User rasmus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:08:37Z http://mathoverflow.net/feeds/user/1291 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/121606/explicit-indecomposable-monomorphism-of-finitely-generated-non-indecomposable-abe Explicit indecomposable monomorphism of finitely generated non-indecomposable Abelian groups. Rasmus 2013-02-12T14:23:51Z 2013-02-13T08:30:44Z <p>This question is a follow-up of the question I asked <a href="http://mathoverflow.net/questions/104827/classification-of-long-exact-sequences" rel="nofollow">here</a>.</p> <p>Can you write down an explicit example of a monomorphism of finitely generated Abelian groups which is an indecomposable object in the category of two-step complexes of Abelian groups such that one of the two entries is not an indecomposable Abelian group?</p> http://mathoverflow.net/questions/116060/inductive-vs-projective-limit-of-sequence-of-split-surjections Inductive vs projective limit of sequence of split surjections Rasmus 2012-12-11T09:43:30Z 2012-12-11T11:21:09Z <p>Let $$A_1\twoheadrightarrow A_2\twoheadrightarrow A_3\twoheadrightarrow A_4\twoheadrightarrow \cdots$$ be an inductive sequence of <strong>countable</strong> abelian groups, the connecting homomorphisms of which are surjective and <strong>split</strong>, that is, we have embeddings $A_{n+1}\rightarrowtail A_n$ such that the composition $A_{n+1}\rightarrowtail A_n\twoheadrightarrow A_{n+1}$ is the identity for every $n$. This means that $A_{n+1}$ is a direct summand of $A_n$.</p> <p>Let $\varinjlim A_n$ denote the inductive limit of the system $$A_1\twoheadrightarrow A_2\twoheadrightarrow A_3\twoheadrightarrow A_4\twoheadrightarrow \cdots$$ and let $\varprojlim A_n$ denote the projective limit of the system $$A_1\leftarrowtail A_2\leftarrowtail A_3\leftarrowtail A_4\leftarrowtail \cdots.$$ We get an induced map $$\varprojlim A_n\to\varinjlim A_n.$$ As <a href="http://math.stackexchange.com/users/5191/zhen-lin" rel="nofollow">Zhen Lin</a> has shown in <a href="http://math.stackexchange.com/a/254702/367" rel="nofollow">over here</a>, this map need not be surjective. Here is a weaker question:</p> <p><strong>Question:</strong> If we have $\varinjlim A_n=0$, then can we conclude that $\varprojlim A_n=0$?</p> <p>This would, of course, follow if the map $\varprojlim A_n\to\varinjlim A_n$ was always injective. Is there any reason to expect this?</p> <p>[Earlier versions of this question were posted <a href="http://math.stackexchange.com/questions/254649" rel="nofollow">here</a> and <a href="http://math.stackexchange.com/questions/254822" rel="nofollow">here</a> on MSE.]</p> http://mathoverflow.net/questions/104827/classification-of-long-exact-sequences Classification of long exact sequences Rasmus 2012-08-16T12:05:52Z 2012-08-16T20:21:17Z <p>Let $\mathcal C$ be the category of long exact sequences of finitely generated abelian groups almost all of whose entries vanish.</p> <p>The category $\mathcal C$ is naturally additive as a subcategory of complexes of abelian groups.</p> <blockquote> <p>Question: Can we write down a complete list of <strong>isomorphism classes</strong> (up to translation) of <strong>indecomposable objects</strong> of $\mathcal C$?</p> </blockquote> <p>It is easy to see that the number of such isomorphism classes is countably infinite.</p> <p>Here are some indecomposable objects: $$\cdots\to 0\to\mathbb Z\overset{m}{\to}\mathbb Z\to\mathbb Z_m\to 0\to\cdots,$$</p> <p>$$\cdots\to 0\to\mathbb Z_{(m,n)}\to\mathbb Z_n\overset{m}{\to}\mathbb Z_n\to\mathbb Z _{(m,n)}\to 0\to\cdots,$$ where $m$ is a natural number, $n$ is a prime power and $(m,n)$ denotes the greatest common divisor.</p> <p>But there is more; for instance, if $p$ is prime then the indecomposable object $$\cdots\to0\to\mathbb Z_p\to\mathbb Z_{p^2}\overset{p}{\to}\mathbb Z_{p^2}\overset{p}{\to}\cdots\overset{p}{\to}\mathbb Z_{p^2}\overset{p}{\to}\mathbb Z_{p^2}\to\mathbb Z_p\to 0\to\cdots$$ can have any finite "length."</p> <p>If this classification problem has been solved, a reference would be great. Otherwise I would very much appreciate any idea/hint towards a general solution.</p> <p>(I've added the noncommutative-algebra tag because chain complexes can be considered as modules over a certain non-commutative ring. The question I am asking is a sub-problem of classifying all finitely generated indecomposables for this ring.)