User ralph - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T20:31:43Z http://mathoverflow.net/feeds/user/10194 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/132037/group-actions-with-finite-stabilizers-and-compact-quotients Group actions with finite stabilizers and compact quotients Ralph 2013-05-27T19:32:36Z 2013-06-17T15:05:29Z <p>Let $G$ be a discrete group that acts on a contractible finite dimensional $G$-complex $X$ with the following properties: </p> <ul> <li>$X/G$ is compact (i.e. the action is <em>cocompact</em>) </li> <li>Each stabilizer $G_\sigma$ admits a cocompact action on a contractible finite dimensional $G_\sigma$-complex with finite stabilizers </li> </ul> <p><strong>Question:</strong> Is there a finite dimensional contractible $G$-complex $Y$ with <em>finite</em> stabilizers such that $Y/G$ is compact ? </p> <p>The question can be stated more conceptually by help of the following definition: Let $\mathscr{F}$ be the class of all finite groups and define the class $K_i\mathscr{F},\;i\ge 0$ inductively by </p> <ul> <li>$K_0\mathscr{F} := \mathscr{F}$ </li> <li>$K_i\mathscr{F}$ includes all groups $G$ that admit a finite dimensional contractible $G$-complex $X$ such that (1) $X/G$ is compact and (2) the stabilizers are in $K_{i-1}\mathscr{F}$. </li> </ul> <p>Then the question is equivalent to </p> <p><strong>Question:</strong> Is $K_1\mathscr{F} \subsetneqq K_2\mathscr{F}$ ? </p> <p>Remark: If condition (1) is dropped, we get the classical Kropholler classes $H_i\mathscr{F}$. There (among many other results) $H_1\mathscr{F} \subsetneqq H_2\mathscr{F}$ is known: For example, the free abelian group of countably infinite rank belongs to $H_2\mathscr{F}\setminus H_1\mathscr{F}$. </p> http://mathoverflow.net/questions/132111/transgression-in-terms-of-cup-product-in-case-of-non-trivial-action-of-the-group/132450#132450 Answer by Ralph for transgression in terms of cup product in case of non-trivial action of the group on the coeffecients module Ralph 2013-05-31T18:09:20Z 2013-06-01T12:06:50Z <p>I can offer the following generalization for non-trivial coefficients $A$: Write $Q = G/H$. The cap product $$H_1(H;\mathbb{Z}) \otimes H^1(H;A) \to \mathbb{Z}\otimes_H A = A_H$$ is $G$-linear with trivial $H$-action and induces a cup product $$\cup: H^p(Q;H_1(H;\mathbb{Z})) \otimes H^q(Q;H^1(H;A))\xrightarrow{} H^{p+q}(Q;A_H).$$ Let $u \in H^2(Q;H_1(H;\mathbb{Z}))$ be the class that corresponds to the extension $$1 \to H^{ab} \to G/H' \to Q \to 1$$ and let $\kappa: A^H \to A_H$ be the canonical homomorphism. </p> <blockquote> <p><strong>Theorem:</strong> The compostion $$H^p(Q;H^1(H;A)) \xrightarrow{d_2}H^{p+2}(Q;A^H)\xrightarrow{\kappa^\ast}H^{p+2}(Q;A_H)$$ is (up to sign) cup product with $u$, i.e. $(\kappa^\ast \circ d_2^{p,1})(x) = - u \cup x$. </p> </blockquote> <p><em>Proof:</em> By abuse of notation let $\kappa: A \to A_H$. Since $\kappa$ is $G$-linear, it induces a map of spectral sequences $\kappa_r^{pq}: E_r^{pq}(A) \to E_r^{pq}(A_H)$. In particular, $\kappa_2^{p+2,0}\circ d_2^{p,1}(A)=d_2^{p,1}(A_H)\circ \kappa_2^{p,1}$. Clearly, $\kappa_2^{p+2,0}$ is the map $\kappa^\ast$ in the theorem and since $H$ acts trivially on $A_H$, we obtain $(\kappa^\ast \circ d_2^{p,1}(A))(x) = d_2^{p,1}(A_H)(\kappa_2^{p,1}x)=-u \cup \kappa_2^{p,1}x$. Hence, it remains to show $u \cup x = u \cup \kappa_2^{p,1}x$ (note the cup products are w.r.t. different pairings). But this follows immediately by applying $H^\ast(Q;-)$ to the following commutative diagram of pairings: $$\begin{array}{ccc} H_1(H;\mathbb{Z}) \otimes H^1(H;A) \;\; &amp; \xrightarrow{\cap} &amp; A_H \newline \scriptstyle id\otimes \kappa^\ast \displaystyle\downarrow\qquad\quad &amp; &amp; \downarrow\scriptstyle id \newline<br> H_1(H;\mathbb{Z}) \otimes H^1(H;A_H) &amp; \xrightarrow{\cap} &amp; A_H. \end{array}$$</p> http://mathoverflow.net/questions/132281/is-there-an-a-such-that-b-injective-iff-1st-ext-functor-vanishes/132282#132282 Answer by Ralph for Is there an $A$ such that $B$ injective iff 1st Ext functor vanishes? Ralph 2013-05-29T21:56:07Z 2013-05-29T22:07:26Z <p>Yes, such a module exists for each unital ring $R$: Take $A= \bigoplus_I R/I$ where $I$ runs through the left ideals of $R$. This is because $B$ is injective iff $\text{Ext}^1_R(R/I,B)=0$ for each $I$ (Weibel, Lemma 4.1.11). Also note that this is a straightforward generalization of your example $R=\mathbb{Z}$. </p> <p><strong>Added:</strong> If $R$ is a Noetherian commutative ring, you can also take $A=\bigoplus_P R/P$ where $P$ runs through the prime ideals of $R$ (Bruns, Herzog: Cohen-Macaulay rings, 3.1.12). </p> http://mathoverflow.net/questions/132073/homomorphisms-from-powers-of-z-to-z/132083#132083 Answer by Ralph for Homomorphisms from powers of Z to Z Ralph 2013-05-28T07:09:11Z 2013-05-29T00:38:05Z <p>This runs under the name Łoś–Eda Theorem. A reference is the book Paul C. Eklof, Alan H. Mekler: Almost free modules (2002): </p> <p>Call a set $I$ $\omega$-measurable, if its cardinality is greater or equal to the first measurable cardinal. This is equivalent to $I$ being uncountable and supporting a non-principal countably complete ultrafilter. </p> <p>First note that $\mathbb{Z}$ is slender (Cor. III.2.4). Then, by Cor. III.3.6 (and the discussion before Lemma III.3.5), if $I$ is not $\omega$-measurable, the natural map $$\phi: \bigoplus_{i \in I} Hom(\mathbb{Z},\mathbb{Z}) \to Hom(\prod_{i \in I}\mathbb{Z},\mathbb{Z}),\; (g_i)_i \mapsto \big(\; (m_i)_i \mapsto \sum_i g_i(m_i)\;\big)$$ is an isomorphism. </p> <p><strong>Remarks:</strong> 1) If $I$ is $\omega$-measurable, not all homomorphisms $\prod_I \mathbb{Z} \to \mathbb{Z}$ factor through a finite subset of $I$. For, let $D$ be a non-principal countably complete ultrafilter on $I$ and let $K_D = \lbrace x \in \prod_I \mathbb{Z} \mid I \setminus \sup(x) \in D\rbrace$. Then it's not hard to show that the composition $\prod_I \mathbb{Z} \twoheadrightarrow \prod_I \mathbb{Z}/K_D \cong \mathbb{Z}$ doesn't factor through a finite subset of $I$ (the latter isomorphism uses II.3.3). </p> <p>2) Irrespective whether $I$ is $\omega$-measurable or not, there is a canonical isomorphism $$Hom(\prod_{i \in I}\mathbb{Z},\mathbb{Z}) \cong \bigoplus_D Hom(\mathbb{Z},\mathbb{Z})$$ where $D$ runs through all countably complete ultrafilters on $I$ (Cor. III.3.7). </p> http://mathoverflow.net/questions/130802/a-group-3-cocycle-trivial-on-a-pair-of-generating-subgroups/130810#130810 Answer by Ralph for A group 3-cocycle, trivial on a pair of generating subgroups? Ralph 2013-05-16T10:15:11Z 2013-05-16T10:15:11Z <p>If $\mathbb{k}^\times$ is the multiplicative group of some field, the following works with $\mathbb{k}=\mathbb{F}_3$: </p> <p>The Quaternion group $Q_8$ is generated by two cyclic subgroups $H,K$ of order 4 and $$H^\ast(H;\mathbb{F}_2)\cong H^\ast(K;\mathbb{F}_2)\cong \mathbb{F}_2[a,b]/(a^2)\; ,\quad|a|=1, |b|=2$$ $$H^\ast(Q_8;\mathbb{F}_2) = \mathbb{F}_2[x,y,z]/(xy,x^3-y^3)\;,\quad |x|=|y|=1,|z|=4$$ Thus $x$ restricts on $H,K$ to some class $c$ having $c^2=0$ and consequently $x^3\neq 0$ restricts to zero on $H,K$. </p> http://mathoverflow.net/questions/129143/verifying-the-correctness-of-a-sudoku-solution Verifying the correctness of a Sudoku solution Ralph 2013-04-29T19:15:11Z 2013-05-07T16:21:02Z <p>A Sudoku is solved correctly, if all columns, all rows and all 9 subsquares are filled with the numbers 1 to 9 without repetition. Hence, in order to verify if a (correct) solution is correct, one has to check by definition 27 arrays of length 9. </p> <p><strong>Q1:</strong> Are there verification strategies that reduce this number of checks ?