User pasha zusmanovich - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T06:46:49Z http://mathoverflow.net/feeds/user/1223 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/116698/if-m-nr-and-m-mr-satisfy-the-same-polynomial-identities-is-it-true-that/130327#130327 Answer by Pasha Zusmanovich for If $M_n(R)$ and $M_m(R)$ satisfy the same polynomial identities is it true that $m=n$? Pasha Zusmanovich 2013-05-11T09:17:06Z 2013-05-13T09:00:15Z <p>This is not true in general. For example, if $R$ is a free associative algebra of rank $>1$, then $M_n(R)$ does not have nontrivial identities for any $n$. </p> <p>Moreover, I believe there are examples of algebras $R$ such that much stronger condition holds: $M_n(R)$ is isomorphic to $M_m(R)$ for some $m \ne n$. This reminds cancellation problems in the commutative setting, though I cannot provide examples of such algebras.</p> <p>This is true if, for example, if $R$ is finite-dimensional and prime (even not necessary associative). Then we can note that $M_n(R)$ is prime too, pass to the algebraic closure of the base field, and invoke theorem of Razmyslov that finite-dimensional prime algebras over algebraically closed fields are determined by their identities (Yu.P. Razmyslov, Identities of Algebras and Their Representations, AMS, 1994 (translation from Russian), around p. 30).</p> <p><b>EDIT</b>:</p> <p>I was not careful enough when reading the question, sorry. The question explicitly asks for situation when $R$ is a PI algebra, so the example with a free algebra obviously does not qualify. </p> <p>I still think this is not true in the whole generality: probably just the condition of being unitary is to weak, one should demand something like (semi)primeness. </p> <p>I can think of two approaches. First, there is a lot of works about identities of tensor product of algebras, and of $M_n(R)$ in particular (typical results: if $R$ satisfies the standard identity of degree $k$, then $M_n(R)$ satisfies the standard identity of some given degree in terms of $n$ and $k$, see e.g. M. Domokos, Eulerian polynomial identities and algebras satisfying a standard identity, J. Algebra 169 (1994), N3, 913-928 DOI: 10.1006/jabr.1994.1317). Second, perhaps one can do something along the lines of Sections 4 and 5 of arXiv:0911.5414.</p> http://mathoverflow.net/questions/95125/applications-of-algebraic-geometry-commutative-algebra-to-biology-pharmacology/130331#130331 Answer by Pasha Zusmanovich for Applications of algebraic geometry/commutative algebra to biology/pharmacology ? Pasha Zusmanovich 2013-05-11T11:13:13Z 2013-05-12T05:34:25Z <p> 1. Ren&eacute; Thom's theory of morphogenesis involves singularities, unfoldings, perturbations of analytic/geometric structures, etc., which, in its turn, involves (or, rather, should involve, as the whole theory is rather sketchy) a good deal of commutative algebra. </p> <p> 2. A conference "<a href="http://www.ihes.fr/jsp/site/Portal.jsp?document_id=3299&portlet_id=14" rel="nofollow">Moduli spaces and macromolecules</a>". </p> <p><p> 3. Some biological models involve systems of boolean equations, or sentences of propositional calculus, which could be interpreted as polynomials over GF(2), with subsequent application of Gr&ouml;bner basis technique. A (more or less random) sample of possibly relevant papers (I avoid mentioning algebraic statistics which was mentioned many times elsewhere):</p> <ul> <li> G. Boniolo, M. D'Agostino, P.P. Di Fiore, Zsyntax: A formal language for molecular biology with projected applications in text mining and biological prediction, PLoS ONE 5 (2010), N3, e9511 DOI:10.1371/journal.pone.0009511 </li> <li> A.S. Jarrah and R. Laubenbacher, Discrete models of biochemical networks: the toric variety of nested canalyzing functions, Algebraic Biology, Lect. Notes Comp. Sci. 4545 (2007), 15-22 DOI:10.1007/978-3-540-73433-8_2 </li> <li> R. Laubenbacher and B. Stigler, A computational algebra approach to the reverse engineering of gene regulatory networks, J. Theor. Biol. 229 (2004), 523-537 