</p> http://mathoverflow.net/questions/31427/proving-homotopy-invariance-of-cellular-homology-by-constructing-a-chain-homotopy Proving homotopy invariance of cellular homology by constructing a chain homotopy Rasmus 2010-07-11T16:35:24Z 2012-08-12T10:41:29Z <p>I'm trying to follow an argument in Lück's "Algebraische Topologie: Homologie und Mannigfaltigkeiten" (to which there apparently doesn't exist an english translation). The aim is to check homotopy invariance of cellular homology by constructing a chain homotopy.</p> <p>Let me sketch the argument. Let $h\colon (X,A)\times [0,1]\to (Y,B)$ be a cellular homotopy from $f_0$ to $f_1$. Using the CW-structure on $[0,1]$ with the two 0-cells {0}, {1} and one 1-cell, we identify $$C_n((X,A)\times [0,1])=C_n(X,A)\oplus C_n(X,A)\oplus C_{n-1}(X,A).$$ Then $C_n(h)$ is of the form $C_n(f_0)\oplus C_n(f_1)\oplus u_{n-1}$, where $u_{n-1}$ is some map $C_{n-1}(X,A)\to C_n(Y,B)$.</p> <p>Now we would like to compute the $n$th differential of $C_*((X,A)\times [0,1])$ under the above identification. Since it is a map from $C_n(X,A)\oplus C_n(X,A)\oplus C_{n-1}(X,A)$ to $C_{n-1}(X,A)\oplus C_{n-1}(X,A)\oplus C_{n-2}(X,A)$, we can denote it by a 3x3-matrix. My computation yielded $$\begin{pmatrix} c_n &amp; 0 &amp; 0 \newline 0 &amp; c_n &amp; 0 \newline -id &amp; id &amp; c_{n-1} \end{pmatrix},$$ where $c_n$ is the $n$th differential of $C_*(X,A)$. In the book, however, I find $$\begin{pmatrix} c_n &amp; 0 &amp; (-1)^n \newline 0 &amp; c_n &amp; (-1)^{n-1} \newline -id &amp; id &amp; c_{n-1} \end{pmatrix}.$$</p> <blockquote> <p><strong>Question:</strong> What is meant by $(-1)^n\colon C_{n}(X,A)\to C_{n-2}(X,A)$ and how can I understand that the second matrix above is the correct expression?</p> </blockquote> <hr> <p><strong>Edit:</strong> Apparently, the correct form of the the matrix representing the $n$th differential of $C_*((X,A)\times [0,1])$ is $$\begin{pmatrix} c_n &amp; 0 &amp; 0 \newline 0 &amp; c_n &amp; 0 \newline (-1)^{n+1}\cdot id &amp; (-1)^n\cdot id &amp; c_{n-1} \end{pmatrix}$$ or $$\begin{pmatrix} c_n &amp; 0 &amp; 0 \newline 0 &amp; c_n &amp; 0 \newline (-1)^{n}\cdot id &amp; (-1)^{n+1}\cdot id &amp; c_{n-1} \end{pmatrix},$$ depending on the orientation of the 1-cell in $[0,1]$.</p> http://mathoverflow.net/questions/87830/earliest-most-standard-reference-for-derived-categories-of-hereditary-algebras/102366#102366 Answer by Rasmus for Earliest/most standard reference for derived categories of hereditary algebras Rasmus 2012-07-16T16:33:41Z 2012-07-16T16:33:41Z <p>In Krause's <a href="http://www.math.uic.edu/~bshipley/krause.chicago.pdf" rel="nofollow"><em>Derived categories, resolutions, and Brown representability</em></a>, the general version of this result for hereditary abelian categories is proved (Section 1.6).</p> http://mathoverflow.net/questions/23269/non-examples-of-model-structures-that-fail-for-subtle-surprising-reasons/96230#96230 Answer by Rasmus for Non-examples of model structures, that fail for subtle/surprising reasons? Rasmus 2012-05-07T15:26:26Z 2012-05-08T15:03:25Z <p>By a result of Schochet, the category of C*-algebras with homotopy equivalences and "Schochet fibrations" is a <em>pointed category of fibrant objects</em> whose homotopy category is the ordinary homotopy category of C*-algebras.</p> <p>It was observed by Andersen and Grodal that the above pointed category of fibrant objects is not the full subcategory of fibrant objects of a Quillen model category.</p> <p>Here is a recent reference reviewing both results: <a href="http://arxiv.org/abs/1011.2926" rel="nofollow">O. Uuye: Homotopy Theory for C*-algebras</a></p> http://mathoverflow.net/questions/95164/the-derived-category-of-integral-representations-of-a-dynkin-quiver The derived category of integral representations of a Dynkin quiver. Rasmus 2012-04-25T15:12:25Z 2012-04-25T21:31:20Z <p>Let $Q$ be a Dynkin quiver. Let $\mathbb CQ$ be its complex path algebra. It is defined in a way such that modules over $\mathbb CQ$ are the same as representations of the quiver $Q$. Let's write $\mathrm{mod}(\mathbb CQ)$ for the finite-dimensional $\mathbb CQ$-modules. There is a beautiful description of the bounded derived category $D^b(\mathrm{mod}(\mathbb CQ))$ in terms of translation quivers and mesh relations, see <a href="http://www.springerlink.com/content/e7842v7117014614/" rel="nofollow">Happel: On the derived category of a finite-dimensional algebra</a>.