User mathreader - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:59:55Z http://mathoverflow.net/feeds/user/2164 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7523/is-every-finite-dimensional-lie-algebra-the-lie-algebra-of-a-closed-linear-lie-gr Is every finite-dimensional Lie algebra the Lie algebra of a closed linear Lie group? mathreader 2009-12-01T23:43:20Z 2011-07-24T08:24:40Z <p>This question is closely related to <a href="http://mathoverflow.net/questions/124/is-every-finite-dimensional-lie-algebra-the-lie-algebra-of-an-algebraic-group" rel="nofollow">this one</a>.</p> <p>Ado's theorem states that given a finite-dimensional Lie algebra $\mathfrak g$, there exists a faithful representation $\rho\colon\mathfrak g \to \mathfrak{gl}(V)$, with $V$ a finite-dimensional vector space. In the real or complex case one can take the exponent of the image and obtain a (virtual) Lie subgroup $\exp\rho(\mathfrak g)$ in $GL(V)$ having Lie algebra $\rho(\mathfrak g)$. But nothing guarantees that this subgroup will be closed in $GL(V)$. </p> <p>So the question is: is every finite-dimensional Lie algebra the Lie algebra of some closed linear Lie group? I am primarily interested in the real and complex case, but it might be interesting to ask what happens in the ultrametric case as well.</p> http://mathoverflow.net/questions/62624/complex-lie-group-without-faithful-real-representations Complex Lie group without faithful real representations? mathreader 2011-04-22T15:01:35Z 2011-04-22T16:17:08Z <p>Does there exist a complex analytic Lie group which doesn't have faithful representations in $GL(N,\mathbb R)$, viewed as a real Lie group?</p> <p>There are examples of complex Lie groups which do not allow faithful complex representations, like tori $\mathbb C^n/\mathbb Z^{2n}$, but such tori have many faithful real representations.</p> <p>Also there are examples of real Lie groups without faithful linear representations, like the universal cover of $SL(2,\mathbb R)$ (but they are not complex analytic Lie groups).</p> <p>How about complex Lie groups without faithful real representations?</p> http://mathoverflow.net/questions/35713/abelianization-of-a-semidirect-product/62523#62523 Answer by mathreader for Abelianization of a semidirect product mathreader 2011-04-21T11:53:03Z 2011-04-21T11:53:03Z <p>A description of the derived subgroup of a semidirect product, from which the abelianization can be obtained, was published in:</p> <p>Daciberg Lima Gonçalves, John Guaschi The lower central and derived series of the braid groups of the sphere Trans. Amer. Math. Soc. 361 (2009), 3375-3399. <a href="http://www.ams.org/journals/tran/2009-361-07/S0002-9947-09-04766-7/" rel="nofollow">http://www.ams.org/journals/tran/2009-361-07/S0002-9947-09-04766-7/</a> (Proposition 3.3)</p> <p>You may also find it in their preprint: <a href="http://arxiv.org/abs/math/0603701" rel="nofollow">http://arxiv.org/abs/math/0603701</a> (Proposition 29)</p> http://mathoverflow.net/questions/62471/question-on-parabolic-subgroups/62489#62489 Answer by mathreader for question on parabolic subgroups mathreader 2011-04-21T03:35:58Z 2011-04-21T03:35:58Z <p>If I understood your question correct, then the answer is no. I will assume for simplicity that you are talking about parabolic subgroups of complex simple Lie groups. Then your question translates to the corresponding question about closed subsystems of root systems. Recall that the standard parabolic subgroups bijectively correspond to the closed subsystems $R$ of the root system $\Phi$ such that $R$ contains the set of all positive roots $\Phi^+$. (Here `closed' means $\alpha\in R$, $\beta\in R$ implies $\alpha+\beta\in R$.) Your question is equivalent to the following one: given a closed subsystem $R$ containing all positive roots $\Phi^+$, is there an element $w$ of Weyl group such that $w.R=-R$?