DOI:10.1016/j.jtbi.2004.04.037 arXiv:q-bio/0312026 </li> <li> I. Lynce and J.P. Marques Silva, Efficient haplotype inference with boolean satisfiability, AAAI'06, July 2006; SAT in Bioinformatics: making the case with haplotype inference, SAT'06, August 2006; http://sat.inesc-id.pt/~ines/ </li> </ul> http://mathoverflow.net/questions/97843/recovering-a-lie-algebra-from-its-affine-lie-algebra/130329#130329 Answer by Pasha Zusmanovich for Recovering a Lie algebra from its affine Lie algebra Pasha Zusmanovich 2013-05-11T09:45:48Z 2013-05-11T09:45:48Z <p>(Too long for a comment). This is a (vague) comment on Qiaochu Yuan's comment on whether $\widehat{\mathfrak g_1} \simeq \widehat{\mathfrak g_2}$ implies $\mathfrak g_1 \simeq \mathfrak g_2$. This is definitely true. </p> <p>The twisted case could be reduced somehow to the untwisted one, so I will deal with the latter. By taking central quotients and commutants, the question is reduced to whether isomorphism of the corresponding loop algebras -- $\mathfrak g_1 \otimes \mathbb C[t,t^{-1}] \simeq \mathfrak g_2 \otimes \mathbb C[t,t^{-1}]$ -- implies the isomorpism of the underlying simple Lie algebras $\mathfrak g_1 \simeq \mathfrak g_2$. This is true even when the algebra of Laurent polynomials $\mathbb C[t,t^{-1}]$ is replaced by an arbitrary commutative associative algebra $A$ with unit, and probably can be dealt with by looking on some invariants of loop algebars (for example, looking on the second cohomology $H^2(\mathfrak g \otimes A, \mathfrak g \otimes A)$, we can separate the case of $sl(n)$ from the other types). </p> <p>It can be dealt with, however, from a somewhat unusual viewpoint (which is probably an overkill): note that this isomorphism implies that the identities of algebras $\mathfrak g_1$ and $\mathfrak g_2$ are the same (as identities of $\mathfrak g \otimes A$ and $\mathfrak g$ are the same), and by theorem of Kushkulei and Razmyslov (see Yu.P. Razmyslov, Identities of Algebras and Their Representations, AMS, 1994 (translation from Russian), around p. 30), this implies $\mathfrak g_1 \simeq \mathfrak g_2$.</p> http://mathoverflow.net/questions/107782/reference-for-list-of-left-regular-representations-of-real-associative-algebras/130277#130277 Answer by Pasha Zusmanovich for reference for list of left-regular representations of real associative algebras Pasha Zusmanovich 2013-05-10T17:09:19Z 2013-05-10T17:09:19Z <p>(Too long for a comment). More (modern and not-so-modern) references, some of them may (partially) contain list(s) you are interested in:</p> <ul> <li><p>A.A. Albert, Structure of Algebras, AMS, 1939: on page 172 discusses classification of 4-dimensional algebras.</p></li> <li><p>S.C. Althoen, K.D. Hansen, L.D. Kugler, <a href="http://hofprints.hofstra.edu/23/01/Althoen,_S._C.,_Hansen,_K._D._and_Kugler,_L._D._(2001)_A_survey_of_four-dimensional_C-associative_algebras._In_Proceedings_Robert_J._Bumcrot_Festschrift,_Hofstra_University.pdf" rel="nofollow">A survey of four-dimensional C-associative algebras</a>, Algebras Groups Geom. 21 (2004), N1, 9-27</p></li> <li><p>R. Ballieu, Anneaux finis; systèmes hypercomplexes de rang trois sur un corps commutatif, Ann. Soc. Sci. Bruxelles Sér. I. 61 (1947), 222-227: presumably contains classification of complex 3-dimensional algebras.</p></li> <li><p>W.A. de Graaf, Classification of nilpotent associative algebras of small dimension, arXiv:1009.5339.</p></li> <li><p>D. Happel, Klassifikationstheorie endlich-dimensionaler Algebren in der Zeit von 1880 bis 1920, Enseign. Math. 26 (1980), 91-102 DOI:10.5169/seals-51060 : A nice historical survey from the modern viewpoint, with a large bibliography.</p></li> <li><p>O.C. Hazlett, On the classification and invariantive characterization of nilpotent algebras Amer. J. Math. 38 (1916), N2, 109-138 <a href="http://www.jstor.org/stable/2370262" rel="nofollow">http://www.jstor.org/stable/2370262</a></p></li> <li><p>G. Pickert, Dreidimensionale assoziative nichtkommutative Algebren, J. Algebra 234 (2000), N2, 280-290 DOI:10.1006/jabr.2000.8550</p></li> <li><p>Scorza, Atti Acad. Sci. Fis. Mat. Napoli 20 (1935), N13 and N14: Classification of 3- and 4-dimensional algebras.