</p> <p>I would be interested in a similar description of the bounded derived category $D^b(\mathrm{mod}(\mathbb ZQ))$ of finitely generated modules over the path <em>ring</em> $\mathbb ZQ$. More precisely, I am looking for a result along the lines "There is a nicely characterized full subcategory in $D^b(\mathrm{mod}(\mathbb ZQ))$ which, after tensoring with $\mathbb C$, becomes equivalent to $D^b(\mathrm{mod}(\mathbb CQ))$." Since Happel's paper is from 1987, I believe someone must have tried to generalise his results to integer representations since then.</p> http://mathoverflow.net/questions/35778/what-does-the-representation-theory-of-the-reduced-c-algebra-correspond-to What does the representation theory of the reduced C*-algebra correspond to? Rasmus 2010-08-16T16:38:07Z 2011-05-06T20:22:00Z <p>Let $G$ be a locally compact group. The group C*-algebra $C^* (G)$ is designed to come with a natural bijection between its (nondegenerate) representations and the (strongly continuous, unitary) representations of $G$.</p> <p><em>Question</em>: Is there a similar statement for the reduced group C*-algebra $C^*_r (G)$?</p> <p>If the answer is no, I'll probably end up asking for the actual purpose of defining $C^*_r (G)$. So far, I know that its isomorphic to $C^* (G)$ in important cases, and that its construction is in some sense simpler than the one of $C^* (G)$.</p> <p>(The definitions and the claims used above can be found in Blackadar's <em>Operator Algebras</em>.) </p> http://mathoverflow.net/questions/56163/what-is-the-group-of-additive-operations-on-topological-k-theory What is the group of additive operations on topological K-theory? Rasmus 2011-02-21T10:59:19Z 2011-03-01T16:08:35Z <p>Let us view topological K-theory as a functor $K$ from the cateory of compact pairs (that is, a compact Hausdorff spaces with a distinguished closed subset) to the category of $\mathbb Z/2$-graded Abelian groups. We could also restrict to second countable spaces and thus countable groups.</p> <p>An additive operation on topological K-theory is just a natural transformation from $K$ to $K$. These natural transformations form an Abelian group under addition.</p> <p><em>Question:</em> What is the isomorphism class of this group?</p> <p>Examples of operations are discussed in Efton Park's book and in Max Karoubi's book, but I cannot find discussion of the collection of all (additive) operations.</p> <p><em>Edit:</em> After looking at Karoubi's book again, I have to state that it actually contains a very satisfactory treatment of operations in K-theory.</p> http://mathoverflow.net/questions/56267/does-every-exact-six-term-sequence-arise-as-the-k-theory-of-a-locally-compact-pai Does every exact six-term sequence arise as the K-theory of a locally compact pair? Rasmus 2011-02-22T10:59:58Z 2011-02-22T10:59:58Z <p>Consider six countable Abelian groups and six group homomorphims as in the following diagram</p> <pre> G &rarr; H &rarr; I &uarr; &darr; L &larr; K &larr; J </pre> <p>Assume that the resulting sequence is exact at all six entries.</p> <p>Question: Is there a (second countable) locally compact Hausdorff space X with a closed subspace A, such that the resulting six-term sequence in K-theory</p> <pre> K<sup>0</sup>(X,A) &rarr; K<sup>0</sup>(X) &rarr; K<sup>0</sup>(A) &uarr; &darr; K<sup>1</sup>(A) &larr; K<sup>1</sup>(X) &larr; K<sup>1</sup>(X,A) </pre> <p>is isomorphic to the above one?</p> <p>An answer in the finitely generated case would also be interesting.</p> http://mathoverflow.net/questions/55954/what-is-the-spectrum-of-the-commutative-c-algebra-i-have-constructed-here What is the spectrum of the commutative C*-algebra I have constructed here? Rasmus 2011-02-19T04:02:16Z 2011-02-19T04:02:16Z <p>Let $B$ and $F$ be compact Hausdorff spaces.</p> <p>Let $E\to B$ be a fiber bundle with fibre $F$ and structure group $\mathrm{Homeo}(F)$, the group of homeomorphisms of $F$.</p> <p>I think this induces a fiber bundle $E'$ over $B$ with fiber $C(F,\mathbb C)$, the C*-algebra of continuous functions on $F$, with structure group $\mathrm{Aut}(C(F,\mathbb C))\cong\mathrm{Homeo}(F)$, the group of *-automorphisms of $C(F,\mathbb C)$.</p> <p>(To be more explicit at this point: my idea is: take a covering of $B$ which trivialises $E$. The transition functions give me a cocycle with values in the structure group $\mathrm{Homeo}(F)$. But, since $\mathrm{Homeo}(F)\cong\mathrm{Aut}(C(F,\mathbb C))$, I get a cocycle with values in $\mathrm{Aut}(C(F,\mathbb C))$, which I'd like to use to glue my new bundle $E'$.)