User zuriel - MathOverflowmost recent 30 from http://mathoverflow.net2013-06-19T07:02:53Zhttp://mathoverflow.net/feeds/user/15770http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/98248/list-of-commutator-identities-and-equivalencesList of commutator identities and equivalencesZuriel2012-05-29T06:24:27Z2012-05-29T06:24:27Z
<p>Let $G$ be a group and let $[a,b]=a^{-1}b^{-1}ab$ be the commutator of $a$ and $b$ in $G$. There are several well-known commutator identities such as</p>
<p>$[x, z y] = [x, y]\cdot [x, z]^y$</p>
<p>and</p>
<p>$[[x, y^{-1}], z]^y\cdot[[y, z^{-1}], x]^z\cdot[[z, x^{-1}], y]^x = 1$.</p>
<p>I wonder if there is a more complete list of commutator identities and commutator equivalences of the form</p>
<p>$\alpha(x_1,\cdots,x_n)\equiv\beta(x_1,\cdots,x_n)\mod{N}$,</p>
<p>where $\alpha$ and $\beta$ are expressions involving commutators and $N$ is some subgroup of $G$, for example the $k$-th term in the lower central series of $G$. Anyone knows such a list? Thank you!</p>
http://mathoverflow.net/questions/97533/automorphism-group-of-factor-groupsAutomorphism group of factor groupsZuriel2012-05-21T06:09:30Z2012-05-21T07:09:36Z
<p>Let $G$ be a group and let $H$ be a factor group of $G$. Is there any result that relates $\operatorname{Aut}(G)$ (the automorphism group of $G$) and $\operatorname{Aut}(H)$?</p>
<p>As a very special case of the question, let $F_2$ be the free group with two generators $x$ and $y$. Let $G_2$ be the factor group of $F_2$ by adding relations such that $[x,[x,y]]=[y,[x,y]]=1$; that is, $G_2$ is the discrete Heisenberg group. Then is there any relation between $\operatorname{Aut}(F_2)$ and $\operatorname{Aut}(G_2)$? Hence or otherwise, how to find out $\operatorname{Aut}(G_2)$?</p>
http://mathoverflow.net/questions/95126/profiniteness-condition-for-hochschild-serre-spectral-sequenceProfiniteness Condition for Hochschild-Serre Spectral Sequence?Zuriel2012-04-25T07:46:32Z2012-04-27T08:56:01Z
<p>This question may seem elementary to experts but I am quite confused about it:</p>
<p>According to the entry of Lyndon–Hochschild–Serre spectral sequence on wikipedia, for a group extension $1\to N\to G\to Q\to1$, there is a Lyndon–Hochschild–Serre spectral sequence if $G$ is a profinite group and $N$ is a closed normal subgroup of $G$:</p>
<p><a href="http://en.wikipedia.org/wiki/Lyndon-Hochschild-Serre_spectral_sequence" rel="nofollow">http://en.wikipedia.org/wiki/Lyndon-Hochschild-Serre_spectral_sequence</a></p>
<p>Also in the book Neukirch, Jürgen; Schmidt, Alexander; Wingberg, Kay (2000), <em>Cohomology of Number Fields</em>, the same condition was assumed.</p>
<p>However, according to some other books, for example Brown, Kenneth S. (1972), <em>Cohomology of Groups</em>, and <em>A User's Guide to Spectral Sequences</em> by John McCleary, the profiniteness condition on the group $G$ was NOT assumed. </p>
<p>Why there is such a difference? Do we really need the condition that $G$ is a profinite group and $N$ is a closed normal subgroup of $G$ to construct the Lyndon–Hochschild–Serre spectral sequence?</p>
http://mathoverflow.net/questions/94424/how-to-compute-transgressions-in-a-serre-spectral-sequenceHow to Compute Transgressions in a Serre Spectral Sequence?Zuriel2012-04-18T16:23:32Z2012-04-19T10:19:24Z
<p>For a short exact sequence of groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$ there is an associated fibration $K(A,1)\rightarrow K(B,1)\rightarrow K(C,1)$, which can be constructed by realizing the homomorphism $B\rightarrow C$ by a map $K(B,1)\rightarrow K(C,1)$ and the convert it into a fibration. The fiber is $K(A,1)$ (from the associated long exact sequence of homotopy groups). </p>
<p>For a fibration $F\rightarrow X\rightarrow B$, the differential $d_n\colon E_{n,0}^n\to E_{0,n-1}^n$ in the Serre spectral sequence was shown to be equal to the transgression in Hatcher's book on Spectral Sequences (Proposition 1.13). The transgression was defined using (relative) homology groups.</p>