</p> <p><strong>Q2:</strong> What is the minimal number of checks that verfify the correctness of a (correct) solution ? </p> <p><img src="http://imageshack.us/a/img542/1809/sudoku1.jpg"/> <img src="http://imageshack.us/a/img607/9485/sudoku2.jpg"/></p> <p>The following simple observation yields an improved verfication algorithm: At first enumerate rows, columns and subsquares as indicated in pic 2. Suppose the columns $c_1,c_2,c_3$ and the subsquares $s_1, s_4$ are correct (i.e. contain exactly the numbers 1 to 9). Then it's easy to see that $s_7$ is correct as well. This shows: </p> <p>(A1) If all columns, all rows and 4 subsquares are correct, then the solution is correct. </p> <p>Now suppose all columns and all rows up to $r_9$ and the subsquares $s_1,s_2,s_4,s_5$ are correct. By the consideration above, $s_7,s_8,s_9$ and $s_3,s_6$ are correct. Moreover, $r_9$ has to be correct, too. For, suppose a number, say 1, occurs twice in $r_9$. Since the subsquares are correct, the two 1's have be in different subsquares, say $s_7,s_8$. Hence the 1's from rows $r_7, r_8$ both have to lie in $s_9$, i.e. $s_9$ isn't correct. This is the desired contradiction. </p> <p>Hence (A1) can be further improved to </p> <p>(A2) If all columns and all rows up to one and 4 subsquares are correct, then the solution is correct. </p> <p>This gives as upper bound for <strong>Q2</strong> the need of checking 21 arrays of length 9. </p> <p><strong>Q3:</strong> Can the handy algorithm (A2) be further improved ? </p> http://mathoverflow.net/questions/127769/applications-of-govorov-lazard-theorem Applications of Govorov-Lazard Theorem? Ralph 2013-04-16T23:16:48Z 2013-05-05T08:08:22Z <p>I asked <a href="http://math.stackexchange.com/questions/201705/applications-of-govorov-lazard-theorem" rel="nofollow">this question on SE</a> a long time ago, but never received an answer: </p> <p>The Govorov-Lazard Theorem states that a (left) module over an unital ring is flat iff it is a direct limit of finitely generated free (left) modules. </p> <p>The theorem is contained in many textbooks like Eisenbud (Commutative Algebra) or Rotman (Introduction to Homological Algebra). However, no applications are given there. </p> <p>Question: Are there interesting applications of the Govorov-Lazard Theorem ? </p> <p>N.b.: The only application I've seen so far, was in a question on SE, where someone remarked that if $A,B$ are commutative $R$-algebras and $M$ is a flat $A$-module and $N$ a flat $B$-module, then it follows from Govorov-Lazard that $M\otimes_R N$ is a flat $A\otimes_R B$-module. But, of course, this follows more easily from standard properties of the tensor product. </p> http://mathoverflow.net/questions/129399/representations-over-mathbbz-p/129427#129427 Answer by Ralph for Representations over $\mathbb{Z}_p$ Ralph 2013-05-02T14:07:36Z 2013-05-02T14:24:01Z <p>Concerning 2: Yes, the decomposition is unique, since the Krull-Schmidt theorem applies to finitely generated modules over $\mathbb{Z}_pG$. This follows by a theorem of Swan (Induced Representations and Projectives. Ann. of Math. 71(1960), 552-578. Remark after Prop. 6.1): </p> <blockquote> <p>Let $R$ be a commutative complete local ring and let $A$ be an $R$-algebra that is finitely generated as $R$-module. Then the Krull-Schmidt theorem holds for finitely generated $A$-modules. </p> </blockquote> <p>An alternative reference is Curtis, Reiner: Representation Theory of Finite Groups and Associative Algebras. Theorem 76.26. </p> http://mathoverflow.net/questions/128768/truncation-of-bg/128774#128774 Answer by Ralph for Truncation of BG? Ralph 2013-04-25T22:58:46Z 2013-04-26T00:10:28Z <p>There are several functorial models for $BG$, see for example [Adem, Milgram: Cohomology of Finite Groups, Chapter II] where $$BG = \coprod_{i=0}^\infty \sigma^i \times G^i/(\text{relations})$$ with $\sigma^i=\lbrace (x_1,...,x_i)\mid 0\le x_1 \le \cdots \le x_i \le 1\rbrace$ the standard $i$-simplex. </p> <p>Now assume $G$ is discrete. Then $$B_nG := \coprod_{i=0}^n \sigma^i \times G^i/(\text{relations})$$ is the n-skeleton of $BG$ and depends functorially on $G$. Since the cohomology in degree $\lt n$ of a CW complex is determined by the $n$-skeleton, we obtain $H^k(B_nG;M)=H^k(BG;M)$ for $0 \le k &lt; n$ and $H^k(B_nG;M)=0$ for $k> n$ and all (local) coefficients $M$. </p> <p>In general $H^n(B_nG;M)\neq H^n(BG;M)$, but there is an exact sequence: Write $BG=EG/G$ $(EG$ is described explicitely in Adem-Milgram) and let $E_nG$ be the $n$-skeleton of $EG$. Then the following sequence is exact (Cartan, Eilenberg: Homologica Algebra, XVI, §9, Appl. 1): $$0 \to H^n(BG;M)\to H^n(B_nG;M)\to H^n(E_nG;M)^G \to H^{n+1}(BG;M)\to 0$$</p> <p><strong>Remark:</strong> In case that $G$ is not discrete, the above has to be adjusted accordingly. For example, if $G$ is (topologically) a $m$-dimensional CW complex then $B_nG$ is the $n(m+1)$-skeleton of $BG$. </p> http://mathoverflow.net/questions/128613/support-of-a-module-over-a-polynomial-algebra/128632#128632 Answer by Ralph for Support of a module over a polynomial algebra Ralph 2013-04-24T16:21:17Z 2013-04-24T16:21:17Z <p>If I understand the definitions properly, the following should work: </p> <p>Let $D \xrightarrow{\alpha} E \xrightarrow{\beta} F$ be exact and suppose $\text{Supp} E \not\subseteq \text{Supp}D \cup \text{Supp}F$. Hence there is $a \in \text{Supp} E$ such that $a \notin \text{Supp}D$ and $a \notin \text{Supp} F$. This means: </p> <ol> <li>$\forall f: f\cdot E = 0 \Rightarrow f(a)=0$</li> <li>$\exists g: g \cdot D=0 \wedge g(a)\neq 0$</li> <li>$\exists h: h \cdot F=0 \wedge h(a)\neq 0$</li> </ol> <p>For $x\in E$ we have $\beta(hx)=h\beta(x)=0$, i.e. $hx\in \ker \beta=\text{im}\;\alpha$. Thus there is $y\in D$ such that $hx=\alpha(y)$. So $ghx=g\alpha(y)=\alpha(gy)=\alpha(0)=0$ and consequently $ghE=0$. Now, by 1., $0=(gh)(a)=g(a)h(a)$, contradicting 2. and 3. q.e.d. </p> http://mathoverflow.net/questions/127289/homology-groups-of-divisible-and-powered-nilpotent-groups/127562#127562 Answer by