User trew - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T23:07:53Z http://mathoverflow.net/feeds/user/9514 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/125018/classification-of-hopf-algebra-with-exactly-two-1-dimensional-modules Classification of Hopf algebra with exactly two 1-dimensional modules trew 2013-03-19T20:56:35Z 2013-03-26T22:06:20Z <p>Is there a classification of indecomposable non-semisimple finite dimensional Hopf algebras with exactly two 1 dimensional modules? If not, is there one when all simple modules are 1 dimensional and there are only two simple modules? If there is no full answer, I'm also interested in some examples of such Hopf algebras. By the way: is there a formula for the number of simple modules involving the structure of the group of grouplike elements like in the case of a group algebra? Thank you for answers. edit: Im also interested for non-semisimple indecomposable k-algebras whose basic algebra is a hopf algebra(in general or more special with only 2 simple modules).maybe there is some kind of criteria?</p> http://mathoverflow.net/questions/125017/an-injective-smooth-function-with-injective-differential-must-have-a-continuous-i/125021#125021 Answer by trew for An injective smooth function with injective differential must have a continuous inverse? trew 2013-03-19T21:04:17Z 2013-03-19T21:04:17Z <p>Yes this is true since your map is open by the local diffeomorphism theorem. We can find for every $x \in U$ an open $U_x$ such that f is a diffeomorphism from $U_x$ to $f(U_x)$.(i think we need that f is in $C^{1}$ not just differentiable)</p> http://mathoverflow.net/questions/87683/heller-functor-of-finite-order heller functor of finite order trew 2012-02-06T16:31:40Z 2012-02-07T15:59:09Z <p>Hi, let A be a finite dimensional selfinjective algebra. Assume mod A is periodic.That is for every finite dimensional module with no projective summand we have: $\Omega^{n} (M) =M$ for some n ,where $\Omega$ is the Hellerfunctor(giving the kernel of a projective cover of M ). When is it true that $\Omega$ is of finite order,that is $\Omega^{n}$ is isomorphic as a functor to the identity functor of the stable category of modules ?(Assume there is a common biggest period for all module if necassary)</p> <p>Thanks for help</p> http://mathoverflow.net/questions/69653/a-question-about-the-existence-of-a-specific-extension-of-a-character A question about the existence of a specific extension of a character. trew 2011-07-06T18:21:41Z 2012-01-31T12:45:18Z <p>general situation:<br> Let $N \leq G$ be a subgroup,and let $\chi \in Irr(G)$ be an irreducible character of G such that $\chi_N$ is not irreducible( i dont think that this is really needed) and let $\psi \in Irr(N)$ be an irreducible constituent of $\chi_N$. Assume there is an subgroup H of G with $N \leq H \leq G$ to which $\psi$ is extendible. Then there is a character $\mu \in Irr(H)$ with $\mu_N =\psi$ and $[\chi_H , \mu] \neq 0.$ NOTE: the problem here is to find such an extension with $[\chi_H , \mu] \neq 0.$ Is this true in general?<br> I have the following specific situation: Let J be a nilpotent finitedimensional algebra over a finite field. Then G=1+J(called finite algebra group) is a p-group,$N=1+J^{2}$ is a normal subgroup. In the paper "On characters and commutators of finite algebra groups" written by Halasi,he writes:<br> Lemma 3.1: "Let G=1+J be a finite Algebra group and $\chi \in$ Irr(G).Then the following properties are equivalent:<br> 1.There exists a proper algebra group H and $\varphi \in Irr(H)$ such that $\varphi^{G}=\chi$.<br> 2.$\chi_{1+J^{2}}$ is not irreducible.<br> Proof:...<br> Assume now that $\chi_{1+J^{2}}$ is not irreducible and let $\psi \in Irr(1+J^{2})$ be a constituent of $\chi_{1+J^{2}}$.Let H be a maximal algebra subgroup such that $\psi$ is extendible to H.Then H $\neq$G.We choose a $\varphi \in Irr(H)$such that $\varphi$ is an extension of $\psi$ and $\varphi$ is a constituent of $\chi_H$...."