User joseph - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:14:48Z http://mathoverflow.net/feeds/user/17456 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84510/irreducibility-of-induced-representation-over-arbitrary-field Irreducibility of Induced Representation over arbitrary field joseph 2011-12-29T12:29:21Z 2012-01-11T04:37:58Z <p>In representation theory of finite groups, there is Mackey test for irreducibility of an induced representation (Serre- Linear Representations of Finite Groups - #7.4). The author has stated it for field $\mathbb{C}$. Is it true for representations over arbitrary field whose characteristic is zero or co-prime to the order of group ?</p> http://mathoverflow.net/questions/77812/uniqueness-of-splitting-field-for-linear-representations-of-finite-groups Uniqueness of splitting field for linear representations of finite groups joseph 2011-10-11T13:09:21Z 2011-10-12T06:10:22Z <p>If $F$ is any field, $\bar{F}$ its algebraic closure, then it is well-known that all irreducible (indecomposable) $\bar{F}$-representations of a finite group $G$ are realizable over some <em>finite</em> extension $E$ of $F$, and we call $E$, a splitting field for $G$. Any extension of $E$ is also a splitting field of $G$. By finiteness of $[E\colon F]$, we can have a <em>minimal</em> extension of $F$ which is splitting field for $G$.</p> <p>I puzzled by the following question:</p> <p>Is such minimal extension unique?</p> <p>(For example, consider a finite group $G$, $\mathbb{F}_p$ a field of order $p$ with $(|G|,p)=1$. Then there is $n\in \mathbb{N}$ such that $\mathbb{F}_q$ ($q=p^n$) is a splitting field field of $G$.</p> <p>Is it possible that there are subfields of $\mathbb{F}_q$ of order $p^{m_1}$ and $p^{m_2}$, $m_1\nmid m_2$, and $m_2\nmid m_1$ such that they are splitting fields of $G$, but their subfields are not splitting fields?)</p> http://mathoverflow.net/questions/74558/coverings-of-a-graph-of-groups Coverings of a graph of groups joseph 2011-09-05T06:50:51Z 2011-09-18T15:00:30Z <p>For topological space $X$ (connected, path connected etc.), there is classification of coverings of $X$ : for fixed $x_0\in X$, consider $\pi_1(X,x_0)$. Then there is a $1-1$ correspondance between conjugacy class of subgroups of $\pi_1(X,x_0)$ and covering spaces of $X$ (upto isomorphism). We define universal cover of $X$ to be a cover $\tilde{X}$ which is a cover of every cover $Y$ of $X$. This exist and unique up to isomorphism.</p> <p>Now consider a <em>graph of groups</em> $(X,A)$, where $X$ is a graph and $A=(A_v,A_e)$ is a family of groups attached to vertices $v\in V(X)$ and edges $e\in E(X)$ of $X$ with injections from edge groups to end-vertex groups.</p> <p>In the book <em>"Trees"-Serre</em>, the universal cover of $(X,A)$ is defined to be a connected graph $\tilde{X}$ with:</p> <p>1) a morphism $p\colon \tilde{X}\rightarrow X$ of graphs;</p> <p>2) an action of $\pi_1(X,x_0)$, $(x_0\in V(X))$ on $\tilde{X}$ such that stabilizer of $\tilde{v}\in p^{-1}(v)$ is isomorphic to vertex group $A_v$, $v\in V(X)$</p> <p>(In other words, it is a graph, with action of $\pi_1(X,x_0)$, $(x_0\in V(X)$, such the quotient graph of groups $X/\pi_1(X,x_0)$ is isomorphic to given graph of groups).</p> <p><em><strong>Question</em></strong>: Is there a construction of universal cover of a graph of groups analogous to the construction of universal cover of topological spaces ( i.e. a cover of every cover of given graph of groups).</p> <p>As an illustration, how can we obtain all coverings of $(X,A)$ where $X$ is the graph $\circ --\circ$ with vertex groups $\mathbb{Z}/l$, and $\mathbb{Z}/n$ and edge group trivial.