Ralph for Homology groups of divisible and powered (nilpotent) groups Ralph 2013-04-14T23:10:53Z 2013-04-14T23:10:53Z <p>For $G=UT(3,\mathbb{Q})/Z$, I computed $H_2(G,\mathbb{Z})=\mathbb{Z}$, which confirms that $G$ is a counterexample to (1). (This doesn't contradict Adam's answer, since $G\supseteq \mathbb{Q}/\mathbb{Z}$ isn't uniquely divisible)</p> <p>The essential steps in my computation are (coefficients are always $\mathbb{Z}$ which are suppressed): </p> <ul> <li><p>The LHS spectral sequence of the central extension $\mathbb{Q}/\mathbb{Z} \hookrightarrow G \twoheadrightarrow \mathbb{Q}^2$ yields $H_3(G)=0$ and $H_2(G) = E^3_{2,0} \le H_2(\mathbb{Q}^2)=\mathbb{Q}$. </p></li> <li><p>The LHS of the central extension $\mathbb{Q} \hookrightarrow UT(3,\mathbb{Q}) \twoheadrightarrow \mathbb{Q}^2$ yields $H_2(UT(3,\mathbb{Q}))=\mathbb{Q}^2$ </p></li> <li><p>The LHS of the central extension $\mathbb{Z} \hookrightarrow UT(3,\mathbb{Q}) \twoheadrightarrow G$ has $$E^2=\begin{array}{cccc} |0 &amp; &amp; &amp; &amp; \newline |\mathbb{Z} &amp; \mathbb{Q}^2 &amp; &amp; \newline |\underline{\mathbb{Z}} &amp; \underline{\mathbb{Q}^2} &amp; \underline{H_2(G)} &amp; \underline{0} \newline \end{array}$$ Since $UT(3,\mathbb{Q}) \twoheadrightarrow G$ induces an isomorphism on the abelianization, we have $E^3_{01}=0$. Hence $d^2: H_2(G)\to \mathbb{Z}$ is epi. For $K = \ker d^2$ there is a s.e. sequence $\mathbb{Q}^2 \hookrightarrow H_2(UT(3,\mathbb{Q})) \twoheadrightarrow K$. Tensoring yields $K\otimes \mathbb{Q}=0$. If $K \neq 0$ there is $\mathbb{Z} \hookrightarrow K$ (since $K\le H_2(G)\le \mathbb{Q})$ and tensoring gives the contradiction $\mathbb{Q} \hookrightarrow K\otimes \mathbb{Q}=0$. Hence $K=0$ and $H_2(G)\cong \mathbb{Z}$ as claimed. </p></li> </ul> http://mathoverflow.net/questions/125822/morphism-of-injective-objects/125854#125854 Answer by Ralph for morphism of injective objects Ralph 2013-03-28T20:48:35Z 2013-03-28T20:48:35Z <p>Note that there is in general <em>no</em> morphism $g: I \to J$ such that the following diagram commutes: $$\begin{array}{ccc} I &amp; \xrightarrow{g} &amp; J \newline i\uparrow &amp; &amp; \uparrow j \newline A &amp; \xrightarrow[f]{} &amp; B \end{array}\tag{\ast}$$ For, the commutativity of the diagram forces $\ker(i) \subseteq \ker(j\circ f)$, but the three maps $i,j,f$ are completely independent from each other. <em>However</em> $g$ can be choosen such that the diagram commutes <em>up to homotopy</em>. Moreover, $g$ is unique up to homotopy. The proof uses the </p> <blockquote> <p><strong>Comparison Theorem:</strong> If $I$ is a bounded below complex of injectives and $\alpha: X \to Y$ is a weak equivalence of two arbitrary complexes, then $$[Y,I] \to [X,I],\; [h] \mapsto [h \circ \alpha]$$ is an isomorphism of homotopy classes. </p> </blockquote> <p>A reference for (the projective version of) the comparison theorem is Brown: Cohomology of Groups, Theorem I.8.5. </p> <p>We are now ready to construct $g$ above: $i: A \to I$ is a weak equivalence and $J$ is injective, bounded below. Hence $[I,J] \to [A,J],\; [h] \mapsto [h \circ i]$ is an isomorphism. Because of $[j \circ f] \in [A,J]$ there is consequently $g: I \to J$ with $[g \circ i]=[j \circ f]$, i.e. $g\circ i \simeq j \circ f$. The uniqueness of $g$ is shown in the same way using the injectivity of the comparison map. </p> http://mathoverflow.net/questions/124182/applications-of-n-dimensional-crystallographic-groups/125623#125623 Answer by Ralph for Applications of n-dimensional crystallographic groups Ralph 2013-03-26T13:17:52Z 2013-03-26T13:17:52Z <p>The following are applications in the theory of $p$-groups: </p> <p>Space groups have been used by </p> <ul> <li>Felsch, Neubüser, Plesken: Space groups and groups of prime-power order. IV: Counterexamples to the class-breadth conjecture. Journal London Math. Soc. (2), 24 (1981) 113-122 </li> </ul> <p>to construct counterexamples to the class-breadth conjecture for $p=2$. Recall the the conjecture claims $\text{class} \le \text{breath} + 1$ for $p$-groups $P$ where the breath $b$ is defined such that $p^b$ is the maximal size of the conjugacy classes of $P$. In their counterexamples $P=S/2^kT$ where $S$ is a space group, $T$ the translation subgroup and $k$ a carefully choosen integer. </p> <p>Space groups those point groups are $p$-groups are also the core in proving the celebrated coclass conjectures of Leedham-Green and Newman (see the book Leedham-Green, McKay: The structure of groups of prime power order, 2002). I don't know enough to tell details, but it's striking that the series of papers that contain the proof are titled "Space groups and groups of prime-power order" (I-VIII). </p> http://mathoverflow.net/questions/125276/good-uses-of-siegel-zeros/125324#125324 Answer by Ralph for Good uses of Siegel zeros? Ralph 2013-03-22T20:40:59Z 2013-03-22T20:40:59Z <p>There is an asymptotic formula for the relative class number of $p$th-cyclotomic fields where a Siegel zero $\beta$ occurs: $$h^-(p)=\frac{p+3}{4}\log p-\frac{p}{2}\log 2\pi + \log (1-\beta) +O(\log^2_2 p)$$ If no Siegel zero exists or if $p\equiv 1(4)$, the term $\log (1-\beta)$ disappears in the formula. </p> <p>This is Theorem 1 from Puchta: On the class number of $p$-th cyclotomic field. Arch. Math. 74 (2000), 266-268. But note the misprint of the sign in the theorem: $\frac{p-3}{4}$ should be $\frac{p+3}{4}$. </p> <p>Compare also with a classical formula of Masley, Montgomery (Cyclotomic fields with unique factorization. J. Reine Angew. Math. , 286/287 (1976), 248–256. Theorem 1): $$h^-(p)=\frac{p+3}{4}\log p-\frac{p}{2}\log 2\pi+O(\log p)$$</p> http://mathoverflow.net/questions/125144/on-the-divisibility-of-the-special-linear-group-of-degree-n-over-an-algebraical/125198#125198 Answer by Ralph for On the divisibility of the special linear group of degree $n$ over an algebraically closed field Ralph 2013-03-21T19:59:54Z 2013-03-21T20:06:48Z <p>$SL_n(k)$ is not divisible over any algebraically closed field $k$ and $n>1$. </p> <p>This would be indeed a nice exercise in a Linear Algebra course. </p> <p><em>Proof:</em> Let $\zeta\neq 1$. I'll show that there is no $X\in SL_n(k)$ such that $X^n=A$ for<br> $$A=\scriptstyle\begin{pmatrix} \zeta &amp; 1 &amp; &amp; &amp; \newline &amp; \zeta &amp; 1 &amp; &amp; \newline &amp; &amp; \ddots &amp; \ddots &amp; \newline<br> &amp; &amp; &amp; \zeta &amp; 1 \newline &amp; &amp; &amp; &amp; \zeta \end{pmatrix}.