</p> <p>Let $\Gamma(B,E')$ denote the continuous sections of $E'$. I think pointwise operations turn this into a C*-algebra. Since the fiber $C(F,\mathbb C)$ is commutative, $\Gamma(B,E')$ is commutative as well.</p> <p><em>Question:</em> What is the spectrum of $\Gamma(B,E')$?</p> <p>Example: If $E\cong B\times F$ is the trivial bundle, then $E'\cong B\times C(F,\mathbb C)$ and thus $$\Gamma(B,E')\cong C(B,C(F,\mathbb C))\cong C(B\times F,\mathbb C).$$ This suggests that the spectrum of $\Gamma(B,E')$ is actually $E$.</p> <p>(This question remained unanswered up to now at <a href="http://math.stackexchange.com/questions/22385" rel="nofollow">math.se</a>.)</p> http://mathoverflow.net/questions/52032/examples-of-non-metrizable-spaces/52050#52050 Answer by Rasmus for Examples of non-metrizable spaces Rasmus 2011-01-14T07:37:32Z 2011-01-14T07:37:32Z <p>Every <a href="http://en.wikipedia.org/wiki/Finite_topological_space" rel="nofollow">finite topological space</a> which is not discrete is not Hausdorff and hence not metrizable. Yet, there are lots of finite topological spaces: they are in one-to-one correspondence with finite preordered sets.</p> http://mathoverflow.net/questions/43069/is-there-a-list-of-all-connected-t-0-spaces-with-5-points Is there a list of all connected T_0-spaces with 5 points? Rasmus 2010-10-21T16:57:41Z 2010-10-21T17:27:34Z <p>Is there some place (on the internet or elsewhere) where I can find the number and preferably a list of all (isomorphism classes of) finite connected $T_0$-spaces with, say, 5 points?</p> <p>In know that a $T_0$-topology on a finite set is equivalent to a partial ordering, and <a href="http://en.wikipedia.org/wiki/Partial_order" rel="nofollow">wikipedia</a> tells me that there are, up to isomorphism, 63 partially ordered sets with precisely 5 elements. However, I am only interested in connected spaces, and I'd love to have a list (most preferably in terms of Hasse diagrams).</p> http://mathoverflow.net/questions/37417/characterisation-of-positive-elements-in-lz Characterisation of positive elements in l¹(Z) Rasmus 2010-09-01T20:19:27Z 2010-09-02T19:14:05Z <p>Consider the Banach $^*$-algebra $\ell^1(\mathbb Z)$ with multiplication given by convolution and involution given by $a^*(n)=\overline{a(-n)}$.</p> <p>I would like to find nice necessary and sufficient conditions for an element $b\in\ell^1(\mathbb Z)$ to be positive, that is, to be of the form $a^* * a$ for some $a\in\ell^1(\mathbb Z)$.</p> <p>By now, I have found two necessary conditions. Namely, if $b\in\ell^1(\mathbb Z)$ is positive, then $$b(-n)=\overline{b(n)}$$ and $$\lvert b(n)\rvert\leq b(0)$$ for every $n\in\mathbb Z$.</p> <p><strong>Edit:</strong> As t3suji states in his comment below both conditions follow from the more general fact that $a$ is a <a href="http://en.wikipedia.org/wiki/Positive_definite_function" rel="nofollow">positive-definite function</a>.</p> <blockquote> <p>Question: Is this condition also sufficient for positivity? If not, what to I have to add?</p> </blockquote> <p>Good references would also be great.</p> <p>Motivation: In the end I want to investigate the (failure of) the Gelfandâ€“Naimark theorem for the above non-C*-algebra.</p> http://mathoverflow.net/questions/37182/representation-of-automorphism-on-finite-dimensional-matrix-algebras/37185#37185 Answer by Rasmus for Representation of $*$-automorphism on finite dimensional matrix algebras Rasmus 2010-08-30T18:38:38Z 2010-08-30T18:47:05Z <p>As an alternative to Robin Chapman's solution, I would like to state Exercise 7.8 from Rørdam's, Larsen's and Laustsen's "Introduction to the K-theory of C*-algebras":</p> <blockquote> <p>For every unital AF-algebra $A$ there is a short exact sequence $$1\to\overline{\mathrm{Inn}}(A)\to\mathrm{Aut}(A)\to\mathrm{Aut}(K_0(A))\to 1,$$ where $\overline{\mathrm{Inn}}(A)$ denotes <em>approximately</em> inner automorphisms and $\mathrm{Aut}(K_0(A))$ denotes group automorphisms preserving the unit class and the positive cone in $K_0(A)$.</p> </blockquote> <p>If $A$ is the matrix ring, then $\mathrm{Aut}(K_0(A))$ is trivial and hence every automorphism of $A$ is approximately inner. Since $A$ is separable, every approximately inner automorphism is the pointwise limit of a sequence of inner automorphisms. And I think the finite-dimensionality of $A$ implies that the pointwise limit of a sequence of inner automorphisms is again inner.