>To conclude, the answer to your question is: the dual of the $p$-torsion subgroup is the maximal pro $p$-finite quotient of the dual group.</p> <p>As you are asking for references, I think that any standard book with an exposition of the Pontryagin-Van Kampen's duality, if not the above discussion, will certainly give you proofs of all the facts I used in my answer. I personally like Dikran Dikranjan's approach to the duality theorem: <a href="http://users.dimi.uniud.it/~dikran.dikranjan/ITG.pdf" rel="nofollow">http://users.dimi.uniud.it/~dikran.dikranjan/ITG.pdf</a></p> http://mathoverflow.net/questions/116467/finite-uniquely-divisible-abelian-groups/116472#116472 Answer by Simone Virili for Finite / uniquely divisible abelian groups Simone Virili 2012-12-15T18:15:33Z 2012-12-15T18:15:33Z <p>well... as the Prüfer groups are not uniquely divisible, by the structure of divisible abelian groups you get that $Q$ is a rational vector space. In particular, $Q$ is flat (i.e., torsion-free as $\mathbb Z$ is a PID). This means that $F$ is pure in $A$. Now, it is well-known that finite pure subgroups split.</p> http://mathoverflow.net/questions/115409/torsion-theories-localizing-the-base-ring-to-the-same-ring/116432#116432 Answer by Simone Virili for torsion theories localizing the base ring to the same ring Simone Virili 2012-12-15T09:00:41Z 2012-12-15T13:02:57Z <p>As you mention Golan, I guess that all your torsion theories are hereditary. Let $\tau_1$ and $\tau_2$ be t.t. on Mod$(R)$ and $\phi_1:R\to R_1$, $\phi_2:R\to R_2$ the two loc. of $R$. The fact that there exists an isomorphism $\phi:R_1\to R_2$ s.t. $\phi\phi_1=\phi_2$, means that $-\otimes_RR_1$ is naturally eq. to $-\otimes_RR_2$. If $\tau_1$ and $\tau_2$ are perfect then these functors coincide with the localization functors. Thus, in such case, $M\in \mathcal T_{\tau_1}$ (the torsion class of $\tau_1$) iff $M\otimes_RR_1=0$ iff $M\otimes_RR_2=0$ iff $M\in \mathcal T_{\tau_2}$. So $\mathcal T_{\tau_1}=\mathcal T_{\tau_2}$, that is, $\tau_1=\tau_2$.</p> <p>If your torsion theories are not perfect I do not remember if $\ker(-\otimes_RR_1)=\mathcal T_1$ holds true, if so you should be able to proceed as above...</p> http://mathoverflow.net/questions/114494/ring-with-a-condition/114649#114649 Answer by Simone Virili for ring with a condition Simone Virili 2012-11-27T11:11:12Z 2012-11-27T16:04:08Z <p>After answering, the question changed so I adapt my answer to the new question.</p> <p>As I was remarking in the comments to your question, it is impossible to construct such ring. In fact, as you want that, for all $x,y\in R\setminus\{0\}$, the intersection $\{r\in R:xr\in (I:I)\}\cap\{r\in R:xry\notin I\}\neq \emptyset$, in particular you need that $\{r\in R:xry\notin I\}\neq \emptyset$ for all $x,y\in R\setminus\{0\}$. Notice that, if $x\in I$, then, as $I$ is a right ideal, $xry=x(ry)\in I$ for all $r$ and $y\in I$, so this set is always empty.</p> <p>Furthermore, if you have $r$ such that $xr\in (I:I)$, then, by definition of $(I:I)$, $xry\in I$ provided $y\in I$. So, if $y\in I$, then $\{r\in R:xr\in (I:I)\}\cap\{r\in R:xry\notin I\}= \emptyset$.</p> <p>Maybe you have some hope looking in a non-commutative ring and taking $I$ to be a right <strong>but not left</strong> ideal and imposing your second condition just for $x,y\in R\setminus I$. Furthermore, another necessary condition is that $I\neq (I:I)$. Anyway, this is another question! </p> <p>Finally, let me also add that the first condition $(I:I)\cap \{t\in R:t(I:I)t\subseteq I\}\neq \{0\}$ is implied by $I\neq\{0\}$ in fact $I$ is always contained in this intersection.</p> http://mathoverflow.net/questions/112660/free-direct-summand-of-a-module/112667#112667 Answer by Simone Virili for Free direct summand of a module Simone Virili 2012-11-17T10:03:51Z 2012-11-17T10:03:51Z <p>The answer to your question is: no, it is not true. Here is a counterexample.