Ralph
|
Registered User
|
|
|
May 17 |
comment |
Verifying the correctness of a Sudoku solution @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. |
|
May 17 |
comment |
Verifying the correctness of a Sudoku solution @François: You managed to algebraicify the problem. Great. Thanks. |
|
May 17 |
comment |
Verifying the correctness of a Sudoku solution This is great work. Thanks a lot. The direction proved in Lemma 1 by using François' linear map is particularly elegant. I tried hard to find a similar approach for the other direction, but didn't succeed yet. |
|
May 17 |
comment |
Verifying the correctness of a Sudoku solution @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. |
|
May 17 |
comment |
A group 3-cocycle, trivial on a pair of generating subgroups? 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 $H^2(H;\mathbb{C}^\times)\neq 0$. |
|
May 17 |
accepted | A group 3-cocycle, trivial on a pair of generating subgroups? |
|
May 16 |
comment |
A group 3-cocycle, trivial on a pair of generating subgroups? No, I'm afraid, but it doesn't: For $\mathbb{k}=\mathbb{C}$ the question is equivalent to finding $\omega \in H^4(G;\mathbb{Z})\cong H^3(G;\mathbb{C}^\times)$ that restricts to zero. Since the cohomology of $Q_8$ is 4-periodic, $H^4(Q_8;\mathbb{Z})$ is generated by a cohomology class z (of order 8) that restricts non-zero on both, H and K. |
|
May 16 |
answered | A group 3-cocycle, trivial on a pair of generating subgroups? |
|
May 7 |
awarded | ● Good Question |
|
May 4 |
awarded | ● Popular Question |
|
May 2 |
comment |
Truncation of BG? I think the point is that the assumption $H^p(G^q,M)=0$ for $p,q>0$ from your 3rd comment implies that $H^p(G,M)=H^p\Gamma(G^\ast,M)$ (Prop. 5.1), but the latter agrees with $H^p_{top}(BG,M)$ in general only if $M$ is discrete with trivial G-action. Anyway, it has been interesting to learn about Grothendieck's cohomology theory of topological groups. Thanks for this. |
|
May 2 |
comment |
Applications of Govorov-Lazard Theorem? Very interesting! Thanks. |
|
May 2 |
comment |
Verifying the correctness of a Sudoku solution Emil, thanks. I'll work through the details of your answer (and the other answers and comments) at the weekend and reply at the beginning of next week. |
|
May 2 |
revised |
Representations over $\mathbb{Z}_p$ added 133 characters in body |
|
May 2 |
comment |
Representations over $\mathbb{Z}_p$ Q1: If the theorems you know give you conditions for the indecomposability of an induced representation, then didn't they solve your problem ? |
|
May 2 |
answered | Representations over $\mathbb{Z}_p$ |
|
Apr 30 |
accepted | Support of a module over a polynomial algebra |
|
Apr 30 |
comment |
Verifying the correctness of a Sudoku solution Emil, that's very interesting. It would be great if you still have your notes. BTW: Have you specified a formal definition of what a "check" is ? (e.g. did you only consider algorithms that always check full columns, rows, etc. or did you also take algorithms into account that check only parts of them ?) |
|
Apr 30 |
awarded | ● Nice Question |
|
Apr 29 |
comment |
Verifying the correctness of a Sudoku solution @Denis: Thanks. I'll correct it later. |
|
Apr 29 |
comment |
Verifying the correctness of a Sudoku solution @François: Exactly! I think your conclusion in the 2nd comment is just the row-version of what I described for columns in the paragraph after the pics. |
|
Apr 29 |
comment |
Truncation of BG? I don't think that the statement in your 3rd comment is true as stated, irregardless of the notion of group cohomology: Take in my example above $R$ as trivial coefficients. Then still $H^1_{top}(BR,R)=0$ while either $Hom(R,R)\neq 0$ and $Hom_{continuous}(R,R)\neq 0$ since both contain $id_R$. |
|
Apr 29 |
asked | Verifying the correctness of a Sudoku solution |
|
Apr 26 |
comment |
Truncation of BG? ... Since $BR=\{\ast\}$, $H^1_{top}(BR,Q)=0$ while $H^1_{gr}(R,Q)=Hom(R,Q)=Hom_Q(R,Q)=Q^R$. |
|
Apr 26 |
comment |