$$</p> <p>Let $X^n=A$ and let $\mu_1,...,\mu_n$ be the eigenvalues of $X$. Clearly $\mu_i^n=\zeta$. Each eigenvector of $X$ is also an eigenvector of $A$ and since $A$ has exactly one eigenvector (up to scalar multiples), we conclude $\mu_1=\ldots=\mu_n$. Hence $\det(X)=\mu_1^n=\zeta\neq 1$. qed. </p> http://mathoverflow.net/questions/124821/characters-of-p-groups/124849#124849 Answer by Ralph for Characters of p-groups Ralph 2013-03-18T11:08:09Z 2013-03-18T11:08:09Z <p>As already explained by Geoff, the number of conjugacy classes of size $>1$ is divisible by $p-1$. A reference is: </p> <p>A. Mann: Groups with few class sizes and the centralizer equality subgroup. Israel J. Math. 142(2004), 367-380. </p> <p>The result (with a proof quite similar to Geoff's) is on page 376 (after Corollary 18). </p> http://mathoverflow.net/questions/124462/properties-of-ring-epimorphisms-that-are-true-only-over-commutative-rings/124488#124488 Answer by Ralph for Properties of ring epimorphisms that are true only over commutative rings Ralph 2013-03-14T09:26:24Z 2013-03-14T09:26:24Z <p>For Artinian commutative rings $R$ all ring epimorphisms $R\to S$ are surjective. </p> <p>In general, this is false if $R$ is non-commutative. For, Isbell has constructed an epimorphism $R \to S$ where $A$ is finite (hence Artinian) and $S$ infinite. I'll have a look if I can find Isbell's example later on. </p> http://mathoverflow.net/questions/124122/do-exact-functors-commute-with-spectral-sequences Do exact functors commute with spectral sequences ? Ralph 2013-03-10T02:36:45Z 2013-03-10T02:36:45Z <p>Let $F: \mathcal{A} \to \mathcal{B}$ be an exact covariant functor of abelian categories and let $$\mathscr{C}: A \to A \to B \to A$$ be an exact couple in $\mathcal{A}$ with corresponding spectral sequence $E^r(\mathscr{C})$.<br> $$F(\mathscr{C}): F(A) \to F(A) \to F(B) \to F(A)$$ is a exact couple in $\mathcal{B}$ and defines a spectral sequence $E^r(F(\mathscr{C}))$. <em>I am pretty sure</em> there is an isomorphism of spectral sequences that is natural in $\mathscr{C}$: $$E^r(F(\mathscr{C})) \cong F(E^r(\mathscr{C}))$$ Such a isomorphism is equivalent to a natural isomorphism $D(F(\mathscr{C}))\cong F(D(\mathscr{C}))$ for all exact couples where $D(-)$ denotes the derived couple.</p> <p><strong>Question:</strong> Does someone know a reference for such an isomorphism or a place in the literature where such an isomorphism is used (i.e. in this generality, not for particular $F)$ ? </p> http://mathoverflow.net/questions/121468/epimorphisms-and-free-submodules/123579#123579 Answer by Ralph for Epimorphisms and free submodules Ralph 2013-03-05T00:09:03Z 2013-03-10T01:36:10Z <p>A simple counterexample can be constructed using free objects: </p> <p>For a set $X$ let $\mathbb{Z}\langle X\rangle$ be the free ring on $X$ and let $\mathbb{Z}F(X)$ be the group ring of the free group $F(X)$ on $X$. The inclusion $X \hookrightarrow F(X)$ induces a ring homomorphism $i: \mathbb{Z}\langle X\rangle \to \mathbb{Z}F(X)$ which is an epimorphism. If $x\neq y$ are in $X$ then $xy^{-1} \in \mathbb{Z}F(X)$ and there is obviously no "polynomial" $f \in \mathbb{Z}\langle X\rangle$ such that $fxy^{-1} \in \mathbb{Z}\langle X\rangle$. </p> <p>To see that $i$ is an epimorphism, note that $\mathbb{Z}F(X)$ is generated (as a ring) by $x,x^{-1}\;(x \in X)$ and a unitary ring homomorphism $\varphi: \mathbb{Z}F(X) \to T$ satisfies $\varphi(x^{-1})=\varphi(x)^{-1}$. Hence $\varphi$ is determined by its values on $X$. </p> <hr> <p><strong>Update:</strong> The property in question holds true for commutative rings. </p> <p><em>Proof:</em> At first assume this has already been proved for zero-dimensional local commutative rings. Let $R \le S$ be an epimorphic extensions of comm. rings and choose a minimal prime $\mathfrak{p}$ of $R$ (exists by Zorn's lemma). $R_\mathfrak{p}$ is a local ring of dimension $\text{ht}(\mathfrak{p})=0$. Since $R\setminus \mathfrak{p}$ is a multiplicative subset of $S$ we can localize and obtain a comm. diagramm $$\begin{array}{ccc} R &amp; \xrightarrow[\scriptstyle\text{epi}]{i} &amp; S\;\;\; \newline {\scriptstyle\text{epi}}\downarrow &amp; &amp; \downarrow\scriptstyle\text{epi} \newline R_\mathfrak{p} &amp; \xrightarrow[i_\mathfrak{p}]{} &amp; (R\setminus \mathfrak{p})^{-1}S \end{array}$$ Hence $i_\mathfrak{p}$ is epi and it's easy to see that $i_\mathfrak{p}$ is mono as well. </p> <p>Let $s \in S$. By our assumption there are $r_i/t_i \in R_\mathfrak{p},\;r_1/t_1 \neq 0$ such that $\frac{r_1}{t_1}\frac{s}{1}=\frac{r_2}{t_2}$, i.e. there is $t \in R\setminus \mathfrak{p}$ with $(tt_2r_1)s=tt_1r_2 \in R$. Moreover $tt_2r_1 \neq 0$ (because $r_1/t_1 \neq 0$ in $R_\mathfrak{p}$ just says there is no $t \in R\setminus \mathfrak{p}$ such that $tr_1=0$ in $R)$ and we are done. </p> <p>Now suppose $R$ is zero-dimensional local comm. with max. ideal $\mathfrak{m}$. Again from the comm. diagramm $$\begin{array}{ccc} R &amp; \xrightarrow[\scriptstyle\text{epi}]{i} &amp; S\;\;\; \newline {\scriptstyle\text{epi}}\downarrow\;\; &amp; &amp; \downarrow\scriptstyle\text{epi} \newline R/\mathfrak{m} &amp; \xrightarrow[i_\mathfrak{m}]{} &amp; S/\mathfrak{m}S \end{array}$$ we conclude that $i_\mathfrak{m}$ is epi and since $R/\mathfrak{m}$ is a field, $i_\mathfrak{m}$ is an isomorphism (see the link in the OP's question). Hence each $s \in S$ can be written as $$s=r + \sum_{i=1}^l m_is_i\qquad (r\in R, s_i \in S, m_i \in \mathfrak{m}, m_i \neq 0)$$ We want to show that there is $r_0 \in R,\;r_0 \neq 0$ such that $r_0s\in R$. If $l=0$ then $s=r\in R$ and we are done. Otherwise, since $\mathfrak{m}=\sqrt{0}$ there is $n_1> 0$ maximal such that $m_1^{n_1} \neq 0$ and multiplying yields $$m_1^{n_1}s=m_1^{n_1}r + \sum_{i=2}^l (m_1^{n_1}m_i)s_i.$$ If $m_1^{n_1}m_2=0$ ignore the corresponding summand. Otherwise, there is $n_2>0$ maximal such that $m_1^{n_1}m_2^{n_2}\neq 0$. Multiplying again yields $$m_1^{n_1}m_2^{n_2}s=m_1^{n_1}m_2^{n_2}r + \sum_{i=3}^l(m_1^{n_1}m_2^{n_2}m_i)s_i.$$ Proceeding this way, we obtain the required $r_0$ in the form $r_0 = \prod_{j=1}^k m_{i_j}^{n_{i_j}}.\;\;$ qed. </p> http://mathoverflow.net/questions/121499/transgression-maps-in-group-cohomology-and-group-homology-duality-of-spectral-s/124114#124114 Answer by Ralph for Transgression maps in group cohomology and group homology / duality of spectral sequences Ralph 2013-03-10T00:43:41Z 2013-03-10T01:27:34Z <p>We have the following general result: </p> <blockquote> <p><strong>Lemma:</strong> Let $K$ be a filtered complex of ablian groups and let $I$ be an ablian group. Filter the cocomplex $K^\ast := \text{Hom}(K,I)$ by the dual filtration (def. in proof). Then there is a homomorphism $\phi_r: E_r(K^\ast) \to E^r(K)^\ast$ of abelian groups such that the following diagram commutes: $$\begin{array}{ccc} E_r^{ij} &amp; \xrightarrow{d_r} &amp; E_r^{i+r,j-r+1} \newline {\scriptstyle\phi_r}\downarrow\;\; &amp; &amp; \;\;\downarrow{\scriptstyle\phi_r} \newline (E^r_{ij})^\ast &amp; \xrightarrow[(d^r)^\ast]{} &amp; (E^r_{i+r,j-r+1})^\ast \end{array}$$ If $I$ is injective, then $(\phi_r)_{r\ge 0}$ is an isomorphism of spectral sequences.