<br> I wonder why there is such a $\varphi$ .Is it true in the general situation or what properties of 1+J and $1+J^{2}$ or H(being maximal) are used here? Thanks for helping</p> http://mathoverflow.net/questions/85855/modular-representation-theory-of-the-generalized-quaternion-group Modular Representation Theory of the generalized Quaternion group trew 2012-01-16T22:46:21Z 2012-01-16T22:46:21Z <p>Hi, here (http://www.math.ku.dk/english/research/conferences/group.actions2011/problem.session.seattle96-maybe.pdf/ Conjuncture 3.6) i read that the indecomposables of the generalized Quaternion group,which has tame representation type,are still not classified. Is this still the case and if yes what do we know about the representation theory of those groups? Thanks for the help</p> http://mathoverflow.net/questions/83520/what-do-we-know-about-periodic-modules-in-p-groups What do we know about periodic modules in p-groups? trew 2011-12-15T13:40:07Z 2011-12-28T15:46:12Z <p>Hi,</p> <p>a module in KG,where G is a p-group and K a field of characteristic p, is called periodic if $\Omega^{n} M = M$, for a natural n. In general the full subcategory of periodic modules seems to have also wild representation type( <a href="http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.ijm/1255989005" rel="nofollow">http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.ijm/1255989005</a> ). I wonder if there are still some interesting results about periodic modules. So I search for a kind of up-to-date survey paper listing such results. some questions are:</p> <p>In which dimensions can a module of period n occur?(results like in this paper: <a href="http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.ijm/1256048241" rel="nofollow">http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&amp;id=pdf_1&amp;handle=euclid.ijm/1256048241</a> where it is proven that a power of p divides the dimension)</p> <p>Which periods can occur in a given group?</p> <p>Is there any interesting relation of the subcategory of periodic modules and the pure group structure?</p> <p>Thank you</p> <p>edit: Another question: Can we give an example of a periodic module in an arbitrary KG?Maybe there is a canonical construction.</p> <p>edit2: after reading parts of benson im a bit confused.for example in the introduction he says compelextity 1 is equivalent being periodic.But he says something else in a later theorem. Is the following correct?: M has complextity 1 iff</p> <p>$M_E$ has maximal complextity 1 for an elementar abelian subgroup E of G iff</p> <p>M is a direct sum of indecomposable periodcis and projectives iff</p> <p>in the minimal projective resolution the terms have bounded dimension.</p> http://mathoverflow.net/questions/83519/dimension-of-the-radical-of-the-tensor-product Dimension of the radical of the tensor product trew 2011-12-15T13:26:50Z 2011-12-15T13:26:50Z <p>Hi, are there any useful formulas for $dim_{k}(rad(M \otimes_{k} N))$(for example relating it do $dim_{k} M$ and $dim_{k} N$ and $dim_{k} rad(M)$and $dim_{k} radN$) where M and N are finite-dimensional KG modules(Characteristic p) and G is a finite group with p divides the order(or some interesting specialisation like a p-group if you want) Thank you!</p> http://mathoverflow.net/questions/10535/ways-to-prove-the-fundamental-theorem-of-algebra/75033#75033 Answer by trew for Ways to prove the fundamental theorem of algebra trew 2011-09-09T20:05:22Z 2011-09-09T20:05:22Z <p>There seems to be a new elementary proof using only bolzano-weierstraß and an inequality: <a href="http://de.arxiv.org/PS_cache/arxiv/pdf/1109/1109.1459v1.pdf" rel="nofollow">http://de.arxiv.org/PS_cache/arxiv/pdf/1109/1109.1459v1.pdf</a></p> http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group Irreducible representations of the unitriangular group trew 2011-06-19T12:43:07Z 2011-08-14T12:21:29Z <p>Hi, I wonder how much is known about the irreducible representations of the nxn unitriangular group over a finite field with q elements.