</p> <p>$(A,\mathfrak m)=(\mathbb Z_p,p\mathbb Z_p)$ is the (local Noetherian commutative) ring (PID) of $p$-adic integers (you can construct this ring in many ways... probably the standard one is to complete $\mathbb Z$ with respect to the $p$-adic topology -a base for this topology is given by the powers of the maximal ideal $p\mathbb Z$- the result is that $\mathbb Z_p$ is the inverse limit of the groups of the form $\mathbb Z/p^n$ with the canonical transition maps). </p> <p>Now, take $F=A$ and $M=p^2 A$. It follows essentially by the definition of $\mathbb Z_p$ that I gave you, that $F/M\cong \mathbb Z/p^2$ that has composition length $2$, in particular this is finite. Anyway, $M\cong A$ as modules (just consider the morphism $A\to M$ such that $x\mapsto p^2x$... then surjectivity is obvious and injectivity follows by the fact that $A$ is a domain and so multiplication by $p^2$ has trivial kernel). Thus $M$ has not only free summands but it is itself free. Finally, notice that $pA\supsetneq p^2A=M$ so $M$ is properly contained in $\mathfrak m F$.</p> http://mathoverflow.net/questions/111887/tensorproduct-p-adic-groupring/112625#112625 Answer by Simone Virili for tensorproduct, p-adic groupring Simone Virili 2012-11-16T23:49:01Z 2012-11-16T23:49:01Z <p>I do not think this is a research question. Anyway, first of all you should convince yourself that $\mathbb Z_p\otimes_{\mathbb Z} (\mathbb Z[G])\cong(\mathbb Z_p\otimes_{\mathbb Z} \mathbb Z)[G]$. After this I think you will have no difficulty in verifying that $\mathbb Z_p\otimes_{\mathbb Z} \mathbb Z\cong \mathbb Z_p$. </p> <p>This is more an exercise than a theorem so it has no specific name. Anyway, similar operations are usually called "extensions of scalars". </p> http://mathoverflow.net/questions/112564/finitely-generated-resolutions/112565#112565 Answer by Simone Virili for Finitely generated resolutions Simone Virili 2012-11-16T10:33:27Z 2012-11-16T20:35:41Z <p>Let me start recalling the Schanuel's Lemma:</p> <p><em>If $M$ is a module and $P,P'$ are projective modules, then for every short exact sequences $0\to K\to P\to M\to 0$ and $0\to K'\to P'\to M\to 0$, there is an isomorphism $K\oplus P'\cong K'\oplus P$.</em></p> <p>So, if you have a short exact sequence $0\to K\to P\to M\to 0$ with $M$ f.g., $P$ f.g. projective and $K$ not f.g., you will not be able to find any other f.g. projective $P'$ which admits a projection onto $M$ with f.g. kernel.</p> <p>So, your question is equivalent to ask the following: for which class of rings does the class of f.g. left modules coincide with that of finitely presented left modules? The answer to this question is: the class of left Noetherian rings.</p> <p>In some cases, one can find such resolutions outside from the Noetherian context. For example, if a ring is left coherent you can find such resolutions for any finitely generated left ideal of the ring.</p> http://mathoverflow.net/questions/111285/cofibrant-replacement-of-chain-complexes/111303#111303 Answer by Simone Virili for Cofibrant Replacement of chain complexes Simone Virili 2012-11-02T20:49:11Z 2012-11-02T20:49:11Z <p>The "addition" to your own question seems to be a starting point for the answer you were looking for but let me add some comment. In particular, the projective model structure you mention is a very particular model structure which has the good fortune to be an abelian model structure in the sense of [M. Hovey, Cotorsion pairs and model categories, Contemporary Maths. (2006)]. In particular, in the projective model structure on chain complexes, the coﬁbrations are the monomorphisms with coﬁbrant (=DG-projective) cokernel, the ﬁbrations are the epimorphisms, and the weak equivalences are the quasi-isomorphisms. So the notion of "projective" complex you are looking for is that of DG-projective complex, that is, a complex in which each entry is projective, and such that any map from it to an exact complex is chain homotopic to 0 (in the bounded case this is equivalent to level-wise projective so your intuition is correct).