</p> <p>Using the statement above, one immediately sees that, for instance, $\mathbb C\oplus\mathbb C$ possesses an automorphism which is not approximately inner.</p> http://mathoverflow.net/questions/3344/sources-for-exact-triangles-in-triangulated-categories Sources for exact triangles in triangulated categories. Rasmus 2009-10-29T21:45:30Z 2010-08-24T15:34:01Z <p>The other day I came across the statement that in the triangulated category $\mathfrak{KK}$ (of C*-algebras with KK-groups as morphism sets) "there are many other sources of exact triangles besides extensions". Except for mapping cone triangles I don't know what is meant. What can you come up with?</p> <p>I would also appreciate answers focussing on triangulated categories in general.</p> http://mathoverflow.net/questions/18938/triangulations-coming-from-a-poset-or-what-conditions-are-necessary-and-suffici Triangulations coming from a poset. Or: What conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset? Rasmus 2010-03-21T16:57:46Z 2010-08-24T10:50:23Z <p>Every partially ordered set gives a triangulation of (the geometric realisation of) its order complex. (The n-simplices of the order complex are the chains $x_0\leq x_1\leq\cdots\leq x_n$.) However, there are triangulations of topological spaces that do not arise this way.</p> <p>Is there a name for triangulations having this special property of "coming from a poset?"</p> <p>EDIT: Apparently, the following formulation of my question is cleaner: <em>what conditions are necessary and sufficient for a finite simplicial complex to be the order complex of a poset?</em></p> http://mathoverflow.net/questions/33389/justification-for-the-matching-condition-for-the-wave-function-at-potential-jumps Justification for the matching condition for the wave function at potential jumps. Why is it both restrictive enough and sufficiently general? Rasmus 2010-07-26T12:43:31Z 2010-08-19T19:17:53Z <p>Consider Schrödinger's <em>time-independent</em> equation $$-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi=E\psi.$$ In typical examples, the potential $V(x)$ has discontinuities, called <em>potential jumps</em>.</p> <p>Outside these discontinuities of the potential, the wave function is required to be twice differentiable in order to solve Schrödinger's equation.</p> <p>In order to control what happens at the discontinuities of $V$ the following assumption seems to be standard (see, for instance, Keith Hannabus' <em>An Introduction to Quantum Theory</em>):</p> <blockquote> <p><strong>Assumption</strong>: The wave function and its derivative are continuous at a potential jump.</p> </blockquote> <p><strong>Questions</strong>:</p> <p>1) Why is it necessary for a (physically meaningful) solution to fulfill this condition?</p> <p>2) Why is it, on the other hand, okay to abandon twofold differentiability?</p> <p>Edit: One thing that just became clear to me is that the above assumption garanties for a well-defined probability/particle current.</p> http://mathoverflow.net/questions/35507/what-is-the-commutative-analogue-of-a-c-subalgebra What is the commutative analogue of a C*-subalgebra? Rasmus 2010-08-13T16:49:48Z 2010-08-17T10:22:24Z <p>Using the duality between locally compact Hausdorff spaces and commutative $C^*$-algebras one can write down a vocabulary list translating topological notions regarding a locally compact Hausdorff space $X$ into algebraic notions regarding its ring of functions $C_0(X)$ (see Wegge-Olsen's book, for instance). For example, we have the following correspondences: $$\;\;\;\text{open subset of X}\quad \longleftrightarrow\quad\text{ideal in C_0(X)}$$ $$\;\;\;\;\;\quad\text{dense open subset of X}\quad \longleftrightarrow\quad\text{essential ideal in C_0(X)}$$ $$\;\;\;\quad\text{closed subset of X}\quad \longleftrightarrow\quad\text{quotient of C_0(X)}$$ $$\text{locally closed subset of X}\quad \longleftrightarrow\quad\text{subquotient of C_0(X)}$$ $$\;\;\;\quad\qquad\qquad\qquad\qquad\text{???}\qquad\qquad \longleftrightarrow\quad\text{C^*-subalgebra in C_0(X)}$$ By ideal I always mean a two-sided closed (and hence self-adjoint) ideal.</p> <p>Well, I can't quite see <strong>how to reconvert a $C^*$-subalgebra in $C_0(X)$ into something topological involving only the space $X$ (and some data describing the subalgebra in topological terms).</strong> Can you come up with something handy?</p> <hr> <p><em>Example:</em> A simple example of a subalgebra of a commutative $C^*$-algebra not being an ideal is $$\mathbb C\cdot(1,1)\subset \mathbb C\oplus\mathbb C.