</p> http://mathoverflow.net/questions/74985/number-of-subgroups-of-p-groups Number of Subgroups of p-Groups joseph 2011-09-09T09:56:56Z 2011-09-09T15:07:23Z <p>If $G$ is a finite $p$-group of order $p^n$, then it is well known that for ($1\leq m\leq n$), number of subgroups of order $p^m$ is $1$(mod $p$). </p> <p><em><strong>Question:</em></strong> Is it true that number of subgroups of order $p^m$, which are isomorphic within themselves, is $0$(mod $p$) or $1$(mod $p$). </p> <p>It looks to be true for groups of order $p^2$, $p^3$.</p> http://mathoverflow.net/questions/74712/embedding-of-finite-2-groups Embedding of Finite 2-groups joseph 2011-09-07T05:16:47Z 2011-09-07T05:47:52Z <p>Consider the finite 2-groups containing cyclic subgroup of index 2:</p> <blockquote> <p>$C_{2^n}$, $C_{2^{n-1}}\times C_2$, $D_{2^n}$, $SD_{2^n}$, $QD_{2^n}$, $Q_{2^n}$.</p> </blockquote> <p>Can every finite (non-abelian) 2-group be embedded in $H\times K$ where $H$, and $K$ are one of the groups in the list?</p> http://mathoverflow.net/questions/74032/finite-quotients-of-graphs Finite quotients of graphs joseph 2011-08-30T02:36:09Z 2011-08-30T16:45:49Z <p>If $X$ is an infinite graph, $G$ is a group acting on $X$ with finite quotient; make $Y=X/G$ into graph of groups by attaching stabilizers at vertices and edges. Let $Z$ be a graph of groups, with graph equal to graph of $X/G$, all groups at vertices and edged being finite(but not all trivial), and suppose there is morphism of graph of groups $\phi:X/G\rightarrow Z$ , such that $\phi$ is graph isomorphism, and its restriction to vertex groups (and edge groups) is surjective homomorphism. </p> <p>Does there exist a group whose action on $X$ gives the graph of groups $Z$.</p> http://mathoverflow.net/questions/73942/groups-acting-on-graph Groups acting on graph joseph 2011-08-29T05:18:05Z 2011-08-29T07:21:00Z <p>Let $S$ be an infinite graph, $G$ is a group acting (effectively) on $S$ with finite quotient graph $S/G$. Make $S/G$ into graph of groups in obvious way by assigning stabilizers at vertices and edges.</p> <p>Let $\tilde{S}$ be universal cover of $S$ and $H$ be a group acting (effectively) on $\tilde{S}$ with same quotient as graph $S/G$, but the graph of group may be different. </p> <p>Question: Does there exist a subgroup $K$ of $G$ which acts on $S$, with quotient graph of groups $S/K$ and $\tilde{S}/H$ isomorphic?</p> <p>[Here I am failing to use covering space theory directly, because here I am considering quotients spaces with some algebraic structures on them, namely graph of groups. This problem arise when I was studying Serre's "Trees".] </p> http://mathoverflow.net/questions/84510/irreducibility-of-induced-representation-over-arbitrary-field/84518#84518 Comment by joseph joseph 2012-01-11T04:26:59Z 2012-01-11T04:26:59Z As you have pointed out, for splitting fields, the theorem is true; no doubt about it. But I am not sure, whether it holds for non-splitting field. Is the answer &quot;Exactly YES&quot;? http://mathoverflow.net/questions/84510/irreducibility-of-induced-representation-over-arbitrary-field/84513#84513 Comment by joseph joseph 2012-01-11T04:23:53Z 2012-01-11T04:23:53Z The isomorphism of vector spaces $Hom_H(V,V)\rightarrow Hom_G(Ind^G_H V, Ind^G_H V)$ will hold if $V$ is irreducible and $Res^H_{H\cap H_s} V$ and $Res^{H_s}_{H\cap H_s}V_s$ ($s\in G\setminus H$) do not share an irreducible component; what about converse? http://mathoverflow.net/questions/77812/uniqueness-of-splitting-field-for-linear-representations-of-finite-groups/77824#77824 Comment by joseph joseph 2011-10-12T08:24:17Z 2011-10-12T08:24:17Z Thanks! Nice Answer!! http://mathoverflow.net/questions/77812/uniqueness-of-splitting-field-for-linear-representations-of-finite-groups/77824#77824 Comment by joseph joseph 2011-10-12T04:08:50Z 