</p> <p>I suggest you to read the very well written paper by Hovey that I mentioned above and to use the papers in his reference list for further details. Of course, Hovey's book on model categories is a standard reference but the paper I'm suggesting is far easier and faster to read. </p> http://mathoverflow.net/questions/110910/mathbbz-pk-mathbbzg-modules-in-repr-theory/110919#110919 Answer by Simone Virili for $\mathbb{Z}/p^k \mathbb{Z}[G]$-modules in repr. theory Simone Virili 2012-10-28T16:55:09Z 2012-10-28T16:55:09Z <p>well... one can start thinking to the case when $G$ is trivial. You can probably feel how $\mathbb Z/p^k\mathbb Z$-modules are easier then $\mathbb Z_p$-modules. In particular, the torsion $\mathbb Z_p$-modules are the abelian $p$-groups, while the $\mathbb Z/p^k\mathbb Z$-modules are the $p^k$-bounded abelian $p$-groups. It seems more than reasonable to start classifying the indecomposable bounded groups in order to classify the torsion ones... when $G$ is not trivial, the situation is similar.</p> http://mathoverflow.net/questions/109227/how-to-prove-this-equation/109236#109236 Answer by Simone Virili for How to prove this equation Simone Virili 2012-10-09T15:39:10Z 2012-10-09T15:47:01Z <p>Any topological group of the form $K\cong H_0\times U(1)^k$ (with $H_0$ finite and $k$ a positive integer) is a closed subgroup of $U(1)^h$ (for some positive integer $h\geq k$). Furthermore, all the closed subgroups of $U(1)^h$ are of this form.</p> <p>The proof is an easy application of the Pontryagin-Van Kampen duality. In fact, such groups are the duals of the finitely generated Abelian groups (which are quotients of $\mathbb Z^h$). It is well known that such groups are of the form $F\times \mathbb Z^k$ (with $F$ a finite group). </p> <p>To find a general form for a compact abelian group is as difficult as giving a structure theorem for discrete abelian group (which is known to be quite a difficult, and fairly open, problem in general, even if there are nice results for countable torsion groups).</p> <p>EDIT: just to answer also to the comment of Stefan Geschke. Finitely generated groups in $\mathrm{Mod}(\mathbb Z)$ can be characterized as the Noetherian objects of the category. So I guess that (by duality) the objects of the form $K\cong H_0\times U(1)^k$ should be the Artinian objects in the category of compact abelian topological ($T_2$) groups. (N.B.= here by Artinian I mean the category-theoretical notion)</p> http://mathoverflow.net/questions/108255/another-question-about-amenability-and-folner-sequences Another question about amenability and Følner sequences Simone Virili 2012-09-27T15:07:01Z 2012-09-27T21:06:59Z <p>Følner's characterization of Amenability says that a group $G$ is amenable if there exists a directed set $(I,\leq)$ and a net {$F_i:i\in I$} of finite subsets of $G$ such that for every $γ ∈ G$, $$\lim_{i\in I}\frac{|γF_i\Delta F_i|}{|F_i|}\ \ \ \rightarrow 0\ ,$$ where $\Delta$ is the symmetric difference of two sets. It is also known that, if $G$ is countable, the word "net" can be substituted by "sequence" (that is $I=\mathbb N$ with the usual order). </p> <p>Is it true that for countable (or at least finitely generated) groups we can always find a Følner sequence as above, which satisfies the following conditions:</p> <p>(1) $F_{n}\subseteq F_{n+1}$, for all $n\in \mathbb N$;</p> <p>(2) $\bigcup_{n\in\mathbb N}F_n=G$.</p> <p>The motivation for my question comes from the paper "The Abramov-Rokhlin Entropy Addition Formula for Amenable Group Actions" by Ward and Zhang (Mh. Math, 1992). In fact, their Theorem 2.6 (that they attribute to Ornstein and Weiss) is proved for a Følner sequence as the one above but it is applied to actions of arbitrary countable Amenable groups. So... it seems that such sequences always exist.</p> http://mathoverflow.net/questions/104533/colocalization-of-the-derived-category (Co)localization of the derived category Simone Virili 2012-08-12T02:11:28Z 2012-08-12T17:01:45Z <p>Let me start saying that a similar question can be stated for general locally Noetherian Grothendieck categories but I state it for categories of rings as it is simpler. So we fix a right Noetherian ring (non-commutative) and we let $Mod(R)$ be the category of right $R$-modules.</p> <p>Recall the following theorem due to H. Krause (for its proof Brown's representability is used).</p> <p><strong>Theorem 1.</strong> <em>Let $\mathcal T$ be a triangulated category with small coproducts which is well generated. Let $H : \mathcal T \to \mathcal A$ be a cohomological functor into an abelian category $\mathcal A$ which has small coproducts and exact $\alpha$-filtered colimits for some regular cardinal $\alpha$. Suppose also that $H$ preserves small coproducts. Then there exists an exact localization functor $L:\mathcal T \to\mathcal T$ such that for each object $X$ we have $LX=0$ if and only if $H(X[n])=0$ for all $n \in \mathbb Z$.