$$</p> <hr> <p><em>First attempts:</em> Instead of talking about a subalgebra, we should probably talk about the injective $^*$-homomorphism given by the inclusion of this subalgebra. But is this inclusion proper (i.e., does it preserve approximate units) in general? Well, at least when we restrict to compact spaces. Then an injective $^*$-homomorphism $C(Y)\to C(X)$ will induce a surjective continuous map $X\to Y$. How to proceed?</p> <hr> <p><em>Remark:</em> Alternatively, we could think about this question within the duality of affine algebraic varieties and finitely generated commutative reduced algebras or even within the duality between affine schemes and commutative rings.</p> <hr> <p><strong>Disclaimer:</strong> I posted this question yesterday on <a href="http://math.stackexchange.com/questions/2229/what-is-the-commutative-analogue-of-a-c-subalgebra" rel="nofollow">MSE</a>. I also got an interesting answer. However, I'm not yet fully satisfied. If I violate any policy by reposting the question here, please tell me about it.</p> http://mathoverflow.net/questions/24683/relative-k-theory-and-split-exact-sequences-of-c-algebras/33126#33126 Answer by Rasmus for Relative K-theory and split exact sequences of C* algebras Rasmus 2010-07-23T19:04:21Z 2010-07-23T19:04:21Z <p>The unitalization case $A=J^+$ is treated in Blackadar's <em>$K$-theory for Operator Algebras</em>. In Proposition 5.4.1 he <em>directly</em> showes that $$K_0(J^+,J)\cong\ker(K_0(J^+)\to K_0({J^+}/J)).$$</p> http://mathoverflow.net/questions/31714/is-tor-always-torsion Is Tor always torsion? Rasmus 2010-07-13T15:15:46Z 2010-07-13T16:12:47Z <p><strong>Question:</strong> Is the following statement true?</p> <blockquote> <p>Let $R$ be an associative, commutative, unital ring. Let $M$ and $N$ be $R$-modules. Let $n\geq 1$. Then $Tor_n^R(M,N)$ is torsion.</p> </blockquote> <p>By " $Tor_n^R(M,N)$ is torsion" I mean that every of its elements is a torsion element. Maybe I want to assume that $R$ is an integral domain.</p> <p>Remark: The above statement is true if $R$ is a principal ideal domain (then $Tor_n^R$ vanishes for $n\geq 2$) and $M$ and $N$ are finitely generated (then we can apply the <a href="http://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain" rel="nofollow">structure theorem</a>).</p> http://mathoverflow.net/questions/30911/is-hartshornes-definition-of-the-category-of-varieties-natural Is Hartshorne's definition of the category of varieties natural? Rasmus 2010-07-07T16:10:37Z 2010-07-07T18:00:08Z <p>I'm an absolute beginner in algebraic geometry so please excuse if the following question is somewhat superficial.</p> <p>Hartshorne's "Algebraic geometry" begins with the definition of (quasi-)affine and (quasi-)projective varieties over some fixed algebraically closed field. At a first glance, these seem to be quite different, so that I would have expected that one would pose questions <em>either</em> on quasi-affine <em>or</em> on quasi-projective varieties.</p> <p>However, Hartshorne then defines a <em>variety</em> to be either a quasi-affine or a quasi-projective variety. These varieties (together with certain continuous and in some sense regular maps) then form the category of varieties.</p> <blockquote> <p>Here is my question: Is the above definition natural in the sense that we really want to compare quasi-affine and quasi-projective varieties or at least study them both at the same time?</p> </blockquote> <p>For instance, is there a (non-trivial) example of a quasi-affine variety which is isomorphic in the above category to a quasi-projective variety? If not, isn't this "unifying" definition a bit artificial?</p> http://mathoverflow.net/questions/30664/uniform-convergence-of-difference-quotient Uniform convergence of difference quotient Rasmus 2010-07-05T20:14:53Z 2010-07-05T20:48:25Z <p>Let $\phi\in C^\infty_c(\mathbb R)$ be a smooth function with <em>compact support</em>.</p> <p>For $h>0$ define the difference quotient $\phi_h\in C^\infty_c(\mathbb R)$ by $\phi_h(t)=\dfrac{\phi(t+h)-\phi(t)}{h}$.</p> <p>By definition, for fixed $t\in\mathbb R$, we have $\phi_h(t)\to\phi'(t)$ as $h\to 0$.</p> <blockquote> <p>Question: Can we conclude, that $\phi_h\to\phi'$ uniformly on $\mathbb R$ (as $h\to 0)$?</p> </blockquote> <p>Motivation: This is used in a proof of Stone's Theorem on the existence of generators of operator groups I'm trying to understand.