<br> I know that all characterdegrees are a power of q and all degrees which occur are known.But what is known about the irreducible representations or the complete charactertable at least for small values of n?For example is the charactertable for n=3 known? Thanks for helping</p> http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group/72866#72866 Answer by trew for Irreducible representations of the unitriangular group trew 2011-08-14T12:21:29Z 2011-08-14T12:21:29Z <p>Here is what I found out about the characters when n=4.I dont know if thats interesting and how to get the actuall irreducible characters then.Maybe someone has an idea: There are $q^{3}$ linear and from <a href="http://fourier.math.uoc.gr/~marial/uni1.published.pdf" rel="nofollow">http://fourier.math.uoc.gr/~marial/uni1.published.pdf</a> there are $q^{3}-q$ characters of degree q and $q(q-1)$ characters of degree $q^{2}$. Now lets look at the characters of $G/Z(G)=1+J/J^{3}$ ,where Z(G) is the center of the group and J are the lower triangular matrices with zeros on the diagonal.This is again an algebragroup,so we have: $q^{5}=q^{3}+aq^{2}+bq^{4}$,where a is the number of the degree q characters of G/Z(G) and b the number of degree $q^{2}$ characters. Assume $\phi$ is a degree $q^{2}$ character in G/Z(G),then $[ \phi_{Z(G/Z(G))} , \psi] \neq 0$ ,for a linear character $\psi$ of Z(G/Z(G)).But then since $\psi$ is G/Z(G) invariant as a character of the center and using Clifford: $\phi_{Z(G/Z(G))}=q^{2} \psi$ and then $\psi^{G/Z(G)} = q^{2} \phi +...$ which is not possible because of $\psi^{G/Z(G)}(1)=q^{3} &lt; q^{4} =q^{2} \phi(1)$. So we have b=0 and a=q^{3}-q from $q^{5}=q^{3}+aq^{2}$.So all the degree q characters of G are also the degree q characters of G/Z(G). Let now $\chi$ be a character of degree q of G/Z(G) then we can choose a linear character $\psi$ of Z(G/Z(G)) with $[\chi_{Z(G/Z(G))} ,\psi] \neq 0$ and again as above: $\psi^{G/Z(G)} = q^{2} \chi +...$ Since $(1_{Z(G/Z(G))})^{G/Z(G)}$ has all linear characters as constituents $\psi^{G/Z(G)}$ can only have degree q irreducible constituents.So for all(there are $q^{2}-1$ such) the nontrivial linear characters $\psi_k$ of Z(G/Z(G)),we have : $(\psi_k)^{G/Z(G)} = q \sum\limits_{i=1}^{q} {\chi_i}$. Similary one can show that for all nontrivial linear characters $\vartheta_k$ of Z(G) one has: $(\vartheta_k)^{G}=q \sum\limits_{i=1}^{q} {\phi_i}$,where $\phi_i$ are degree-$q^{2}$ characters of G.</p> http://mathoverflow.net/questions/69653/a-question-about-the-existence-of-a-specific-extension-of-a-character/69681#69681 Answer by trew for A question about the existence of a specific extension of a character. trew 2011-07-06T23:24:24Z 2011-07-10T15:22:18Z <p>Hi, I just saw a theorem in a paper of Isaacs which could be useful. Assume additional to the general case(with same notation)that N is normal in H and H/N is abelian. Then the lemma: </p> <p>"Let N be a normal subgroup of H and H/N abelian.Let $\vartheta \in Irr(H)$and $\psi \in Irr(N)$ and $[\psi,\vartheta_N] \neq 0$.Then every Z $\in Irr(H)$ with $[Z_N,\psi] \neq 0$ has the form $Z=\lambda \vartheta$ for a linear $\lambda \in Lin(H/N)$ " </p> <p>tells us that every irreducible constituent of $\psi^{H}$ has the form $\lambda \vartheta$,where $\lambda \in Lin(H/N)$ because if Z is an irreducible constituent of $\psi^{H}$ then $[Z,\psi^{H}]=[Z_N,\psi]$.We can then write $\psi^{H}$ as a sum of all different $\lambda \vartheta$ with multiplicity one,because $[\psi^{H} , \lambda \vartheta]=[\psi, (\lambda \vartheta)_N]=[\psi,\psi]=1$. </p> <p>We have $[\chi_H , \psi^{H}]=[\chi,\psi^{G}]=[\chi_{N},\psi] \neq 0$ Choose an irreducible constituent of $\psi^{H}$,called $\phi$ with $[\phi,\chi_H] \neq 0.$ On the one hand we have: $(\psi^{H})_{N} = |H:(N)| \psi$ and on the other (since $[\phi,\psi^{H}]=[\phi|{N} , \psi]$): </p> <p>$(\psi^{H})_{N} = a \phi +etc$ Comparing these 2 gives: $\phi|{N}=c \psi$ for natural a and c and etc as a combination of other caracters. we have to show that c=1 to finish the proof. $c=[\phi|{N} , \psi]=[\phi,\psi^{H}]$,but we showed above that $\psi^{H}$ has only irreducible constituents with multiplicity one. I am thankful for proofreading this or giving hint to avoid the additional assumptations.