</em></p> <p>Fix now a hereditary torsion theory $\tau$ on $Mod(R)$. We go on defining in three steps an exact localization functor of the derived category $L_\tau:{ D}(R)\to { D}(R)$.</p> <p>(1) Denote by $$H^n:{\bf D}(R)\to Mod (R)$$ the usual $n$-th cohomology, for every $n\in\mathbb Z$. It is clear that each $H^n(-)$ is cohomological and preserves coproducts. </p> <p>(2) Fix a hereditary torsion theory $\tau$ on $Mod(R)$. The $\tau$-localization functor $$Q_\tau:Mod(R)\to \mathcal A_\tau=Mod(R)/\mathcal T_{\tau} ,$$ where $\mathcal T_{\tau}$ is the hereditary torsion class associated to $\tau$, is exact and preserves coproducts. Furthermore, $\mathcal A_\tau$ is a Grothendieck category and so it has small coproducts and exact colimits. </p> <p>(3) For every $n\in\mathbb Z$ denote by $$H_\tau^n:{\bf D}(R)\to \mathcal A_\tau$$ the composition of the above two functors, that is $H_\tau^n(-)=Q_\tau H^n(-)$. By (1) and (2) one can easily derive that $H_\tau^n(-)$ is cohomological and preserves coproducts. </p> <p>Now we have all the instruments to construct the localization functor $L_\tau(-)$:</p> <p><strong>Corollary.</strong> *Let $\tau$ be a hereditary torsion theory on $Mod(R)$. Then there exists an exact localization functor $L_\tau:{\bf D}(R)\to {\bf D}(R)$ such that $L_\tau(X)=0$ if and only if the $n$-th cohomology of $X$ is $\tau$-torsion for every $n\in\mathbb Z$.*</p> <p><em>Proof.</em> Consider the functor $\prod_{n\in\mathbb Z}H^n_\tau:{\bf D}(R)\to \prod_{n\in\mathbb Z}\mathcal A_\tau$. By the above discussion, this functor is a cohomological functor preserving coproducts from ${\bf D}(R)$ to a bicomplete abelian category with exact colimits. Let $X\in {\bf D}(R)$, then $\prod_{n\in\mathbb Z}H^n_\tau(X)=0$ if and only if $H_\tau^n(X)=0$ for every $n\in\mathbb Z$, if and only if $Q_\tau H^n(X)=0$ for every $n\in\mathbb Z$. This is equivalent to say that all the cohomologies of $X$ are in the kernel of $Q_\tau(-)$ that is, they are $\tau$-torsion. Now, to prove the existence of $L_\tau(-)$ it is enough to apply Theorem 1.</p> <hr> <p><strong>Question.</strong> In the above notation, is it possible to prove that for a given object $X\in{\bf D}(R)$, $L_\tau(X)$ belongs to the smallest localizing subcategory of ${\bf D}(R)$ containing $X$?</p> <hr> <p>More specifically, consider an indecomposable injective module $E$ and let $\tau$ be the hereditary torsion theory cogenerated by $E$. In the commutative case, there is a unique prime ideal associated to $E$ and localizing with respect to $\tau$ is the same as localizing at that prime ideal. In particular, localized modules can be constructed as a direct limit of $R$-modules and this construction can be "lifted" to the derived category to answer positively to the above question.</p> <p>In the non-commutative case, there is no prime ideal associate in general but we can suppose it if we suppose our ring to be right FBN (in the general case $E=E(C)$ with $C$ a cocritical module which can be chosen of the form $R/I$ with $I$ an irreducible ideal). But even in the case when the localization at $\tau$ is Ore, the localized modules are constructed as direct limits in the category of Abelian groups and after that they are induced with the structure of modules. This makes no sense (to me) in ${\bf D}(R)$. </p> http://mathoverflow.net/questions/103837/is-there-a-categorical-treatment-of-dynamical-systems/103840#103840 Answer by Simone Virili for Is there a categorical treatment of dynamical systems? Simone Virili 2012-08-03T05:21:48Z 2012-08-11T05:06:22Z <p>For every category $\mathfrak C$ and every monoid $\mathcal S$, you can define the category of $\mathcal S$-flows on $\mathfrak C$ as follows:</p> <p>-- Objects are pairs $(X, \alpha:\mathcal S\to \mathrm{End}(X))$, where $\alpha$ is a monoid homomorphism (that is, multiplication in $\mathcal S$ becomes composition of morphisms in $\mathfrak C$ and the neutral element goes to identity);</p> <p>-- Morphisms in the category between two objects $(X,\alpha)$ and $(X',\alpha')$ are just morphisms $\phi:X\to X'$ that commute with the actions, that is $\alpha'(s)\phi=\phi\alpha(s)$ for all $s\in \mathcal S$. </p> <p>Clearly an isomorphism in this category is just a morphism that is an isomorphism in $\mathfrak C$. </p> <p>This is standard and probably does not answer completely your question as you probably want a weaker form of equivalence than really isomorphism in the category of $\mathcal S$-flows. Anyway I think this is a good starting point.