</p></li> <li><p>D.A. Suprunenko and R.I. Tyshkevich, Commutative Matrices, Acad. Press, 1968 (translation from Russian): On p.61 (of the Russian edition) there is a discussion of commutative nilpotent algebras of dimension 5.</p></li> </ul> http://mathoverflow.net/questions/121633/what-is-an-ideal-supporting-algebra/130266#130266 Answer by Pasha Zusmanovich for What is an ideal-supporting algebra? Pasha Zusmanovich 2013-05-10T14:09:51Z 2013-05-10T14:09:51Z <p>It seems that the standard term for universal algebras whose congruences behave "as good as ideals" is "ideal-determined algebras". This is a more general notion than $\Omega$-group. This notion, along with its numerous particular cases and variations, was studied by Agliano, Chajda, Fichtner, Gr&auml;tzer, Gumm, Slominski, Ursini and others. See, for example, I. Chajda, G. Eigenthaler, and H. L&auml;nger, Congruence Classes in Universal Algebra, Heldermann Verlag, 2003, Chapter 10.</p> http://mathoverflow.net/questions/112715/why-when-classification-of-simple-objects-is-simple-e-g-unknown-classifica/130260#130260 Answer by Pasha Zusmanovich for Why/when classification of simple objects is "simple" ? E.g. (unknown) classification of simple Lie algebras in char =2,3... Pasha Zusmanovich 2013-05-10T13:32:20Z 2013-05-10T13:32:20Z <p>(This is too long for a comment). There is a recent surge of activity around (attempts of) classification of simple finite-dimensional Lie algebras in $p=2,3$. It is my understanding that the common view among experts is that the case $p=3$ might be in sight, while situation in $p=2$ is still chaotic. The main current players in the field are A. Grishkov, M.I. Kuznetsov and his students, and D. Leites and his collaborators. </p> <p>As in Kostrikin-Shafarevich program for large characteristics, deformations (of some initial set of algebras) play a role. Computers are involved a lot.</p> <p>A sample of recent publications:</p> <ul> <li>S. Bouarroudj, P. Grozman, D. Leites, Infinitesimal deformations of symmetric simple modular Lie algebras and Lie superalgebras, arXiv:0807.3054</li> <li>S. Bouarroudj, A. Lebedev, D. Leites, I. Shchepochkina, Deforms of Lie algebras in characteristic 2: semi-trivial for Jurman algebras, non-trivial for Kaplansky algebras, arXiv:1301.2781.</li> <li>B. Eick, Some new simple Lie algebras in characteristic 2, J. Symb. Comput. 45 (2010), N9, 943-951</li> <li>A. Grishkov, On simple Lie algebras over a field of characteristic 2, J. Algebra 363 (2012), 14-18</li> <li>A. Grishkov and M. Guerreiro, On simple Lie algebras of dimension seven over fields of characteristic 2, Sao Paulo J. Math. Sci. 4 (2010), N1, 93--107</li> <li>D. Leites, Towards classification of simple finite dimensional modular Lie superalgebras, arXiv:0710.5638.</li> <li>M. Vaughan-Lee, Simple Lie algebras of low dimension over GF(2), LMS J. Comput. Math. 9 (2006), 174--192</li> </ul> http://mathoverflow.net/questions/36850/journals-for-undergraduates/39206#39206 Answer by Pasha Zusmanovich for Journals for undergraduates Pasha Zusmanovich 2010-09-18T09:00:41Z 2013-01-29T09:18:52Z <p><a href="http://thehcmr.org" rel="nofollow">The Harvard College Mathematics Review</a> was another interesting venture, but it seems to be discontinued as of now. </p> http://mathoverflow.net/questions/99018/are-sg-and-ug-isomorphic-as-g-modules-for-g-lie-algebra-over-f-p-are-sgg/111481#111481 Answer by Pasha Zusmanovich for Are S(g) and U(g) isomorphic as g-modules for g Lie algebra over F_p ? Are S(g)^g and U(g)^g isomorphic as com.algs ? Pasha Zusmanovich 2012-11-04T19:27:57Z 2012-11-04T19:27:57Z <p>A sort of involved and (seemingly very partial) variant of Duflo isomorphism in characretristic $p$ is claimed in the following paper: N.A. Koreshkov, Central elements and invariants in modular Lie algebras, Russ. Math. (Izv. VUZ) 46 (2002), N7, 20-24 (Russian original is available, for example, at <a href="http://www.ksu.ru/journals/izv_vuz/arch/2002/07/05-7.PDF" rel="nofollow">http://www.ksu.ru/journals/izv_vuz/arch/2002/07/05-7.PDF</a> ). Roughly, it is proved there that for a finite-dimensional Lie algebra over a finite field $\mathbb Z_p$, a certain modification of $S(\overline{L})^{\overline L}$ is isomorphic to a certain subring of $U(\overline L)^{\overline L}$, where $\overline L$ is obtained from $L$ by some (infinite) field extension.