</p> http://mathoverflow.net/questions/68101/p-groups-realisable-as-1j-where-j-is-a-nilpotent-finite-f-algebra p-groups realisable as 1+J,where J is a nilpotent finite F-Algebra trew 2011-06-17T22:23:09Z 2011-06-19T00:45:06Z <p>There is a functor F: {finite nilpotent Algebra over a finite field F} -> {finite p-groups} sending J to 1+J.(You can think of J as the Jacobsonradical of 1F+J) and sending f:A->B to F(f):1+A -> 1+B with F(f)(1+a)=1+f(b). I want to know which classes of p-groups are realisable as 1+J and maybe if you see some interesting properties of the functor F. For example if J^2 = 0,then G=1+J is elementary abelian and if J^3=0 then G=1+J is special. Such realisations could be interesting since,we have for example in general G' $\leq$ 1+J^2 and a series 1+J $\leq$ 1+J^2 ... $\leq$ 1+J^k =0 and there it is known that the characters of 1+J are induced from linear characters of a subgroup 1+A where A is a subalgebra of J.</p> http://mathoverflow.net/questions/55983/good-books-in-modular-representation-theory/55985#55985 Answer by trew for Good books in Modular Representation Theory trew 2011-02-19T15:58:20Z 2011-02-19T16:03:33Z <p>Hi, a very elementary written book is Local Representation Theory by Alperin.The second book on finite groups by Huppert has also a big part about modular representation theory.(you should read the first book too,with an long introduction to representation theory in the semisimple case) a more advanced book is that of feit and a recent (2010) book is "Representations of Groups: A Computational Approach" by Lux and Pahlings.It has many nice examples with gap and the sporadic groups.You should also take a look for the books of curtis and reiner.The one from 1962 is easy to read and might be the best introduction.</p> http://mathoverflow.net/questions/52261/applications-of-the-theorem-of-gelfand-naimark Applications of the Theorem of Gelfand-Naimark trew 2011-01-16T20:12:24Z 2011-01-16T21:45:51Z <p>Hi,<br> I am interested in the correspondence of algebraic results about C(X) (the space of continuous functions $X\to {\mathbb C}$(complex numbers) or $X\to {\mathbb R}$(real numbers) and topological properties of X,for example <a href="http://projecteuclid.org/euclid.mmj/1029000524" rel="nofollow">results like this</a> .(Does by the way someone know what's the "deep result by bkouche" mentioned in the paper?)</p> <p>Can you by the way use this result to prove interesting theorems with this translation<br> (like: -a manifold(or even a CW-complex) is paracompact<br> -the theorem of Tietze, etc?)</p> <p>There are many such correspondences which are obtained by using Gelfand-Naimark but I couldn't find literature where you can find full details with all needed definitions and proofs.(I couldn't even find a proof of the categorical Gelfand-Naimark theorem in the nonunital case,only some sketches.) Does such literature exist for a beginner in this topic? The book "Basic Noncommutative Geometry" written by Khalkhali is a good source but omits most details (see <a href="http://books.google.de/books?id=UInc5AyTAikC&amp;printsec=frontcover&amp;dq=Khalkhali&amp;hl=de&amp;ei=w0wzTazYGIb2sgb7rcybCg&amp;sa=X&amp;oi=book_result&amp;ct=result&amp;resnum=1&amp;ved=0CCoQ6AEwAA#v=onepage&amp;q&amp;f=false" rel="nofollow">page 16</a> for a little list).</p> <p>So I would be glad if you can recommend to me a good book/link, or write a nice result here if it's not too complicated.</p> <p>Example: X is connected iff C(X) has no idempotents because direct sums of subalgebras correspond to disjoint union of closed subspaces,since C is an equivalence.</p> http://mathoverflow.net/questions/47952/proving-interesting-theorems-about-s-n-using-its-character-table Proving interesting theorems about S_n using its character table. trew 2010-12-01T21:56:50Z 2010-12-02T06:43:54Z <p>Hi, i wonder if there are interesting proofs about $S_n$ (group theoretic or not) using its character table. Using the Murnaghan-Nakayama rule you can for example prove that for $n>4$ $A_n$ is the only normal subgroup of $S_n$ because there are no nonlinear characters $x$ and $g$(not 1) in $S_n$ with $x(g)=x(1)$, since $x(1)>x(g)$ . Do you know any other nontrivial theorems about $S_n$ with a proof using its charactertable ?