</p> <p>Just to understand what this category is, I can give you some examples. If you consider $\mathbb N$-flows on the category of modules $\mathrm {Mod}(R)$ over a ring $R$, you just obtain the category of modules over the polynomial ring $R[X]$. In fact, it is a classical way of looking at modules over $R[X]$ as $R$-modules with a distinguished $R$-linear endomorphism acting on them, that is, discrete-time dynamical systems. </p> <p>If you take $\mathbb Z$-flows you obtain the ring of Laurent polynomials $R[X^{\pm 1}]$. This is easily generalized to $\mathbb N^k$ and $\mathbb Z^k$, giving rise to polynomials in $k$ commuting variables. This point of view is generally adopted by K. Schmidt in his book "Dynamical Systems of Algebraic Origin". In fact, the general approach there is to study dynamical systems of the form $(G,\phi_1,\dots,\phi_k)$, where $G$ is a compact abelian topological group and $\phi_1,\dots,\phi_k$ are commuting topological automorphisms of $G$. This is the category of $\mathbb Z^k$-flows on the category of compact abelian groups. Via Pontryagin duality, this category can be seen to be dual to the category of $\mathbb Z^k$-flows on discrete Abelian groups, that, by what we said above, is exactly $\mathrm{Mod}(\mathbb Z[X_1^{\pm 1},\dots,X_k^{\pm 1}])$.</p> <p>Generalizing more, you can easily prove that $\mathcal S$-flows on $\mathrm {Mod}(R)$ are the category of modules over the monoid ring $R[\mathcal S]$.</p> http://mathoverflow.net/questions/100042/additive-integer-valued-functions-on-the-module-category/103878#103878 Answer by Simone Virili for Additive integer-valued functions on the module category Simone Virili 2012-08-03T15:46:35Z 2012-08-03T15:46:35Z <p>Let me suggest you some references:</p> <p>Northcott Reufel, "A generalization of the concept of length" (http://qjmath.oxfordjournals.org/content/16/4/297.full.pdf)</p> <p>Vamos, "ADDITIVE FUNCTIONS AND DUALITY OVER NOETHERIAN RINGS" (http://qjmath.oxfordjournals.org/content/19/1/43.full.pdf)</p> <p>Zanardo, "Multiplicative invariants and length functions over valuation domains" (http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jca/1323364358)</p> <p>If your interest goes in this direction I know that people in Padova is working on generalizations of the work of Zanardo to classify length functions of Prufer domains. </p> <p>The classification given by Vamos on Noetherian rings was generalized by him in his (non-pubblished) PhD thesis to a classification for rings with Gabriel-Krull dimension. I recently gave an alternative poof of Vamos' result for Grothendieck categories with Gabriel-Krull dimension based on the formalism of torsion theories, contact me if you are interested.</p> http://mathoverflow.net/questions/69218/if-you-were-to-axiomatize-the-notion-of-entropy/102664#102664 Answer by Simone Virili for If you were to axiomatize the notion of entropy ..... Simone Virili 2012-07-19T13:31:25Z 2012-07-19T13:31:25Z <p>I want just to mention a framework in which we (Dikran Dikranjan, Anna Giordano Bruno and me) are going to redefine many notions of entropy.</p> <p>The idea of prof. Dikranjan was to define the category of normed semigroups. Indeed, a normed semigroup is just a semigroup $S$ with a norm $$v:S\to \mathbb R_{\geq 0}$$ such that $v(xy)\leq v(x)+v(y)$. The morphisms in the category are just semigroup homomorphisms such that the norm of the image is $\leq$ than the norm of the original point. </p> <p>In this category one can define a notion of entropy of an endomorphism $\phi:(S,v)\to (S,v)$. In fact one takes $$h(\phi)=\sup \left( \lim_{n\to\infty}\frac{v(x\phi(x)\dots\phi^{n-1}(x))}{n}: x\in S\right ) .