Truncation of BG? $\mathbb ZG$ is the usual group ring (integer linear combinations of elements of $G)$; it doesn't depend on the topology of $G$. // In the following I'll use $H^\ast_{top}$ to denote the (cellular) cohomology of a topological space and $H^\ast_{gr}$ for the group cohomology defined in my 4th comment. // If we mean by "group cohomology" the same, then I think there are counterexamples to the statement in your 3rd comment: Set $Q := \mathbb{Q},\;R := \mathbb{R}$ and take $G=(R,+)$ with the usual (Euclidean) topology. $R$ is contractible, hence $H^p_{top}(R^q;Q)=0$ for $p,q>0$. |
|
Apr 26 |
comment |
Truncation of BG? By group cohomology I mean $H^i(G;M)=Ext_{\mathbb ZG}^i(\mathbb{Z},M)$. What is $G^q$ ? |
|
Apr 26 |
comment |
Truncation of BG? I get $B_1S^1=\lbrace \ast\rbrace \; \coprod\; [0,1]\times S^1/\sim$ with the relations $$[0,1] \times 1 \sim \lbrace \ast\rbrace,\;0 \times S^1 \sim \lbrace \ast\rbrace,\;1\times S^1\sim \lbrace\ast\rbrace$$ This looks to me like $B_1S^1=S^2\cong CP^1$. |
|
Apr 26 |
comment |
Truncation of BG? Note that $H^\ast(BS^1,-)\neq H^\ast(S^1,-)$ (where the latter is group cohomology). For example, $H^2(BS^1,\mathbb{Q})=\mathbb{Q}$, while $H^2(S^1,\mathbb{Q})=Hom_\mathbb{Q}(\mathbb{R},\mathbb{Q})$ has as $\mathbb{Q}$-dimension the cardinality of the power set of the reals. An interesting paper that compares the homology of $BG$ and the group homology of $G$ for Lie groups $G$ is Milnor: On the homology of Lie groups made discrete. Comment. Math. Helv. 58(1983), 72-85. |
|
Apr 26 |
accepted | Truncation of BG? |
|
Apr 26 |
revised |
Truncation of BG? added exact sequence |
|
Apr 25 |
comment |
Truncation of BG? Thank you, José! Seems our changes overlapped. |
|
Apr 25 |
revised |
Truncation of BG? replaced "<" by "\lt" in math mode |
|
Apr 25 |
comment |
Truncation of BG? What are these cases where $H^\ast(BG,-)=H^\ast(G,-)$ you have in mind ? |
|
Apr 25 |
answered | Truncation of BG? |
|
Apr 25 |
comment |
Truncation of BG? In general the cohomology of the topological space $BG$ doesn't equal the group cohomology of $G$. However, they agree if $G$ is discrete. |
|
Apr 24 |
answered | Support of a module over a polynomial algebra |
|
Apr 24 |
comment |
Support of a module over a polynomial algebra What is $V_f$ ? |
|
Apr 18 |
comment |
Applications of Govorov-Lazard Theorem? Thanks for the example. I think the principle you formulated in the last sentence is good to keep in mind. |
|
Apr 18 |
comment |
Applications of Govorov-Lazard Theorem? Thanks for the examples. Why have you made your answer community wiki ? |
|
Apr 17 |
comment |
Applications of Govorov-Lazard Theorem? @darij: As far as I can see, L-G is neither used in Higgins paper nor does Higgins treat the flat case. If I'm not missing something this is quite surprising because the flat case follows immediately from Higgins results and L-G. Higgins paper was submitted 1968 while the paper from Govorov was published 1965 and that of Lazard 1969. So it unclear, if Higgins should had known L-G when he was writing his paper. In any case L-G is an interesting application on the flat Birkhoff-Witt. Wouldn't you like to post this as an answer ? |
|
Apr 17 |
comment |
Applications of Govorov-Lazard Theorem? @James: What is the equational criterion ? |
|
Apr 17 |
comment |
Applications of Govorov-Lazard Theorem? Thanks! I'll have a look at the paper. |
|
Apr 16 |
asked | Applications of Govorov-Lazard Theorem? |
|
Apr 14 |
revised |
Homology groups of divisible and powered (nilpotent) groups added tag <group-cohomology> |
|
Apr 14 |
answered | Homology groups of divisible and powered (nilpotent) groups |
|
Apr 9 |
comment |
$k[[x]]$ as a $(k[[x]])^p$ module for ugly fields @Graham: Why is it projective if it's a direct limit of free modules ? |
|
Apr 9 |
comment |
cap products and injective abelian groups Chris, I'm sorry, I saw your question 4 comments above just now. The isomorphism of complexes $\hom_G(X,\hom(M,D))\cong \hom(X\otimes_G M,D)$ holds for all $D$. Taking (co)homology yields $H^n(G,\hom(M,D)) \cong H^n(\hom(X\otimes_G M,D))$. If $D$ is divisible, then $\hom(-,D)$ is exact (as explained by Mariano). The point is now that exact functors commute with (co)homology, i.e. $H^n(\hom(X\otimes_G M,D))\cong \hom(H_n(X\otimes_G M),D)=\hom(H_n(G,M),D)$. |
|
Mar 28 |
accepted | morphism of injective objects |
|
Mar 28 |
answered | morphism of injective objects |
|
Mar 26 |
answered | Applications of n-dimensional crystallographic groups |