</p> http://mathoverflow.net/questions/45469/examples-of-nice-families-of-irreducible-polynomials-over-z Examples of nice families of irreducible polynomials over Z trew 2010-11-09T17:41:36Z 2010-11-10T05:31:31Z <p>Hi, i search for irreducible polynomials over Z which have variable coefficients you can "choose". Since I found nearly nothing in books or the internet i hope you can help me. Here 3 examples: Let g a polynomial over Z with degree smaller then n/2 ,then: $g* (\prod_{i=1}^n (x-a_i)) -1$ is irreducible if the a_i are all distinct. Here you can choose n the coefficients of g and the a_n so its a nice example. Another one,I found, is from Furtwängler : $x^4 (\prod_{i=1}^{n-4} (x-b_i)) -(-1)^n *(2x+4)$ where the b_i are strictly increasing.Can you generalize this example ?I think it should work also for some integers other than 2 and 4. Here is another nice example : <a href="http://mathoverflow.net/questions/18094/polynomial-with-the-primes-as-coefficients-irreducible" rel="nofollow">http://mathoverflow.net/questions/18094/polynomial-with-the-primes-as-coefficients-irreducible</a></p> <p>I try to find examples where its easy to control zeros modulo p of the some irreducible polynomials and its derivation for another problem.</p> http://mathoverflow.net/questions/40346/a-result-about-characters-of-f-algebra-groups A result about Characters of F-algebra groups trew 2010-09-28T17:16:21Z 2010-09-28T18:28:57Z <p>Let G be an F-algebra group(G=1+J , where J is the jacobson radical of a finite dimensional F-algebra ,where F is a field of prime characteristic) In a paper of Isaacs ("Characters of groups associated with finite algebras" from 1995) there is a claim of Gutkin with a wrong proof.It says: Let x be an irreducible character of G ,then $x=a^G$ ,where a is a linear character of some subgroup H&lt;=G of the form : H=1+U where U is multiplicativly closed F-subspace of J. This result would be a generalisation of the result of Isaacs paper,but its unclear to me if it has been proven until today.Does someone know if the result is true and can recommand me a paper/link for more information?</p> http://mathoverflow.net/questions/39966/minimal-conditions-for-the-exponential-law-for-compact-open-topologies Minimal conditions for the exponential law for compact-open topologies. trew 2010-09-25T18:37:00Z 2010-09-26T01:26:25Z <p>What are the minimal conditions on three topological spaces $X,Y$ and $Z$ such that with the compact-open topology the map</p> <p>$${(X^Y)}^Z \to X^{Y \times Z}$$</p> <p>given by taking adjoints is a homeomorpism. The map sends $f: Z \to X^Y$ to $g:Y \times Z \to X$ by the relation $g(y,z)=f(z)(y)$. </p> <p>This result is known for $Z$ Hausdorff and $Y$ locally compact. I'm interested in a proposition of the form the adjoint construction is a homeomorphism of mapping spaces if and only if some statement regarding the spaces $X$, $Y$ and $Z$. It would also be interesting to see some counterexamples,for example for Z not Hausdorff,etc. </p> http://mathoverflow.net/questions/125018/classification-of-hopf-algebra-with-exactly-two-1-dimensional-modules Comment by trew trew 2013-03-26T18:37:16Z 2013-03-26T18:37:16Z sorry, i forgot to add that im looking for nonsemisimple and indecomposable algebras of the above type. http://mathoverflow.net/questions/125018/classification-of-hopf-algebra-with-exactly-two-1-dimensional-modules Comment by trew trew 2013-03-19T21:30:58Z 2013-03-19T21:30:58Z no im interested in any field.but if you have some examples over C you can post them too of course http://mathoverflow.net/questions/125017/an-injective-smooth-function-with-injective-differential-must-have-a-continuous-i/125021#125021 Comment by trew trew 2013-03-19T21:17:04Z 2013-03-19T21:17:04Z <a