$$ It is interesting to notice that already at this level, the above entropy function satisfies some good properties. Furthermore, it turns out that many of the usual notions of entropy (topological, algebraic, mesure-theoretic entropy, ...) for endomorphisms or automorphisms can be defined using a suitable functor from the category in which they are defined to the category of normed semigroups (the semigroup can be the set of subset with intersection (or sum in groups), the set of open covers with intersection, ... the norm can be $\log$ of the cardinality, measure, minimal cardinality of a subcover ...).</p> <p>Let me conclude remarking that if you are looking for lists of axioms for entropy functions you should look to the following papers:</p> <p>L. N. Stojanov, Uniqueness of topological entropy for endomorphisms on compact groups, Boll. Un. Mat. Ital. B 7 (1987) no. 3, 829–847. (axiomatic char. of topological entropy on compact groups)</p> <p>D. Dikranjan and A. Giordano Bruno, Entropy on abelian groups, preprint; arXiv:1007.0533. (axiomatic char. of algebraic entropy on discrete groups)</p> <p>L. Salce, P. Vamos, S. Virili, Length functions, multiplicities and algebraic entropy, Forum Math. (2011) (axiomatic char. of a notion of algebraic entropy on modules)</p> <p>The above characterizations are discussed in the following survey,</p> <p>D. Dikranjan, M. Sanchis, S. Virili, New and old facts about entropy in uniform spaces and topological groups, Topology appl. (2012) 1916-1942</p> http://mathoverflow.net/questions/36025/explicit-computations-using-the-haar-measure/102039#102039 Answer by Simone Virili for Explicit computations using the Haar measure Simone Virili 2012-07-12T13:38:00Z 2012-07-12T13:38:00Z <p>There are cases in which computation can be made easier. Consider for example a locally compact abelian group $G$ and denote by $\mu$ a Haar measure.</p> <p>One hypothesis on $G$ that can make life easier is to suppose $G$ totally disconnected. In such case, $G$ has a base of neighborhoods of $0$ consisting of clopen compact subgroups. Notice also that, if we have two such neighborhoods $V_1\subseteq V_2$, then $V_2/V_1$ is finite and $\mu(V_2)=|V_2/V_1|$, provided $\mu(V_1)=1$ (we can always suppose this up to a renormalization of the measure). </p> <p>This seems quite abstract but there are examples in which this could be very useful in concrete computation (see the answer of Davidac897 for a very explicit occurrence of what I'm saying).</p> http://mathoverflow.net/questions/101390/a-bijective-correspondence-induced-by-fourier-transform/101516#101516 Answer by Simone Virili for A bijective correspondence induced by Fourier Transform Simone Virili 2012-07-06T17:06:57Z 2012-07-11T09:09:17Z <p>I try to write down in full detail the answers of Nik and BS (many thanks to both!). </p> <p>The references in the proof are to Folland's book "A Course in Abstract Harmonic Analysis" and Rudin's book "Fourier Analysis on Groups".</p> <hr> <p>Let $G$ be an LCA group, then {$\widehat\phi:\phi\in L^1(G)^+\cap \mathcal P(G)$}$=L^1(\widehat G)^+\cap \mathcal P(\widehat G)$.</p> <p><em>Proof</em>. Let us start proving the inclusion {$\widehat\phi:\phi\in L^1(G)^+\cap \mathcal P(G)$}$\subseteq L^1(\widehat G)^+\cap \mathcal P(\widehat G)$. Indeed, consider $\phi \in L^1(G)^+\cap \mathcal P(G)$, then $\widehat\phi\in L^1(\widehat G)$ by the Fourier Inversion Theorem (see [Rudin, page 22, line 1]) and $\widehat\phi\geq 0$ by [Folland, Corollary 4.23]. Let now $\mu_\phi$ be the non-negative and bounded (as $\phi\in L^1(G)^+$) regular measure defined on a generic Borel subset $E$ of $G$ by $\mu_\phi(E)=\int_{x\in E}\phi(x)d\mu(x)$ (here $\mu$ is a fixed Haar measure on $G$). One can show that $$\widehat\phi(\gamma)=\int_{x\in G}\phi(x)\gamma(-x)d\mu(x)=\int_{x\in G}\gamma(-x)d\mu_\phi(x)\, .$$ By Bochner's Theorem (see [Rudin, page 19]), $\widehat\phi\in \mathcal P(\widehat G)$. </p> <p>On the other hand, let $\phi\in L^1(\widehat G)^+\cap \mathcal P(\widehat G)$ and $\psi$ be the function defined by $\psi(\gamma)=\phi(-\gamma)$ for all $\gamma\in \widehat G$. It is not difficult to see that $\psi\in L^1(\widehat G)^+\cap \mathcal P(\widehat G)$. By the first part of the proof, $\widehat\psi\in L^1( G)^+\cap \mathcal P( G)$ and, using Fourier Inversion Theorem (see [Folland, Theorem 4.32]), one obtains that $\widehat{\widehat\psi}=\phi$, which is therefore the Fourier transform of a function in $L^1( G)^+\cap \mathcal P( G)$.