</p> <p>If the Weyl group contains $-id$, then the question is yes. However, for the root systems of types $A_n$ ($n\ge2$), $D_{2n+1}$ and $E_6$ this is not true: $-id$ does not belong to the Weyl group. Therefore for these groups there may exist parabolic subgroups which are not conjugate to their opposites via the action of Weyl group. And indeed, looking at $A_2$, we see that the two standard maximal parabolics are not conjugate to their opposites via Weyl group: there is no element in Weyl group which transforms the root subsystem $R_1=\Phi^+\cup{-\alpha_1}$ to $-R_1$, and the same is true for $R_2=\Phi^+\cup{-\alpha_2}$.</p> http://mathoverflow.net/questions/53676/explicit-linear-representations-of-fundamental-groups-of-surfaces explicit linear representations of fundamental groups of surfaces mathreader 2011-01-29T03:33:58Z 2011-02-01T20:57:42Z <p>I am looking for an explicit representation of the fundamental group of a closed orientable surface of genus >1. I guess they should be abundant in degree 2. Did anyone see the explicit matrix construction of such a representation? Are there any integral ones? Maybe in higher degrees?</p> http://mathoverflow.net/questions/37853/to-what-extent-can-algorithms-in-undergraduate-linear-algebra-be-made-continuous/38029#38029 Answer by mathreader for To what extent can algorithms in undergraduate linear algebra be made continuous/polynomial/etc.? mathreader 2010-09-08T04:02:27Z 2010-09-08T04:02:27Z <p>If we talk about algorithms in a strict sense, then the data involved should be given constructively. This effectively limits us to rational numbers (represented as pairs of integers) or some 'easy' algebraic numbers (for which we can provide algorithms for basic arithmetic). And for these data types most linear algebra algorithms are polynomial in time and stable in the sense that they do not give any inaccuracy due to rounding/overflow/etc. This philosophy is implemented in the computer algebra system GAP.</p> http://mathoverflow.net/questions/21114/low-dimensional-nilpotent-lie-algebras/21169#21169 Answer by mathreader for Low dimensional nilpotent Lie algebras mathreader 2010-04-13T00:41:47Z 2010-04-13T00:41:47Z <p>Many articles on classification of low-dimensional Lie algebras do contain mistakes. To the best of my knowledge, the full detailed proof is provided in the dissertation of Ming-Peng Gong:</p> <p><a href="http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.16.5538&amp;rep=rep1&amp;type=pdf" rel="nofollow">http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.16.5538&amp;rep=rep1&amp;type=pdf</a></p> <p>where he classifies all algebras up to dimension 7 over algebraically closed fields of any characteristics except 2, and also over reals.</p> http://mathoverflow.net/questions/20471/why-are-free-groups-residually-finite/21050#21050 Answer by mathreader for Why are free groups residually finite? mathreader 2010-04-11T23:48:52Z 2010-04-11T23:48:52Z <p>Actually there is a simple reasoning showing residual finiteness of free groups using coverings. Indeed, consider the free group $F(n)$ as a fundamental group of a wedge $X$ of $n$ circles. Take any nontrivial element $w$ of $F(n)$. It is sufficient to exhibit a finite cover $X_0 \to X$ such that the path $w$ is not covered by a loop in the graph $X_0$. The graph $X_0$ can be built as follows: take a universal cover $\tilde X$ of $X$. It is a Cayley graph of $F(n)$, a regular tree, every vertex of which has in-degree $n$ and out-degree $n$ with edges labeled by generators of $F(n)$. Consider the path W (starting at a fixed origin) in $\tilde X$ corresponding to the word $w$. In order to obtain $X_0$, we declare Vertices($X_0$) = Vertices($W$) and Edges($X_0$) = Edges($W$) plus additional edges added in such a way that $X_0$ becomes an $n$-regular graph: every vertex should have $n$ in-coming and $n$ out-going edges. This is always possible since $W$ was originally taken from an $n$-regular graph.</p> <p>Thus we get a finite $n$-regular graph $X_0$, hence a finite sheeted covering $X_0\to X$ in which the path $w$ is not a loop. Therefore, element $w$ lies outside of a finite index subgroup of $F(n)$, hence it lies outside some finite index normal subgroup, hence it maps nontrivially into some finite group.