</p> http://mathoverflow.net/questions/21899/definition-of-an-algebra-over-a-noncommutative-ring Definition of an algebra over a noncommutative ring Rasmus 2010-04-19T21:54:26Z 2010-04-21T17:59:25Z <p>I've tried in vain to find a definition of an algebra over a <em>noncommutative</em> ring. Does this algebraic structure not exist? In particular, does the following definition from <a href="http://en.wikipedia.org/wiki/Algebra_(ring_theory)" rel="nofollow">http://en.wikipedia.org/wiki/Algebra_(ring_theory)</a> make sense for noncommutative $R$?</p> <blockquote> <p>Let $R$ be a commutative ring. An algebra is an $R$-module $A$ together with a binary operation $$[\cdot,\cdot]: A\times A\to A$$ called $A$-multiplication, which satisfies the following axiom: $$[a x + b y, z] = a [x, z] + b [y, z], \quad [z, a x + b y] = a[z, x] + b [z, y]$$ for all scalars $a$, $b$ in $R$ and all elements $x$, $y$, $z$ in $A$.</p> </blockquote> <p>So, is there a common notion of an algebra over a noncommutative ring?</p> http://mathoverflow.net/questions/10480/gelfand-duality-in-ncg/12979#12979 Answer by Rasmus for Gelfand duality in NCG Rasmus 2010-01-25T22:11:58Z 2010-04-20T06:56:26Z <p>Concerning your last question, I would say you should view your C*-algebra itself as the (coordinate ring on the) "non-commutative topological space." The spaces you suggest are commutative. Your C*-algebra might be considered as a non-commutative substitute of the coordinate ring on one of them.</p> http://mathoverflow.net/questions/20314/good-books-on-theory-of-distributions/20491#20491 Answer by Rasmus for Good books on theory of distributions Rasmus 2010-04-06T11:43:22Z 2010-04-06T11:43:22Z <p>Gel'fand, I. M. and Shilov, G. E.: <em>Generalized Functions</em></p> http://mathoverflow.net/questions/19924/why-is-it-called-spectral-triple Why is it called *spectral* triple? Rasmus 2010-03-31T09:04:14Z 2010-03-31T09:32:50Z <p>I know the definition a spectral triple and that it is some kind of non-commutative generalisation of (the ring of functions on) a compact spin manifold.</p> <p>But, why is it called <em>spectral</em> triple?</p> http://mathoverflow.net/questions/19215/ambiguous-definition-of-nerve-of-an-open-covering-on-wikipedia Ambiguous definition of "nerve of an open covering" on wikipedia? Rasmus 2010-03-24T17:26:16Z 2010-03-24T17:49:06Z <p>Let $(U_i)_{i\in I}$ be an open covering of a topological space $X$.</p> <p>At <a href="http://en.wikipedia.org/wiki/Nerve_of_an_open_covering" rel="nofollow">http://en.wikipedia.org/wiki/Nerve_of_an_open_covering</a>, the nerve of the open covering is defined as follows:</p> <blockquote> <p>the nerve $N$ is the set of finite subsets of $I$ defined as follows:</p> <ul> <li>the empty set belongs to $N$;</li> <li>a finite set $J\subset I$ belongs to $N$ if and only if the intersection of the $U_i$ whose subindices are in $J$ is non-empty.</li> </ul> </blockquote> <p>On the other hand, <a href="http://en.wikipedia.org/wiki/Nerve_(category_theory)" rel="nofollow">http://en.wikipedia.org/wiki/Nerve_(category_theory)</a> states:</p> <blockquote> <p>If $X$ is a topological space with open cover $U_i$, the nerve of the cover is obtained from the above definitions by replacing the cover with the category obtained by regarding the cover as a partially ordered set with relation that of set inclusion.</p> </blockquote> <p>Here, "the above definitions" refers to the usual construction of the nerve of a category: A vertex for each object, and a $k$-simplex for each $k$-tuple of composable morphisms.</p> <p>My question is: Does this categorical construction really yield the previously defined nerve of the open covering?</p> <p>For instance, cover the inverval by two intersecting invervals non of them containing the other one. Then it seems to me that the first construction yields two vertices connected by an edge, while the second construction yields to bare vertices.</p> <p>What am i missing?</p> <p>If the second definition is indeed wrong, what is the right way to obtain the nerve of an open covering as a special case of the nerve of a category?