\\\</p> http://mathoverflow.net/questions/101390/a-bijective-correspondence-induced-by-fourier-transform A bijective correspondence induced by Fourier Transform Simone Virili 2012-07-05T10:23:20Z 2012-07-11T09:09:17Z <p>Let $G$ be a discrete Abelian group and denote by $\widehat G$ the (compact) Pontryagin-Van Kampen dual of $G$. I was reading in a paper of Justin Peters that Fourier Transform induces a bijection between the following sets of functions:</p> <p>(1) $L^1(G)^+\cap \mathcal P(G)$ (continuous, non-negative, positive-definite and absolutely integrable functions $G\to \mathbb C$);</p> <p>(2) $L^1(\widehat G)^+\cap \mathcal P(\widehat G)$ (continuous, non-negative, positive-definite and absolutely integrable functions $\widehat G\to \mathbb C$).</p> <p>Peters says that this is easy to deduce from Fourier Inversion Theorem but it does not seem so elementary to me. Can anyone help me? Is this true for any LCA group? </p> <p>Is there any analog if we start with a compact non-commutative group and we use the Tannaka-Krein duality?</p> http://mathoverflow.net/questions/101301/continuous-positive-definite-function-with-prescribed-support Continuous positive-definite function with prescribed support Simone Virili 2012-07-04T11:15:19Z 2012-07-04T14:19:38Z <p>Consider a (locally) compact Abelian group G and a compact neighborhood $C$ of the identity ($0$) of $G$. Is it possible to find a non-trivial (i.e., $\phi(0)\neq 0$) continuous, positive (with real non-negative image), positive-definite and absolutely integrable function $\phi :G\to \mathbb C$ (in symbols $\phi\in L^1(G)^+\cap P(G)$) whose support (the closure in $G$ of the set of points with non-zero image) is contained in $C$?</p> <p>As I do not need a solution for all the compact neighborhoods but only for a cofinal subset of the family of such neighborhoods (ordered by inclusion) we can restrict to symmetric neighborhoods (i.e., $C=-C$). For such $C$, my idea was to find a compact symmetric neighborhood $C'\subseteq C$ such that $C'+C'\subseteq C$ and take as $\phi$ the convolution of $\chi_{C'}$ with itself. Since $\chi_{C'}(x)=\overline{\chi_{C'}(-x)}$, one easily obtains that $\phi$ is positive-definite. It is also easily verified that $\phi$ is absolutely integrable and positive. What about continuity? Does this work?</p> http://mathoverflow.net/questions/132398/surjectivity-of-the-natural-map-of-injective-module-to-its-localization/132399#132399 Comment by Simone Virili Simone Virili 2013-05-31T13:06:30Z 2013-05-31T13:06:30Z Thanks for the observation, you are right, there was a gap at some point, I've edited. http://mathoverflow.net/questions/122752/existence-of-nice-folner-sequences Comment by Simone Virili Simone Virili 2013-02-24T09:00:50Z 2013-02-24T09:00:50Z @Mark: Thanks for your answer and comment! I'm trying to understand how some dynamical results in dynamical systems of the form $G\times M\to M$ (with $G$ amenable and $M$ an Abelian discrete group) generalize from the case when $G$ is Abelian (e.g. $\mathbb Z^n$) to the general situation. I admit that my intuition is still not that good in the non-commutative case! http://mathoverflow.net/questions/122752/existence-of-nice-folner-sequences Comment by Simone Virili Simone Virili 2013-02-24T00:42:52Z 2013-02-24T00:42:52Z @Misha: well... in that case ${B_n:n\in\mathbb N}$ is a Folner sequence so the interesting case is when you have exponential growth... http://mathoverflow.net/questions/121804/partial-order-relation-on-subsets Comment by Simone Virili Simone Virili 2013-02-14T14:35:22Z 2013-02-14T14:35:22Z Essentially you are taking strictly increasing maps from {1,...,k} to {1,...,n}. Than, given two strictly increasing maps $f,g:\{1,...,k\}\to \{1,\dots,n\}$ you say that $f\prec g$ if $f(i)\leq g(i)$ for all $i\in\{1,\dots,k\}$. I would call it just a pointwise order... I cannot help you with any reference. http://mathoverflow.net/questions/120120/what-is-the-dual-of-a-pre-injective-map/120146#120146 