</p> http://mathoverflow.net/questions/12539/essential-theorems-in-group-cohomology/12605#12605 Answer by mathreader for Essential theorems in group (co)homology mathreader 2010-01-22T04:44:41Z 2010-01-22T04:44:41Z <ol> <li> Interpretation of cohomology of small degree:</li> </ol> <p>$H^1(G,A)$ = crossed homomorphisms $G\to A$ modulo principal ones.</p> <p>$H^2(G,A)$ = equivalence classes of extensions of G by A.</p> <p>$H^3(G,Center(G))$ = obstructions to existence of extensions of G by A.</p> <p>2. Transfer and its applications: If $G$ is finite then</p> <p>1) $H^i(G,M)$ is a torsion group annihilated by multiplication by $|G|$.</p> <p>2) Embedding of $p$-primary component of $H^i(G,M)$ into a subgroup of $H^i(P,M)$, for any $p$-Sylow subgroup $P\subset G$.</p> <p>3. In general, Brown's book "Cohomology of groups" gives a decent overview of what is good to know. </p> http://mathoverflow.net/questions/10419/depressed-graduate-student/10439#10439 Answer by mathreader for Depressed graduate student. mathreader 2010-01-02T01:07:53Z 2010-01-02T01:07:53Z <p>There is an interesting article </p> <p><a href="http://www.scottberkun.com/essays/33-how-to-survive-creative-burnout" rel="nofollow">Scott Berkun, How to survive a creative burnout</a></p> <p>It is not aimed at mathematicians in the first place but still useful in my opinion.</p> http://mathoverflow.net/questions/7389/what-are-the-most-overloaded-words-in-mathematics/7495#7495 Answer by mathreader for What are the most overloaded words in mathematics? mathreader 2009-12-01T20:03:31Z 2009-12-02T01:01:02Z <p><strong>trivial</strong></p> <p>Besides being a synonym to 'obvious', like in 'the proof is trivial', it has the meaning of 'shallow' ('the question is trivial') and moreover denotes a bunch of mathematical notions:</p> <p>trivial group</p> <p>trivial representation</p> <p>trivial topology</p> <p>trivial solution (in ODE/PDE)</p> <p>etc.</p> <p>Sometimes it produces confusion as it is not quite clear which sort of triviality is meant.</p> http://mathoverflow.net/questions/7155/famous-mathematical-quotes/7188#7188 Answer by mathreader for Famous mathematical quotes mathreader 2009-11-29T21:34:00Z 2009-11-29T21:34:00Z <p><strong>In mathematics you don't understand things. You just get used to them.</strong> </p> <p>--John von Neumann, reply to a physicist at Los Alamos who had said "I don't understand the method of characteristics."</p> <p>---- footnote on page 226 of Gary Zukav, The Dancing Wu Li Masters: An Overview of the New Physics, Rider, London, 1990.</p> <p>(taken from Warren Dicks' Home Page)</p> http://mathoverflow.net/questions/16991/what-are-the-connections-between-pi-and-prime-numbers/17041#17041 Comment by mathreader mathreader 2012-10-18T09:44:45Z 2012-10-18T09:44:45Z By the way, do you happen to know what constant appears if we change $\sigma(n)$ into the sum $\sigma'$ of all Gaussian integer divisors of n with positive real parts? I.e. this sequence: <a href="http://oeis.org/A078930" rel="nofollow">oeis.org/A078930</a> Computations show that $\sum_1^n\sigma'(i)\approx Cn^2$, where C=1.7972... http://mathoverflow.net/questions/67355/residual-finiteness-of-fundamental-groups-of-surfaces/67357#67357 Comment by mathreader mathreader 2011-12-29T04:49:23Z 2011-12-29T04:49:23Z Could you possibly clarify, which homomorphism $\pi_1(F)\to S_3$ is meant in this proof? http://mathoverflow.net/questions/60315/modern-reference-for-maximal-connected-subgroups-of-compact-lie-groups/60318#60318 Comment by mathreader mathreader 2011-07-25T08:48:20Z 2011-07-25T08:48:20Z In Memoirs of the AMS, there are two volumes, one of Gary Seitz (for classical groups) and another of Seitz and Donna Testerman (for exceptional groups inclusions). http://mathoverflow.net/questions/7523/is-every-finite-dimensional-lie-algebra-the-lie-algebra-of-a-closed-linear-lie-gr/71112#71112 Comment by mathreader mathreader 2011-07-24T11:28:15Z 2011-07-24T11:28:15Z