</p> </blockquote> <p>As a corollary we obtain: </p> <blockquote> <p>Let $A$ be a $G$-modules, $I$ an injective abelian group (with trivial $G$-action), and denote the homology LHS spectral sequence with coefficients in $A$ by $E^r(A)$ (and respectively for cohomology). Then there is an isomorphism $\phi_r: E_r(A^\ast) \to E^r(A)^\ast$ of spectral sequences. </p> </blockquote> <p><em>Proof:</em> Let $P$ resp. $Q$ be projective resolutions of $\mathbb{Z}$ over $G$ resp. $G/H$ and put $X=Q \otimes P$. Then the cohomology LHS spectral sequence is the spectral sequence of of the filtration of the cocomplex $$\text{Hom}_G(X,A^\ast)=\text{Hom}_G(X,\text{Hom}(A,I))\cong \text{Hom}_G(X\otimes A,I) \cong \text{Hom}(X\otimes_GA,I)$$ (the last iso needs $I$ to be trivial) and the homology LHS spectral sequence is the spectral sequence of the filtration of $X\otimes_G A$. Now the results follows from the lemma. qed. </p> <p><strong>Remark:</strong> If $I$ is injective, then the lemma can be proved easier by using exact couples (in the same manner as in Tyler Lawson's answer to the question linked by Mark Grant). But if there are additional structures in the spectral sequences like products, Steenrod power operations, etc. that one wants to compare, then it's helpful to have an explicit map $\phi_r$ as defined below. For example, this way one can show that $\phi_2^{2,0}: H^2(G/H,(A^\ast)^H)\to H_2(G/H,A_H)^\ast$ is induced by $$\text{Hom}_{G/H}(Q_2,(A^\ast)^H) \to (Q_2 \otimes_{G/H}A_H)^\ast,\; f \mapsto (x \otimes \bar{a} \mapsto f(x)(a)).$$ </p> <p><strong>Proof of the lemma:</strong> Let $C := K^\ast$ with differential $\delta = d^\ast$. The dual filtration is defined by $F^rC^i=\lbrace f \in \text{Hom}(K_i,I)\mid f|F_{r-1}K_i=0\rbrace$. As usual set $$Z^r_{ij}=\lbrace x \in F_iK_{i+j}\mid dx \in F_{i-r}K_{i+j-1}\rbrace\;,\quad B^r_{ij}=dZ_{i+r-1,j-r+2}^{r-1}+Z^{r-1}_{i-1,j+1}$$</p> <p>$$Z_r^{ij}=\lbrace f \in F^iC^{i+j}\mid \delta f \in F^{i+r}C^{i+j+1}\;,\quad B_r^{ij}=\delta Z_{r-1}^{i-r+1,j+r-2}+Z^{i+1,j-1}_{r-1}$$ It's straightforward to show that $Z_r^{ij} \to \text{Hom}(Z^r_{ij},I),\;f \mapsto f|Z^r_{ij}$ induces a hom. $$\phi_r: E_r^{ij}=Z_r^{ij}/B_r^{ij} \to \text{Hom}(Z^r_{ij}/B^r_{ij},I)=(E^r_{ij})^\ast$$ that makes the diagram in the lemma commute. </p> <p>Now suppose $I$ is injective. We want to show that $\phi_r$ is bijective. In case $r=0$, $E^0_{ij}=F_iK_{i+j}/F_{i-1}K_{i+j}, E_0^{ij}=F^iC^{i+j}/F^{i+1}C^{i+j}$ and bijectivity is easy to establish. Assume $\phi_r$ is bijective. Then, by the diagram above, $H(\phi_r)$ is bijective and the bijectivity of $\phi_{r+1}$ follows from the commutativity of the diagram below: </p> <p>$$\begin{array}{ccc} E_{r+1}^{ij} &amp; \xrightarrow{} &amp; \xrightarrow[\alpha]{\sim} &amp; H(E_r^{ij}) \newline {\scriptstyle\phi_{r+1}}\downarrow\;\; &amp; &amp; &amp; {\scriptstyle\cong}\downarrow {\scriptstyle H(\phi_r)} \newline (E_{ij}^{r+1})^\ast &amp; \xleftarrow[\beta^\ast]{\sim} H(E^r_{ij})^\ast &amp; \xleftarrow[\gamma]{\sim} &amp; H( (E^r_{ij})^\ast) \end{array}$$ Here $\alpha$ is induced by $F_{r+1}^{ij}\hookrightarrow F_r^{ij}$, $\beta$ by $F_{ij}^{r+1}\hookrightarrow F^r_{ij}$ and $\gamma$ (which reflects the fact that the ,exact functor $\text{Hom}(-,I)$ commutes with homology) is given by $\bar{f} +\text{im}\ d_r \mapsto (\bar{x} + \text{im}\; d^r \mapsto f(x)\;)$. qed. </p> http://mathoverflow.net/questions/122707/which-topological-spaces-are-topological-groups/122713#122713 Answer by Ralph for Which topological spaces are (topological) groups? Ralph 2013-02-23T11:43:23Z 2013-02-23T11:43:23Z <p>There is a homological criterion that is often helpful, to rule out the possibility for a topological space to admit a continuous group structure (even H-space structure): </p> <p>The rational cohomology ring of a connected topological group (or H-space) $G$ is a connected graded-commutative Hopf algebra over $\mathbb{Q}$ and if $H^i(G;\mathbb{Q})$ is finite dimensional for all $i \ge 0$, then, by a theorem of Borel, $H^\ast(G;\mathbb{Q})$ is the tensor product of an exterior algebra on odd-dimensional generators and a polynomial algebra on even-dimensional generators. </p> <p>For example $H^\ast(\mathbb{C}P^n;\mathbb{Q})=\mathbb{Q}[X]/(X^n),\; \deg x=2$ isn't of this form. Hence $\mathbb{C}P^n$ is no topological group. </p> <p>For the theorem (and variations thereof) and further examples see Hatcher: Algebraic Topology, Section 3.C. </p> http://mathoverflow.net/questions/122553/p-group-with-large-center/122569#122569 Answer by Ralph for p-group with large center Ralph 2013-02-21T19:39:23Z 2013-02-23T08:51:50Z <p>Let $G$ be a group in question. First note that $G/Z(G)\cong C_{p^2}$ isn't possible. Thus $G$ has the presentation $$\langle Z,x,y\mid Z \text{ central}, x^p=a,y^p=b,[x,y]=c\rangle$$ where $Z$ is the center, $a,b, c \in Z$ and $c\neq 1,\; c^p=1$. </p> <p><strong>Added:</strong> Suppose $a=a_0^p,a_0 \in Z$. By replacing $x$ by $xa_0^{-1}$ we have the relation $x^p = 1$. Write $Z=\langle z_1,...,z_n\mid r(Z)\rangle$. Then we obtain the presentations $$\tag{I}G(c)=\langle z_1,...,z_n,x,y\mid r(Z), x^p=y^p=1,[x,y]=c\rangle$$</p> <p>$$G(c,i)= \langle z_1,...,z_n,x,y\mid r(Z), x^p=z_i, y^p=1,[x,y]=c\rangle\tag{II}$$</p> <p>$$G(c,i,j)=\langle z_1,...,z_n,x,y\mid r(Z), x^p=z_i,y^p=z_j,[x,y]=c\rangle\tag{III}$$</p> <p><strong>Added 2:</strong> 1) In case (III) we can assume $i\neq j$ (otherwise, replacing $x$ by $xy^{-1}$ gives case (II)). </p> <p>2) Let $\exp(z_i)=k_i$. Denote by $(k_1,...,k_n)$ the isomorphism type of $C_{p^{k_1}} \times \cdots C_{p^{k_n}}$. Then the groups above have maximal abelian subgroups of the following types: $$\begin{array}{lcl} (I) &amp; : &amp; (k_1,...,k_n,1) \newline (II) &amp; : &amp; (k_1,...,k_n,1), (k_1,..,k_i+1,..,k_n) \newline (III) &amp; : &amp; (k_1,..,k_i+1,..,k_n), (k_1,..,k_j+1,..,k_n) \end{array}$$ Hence (I), (II), (III) belong to different ismorphism types. </p> <p>It remains to check for which parameters $c,i,j$ the groups within (I) resp. (II) resp. (III) are isomorphic. </p> <p><strong>Added 3:</strong> Modulo possible mistakes a complete classification is given by: </p> <hr> <p>Let $p$ be an odd prime, $Z=\langle z_1,...,z_n\rangle \cong C_{p^{l_1}}^{n_1} \times \cdots \times C_{p^{l_m}}^{n_m}$ with $l_1 > \cdots > l_m$ and suppose $$\lbrace 1,...,n\rbrace = \coprod_{i=1}^m \lbrace r_i,...,r_{i+1}-1\rbrace\qquad (n_i=r_{i+1}-r_i)$$ is a decomposition such that $\exp(z_j)=l_i$ for $r_i \le j &lt; r_{i+1}$. Then the groups $G$ with $Z(G)=Z$ and $(G:Z(G))=p^2$ are given (up to isomorphism) by the following non-isomorphic presentations $(c_i := z_{r_i}^{p^{l_i-1}}):$ $$\begin{array}{llcl} G(c_i) &amp; (i=1,...,m) &amp; \qquad &amp; \text{(I)} \newline G(c_i,r_j) &amp; (i,j=1,...,m) &amp; \qquad &amp; \text{(II)} \newline G(c_i,r_j,r_j+1) &amp; (i,j=1,...,m,\;n_j \ge 2) &amp; \qquad &amp; \text{(IIIa)} \newline G(c_i,r_j,r_k) &amp; (i,j,k=1,...,m,\;j \neq k) &amp; \qquad &amp; \text{(IIIb)} \end{array}$$</p> <p>In particular, there are $m^3 +m(1 + |\lbrace n_j \ge 2\rbrace|)$ isomorphism classes. </p> <hr> <p><em>Proof:</em> i) It follows from the structure of the max. abelian subgroups in Add-2, that $G(c,i) \cong G(c,j)$ iff $k_i=k_j$ and $G(c,i,j)\cong G(c,l,q),\;(i\le j,\;l \le q)$ iff $k_i = k_l$ and $k_j=k_q$. </p> <p>ii) We determine the isomorphism types for various $c$. </p> <p><strong>Claim 1:</strong> $\quad G(c,-) \cong G(c',-)$ iff there is $f \in Aut(Z)$ such that $c'=f(c)$. </p> <p>First, $f \in Aut(Z)$ extends to an isomorphism $\varphi_f: G(c,-) \to G(f(c),-)$ by $x \mapsto x, y\mapsto y$. Conversely, if $\varphi: G(c,-) \to G(c',-)$ is an isomorphism, then (easy calculation) there is $q$ coprime to $p$ with $c'=\varphi(c)^q$. Hence $c'=f(c)$ where $f\in Aut(Z)$ is the composition of $\varphi|Z$ and the $q$-power map. $\square$ </p> <p>Using $Z=\prod_{j=1}^m C_{p^{l_j}}^{n_j}$, we have the decomposition $\langle c\in Z \mid c^p=1\rangle = \coprod_{i=1}^m M_i$, where $$M_i = \lbrace (g_1,...,g_i,1,...,1)\mid g_i \neq 1,\;g_j \in C_{p^{l_j}}^{n_j},\; g_j^p=1\;(j=1,...,i)\;\rangle.$$ </p> <p><strong>Claim 2:</strong> $\quad f(M_i)=M_i$ for $f \in Aut(Z)$. Conversely, for $c, c' \in M_i$ there is $f \in Aut(Z)$ such that $f(c)=c'$. </p> <p>I omit the proof which is in essential based on linear algebra. Since $c_i = z_{r_i}^{p^{l_i-1}} \in M_i$, the claim implies $G(c,-)\cong G(c_i,-)$ for all $c \in M_i$ and $G(c_i,-) \not\cong G(c_j,-)$ for $i\neq j$. This completes the classification.</p> <p><strong>Remark:</strong> For $p=2$ (in contrast to $p$ odd) we have in addition the case $G(c,i,i)$ (i.e. $x^2=y^2=c\neq 1$). Example: Quaternion group of order 8. The reason is that $(xy^{-1})^p=c^{p(p+1)/2}$ is $1$ only if $p$ is odd. </p> http://mathoverflow.net/questions/122424/taking-invariants-under-pro-p-group-is-exact/122433#122433 Answer by Ralph for Taking invariants under pro-p-group is exact? Ralph 2013-02-20T18:05:25Z 2013-02-21T00:00:49Z <p>Yes, I think taking invariants is exact in your context. </p> <p><strong>Proof:</strong> Let be $G$ a pro-p group and $0 \to A \to B \to C\to 0$ a short exact sequence (s.e.s) of finite dim. $\mathbb{Q}_l$-vector spaces on which $G$ acts continuously and $\mathbb{Q}_l$-linearly. There is a long exact sequence (l.e.s.) [N, 2.3.2]</p> <p>$$0 \to A^G \to B^G \to C^G \to H^1_c(G,A) \to H^1_c(G,B) \to \cdots$$</p> <p>of $\mathbb{Q}_l$-vector spaces ($H^1_c$ benotes continuous cohomology). Hence it's enough to show $H^1_c(G,A)=0$. </p> <p>Write $A= \mathbb{Q}_l^n$. By the OP's comment above, there is a $G$-submodule $T = \mathbb{Z}_l^n$ of $A$ and $G$ acts continously and $\mathbb{Z}_l$-linearly. With $W := A/T = (\mathbb{Q}_l/\mathbb{Z}_l)^n$ we have the s.e.s. of $\mathbb{Z}_l$-modules $$0 \to T \to A \to W\to 0.\tag{1}$$<br> Since $W$ is a discrete $G$-module, $H^i_c(G,W)=H^i(G,W)=0$ $(i>0)$ by the discrete case below and the l.e.s. of $(1)$ yields the surjection of $\mathbb{Z}_l$-modules </p> <p>$$H^1_c(G,T) \twoheadrightarrow H^1_c(G,A).\tag{2}$$ </p> <p>[N,2.3.9] states: </p> <blockquote> <p>Assume that the cohomology groups of $G$ with coefficients in finite $l$-primary modules are finite. Then $H^i_c(G,T)$ is a finitely generated $\mathbb{Z}_l$-module for all $i$. </p> </blockquote> <p>By the discrete case below we can apply this theorem and find that $H^1_c(G,T)$ is a f.g. $\mathbb{Z}_l$-module. Therefore, by $(2)$, the $\mathbb{Q}_l$-vector space $H^1_c(G,A)$ is f.g. as $\mathbb{Z}_l$-module what is only possible if $H^1_c(G,A)=0$. q.e.d.</p> <hr> <p><strong>Discrete case:</strong> Let $G$ be a pro-p group and $A$ a discrete $G$-module such that $A \to A, x \mapsto px$ is an automorphism. Then $H^i(G,A)=0$ for $i>0$. In particular, for each short exact sequence of discrete $G$-modules $0 \to A \to B \to C\to 0$, the induced sequence $$0 \to A^G \to B^G \to C^G\to 0$$ is exact. </p> <p><strong>Proof:</strong> By the long exact cohomology sequence [RZ, 6.6.1] the latter follows from $H^1(G,A)=0$. Let $i>0$ and $x\in H^i(G,A)$. Since $H^i(G,A)= \varinjlim_U H^i(G/U,A^U )$ [RZ, 6.5.6], there is an open normal subgroup $U\le G$ and $y \in H^i(G/U,A^U)$ such that $x=\text{inf}(y)$. Since $G/U$ is a finite p-group, $y$ and hence $x$ is annulated by a power of p. But multiplication with p is an automorphism on $H^i(G,A)$. Thus $x=0$ and $H^i(G,A)=0$ follow. q.e.d. </p> <hr> <p>[N]$\;\;$ Neukirch, et. al.: Cohomology of Number Fields. </p> <p>[RZ] Ribes, Zalesskii: Profinite Groups, 2nd Edition. </p> http://mathoverflow.net/questions/121820/transfer-map-for-group-homology/121839#121839 Answer by Ralph for Transfer map for group homology. Ralph 2013-02-14T21:25:29Z 2013-02-14T21:25:29Z <p>In case of trivial coefficients there are nice formulas: Let $G=\coprod_h Hh$ and<br> $$t: G \to H,\; g \mapsto \prod_h hgh_g^{-1}$$ where the representative $h_g$ is defined by $Hhg=Hh_g$. Under the identification $H_1(G,A)=G_{ab}\otimes A$ we get $$tr_1: G_{ab}\otimes A \to H_{ab}\otimes A,\; g[G,G]\otimes a \mapsto t(g)[H,H] \otimes a$$ (cf. Brown, Cohomology of Groups, III.9, Ex. 2). In case $A=\mathbb{Z}$ this is just the usual transfer homomorphism from group theory (cf. Robinson, Theory of Groups, 10.1). </p> <p>In cohomology, using the identification $H^1(G,A)=Hom(G,A)$ we have $$tr^1: Hom(H,A) \to Hom(G,A),\; f \mapsto f\circ t.$$</p> http://mathoverflow.net/questions/121755/categories-where-projective-objects-are-flat Categories where projective objects are flat Ralph 2013-02-13T22:59:16Z 2013-02-14T19:57:31Z <p>In many categories arising in the theory of algebras or modules, projective objects are flat. For example each projective module over a ring with identity is flat. However, there are categories where this principle is violated: <a href="http://hopf.math.purdue.edu/LewisG/PrjFlt.pdf" rel="nofollow">Lewis</a> has shown that the category of Mackey functors for the orthogonal group $O(n)$ has non-flat projectives. </p> <p><strong>Q1:</strong> Is there a classification of symmetric monoidal abelian categories $\mathcal{A}$ where all projectives are flat ? BTW: Do such categories have a particular name ? </p> <p>If a classification is too strong to establish, I would be interested in </p> <p><strong>Q2:</strong> Are there criteria that ensure that projectives are flat ? </p> <p>An example for Q2 is, when $\mathcal{A}$ has direct sums and the unit is a progenerator of $\mathcal{A}$ (this is just the categorification of the module case mentioned above). Note that this criteria doesn't work for instance in the category of $RG$-modules ($R$ commutative ring, $G$ a group) where the tensor product is given by $M \otimes_R N$ with diagonal $G$-operation. </p> http://mathoverflow.net/questions/120918/are-epimorphisms-from-a-division-ring-isomorphisms/120974#120974 