<p>My questions is: From the short exact sequence of groups $1\rightarrow A\rightarrow B\rightarrow C\rightarrow 1$, is there any method to directly compute the transgression of the associated fibration $K(A,1)\rightarrow K(B,1)\rightarrow K(C,1)$, at least for the case $n=2$, without constructing $K(G,1)$'s and considering their homologies?</p>
http://mathoverflow.net/questions/94465/homology-of-covering-spacesHomology of Covering SpacesZuriel2012-04-18T21:18:23Z2012-04-18T22:24:55Z
<p>Let $A$ be a subgroup of a group $G$. Then since $A$ is a subgroup of the fundamental group $\pi_1(K(G,1))=G$, there is a covering space $p\colon Y\to K(G,1)$ with $p_*(\pi_1(Y))=A$. So the homology of $Y$ should be completely determined by $A$ and $G$. Suppose that $A$ and $G$ is known, how can one compute $H_*(Y)$, the homology groups of $Y$? </p>
http://mathoverflow.net/questions/93621/triply-graded-spectral-sequenceTriply graded spectral sequence?Zuriel2012-04-10T02:34:00Z2012-04-11T16:19:02Z
<p>As we know, most of the spectral sequences are doubly graded. However, this "doubly graded" condition is not a part of the formal definition of spectral sequence. Is there any useful triply (quadruply, quintuply, etc.) graded spectral sequences? If not, is there a hope that some meaningful work can be done with this topic?</p>
http://mathoverflow.net/questions/93603/why-the-underlying-function-of-a-monomorphism-may-not-be-an-injectionWhy the underlying function of a monomorphism may not be an injectionZuriel2012-04-09T20:48:05Z2012-04-10T08:11:06Z
<p>In category theory, a monomorphism (also called a monic morphism or a mono) is defined to be a left-cancellative morphism. It seems that this definition generalizes the definition of injections. However, even in a concrete category, a monomorphism may not be an injection. Why could this happen? I know examples of monomorphisms which are not injections but what is the reason behind the existence of such examples? Isn't monomorphism a generalised concept of injection? Similar questions can be asked about epimorphisms and surjections.</p>
http://mathoverflow.net/questions/93602/a-question-on-mcclearys-book-on-spectral-sequencesA Question on McCleary's book on Spectral SequencesZuriel2012-04-09T20:40:07Z2012-04-10T07:09:43Z
<p>I am reading John McCleary's A User's Guide to Spectral Sequence and was quite confused about one result: On page 15 of the version I was reading, it says that if $E^{\star,\star}_2$ is the bigraded vector space in Example 1.E, then $P(E^{\star,\star}_2,t)=(1+t^{11})(1+t^4+t^8+t^{12})(1+t^3)$. I am quite confused on how to obtain this result from Example 1.E. It seems to me that $P(E^{\star,\star}_t)$ has a term $t^{11+12+3}=t^{26}$, which by definition of $P(E^{\star,\star}_2,t)$ implies that $\text{dim}_k(\bigoplus _{p+q=26}E^{p,q})=1$. Why is that? I am not sure if I have understood Example 1.E wrongly. Any explanation will be greatly appreciated.</p>
<p>Update: Thank you @Neil Strickland for reminding me. The conditions of Example 1.E are: Suppose $E^{\star,\star}_2$ is given as an algebra by </p>
<p>$E^{\star,\star}_2\cong\mathbb{Q}[x,y,z]/(x^2=y^4=z^2=0)$,</p>
<p>where the bidegree of each generator is given by $\text{bideg}x=(7,1)$, $\text{bideg}y=(3,0)$ and $\text{bideg}z=(0,2)$. Furthermore, suppose $d_2(x)=y^3$ amd $d_3(z)=y$. In this case, the spectral sequence collapses at $E_4$ and, though $x$ and $y$ do not survive to $E_{\infty}$, the product $xy$ does.</p>
http://mathoverflow.net/questions/89626/rational-homology-of-a-covering-spaceRational Homology of a Covering SpaceZuriel2012-02-27T02:00:48Z2012-02-27T03:04:48Z
<p>I have heard that the rational homology of a covering space is easy to compute, compared with the ordinary homology. However, I don't know any details about that. Can anyone help me? Any reference will be greatly appreciated.</p>
http://mathoverflow.net/questions/88481/computing-the-homology-groups-of-spaces-in-a-fibrationComputing the homology groups of spaces in a fibrationZuriel2012-02-15T02:50:52Z2012-02-15T09:03:27Z