</p> http://mathoverflow.net/questions/94907/applications-of-group-theory-to-math-biology-pharmacology/104671#104671 Answer by Pasha Zusmanovich for Applications of group theory to math. biology (pharmacology) ? Pasha Zusmanovich 2012-08-14T07:17:14Z 2012-08-14T07:17:14Z <p>1. There are a few old papers by Robert Rosen where he (seemingly) applies free semigroups to DNA-protein coding problem and argues about biological significance of the notion of freeness. These papers seem to be forgotten by now.</p> <ul> <li>The DNA-protein coding problem, Bull. Math. Biophysics 21 (1959), N1, 71-95 DOI:10.1007/BF02476459</li> <li>Some further comments on the DNA-protein coding problem, Bull. Math. Biophysics 21 (1959), N3, 289-297 - DOI:10.1007/BF02477917</li> <li>Some further comments on the DNA-protein coding problem: A correction and a note, Bull. Math. Biophysics 22 (1960), N2, 199-205 DOI:10.1007/BF02478006</li> <li>An hypothesis of Freese and the DNA-protein coding problem, Bull. Math. Biophys. 23 (1961), 305--318 DOI:10.1007/BF02476743</li> </ul> <p>2. There were some efforts to describe evolution and symmetry breaking of genetic code in terms of symmetries of some algebraic structures, including Lie groups and quantum groups, see Bashford, J.D. and Jarvis, P.D., The genetic code as a periodic table: algebraic aspects, arXiv:physics/0001066, and references therein.</p> 3. (Elementary) representation theory of symmetric group was used in some classical genetics: see, for example, Bennett, J.H., A general class of enumerations arising in genetics, Biometrics 23 (1967), No.3, 517-537 [<a href="http://links.jstor.org/sici?sici=0006-341X(196709)23%3A3%3C517%3AAGCOEA%3E2.0.CO%3B2-0" rel="nofollow">JSTOR link</a>]. There are probably more current treatements in the same direction, of which I am unaware of. http://mathoverflow.net/questions/90045/schur-multipliers-for-lie-algebras/93305#93305 Answer by Pasha Zusmanovich for Schur `multipliers' for Lie algebras Pasha Zusmanovich 2012-04-06T11:27:15Z 2012-06-23T14:19:38Z <p>Throwing a bunch of more references, for what it's worth (haven't looked thoroughly at them). But frankly, for me the "Schur multiplier", at least in the Lie-algebraic context, was always a synonym for the "second (co)homology with trivial coefficients", just defined via the Hopf formula, though probably I am missing some additional data coming by analogy from group theory.</p> <ul> <li>J. Bichon, G. Carnovale, Lazy cohomology: an analogue of the Schur multiplier for arbitrary Hopf algebras, J. Pure Appl. Algebra 204 (2006), no.3, 627-665</li> <li>L.R. Bosko, On Schur multipliers of Lie algebras and groups of maximal class, Intern. J. Algebra Comput. 20 (2010), N6, 807-821; DOI:10.1142/S0218196710005881</li> <li>L.R. Bosko, E.L. Stitzinger, Schur multipliers of nilpotent Lie algebras, arXiv:1103.1812</li> <li>H. Mohammadzadeh, B. Edalatzadeh, Some properties on Schur multiplier and cover of a pair of Lie algebras, arXiv:1105.0077</li> <li>P. Niroomand, On dimension of the Schur multiplier of nilpotent Lie algebras, Centr. Eur. J. Math. 9 (2011), no.1, 57-64 MR:2011m:17031</li> <li>P. Niroomand, F. Russo, A note on the Schur multiplier of a nilpotent Lie algebra, Comm. Algebra 39 (2011), N4, 1293-1297; arXiv:1001.0176; DOI:10.1080/00927871003652660</li> <li>P. Niroomand, F. Russo, A restriction on the Schur multiplier of nilpotent Lie algebras, Electron. J. Lin. Algebra 22 (2011), 1-9 <a href="http://www.math.technion.ac.il/iic/ela/ela-articles/22.html#1" rel="nofollow">http://www.math.technion.ac.il/iic/ela/ela-articles/22.html#1</a></li> <li>F. Saeedi, A. Salemkar, B. Edalatzadeh, The