2011-10-12T04:08:50Z @Torsten: Can you explain second paragraph in answer, please? (I know that Q8 has four 1-dim. irrep, and one 4-dim. irrep over Q; and over C (or Q(i)), it splits as sum of two irrep. of dim. 2 (i.e. twice the irrep. of dim 2 over $\mathbb{C}$)) http://mathoverflow.net/questions/77812/uniqueness-of-splitting-field-for-linear-representations-of-finite-groups Comment by joseph joseph 2011-10-11T13:38:30Z 2011-10-11T13:38:30Z I am considering representations only on <i>splitting fields</i>, and looking whether there is <i>smallest</i> subfield (whether exist) in the algebraic closure $\bar{F}$. Not fixing any representation, since <i>every</i> irreducible representation over a splitting field remains irreducible even after extension of that splitting field, so not considering only one representation, but all <i>irreducible representations</i>. Is this the thing you want to know? http://mathoverflow.net/questions/37136/classification-of-finite-groups-of-isometries/62734#62734 Comment by joseph joseph 2011-09-10T07:44:55Z 2011-09-10T07:44:55Z ***@Richard and Geoff***: Why should we consider &quot;irreducible representations&quot; of groups? In language of representation theory, the question will be simply &quot;To find faithfull representations of finite groups over $\mathbb{R}$&quot;; not necessarily irreducible. http://mathoverflow.net/questions/74985/number-of-subgroups-of-p-groups Comment by joseph joseph 2011-09-09T13:45:41Z 2011-09-09T13:45:41Z Given a subgroup $L$ of $G$, consider the set $X=\{H\leq G \colon H\cong L\}$. Does the cardinality of this set is $0$ (mod $p$) or $1$ (mod $p$)? For groups of order $p^2$, and $p^3$, it looks to be true. http://mathoverflow.net/questions/74558/coverings-of-a-graph-of-groups/74569#74569 Comment by joseph joseph 2011-09-05T10:28:16Z 2011-09-05T10:28:16Z Also the Bass-Serre tree is just a tree; why we do not construct universal cover of graph of groups as again certain graph of groups with obvious universal property? http://mathoverflow.net/questions/74558/coverings-of-a-graph-of-groups/74569#74569 Comment by joseph joseph 2011-09-05T10:28:12Z 2011-09-05T10:28:12Z @HW: The universal cover (Bass-Serre tree) of graph of groups is a tree on which the fundamental group of graph of group acts, with quotient graph of groups isomorphic to given graph of groups. But can we define it as a cover of given graph of groups which is also cover of any other cover of given graph of groups (similar to &quot;topological universal cover&quot;)? I want to get all covers of graph of groups $\circ --\circ$, with vertex groups finite cyclic, edge group trivial. http://mathoverflow.net/questions/74558/coverings-of-a-graph-of-groups Comment by joseph joseph 2011-09-05T07:02:18Z 2011-09-05T07:02:18Z @David: Thanks! I was adding tag of &quot;geometric group theory&quot;, but didn't get it. so added just group theory. Anyway!. http://mathoverflow.net/questions/74558/coverings-of-a-graph-of-groups Comment by joseph joseph 2011-09-05T06:51:37Z 2011-09-05T06:51:37Z As per the construction of Serre, the universal cover of a graph of groups is (just) a graph; so are we not moving away from category of graph of groups to category of graphs? Why we do not construct a universal cover as a graph of groups again? http://mathoverflow.net/questions/73942/groups-acting-on-graph/73948#73948 Comment by joseph joseph 2011-08-29T08:03:40Z 2011-08-29T08:03:40Z @HW: True! To avoid countably branching in $S$ or $\tilde{S}$, we can consider $S$ to be locally finite. Then can we hope for positive answer? http://mathoverflow.net/questions/73942/groups-acting-on-graph Comment by joseph joseph 2011-08-29T07:09:25Z 2011-08-29T07:09:25Z @HW: Thanks! I have changed it.