Comment by Simone Virili Simone Virili 2013-01-28T21:16:29Z 2013-01-28T21:16:29Z no problem, it is nice to have this motivation for my question here http://mathoverflow.net/questions/120120/what-is-the-dual-of-a-pre-injective-map/120146#120146 Comment by Simone Virili Simone Virili 2013-01-28T20:36:05Z 2013-01-28T20:36:05Z Yes I did know such connection, which is in fact part of the motivation for my question... http://mathoverflow.net/questions/116771/on-topology-of-p-adic-numbers/116772#116772 Comment by Simone Virili Simone Virili 2012-12-19T11:33:52Z 2012-12-19T11:33:52Z actually this happens for any topological group admitting a base of neighborhoods given by subgroups http://mathoverflow.net/questions/116719/localization-of-a-pure-injective-module-is-pure-injective Comment by Simone Virili Simone Virili 2012-12-18T18:33:51Z 2012-12-18T18:33:51Z what is your setting? What do mean by &quot;localization&quot;? This could be an interesting question, it would be nice to have some more details (do you refer to a left or right Ore localization? are you thinking to modules over a commutative ring? ...). http://mathoverflow.net/questions/115409/torsion-theories-localizing-the-base-ring-to-the-same-ring/116432#116432 Comment by Simone Virili Simone Virili 2012-12-15T14:16:50Z 2012-12-15T14:16:50Z yes, I was not saying that the localization is equivalent to $-\otimes_RR_1$, I was just saying that maybe, when these two functors are different, maybe they have the same kernel... do you have some easy counterexample? http://mathoverflow.net/questions/116155/how-should-one-look-at-the-set-of-compatible-ring-structures-on-a-given-group Comment by Simone Virili Simone Virili 2012-12-12T10:19:27Z 2012-12-12T10:19:27Z Even if the point of view is completely different, in the second volume (if I remember correctly) of &quot;Infinite Abelian Groups&quot; by Laszlo Fuchs, there is a nice discussion about the possible ring structures on a given Abelian group $G$. (probably you are aware of this, but it is not always possible to find such a ring structure on a given group) http://mathoverflow.net/questions/103781/does-mittag-lefflerness-descend Comment by Simone Virili Simone Virili 2012-12-04T19:16:47Z 2012-12-04T19:16:47Z It does not answer completely your question but I suggest you to have a look to this paper <a href="http://arxiv.org/pdf/0704.3690v1.pdf" rel="nofollow">arxiv.org/pdf/0704.3690v1.pdf</a> In particular Example 1.6 and Proposition 1.7 discuss similar matters http://mathoverflow.net/questions/114688/are-amenable-groups-topologizable Comment by Simone Virili Simone Virili 2012-12-03T00:30:22Z 2012-12-03T00:30:22Z probably Corollary 3 on page 213 in <a href="http://www.mscand.dk/article.php?id=2006" rel="nofollow">mscand.dk/article.php?id=2006</a> will help... http://mathoverflow.net/questions/97555/noncommutative-localization-from-scratch Comment by Simone Virili Simone Virili 2012-12-02T17:55:29Z 2012-12-02T17:55:29Z @Martin: the commutative setting is really different!!! The common point of view for localizing a module $M$ on a commutative ring $R$ with respect to a multiplicative subset $\Sigma\subseteq R$ is to view the elements of $\Sigma$ as endomorphisms of $M$ and so take the direct limit of $|\Sigma|$-many copies of $M$ with these transition maps. When $R$ is not commutative you cannot do so if $\Sigma$ is not included in the center $Z(\Sigma)$. This is a non-trivial complication (in concrete situations it makes really a lot of difference, I can give you concrete examples)! http://mathoverflow.net/questions/114688/are-amenable-groups-topologizable Comment by Simone Virili Simone Virili 2012-12-02T13:48:55Z 2012-12-02T13:48:55Z unfortunately it seems that the bohr topology is T2 only for MAP groups... is there any known example of amenable non-MAP group (with the discrete topology)? http://mathoverflow.net/questions/115028/when-is-a-product-of-a-simple-not-semisimple Comment by Simone Virili Simone Virili 2012-12-02T02:02:44Z 2012-12-02T02:02:44Z well... it is not true that a product of simple modules is semisimple... this is true if you take the product of copies one given simple for the exact reason I explained in the previous comment.