commutator subalgebra and Schur multiplier of a pair of nilpotent Lie algebras, J. Lie Theory 21 (2011), No.2, 491-498 <a href="http://www.heldermann.de/JLT/JLT21/JLT212/jlt21021.htm" rel="nofollow">http://www.heldermann.de/JLT/JLT21/JLT212/jlt21021.htm</a></li> <li>A. Salemkar, V. Alamian, H. Mohammadzadeh, Some properties of the Schur multiplier and covers of Lie algebras, Comm. Algebra 36 (2008), no.2, 697-707; DOI:10.1080/00927870701724193</li> </ul> http://mathoverflow.net/questions/92972/homomorphic-images-of-a-cartesian-product-of-finite-groups Homomorphic images of a Cartesian product of finite groups Pasha Zusmanovich 2012-04-03T07:23:40Z 2012-04-05T19:14:06Z <p>What can be said about the class of groups which can be represented as a homomorphic image of an (infinite) Cartesian product (=unrestricted direct product) of finite groups? What would be simple groups in this class? Which "classical" groups, like various matrix groups, belong to this class?</p> <p>The question can be (trivially) reformulated as a description of the smallest class of groups containing all finite groups and closed under taking Cartesian products and homomorphic images.</p> <p>Remarks:</p> <ol> <li>This class includes ultraproducts of finite groups, or, what is the same, groups having the same first-order theory as a set of finite groups (called pseudo-finite in the literature).</li> <li>A countable Cartesian product of finite groups is a profinite group. I don't know how beneficial it would be to consider the question in the profinite context.</li> <li>A finitely-generated group from this class is necessarily finite (Nikolov and Segal, see arXiv:1108.5130, Theorem 32). </li> </ol> http://mathoverflow.net/questions/90023/equivariant-levi-subalgebras/93095#93095 Answer by Pasha Zusmanovich for Equivariant Levi subalgebras. Pasha Zusmanovich 2012-04-04T11:15:24Z 2012-04-04T11:15:24Z <p>Yes. This results and its variations are contained in papers by E.J. Taft:</p> <ul> <li>Invariant Wedderburn factors, Illinois J. Math. 1 (1957), N4, 565-573 <a href="http://projecteuclid.org/euclid.ijm/1255380679" rel="nofollow">http://projecteuclid.org/euclid.ijm/1255380679</a> .</li> <li>Invariant Levi factors, Michigan Math. J. 9 (1962), N1, 65-68 DOI:10.1307/mmj/1028998623</li> <li>Orthogonal conjugacies in associative and Lie algebras, Trans. AMS 113 (1964), No.1, 18-29 DOI:10.2307/1994088</li> </ul> http://mathoverflow.net/questions/88802/lie-algebras-with-abelian-cartan-subalgebras/93072#93072 Answer by Pasha Zusmanovich for Lie algebras with abelian Cartan subalgebras. Pasha Zusmanovich 2012-04-04T05:14:26Z 2012-04-04T07:40:54Z <p>If $H$ is a Cartan subalgebra of a Lie algebra $L$, and $A$ is an associative commutative algebra, then $H \otimes A$ is a Cartan subalgebra of $L \otimes A$. Specializing this to the case $L$ classical simple, we get another positive answer to the first question (but not to the second one). This example can be, probably, varied, by taking subalgebras of $L \otimes A$, or adding "tails" of derivations, etc.</p> http://mathoverflow.net/questions/87134/references-infinite-dimensional-lie-algebras/93066#93066 Answer by Pasha Zusmanovich for References: Infinite dimensional Lie algebras Pasha Zusmanovich 2012-04-04T03:13:37Z 2012-04-04T03:13:37Z <p>Lie algebras of vector fields were treated in some 1970-1980 works of Kac and Rudakov:</p> <p>Kac: </p> <ul> <li>Simple irreducible graded Lie algebras of finite growth, Math. USSR Izv. 2 (1968), N6, 1271-1311 DOI:10.1070/IM1968v002n06ABEH00072 (this is the (famous) paper where the Kac-Moody algebras were introduced, but it contains also material about Lie algebras of vector fields)</li> <li>Description of filtered Lie algebras with which graded Lie algebras of Cartan type are associated, Math. USSR Izv. 8 (1974), 801-835 MR:51#5685 DOI:10.1070/IM1974v008n04ABEH002128 ZBL:0317.17002</li> </ul> <p>Rudakov:</p> <ul> <li>Groups of automorphisms of infinite-dimensional simple Lie algebras, Math. USSR Izv. 3 (1969), 707-722 ZBL:0222.17014 DOI:10.1070/IM1969v003n04ABEH000798 </li> <li>Subalgebras and automorphisms of Lie algebras of Cartan type, Funct. Anal. Appl. 20 (1986), 72-73 ZBL:0594.17015</li> </ul> <p>I think lot of information you are interested in contained there, albeit maybe not in the most explicit form.