Consider the simplest case: the irrational winding of two-dimensional torus. Here $\mathfrak{g}$ is one-dimensional real Lie algebra isomorphic to $\mathbb{R}$. The representation is given to us by $\rho: \mathfrak{g}\to \diag(i\mathbb{R},i\mathbb{R})$, $x\mapsto \diag(ix, i\alpha x)$, where $\alpha$ is irrational. Then $\exp\rho(\mathfrak{g})$ is a non-closed, or virtual, Lie subgroup of $U(1)\times U(1)$. http://mathoverflow.net/questions/43726/the-free-group-f-2-has-index-12-in-sl2-mathbbz/43741#43741 Comment by mathreader mathreader 2011-04-23T07:08:49Z 2011-04-23T07:08:49Z @Mark Sapir: It looks that the Sanov subgroup is NOT normal in SL(2,Z). I did some computations in GAP, and it turned out that its normalizer has index 3 in SL(2,Z), and its normal closure has index 6. The normal subgroup of index 12, different from the derived subgroup, is generated by the matrices: $$\begin{pmatrix} -1 &amp; 2\\ 0 &amp; -1 \end{pmatrix} \text{ and } \begin{pmatrix} -1 &amp; 0\\ 2 &amp; -1 \end{pmatrix} .$$ http://mathoverflow.net/questions/62624/complex-lie-group-without-faithful-real-representations/62631#62631 Comment by mathreader mathreader 2011-04-22T23:52:37Z 2011-04-22T23:52:37Z @Tom: Thanks! This is a really nice explanation! http://mathoverflow.net/questions/62624/complex-lie-group-without-faithful-real-representations/62631#62631 Comment by mathreader mathreader 2011-04-22T19:41:23Z 2011-04-22T19:41:23Z Thank you for your answer. Could you please explain why this group will not admit faithful representations? http://mathoverflow.net/questions/21114/low-dimensional-nilpotent-lie-algebras/21169#21169 Comment by mathreader mathreader 2011-04-21T04:00:15Z 2011-04-21T04:00:15Z @Jim: In the article quoted by Pasha below, Willem de Graaf claims that M.-P. Gong lost a one-parameter family of 6-dimensional algebras. If this is true, then his results on 7-dimensional algebras may be also incomplete, unfortunately. http://mathoverflow.net/questions/53676/explicit-linear-representations-of-fundamental-groups-of-surfaces/53684#53684 Comment by mathreader mathreader 2011-01-30T01:13:54Z 2011-01-30T01:13:54Z Thank you for the link, Igor! As far as I see, the author considers projective representations. I hope they lift to the linear ones. http://mathoverflow.net/questions/53676/explicit-linear-representations-of-fundamental-groups-of-surfaces Comment by mathreader mathreader 2011-01-29T04:22:52Z 2011-01-29T04:22:52Z @John: Yes, I meant faithful representations. @Igor: Yes, degree 2 means $2 \times 2$ matrices http://mathoverflow.net/questions/44269/commutator-subgroup-does-not-consist-only-of-commutators/44276#44276 Comment by mathreader mathreader 2010-10-31T23:33:07Z 2010-10-31T23:33:07Z How can one prove that [x,y][z,w] is not [a,b] in a free group? http://mathoverflow.net/questions/8846/proofs-without-words/8847#8847 Comment by mathreader mathreader 2010-10-17T17:07:25Z 2010-10-17T17:07:25Z Am I the only one who doesn't understand this &quot;proof&quot; at all? http://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples/38898#38898 Comment by mathreader mathreader 2010-09-16T09:20:30Z 2010-09-16T09:20:30Z This is not a joke, you CAN cancel those du's, if understood properly. Namely, if we have graph of f(x), and a tangent vector v at point (c, f(c)), then dy = dy(v) = projection of v to y-coordinate, dx = dx(v) = projection of v to x-coordinate. Their quotient dy(v)/dx(v) is equal to f'(c). http://mathoverflow.net/questions/9716/whats-the-status-of-the-following-relationship-between-ramanujans-tau-functi Comment by mathreader mathreader 2009-12-25T07:27:58Z 2009-12-25T07:27:58Z Interestingly, the number 26 appears also as a number of 'sporadic' finite simple groups (i.e. not fitting into the infinite series of groups of Lie type, cyclic and alternative). http://mathoverflow.net/questions/8097/number-theory-textbook-with-an-algebraic-perspective/8099#8099 Comment by mathreader mathreader 2009-12-07T11:21:05Z 2009-12-07T11:21:05Z I like Ireland-Rosen book very much too.