href="http://en.wikipedia.org/wiki/Inverse_function_theorem" rel="nofollow">en.wikipedia.org/wiki/Inverse_function_theorem</a> the fact that the dimension must be the same follows from you assumption that Df(x):R^n -&gt; R^m is invertible(so you can use linear algebra to see that n=m) http://mathoverflow.net/questions/125017/an-injective-smooth-function-with-injective-differential-must-have-a-continuous-i/125021#125021 Comment by trew trew 2013-03-19T21:12:52Z 2013-03-19T21:12:52Z I think its called Inverse function theorem in english http://mathoverflow.net/questions/87683/heller-functor-of-finite-order Comment by trew trew 2012-02-06T17:49:01Z 2012-02-06T17:49:01Z thats right i guess.thanks http://mathoverflow.net/questions/83520/what-do-we-know-about-periodic-modules-in-p-groups/83525#83525 Comment by trew trew 2011-12-17T14:20:21Z 2011-12-17T14:20:21Z thank you for your edit!But I think it has period 1 for p=2. http://mathoverflow.net/questions/72876/why-did-gabriel-invent-the-term-quiver Comment by trew trew 2011-08-14T17:45:56Z 2011-08-14T17:45:56Z I dont know what Gabriel thought but here is an answer why the term is used from: <a href="http://www.amazon.de/Elements-Representation-Theory-Associative-Algebras/dp/0521586313/ref=sr_1_1?ie=UTF8&amp;qid=1313343784&amp;sr=8-1" rel="nofollow">amazon.de/&hellip;</a> page 42 says:&quot;...There are two main reasons for using the term quiver rather than graph:the first one is that the former has become generally accepted by specialists;the second is that the latter is used in so many different contexts and even senses( a graph can be oriented or not,with or without multiple arrows or loops) that it may lead,for our purposes at east to certain ambiguities. http://mathoverflow.net/questions/68207/irreducible-representations-of-the-unitriangular-group/68227#68227 Comment by trew trew 2011-06-19T19:24:32Z 2011-06-19T19:24:32Z thanks,i know that result.it would be interesting to know what has been done for n=4,before i try ;) I know for example that all characterdegrees with their multiplicity is known.can someone by the way say what exactly &quot;wild&quot; means? http://mathoverflow.net/questions/68101/p-groups-realisable-as-1j-where-j-is-a-nilpotent-finite-f-algebra Comment by trew trew 2011-06-18T15:54:22Z 2011-06-18T15:54:22Z thank you.do you know a paper which contains most of the known results about those groups?by the way: if someone has an interesting example of a p-group not realisable as 1+J,I would be glad if you post it. http://mathoverflow.net/questions/55983/good-books-in-modular-representation-theory/55985#55985 Comment by trew trew 2011-02-19T16:08:08Z 2011-02-19T16:08:08Z Ok,that sounds strange so i changed that.I dont know how to explain.Maybe just try to read it. http://mathoverflow.net/questions/54798/minimal-distance-between-a-point-and-a-manifold-m-m-is-the-intersection-of-a-bal Comment by trew trew 2011-02-08T20:58:33Z 2011-02-08T20:58:33Z I think its an interessesting question wheter there is an exact solution http://mathoverflow.net/questions/52261/applications-of-the-theorem-of-gelfand-naimark Comment by trew trew 2011-01-17T08:57:42Z 2011-01-17T08:57:42Z sorry,i forgot about your article ;) http://mathoverflow.net/questions/52261/applications-of-the-theorem-of-gelfand-naimark/52266#52266 Comment by trew trew 2011-01-16T21:54:43Z 2011-01-16T21:54:43Z Thank you!I know this from the book of pedersen.Its a very nice application and there is even a one-to-one corresponende of the compactifications of a locally compact space and essential extensions of $C_0 (X)$ described in the book of Khalkhali. http://mathoverflow.net/questions/52261/applications-of-the-theorem-of-gelfand-naimark Comment by trew trew 2011-01-16T20:36:54Z 2011-01-16T20:36:54Z Thank you very much! http://mathoverflow.net/questions/47952/proving-interesting-theorems-about-s-n-using-its-character-table/47953#47953 Comment by trew trew 2010-12-01T22:22:23Z 2010-12-01T22:22:23Z thank you!does someone know if you need to use that result to prove Murnaghan-Nakayama,which also gives you that the character table entries are integers?I think so.