</p> <p>Also, some works about finite-dimensional characteristic $p$ counterparts of these algebras (so-called Lie algebras of Cartan type) - notably papers by Skryabin in Comm. Algebra in mid 1990s, book by Strade - start with a pretty much general context (arbitrary field, arbitrary dimension) and may be also relevant for your purpose.</p> http://mathoverflow.net/questions/90402/rational-forms-of-simple-lie-algebras/93063#93063 Answer by Pasha Zusmanovich for Rational forms of simple Lie algebras Pasha Zusmanovich 2012-04-04T02:36:58Z 2012-04-04T02:36:58Z <p>(Disclaimer: I am not an expert and easily may overlook something).</p> <p>Results about forms of exceptional types are technically difficult and scattered over the literature. For example, the only known to me full description of forms of $D_4$ is buried (somewhat implicitly) inside the book: Knus, Merkurjev, Rost, Tignol, The Book of Involutions, AMS Colloq. Publ, Vol. 44., 1998, <a href="http://www.mathematik.uni-bielefeld.de/~rost/BoI.html" rel="nofollow">http://www.mathematik.uni-bielefeld.de/~rost/BoI.html</a> . The works of Skip Garibaldi (using the language of algebraic groups) are also highly relevant.</p> http://mathoverflow.net/questions/84255/how-about-the-lie-algebra-over-commutative-ring/84698#84698 Answer by Pasha Zusmanovich for How about the Lie algebra over commutative ring? Pasha Zusmanovich 2012-01-01T20:07:50Z 2012-01-01T21:08:06Z <ol> <li><p>J.F. Hurley in a series of papers studied Lie algebras obtained by taking the multiplication table (with integer coefficients, due to Chevalley) of simple Lie algebras of classical or exceptional type and considering them over a commutative ring. The results describe center, ideal structure, etc. of such algebras in terms of the underlying ring. See, for example: Ideals in Chevalley algebras, <a href="http://www.jstor.org/stable/1994801" rel="nofollow">Trans. Amer. Math. Soc. 137 (1969), 245-258</a>; Composition series in Chevalley algebras <a href="http://projecteuclid.org/euclid.pjm/1102977369" rel="nofollow">Pacific J. Math. 32 (1970), 429-434</a>; Centers of Chevalley algebras, <a href="http://www.journalarchive.jst.go.jp/english/jnlabstract_en.php?cdjournal=jmath1948&amp;cdvol=34&amp;noissue=2&amp;startpage=219" rel="nofollow">J. Math. Soc. Japan 34 (1982), No.2, 219-222</a>. In the joint paper with J. Morita (Affine Chevalley algebras, <a href="http://dx.doi.org/10.1016/0021-8693(81)90299-4" rel="nofollow">J. Algebra 72 (1981), N2, 359-373</a>) he does something similar for some Kac--Moody algebras.</p></li> <li><p>Some questions in free Lie algebras were considered over commutative rings, for example: D.Z. Djokovic, On some inner derivations of free Lie algebras over commutative rings, J. Algebra 119 (1988), 233-245, where centralizers of a member of a free generating set are studied. The latter reference is more or less random, probably more can be found in some books (Reutenauer?). </p></li> </ol> <p>There are more instances of considering Lie algebras over commutative rings (for example, plenty of papers about automorphisms of some triangular or close to them algebras), but, unlike in the case of Lie algebras over fields, all these are some isolated examples, rather than a coherent theory. The book(s) of Bourbaki recommended by Anatoly Kochubei are, probably, interesting in that regard. Bourbaki tend to state things in the utmost generality, and it is educational to see how quickly they have to give up considering Lie algebras over rings and have to "throw around properties of vector spaces" (quoting Darij Grinberg).