Answer by Ralph for Are epimorphisms from a division ring isomorphisms ? Ralph 2013-02-06T14:35:47Z 2013-02-06T14:35:47Z <p>I think the question can be answered affirmatively. </p> <p>Since $0,R$ are the only ideals of $R$, $\alpha$ is injective (I assume $\alpha(1_R)=1_S$) and $R$ can be considered as subring and as left $R$-submodule of $S$. Let $\mu:S \otimes_R S \to S$ be the multiplication. A splitting of $\mu$ is given by $j: S \to S \otimes_R S,\; s \mapsto s \otimes 1$. Hence $S \otimes_R S = \ker(\mu) \oplus \text{im}(j)$ and as the image of $j$ equals the image of $S \otimes_R R \to S \otimes_R S$, we have<br> $$\ker(\mu) = \frac{S \otimes_R S}{\text{im}(j)}=\frac{S \otimes_R S}{S \otimes_R R}=S \otimes_R (S/R)$$ Since $R$ is a division ring, $S$ is a free right $R$-module, say, $S=\bigoplus_i R$. Thus $\ker(\mu) = \bigoplus_i (S/R)$ and in particular $\ker(\mu)=0 \Leftrightarrow R=S$. </p> <p>But, by Prop. 1.1 (cited in the OP's question), $\ker(\mu)=0$ since $\alpha$ is epi and the assertion follows. </p> http://mathoverflow.net/questions/120445/non-commutative-finite-rings/120479#120479 Answer by Ralph for non-commutative finite rings Ralph 2013-02-01T01:04:15Z 2013-02-01T17:33:24Z <p>A "natural" example is given by the group ring $\mathbb{F}_2[Q]$ of the Quaternion group of order 8. </p> <p>For, we have to show that each left ideal is also a right ideal, and conversely, each right ideal is also a left ideal. The first half (i.e. left is right) is shown in <a href="http://www.ams.org/journals/proc/1979-076-02/S0002-9939-1979-0537074-2/S0002-9939-1979-0537074-2.pdf" rel="nofollow">this paper</a>. </p> <p>Let $i:\mathbb{F}_2[Q] \to \mathbb{F}_2[Q],\;g \mapsto g^{-1}$ be the antipode. It's a general fact that for a left (right) ideal $I$, $i(I)$ is a right (left) ideal. </p> <p>Now suppose $I$ is a right ideal. Hence $i(I)$ is a left ideal and by the above, it's also a right ideal. Consequently, $I=i(i(I))$ is a left ideal and we are done. </p> <p><strong>Added:</strong> The comment asks for a modular representation of $Q$. Using GAP I found that $\mathbb{F}_2[Q]$ can be embedded into the matrix ring $M_4(\mathbb{F}_2)$. Write $Q=\langle x,y\mid x^4=y^4=1, yxy^{-1}=x^{-1}\rangle$. Then a faithful representation $Q\hookrightarrow GL(4,2)$ is given by $$x \mapsto \begin{pmatrix}1 &amp; 0 &amp; 1 &amp; 0 \newline 0 &amp; 1 &amp; 0 &amp; 0 \newline 0 &amp; 0 &amp; 1 &amp; 1 \newline 0 &amp; 0 &amp; 0 &amp; 1 \end{pmatrix}\qquad y \mapsto \begin{pmatrix}1 &amp; 1 &amp; 1 &amp; 1 \newline 0 &amp; 1 &amp; 0 &amp; 1 \newline 0 &amp; 0 &amp; 1 &amp; 0 \newline 0 &amp; 0 &amp; 0 &amp; 1 \end{pmatrix}$$ Since the Sylow 2-subgroup of $GL(3,2)$ is the Dihedral group $D_8$, four is the smallest degree of a faithfull representation of $Q$ over $\mathbb{F}_2$. </p> http://mathoverflow.net/questions/120351/equivariant-cohomology-for-actions-with-finite-stabilizers/120378#120378 Answer by Ralph for Equivariant Cohomology for actions with finite stabilizers Ralph 2013-01-31T03:58:12Z 2013-01-31T11:32:08Z <p>For ease of referencing I'll prove the stated isomorphism under the hypothesis that $G$ is a compact Lie group and $X$ is a finite dimensional compact topological $G$-manifold. </p> <p>By the hypothesis on $G$ we can choose $EG$ to be a dimension-wise finite CW complex. The $(n+1)$-skeleton $E$ is compact, $n$-connected and for fixed $m$ and $n$ large enough, the inclusion $E \hookrightarrow EG$ induces an isomorphism $$H^m(EG\times_G X,\mathbb{Q}) \xrightarrow{\cong} H^m(E\times_G X,\mathbb{Q})$$ (that's 4) on p. 4 of the linked paper in the question). Hence it suffices to show that the map $E\times_G X \to X/G$ induces an isomorphism in rational cohomology in degree $m$. </p> <p>But this follows from the Vietoris-Begle theorem [Q, Corollary A.7] that applies if we have shown that $f$ is closed and the fibres of $f$ are compact, n-acyclic and relative Hausdorff in $E\times_G X$: </p> <p>Closedness of $f$ and compactness of the fibres $E/G_x$ are obvious, $n$-acyclity follows from $H^i(E/G_x,\mathbb{Q})=H^i(G_x,\mathbb{Q})=0$ for $0&lt; i&lt; n$. Relative Hausdorff means different points in the fibre have disjoint neighborhoods in $E \times_G X$. This holds since $E \times_G X$ is Hausorff. QED</p> <p>[Q] Quillen: The Spectrum of an Equivariant Cohomology ring: I </p> <hr> <p>The cited Vietoris-Begle theorem states: </p> <blockquote> <p>If $f: X \to Y$ is closed and its fibres are compact, relatively Hausdorff in $X$ and $n$-acyclic, i.e. $H^i(f^{-1}(y),k)=H^i(\ast,k)$ for $i&lt; n$ ($k$ any constant coefficients), then $f^\ast:H^i(Y,k) \xrightarrow{\cong} H^i(X,k)$ for $i \lt n$. </p> </blockquote> http://mathoverflow.net/questions/120014/eigenvalues-of-a-diagonal-matrix-times-a-matrix/120064#120064 Answer by Ralph for eigenvalues of a diagonal matrix times a matrix Ralph 2013-01-27T22:50:17Z 2013-01-27T22:50:17Z <p>I suppose $A=(a_{ij}) \in M_n(\mathbb C)$. By <a href="http://en.wikipedia.org/wiki/Gershgorin_circle_theorem" rel="nofollow">Gerschgorin's circle theorem</a> the eigenvalues of $A$ lie in the union of the discs $$|z-a_{ii}| \le \sum_{j\neq i} |a_{ij}|\qquad (i=1,...,n)$$ Hence the eigenvalues of $DA$ lie in the union of the discs $$|z-\frac{a_{ii}}{2}| \le \frac{1}{2}\sum_{j\neq i}|a_{ij}|\qquad(i=1,n)$$ $$\qquad\quad |z-\frac{a_{ii}}{3}| \le \frac{1}{3}\sum_{j\neq i}|a_{ij}|\qquad(i=2,\ldots,n-1)$$</p> http://mathoverflow.net/questions/118549/depth-for-non-commutative-rings Depth for non-commutative rings Ralph 2013-01-10T16:19:53Z 2013-01-10T16:19:53Z <p>The depth of a ring or module is one of the most basic invariants in commutative ring theory. </p> <p><strong>Q1:</strong> Is there also a powerful notion of depth for non-commutative rings ? </p> <p>By a search in mathscinet, I only found the papers [BT], [R]. In the latter the $I$-depth of a module is defined (in analogue to the comm. case) as the minimal $i$ such that $Ext_R^i(R/I,M)\neq 0$ ($R$ a ring, $I$ a left ideal in $R$ and $M$ a left $R$-module). But there are no applications of the depth in the paper. In particular, I wonder: </p> <p><strong>Q2:</strong> Does the Auslander-Buchsbaum formula holds for this depth (for an approriate definition of the dimension of $R$) ? </p> <p>According to the cited papers, there seems to be no regular sequences defined for non-comm. rings. However, as far as I can see, the usual definition that $x_1,...,x_n \in R$ are $M$-regular, if $x_i$ is regular on $M/(\sum_{j=1}^iRx_i)M$ makes sense in the non-comm. case, too. </p> <p><strong>Q3:</strong> Why doesn't this definition work well resp. dosn't have nice properties ? (I believe it doesn't have nice properties because it isn't considered in the literature). </p> <p>PS: There are more results for the depth of non-comm. <em>graded</em> rings in the literature, but I want