<p>Let $F\rightarrow X\rightarrow B$ be a fibration. If we know very well the spaces $F$ and $B$ and wish to compute the homology of $X$. One possible tool is the Serre Spectral Sequence. However, it works under the condition that $\pi_1(B)$ acts trivially on $H_*(F;G)$. If this condition ($\pi_1(B)$ acts trivially on $H_*(F;G)$) does not hold, what other tools can one use to compute the homology of the homology of $X$?</p>
<p>In fact I am interested in the special case that all spaces in the fibration are $K(\pi,1)$ spaces. If any approach works for this particular case it would be wonderful.</p>
<p>Thank you! </p>
http://mathoverflow.net/questions/86082/homotopy-fibre-sequence-of-classifying-spacesHomotopy-Fibre Sequence of Classifying SpacesZuriel2012-01-19T08:44:29Z2012-01-19T08:44:29Z
<p>Let $G$ be a topological group and $H$ be a normal subgroup of $G$ (I think $H$ is required to be admissible in the sense that the quotient map $G\to G/H$ is a principal $H$-bundle, am I right?). Then there exists a homotopy fibre sequence $BH\to BG\to B(G/H)$, where $BG$ denotes the classifying space of $G$. </p>
<p>My questions is: suppose that we already know the groups $G$ and $H$ and suppose that we know the classifying space of $G/H$ and the classifying space of $H$, to what extent can we decide the classifying space of $G$ from these information? How to find the classifying space of $G$ if we know the classifying space of $G/H$ and the classifying space of $H$?</p>
<p>Any reference will be greatly appreciated. </p>
http://mathoverflow.net/questions/84668/the-classifying-space-of-the-discrete-heisenberg-groupThe Classifying Space of the Discrete Heisenberg GroupZuriel2012-01-01T01:48:31Z2012-01-01T09:00:35Z
<p>What is the Classifying Space of the Discrete Heisenberg Group? Which paper/book contains a detailed proof?</p>
<p>Thank you for your time.</p>
http://mathoverflow.net/questions/72913/centerlessness-of-reduced-free-groupCenterlessness of reduced free groupZuriel2011-08-15T08:39:01Z2011-08-15T18:20:38Z
<p>Let F_n be the free group with n generators, where n is an interger greater than 1. Let RF_n be the reduced free group, which is defined to be the quotient group of F_n obtained from F_n by adding the relations that each generator of F_n commutes with all its conjugates.</p>
<p>Can anyone help to prove or disprove that RF_n is centerless, that is, the center of RF_n is trivial?</p>
<p>Your help will be much appreciated.</p>
http://mathoverflow.net/questions/67757/where-can-one-find-reference-proving-that-braid-group-induces-isomorphism-betweenWhere can one find reference proving that Braid group induces isomorphism between punctured disk and the complement of the braid?Zuriel2011-06-14T13:04:23Z2011-06-15T05:23:56Z
<p>It is a known result that if $B$ is an $n$ braid over a disk, then $B$ naturally induces an isomorphism between the fundamental group of a disk with n points removed and the fundamental group of the space $D\times [0,1]-B$, where $D$ is a disk. My question is, in which book/paper can I find a proof of this result?</p>
http://mathoverflow.net/questions/67756/question-about-some-element-in-a-group-commutes-with-its-all-conjuagatesQuestion about some element in a group commutes with its all conjuagates.Zuriel2011-06-14T12:56:59Z2011-06-14T14:56:36Z
<p>Let G be a group and g and h be elements in G. If g commutes with all conjugates of g and h commutes with all conjugates of h, can one conclude that gh commutes with all conjugates of gh?</p>
<p>Thanks!!</p>
http://mathoverflow.net/questions/97533/automorphism-group-of-factor-groups/97537#97537Comment by ZurielZuriel2012-05-21T07:34:02Z2012-05-21T07:34:02Z@Mark Sapir, Thank you for your answer! Just to clarify: did you mean that in the free group with generators $x$ and $y$, the normal subgroup generated by $[x,[x,y]]$ and $[y,[x,y]]$ is a fully invariant group? If yes, why is it so?http://mathoverflow.net/questions/94465/homology-of-covering-spaces/94470#94470Comment by ZurielZuriel2012-04-19T11:05:50Z2012-04-19T11:05:50Z@Igor Rivin, for your long answer, may I know which part(s) of the book should I read to find answers to my question?http://mathoverflow.net/questions/94465/homology-of-covering-spacesComment by ZurielZuriel2012-04-18T22:16:42Z2012-04-18T22:16:42ZMany thanks @Aru Ray! I will