</p> <p>Perhaps the question could be augmented slightly by asking what is the reason for the absence of such a theory for Lie algebras (as opposed, for example, for associative algebras). Perhaps this is related somehow to the fact that classifying some natural (e.g., simple) classes of associative algebras is easier then that of Lie algebras (e.g., root space decomposition technique for Lie algebras which works over algebraically closed fields vs. "idempotent" technique for associative algebras which works over arbitrary fields and even over rings), but I venture into a sheer speculation here.</p> http://mathoverflow.net/questions/79528/embedding-of-relatively-free-groups-of-bigger-rank-into-ones-of-smaller-rank Embedding of relatively free groups of bigger rank into ones of smaller rank Pasha Zusmanovich 2011-10-30T16:42:45Z 2011-10-31T07:45:24Z <p>This question is prompted by <a href="http://mathoverflow.net/questions/78787/varieties-of-groups-with-infinite-relatively-free-group-of-rank-2-finite-infinit" rel="nofollow">this one</a> by Arturo Magidin: whether there exist varieties of groups in which the relatively free group of rank 2 is finite, and the relatively free group of rank 3 is infinite.</p> <p>My question is: in which varieties of groups is it true that relatively free groups of bigger, countable or finite, rank embed into relatively free groups of smaller rank? This is so, for example, for the variety of all groups, and also for Burnside varieties (defined by the identity $x^n = 1$). On the other hand, this is not so for solvable or nilpotent varieties.</p> <p>My knowledge on this is limited by the (old) book of H. Neumann "Varieties of Groups", and by paper of Shirvanyan about free Burnside groups. Probably more is known nowadays.</p> <p>Of course, one can pose the same question also for other algebraic systems, for example, for algebras.</p> http://mathoverflow.net/questions/71144/algebraic-axiomatization-for-abbat-operation-on-matrices/79530#79530 Answer by Pasha Zusmanovich for Algebraic axiomatization for AB+BA^T operation on matrices Pasha Zusmanovich 2011-10-30T17:24:55Z 2011-10-30T17:24:55Z <p>(This is an answer to the updated question). I have a bit trouble with your terminology (what you are talking about are definitely not direct sums of a Lie and a Jordan algebra), but I hope I understand the spirit of your question, without going to much into nitpicking. Again, being put, probably, in a bit more general framework, the question is whether the algebras with a few binary operations, one of them, for example, satisfies the Jacobi identity, another one satisfies the Jordan identity, with some compatibility conditions between them, or something like that, were considered in the literature. </p> <p>The answer is "yes". In fact, there is so much papers about such sort of objects, that it is a bit difficult to point just on a few of them. Lot of such objects are studied nowadays in the framework of operadic theory, see a very interesting compendium arXiv:1101.0267 . Some authors who wrote a lot on the topic: Dzhumadil'daev (e.g., <a href="http://dx.doi.org/10.1142/S0219498809003230" rel="nofollow">http://dx.doi.org/10.1142/S0219498809003230</a> , <a href="http://dx.doi.org/10.1007/s10958-009-9532-x" rel="nofollow">http://dx.doi.org/10.1007/s10958-009-9532-x</a> , <a href="http://www.mpim-bonn.mpg.de/preblob/2920" rel="nofollow">http://www.mpim-bonn.mpg.de/preblob/2920</a> , <a href="http://dx.doi.org/10.1081/AGB-200060504" rel="nofollow">http://dx.doi.org/10.1081/AGB-200060504</a> ), M. Goze, Markl and Remm (look them on arXiv).</p> http://mathoverflow.net/questions/76253/where-did-the-multigraded-segre-product-appear-in-the-literature/78799#78799 Answer by Pasha Zusmanovich for Where did the multigraded Segre product appear in the literature? Pasha Zusmanovich 2011-10-21T21:50:26Z 2011-10-21T21:50:26Z <p>I believe that Segre product, at least in the particular case of grading over $\mathbb Z$ or $\mathbb N$ (I don't know what a vector configuration is), appears in literature in many places. One such place I am aware of is the book by Polishchuk and Positselski "Quadratic Algebras", AMS, 2005, where it is used to construct non-Koszul algebras with mutually inverse Hilbert series (p. 59 onwards). It is used also in the operadic environment (see, for example, a draft of the book by Loday and Vallette at <a href="http://math.unice.fr/~brunov/Operades.html" rel="nofollow">http://math.unice.fr/~brunov/Operades.html</a> ) what encompasses the above-mentioned algebra case and, probably, many other cases, including yours.