to restrict to the ungraded case here. </p> <hr> <p>[BT] J. Bueso, B. Torrecillas: Noncommutative local cohomology. Comm. Alg. 11(1983), 681-693 </p> <p>[R] J. Raynaud: Profondeur, hauteur et localisations en algebre non commutative. J. of Pure Appl. Alg. 31(1984), 199-215 </p> http://mathoverflow.net/questions/132111/transgression-in-terms-of-cup-product-in-case-of-non-trivial-action-of-the-group/132450#132450 Comment by Ralph Ralph 2013-06-03T09:49:31Z 2013-06-03T09:49:31Z Yes, it's OK. Thank you Mark. http://mathoverflow.net/questions/132320/find-all-faithful-irreducible-representations-of-the-dihedral-groups-d-2n-ove Comment by Ralph Ralph 2013-05-30T11:22:23Z 2013-05-30T11:22:23Z Same question (with an answer): <a href="http://math.stackexchange.com/questions/173811/faithful-irreducible-representations-of-cyclic-and-dihedral-groups-over-finite-f" rel="nofollow" title="faithful irreducible representations of cyclic and dihedral groups over finite f">math.stackexchange.com/questions/173811/&hellip;</a> http://mathoverflow.net/questions/132005/on-avoiding-a-linear-subspace-of-an-algebra Comment by Ralph Ralph 2013-05-30T11:15:56Z 2013-05-30T11:15:56Z What is a linear form in this context ? Is it just a homogeneous element ? http://mathoverflow.net/questions/132222/a-question-on-nilpotent-matrix/132236#132236 Comment by Ralph Ralph 2013-05-29T18:00:51Z 2013-05-29T18:00:51Z I see. I missed this condition. Thanks for the hint. Deleted the answer. http://mathoverflow.net/questions/12586/dual-of-zi-for-uncountable-i Comment by Ralph Ralph 2013-05-29T00:57:04Z 2013-05-29T00:57:04Z To address the question in the title: The $\mathbb{Z}$-dual of $\mathbb{Z}^I$ is the free abelian group whose rank equals the cardinality of the set $D$ of all countably complete ultrafilters on $I$. Moreover, $|I| \le |D|$ and if the cardinality of $I$ is less than the first measurable cardinal, then $|I|=|D|$. For references see my answer to this question: <a href="http://mathoverflow.net/questions/132073/homomorphisms-from-powers-of-z-to-z" rel="nofollow" title="homomorphisms from powers of z to z">mathoverflow.net/questions/132073/&hellip;</a> http://mathoverflow.net/questions/132111/transgression-in-terms-of-cup-product-in-case-of-non-trivial-action-of-the-group Comment by Ralph Ralph 2013-05-28T21:57:49Z 2013-05-28T21:57:49Z Is the upper $G/H$ in $H^{p-1}(...)^{G/H}$ correct ? http://mathoverflow.net/questions/132073/homomorphisms-from-powers-of-z-to-z/132083#132083 Comment by Ralph Ralph 2013-05-28T21:03:08Z 2013-05-28T21:03:08Z ... direct product. This may also be the reason why the title of their paper includes the words &quot;non-commutative&quot;. In particular, they refer to the Eklof-Mekler book for generalizations of the Specker phenomenon on uncountable <i>direct products</i> (p. 420, before Def. 2.1). To summarize: 1. The isomorphism in my 1st comment is correct (reference: Eklof-Mekler, Cor. III.3.7). 2. $\phi$ is an isomorphism if $I$ is not $\omega$-measurable (references in my answer). http://mathoverflow.net/questions/132073/homomorphisms-from-powers-of-z-to-z/132083#132083 Comment by Ralph Ralph 2013-05-28T20:34:13Z 2013-05-28T20:34:13Z Todd, if it seemed that I didn't take your comment/question seriously, I apologize. I just had a look into the Shelah-Str&#252;ngmann paper (it's freely available on <a href="http://www.degruyter.com/view/j/jgt.2013.16.issue-3/issue-files/jgt.2013.16.issue-3.xml" rel="nofollow">degruyter.com/view/j/jgt.2013.16.issue-3/&hellip;</a>). Unfortunately Mariano didn't give a precise reference within the paper where it is shown that $\phi$ (the map from my answer) fails to be an isomorphism for uncountable cardinals. But it seems to me that the point is that Shelah-Str&#252;ngmann consider homomorphisms from the <i>free complete product</i> of groups into the integers while I take the ... http://mathoverflow.net/questions/132073/homomorphisms-from-powers-of-z-to-z/132083#132083 Comment by Ralph Ralph 2013-05-28T19:09:32Z 2013-05-28T19:09:32Z @Todd: As explained by Andreas Blass, not all homs factor when I isn't measurable (I'll correct my remark above later). I guess the Shelah-Str&#252;ngmann paper in the question you linked, is based on a non-measurable index set. However, it should be pointed out that in the measurable as well as in the non-measurable case $Hom(\prod_I \mathbb{Z},\mathbb{Z})\cong \bigoplus_D Hom(\mathbb{Z},\mathbb{Z})$ is a free abelian group with basis a set $D$ of ultrafilters on I. http://mathoverflow.net/questions/132037/group-actions-with-finite-stabilizers-and-compact-quotients/132040#132040 Comment by Ralph Ralph 2013-05-27T20:40:09Z 2013-05-27T20:40:09Z Dieter, thanks, but I won't assume further restrictions on the fixed point spaces. Actually, I wonder, if the strict inclusion $H_1\mathscr{F} \subsetneqq H_2\mathscr{F}$ of the Kropholler classes also holds for cocompact actions. http://mathoverflow.net/questions/129143/verifying-the-correctness-of-a-sudoku-solution/129164#129164 Comment by Ralph Ralph 2013-05-17T10:09:43Z 2013-05-17T10:09:43Z @Tony: Very interesting. You already showed that one needs to check $4 \le s \le 9$ squares and that in the cases $s=9,8,7$ 21 checks are necessary. I think, with some more work, the other cases for $s$ can be treated, too. In particular, this gives a nice systematic for the resulting case-by-case analysis. http://mathoverflow.net/questions/129143/verifying-the-correctness-of-a-sudoku-solution/129599#129599 Comment by Ralph Ralph 2013-05-17T10:06:45Z 2013-05-17T10:06:45Z @Fran&#231;ois: You managed to algebraicify the problem. Great. Thanks. http://mathoverflow.net/questions/129143/verifying-the-correctness-of-a-sudoku-solution/129877#129877 Comment by Ralph Ralph 2013-05-17T10:05:49Z 2013-05-17T10:05:49Z This is great work. Thanks a lot. The direction proved in Lemma 1 by using Fran&#231;ois' linear map is particularly elegant. I tried hard to find a similar approach for the other direction, but didn't succeed yet. http://mathoverflow.net/questions/129143/verifying-the-correctness-of-a-sudoku-solution/129425#129425 Comment by Ralph Ralph 2013-05-17T09:54:23Z 2013-05-17T09:54:23Z @Emil: Thank you very much for writing down your original solution. I would like to accept both of your answers, but unfortunately, this isn't possible on MO. http://mathoverflow.net/questions/130802/a-group-3-cocycle-trivial-on-a-pair-of-generating-subgroups/130810#130810 Comment by Ralph Ralph 2013-05-17T07:57:47Z 2013-05-17T07:57:47Z Yes: Let $H$ resp. $K$ be any finite group those integral cohomology has a non-zero class x resp. y of degree 2. Then $xy\in H^4(H\times K;\mathbb{Z})=H^3(H\times K;\mathbb{C}^\times)$ is non-zero and restricts to zero on $H$ and $K$. If you take $H=K=\mathbb{Z}/p\oplus \mathbb{Z}/p$ then you get an example with $H^2(H;\mathbb{C}^\times)\neq 0$.