find the book and look for related results.http://mathoverflow.net/questions/94465/homology-of-covering-spacesComment by ZurielZuriel2012-04-18T21:48:37Z2012-04-18T21:48:37Z@Aru Ray, so you mean that the result has nothing to do with $G$?http://mathoverflow.net/questions/94424/how-to-compute-transgressions-in-a-serre-spectral-sequenceComment by ZurielZuriel2012-04-18T19:38:59Z2012-04-18T19:38:59Z@Josh, it seems to be quite a big book. May I know which page/section should I read for this question?http://mathoverflow.net/questions/94424/how-to-compute-transgressions-in-a-serre-spectral-sequenceComment by ZurielZuriel2012-04-18T19:32:12Z2012-04-18T19:32:12Z@Ralph, yes, it is. However, the name "Lyndon-Hochschild-Serre spectral sequence" was not explicitly mentioned in Hatcher's book.http://mathoverflow.net/questions/93621/triply-graded-spectral-sequenceComment by ZurielZuriel2012-04-10T04:08:29Z2012-04-10T04:08:29ZI am still new to this website and am not sure about many of its functions. Will wait more as you have suggested; thank you!http://mathoverflow.net/questions/93602/a-question-on-mcclearys-book-on-spectral-sequencesComment by ZurielZuriel2012-04-10T02:47:08Z2012-04-10T02:47:08Z@Ben Williams, this change is not mentioned in the book. Does this mean that the case is that this part of the book contains some flaws, not I understood it wrongly?http://mathoverflow.net/questions/86082/homotopy-fibre-sequence-of-classifying-spacesComment by ZurielZuriel2012-04-02T00:51:37Z2012-04-02T00:51:37Z@jim stasheff, may I know which book you are refering to by "Borel"? It seems to me that Eilenberg−Moore spectral sequences require the base space B to be simply connected; what if this condition does not hold? Thanks in advance!! http://mathoverflow.net/questions/88481/computing-the-homology-groups-of-spaces-in-a-fibrationComment by ZurielZuriel2012-02-15T03:59:21Z2012-02-15T03:59:21Z@Ryan, May I know where I can find any reference on the Serre Spectral sequence for any fibration mentioned in your comment? Thank you!http://mathoverflow.net/questions/84668/the-classifying-space-of-the-discrete-heisenberg-group/84676#84676Comment by ZurielZuriel2012-01-02T02:57:36Z2012-01-02T02:57:36Z@Alain Valette, you have mentioned that $\Gamma$ can be viewed as the semi-direct of $\mathbb{Z}^2$ and $\mathbb{Z}$; did you mean that $n\in\mathbb{Z}$ is mapped to the element $f_n\in Aut(\mathbb{Z}^2)$, such that $f_n(m_1,m_2)=m_1+nm_2$? Then why is $\Gamma$ isomorphic to the semi-direct product? It seems that the central extension $0\rightarrow\mathbb{Z}\rightarrow\Gamma\rightarrow\mathbb{Z}^2\rightarrow0$ does not split; then is there any nature semi-direct structure from this central extension? Thank you!http://mathoverflow.net/questions/72913/centerlessness-of-reduced-free-group/72915#72915Comment by ZurielZuriel2011-08-15T10:15:07Z2011-08-15T10:15:07Z@Mark Sapir: as you mentioned in your previous answer, [a, b] is in the center of RF_2 and thus the group has a center. But do we also need to prove [a, b] is not equal to the identity element of the group? I know that it should be true; but how to prove it?
Thank you again for your answer.http://mathoverflow.net/questions/72913/centerlessness-of-reduced-free-group/72915#72915Comment by ZurielZuriel2011-08-15T09:21:00Z2011-08-15T09:21:00ZThank you Mark for your answer!!
Then can we conclude that the center of RF_n is the intersection of all N_i, as is defined in your answer?http://mathoverflow.net/questions/67756/question-about-some-element-in-a-group-commutes-with-its-all-conjuagates/67761#67761Comment by ZurielZuriel2011-06-15T10:57:15Z2011-06-15T10:57:15ZThank you for your answer! May I know more details (title, etc.) of those classical papers by Baer and others? I would like to read them.http://mathoverflow.net/questions/67756/question-about-some-element-in-a-group-commutes-with-its-all-conjuagates/67759#67759Comment by ZurielZuriel2011-06-15T10:54:32Z2011-06-15T10:54:32ZThank you so much for your answer! May I know how $G_1$ and $G_2$ are defined and how do you know that there are exactly 15 elements in each group with the given property?