</p> http://mathoverflow.net/questions/127971/can-nuclearity-be-determined-by-tensoring-with-a-single-c-algebra/128073#128073 Comment by Rasmus Rasmus 2013-04-19T13:05:04Z 2013-04-19T13:05:04Z In the second sentence, you might want to restrict to one representant from each isomorphism class, so that you get a set. http://mathoverflow.net/questions/95164/the-derived-category-of-integral-representations-of-a-dynkin-quiver Comment by Rasmus Rasmus 2013-03-23T07:55:46Z 2013-03-23T07:55:46Z @Benjamin: I think you are right. I was probably aiming for something like a full subcategory $\mathcal C\subset D^b(\mathrm{mod}(\mathbb ZQ))$ such that the obvious functor from $\mathcal C\otimes\mathbb C$ to $D^b(\mathrm{mod}(\mathbb CQ))$ is not only an equivalence but also induces a bijection of isomorphism classes. For $Q=\bullet$, I suppose we could choose the subclass of complexes with at most one non-vanishing entry which is free. http://mathoverflow.net/questions/123283/morphisms-between-k-0/123313#123313 Comment by Rasmus Rasmus 2013-03-01T10:54:38Z 2013-03-01T10:54:38Z One might add that this works because every element in $K_0(A)$ is positive if $A$ is purely infinite simple, so that the obstruction used in Fernando's first counterexample disappears. http://mathoverflow.net/questions/122000/subgroups-of-mathbbzn Comment by Rasmus Rasmus 2013-02-23T13:45:01Z 2013-02-23T13:45:01Z @Aaron: Another instance where the Smith normal form is used and which I think is interesting to you is the computation of the K-theory of a Cuntz--Krieger algebra. There, you have to compute the kernel and cokernel of some integer matrix. The result can be read off from the Smith normal form of the matrix. http://mathoverflow.net/questions/104827/classification-of-long-exact-sequences/104862#104862 Comment by Rasmus Rasmus 2013-02-13T12:17:14Z 2013-02-13T12:17:14Z As it turns out, this is not so hard and $\mathbb Z/8$-modules are enough: <a href="http://mathoverflow.net/questions/121606/" rel="nofollow">mathoverflow.net/questions/121606</a> http://mathoverflow.net/questions/17468/derivators-in-english/17502#17502 Comment by Rasmus Rasmus 2013-02-07T16:16:35Z 2013-02-07T16:16:35Z Irakli Patchkoria's preprint: <a href="http://arxiv.org/abs/1108.6309" rel="nofollow">arxiv.org/abs/1108.6309</a> http://mathoverflow.net/questions/120338/general-recipe-for-building-c-algebras-out-of-combinatorial-object/120347#120347 Comment by Rasmus Rasmus 2013-01-30T20:21:08Z 2013-01-30T20:21:08Z @Mariano: The book &quot;Graph Algebras&quot; by Iain Raeburn. http://mathoverflow.net/questions/116060/inductive-vs-projective-limit-of-sequence-of-split-surjections/116068#116068 Comment by Rasmus Rasmus 2012-12-12T07:23:45Z 2012-12-12T07:23:45Z Thank you very much for your enlightening reply. http://mathoverflow.net/questions/1890/describe-a-topic-in-one-sentence/5784#5784 Comment by Rasmus Rasmus 2012-10-07T18:57:19Z 2012-10-07T18:57:19Z I don't understand. Could you please explain? http://mathoverflow.net/questions/107101/is-the-category-of-vector-bundles-over-a-topological-space-abelian Comment by Rasmus Rasmus 2012-09-14T08:04:14Z 2012-09-14T08:04:14Z Intuitively, the problem is that taking fiberwise kernels or cokernels will not, in general, give you vector bundles because the rank can jump. http://mathoverflow.net/questions/106333/maps-induced-on-cohomology-and-change-of-coefficients/106344#106344 Comment by Rasmus Rasmus 2012-09-09T18:45:25Z 2012-09-09T18:45:25Z @Judy: You can post replies to Ralph's answer as comments below his answer. Just click &quot;add comment&quot;. What you posted is formally a second answer. http://mathoverflow.net/questions/104827/classification-of-long-exact-sequences/104862#104862 Comment by Rasmus Rasmus 2012-08-16T20:04:08Z 2012-08-16T20:04:08Z @tweetie-bird: In arxiv.org/abs/math/0409417 Schmidmeier and Ringel show that the category of finitely generated $\mathbb Z/p^n$-submodule inclusions is <i>controlled $\mathbb Z/p$-wild.</i> This seems to mean roughly that the classification of its objects is at least as complication as the classification of finitely generated modules over the free $\mathbb Z/p$-algebra on two generators. http://mathoverflow.net/questions/104827/classification-of-long-exact-sequences Comment by Rasmus Rasmus 2012-08-16T18:30:55Z 2012-08-16T18:30:55Z @algori, Florian Eisele: Thank you for your help in figuring out what $?$ has to be! http://mathoverflow.net/questions/104827/classification-of-long-exact-sequences Comment by Rasmus Rasmus 2012-08-16T17:55:52Z 2012-08-16T17:55:52Z @algori: But for $m=1$ this would give me $n$, though it should give $1$. http://mathoverflow.net/questions/104827/classification-of-long-exact-sequences Comment by Rasmus Rasmus 2012-08-16T16:40:29Z 2012-08-16T16:40:29Z @Dustin Cartwright: You are right. Thank you for catching this. I will correct the mistake.