</p> http://mathoverflow.net/questions/68268/dissimilarity-between-groups-and-lie-algebras (Dis)similarity between groups and Lie algebras Pasha Zusmanovich 2011-06-20T04:56:22Z 2011-10-11T02:25:19Z <p>There are many questions which sound similar or the same for groups and Lie algebras. Some (very few, actually) of those questions have identical solutions and answers. Some have identical answers but very different solutions. Some have different answers and different solutions.</p> <p>I am not talking about situations where one have a strictly defined (functorial) correspondence of some sort which allows to reduce almost automatically some questions from groups to Lie algebras, like Lie groups-Lie algebras, or Malcev(-like) correspondence. </p> <p>For simplicity, let us confine to finite groups and finite-dimensional Lie algebras over a field. A few examples of what I have in mind:</p> <ol> <li><p>Suppose a group (resp. Lie algebra) is represented as a (not necessarily direct) product (resp. sum) of two nilpotent subgroups (resp. subalgebras). Is it true that the group (resp. Lie algebra) is solvable?</p></li> <li><p>Suppose a group (resp. Lie algebra) is commutative-transitive: $[x,y] = 1$ (resp. 0), $[y,z] = 1$ (resp. 0), and $y \ne 1$ (resp. 0) implies $[x,z] = 1$ (resp. 0). What can be said about its structural properties?</p></li> <li><p><b>(edit: added a couple of hours later)</b> (Lyndon-)Hochschild-Serre spectral sequence connecting group (resp. Lie algebra) (co)homology with (co)homology of its normal subgroup (resp. ideal) and the respective quotient.</p></li> </ol> <p>As far as I know, both questions 1. and 2. have very similar answers for groups and Lie algebras, in some particular situations admit (almost) identical proofs or fragments of proofs, but in the whole generality the proofs are very different, with group-theoretic proof not utilizing in any way a Lie-algebraic one, and vice versa. As for 3., initially the group- and Lie-algebraic cases were given very similar, but disjoint proofs, and later were interpreted as particular instances of the Grothendieck (?) spectral sequence. </p> <p><b>Question.</b> Is it possible, looking on a property of groups/Lie algebras expressed as a formula in the appropriate first- or second-order theory, to predict solely on syntaxical/logical ground, whether this property would convey similar or different results in respective categories? On the other hand, is there some (metamathematical, I dare to say) principle which would make this impossible? </p> <p>I realize that the question is probably too vague, and I would appreciate any help in making it more precise and interesting.</p> http://mathoverflow.net/questions/68429/cohomology-of-generalized-verma-modules-and-invariant-operators/72466#72466 Answer by Pasha Zusmanovich for cohomology of generalized Verma modules and invariant operators Pasha Zusmanovich 2011-08-09T10:10:39Z 2011-08-09T10:10:39Z <p>In: A. Tolpygo, Estimation of the cohomology of Verma modules, Russ. Math. Surv. 48 (1993), N1, 193-194, some estimations on the dimension of cohomology in question are given. In the later paper: Lie algebra cohomology and generating functions, Homology Homotopy Appl. 6 (2004), 59-85 <a href="http://www.intlpress.com/HHA/v6/n1/a6/" rel="nofollow">http://www.intlpress.com/HHA/v6/n1/a6/</a> , Tolpygo elaborates on the technique used in the former shorter note, but in the context of finite-dimensional modules. But probably something could be inferred from the latter paper also in the case of Verma modules. </p> http://mathoverflow.net/questions/70569/system-of-weights-for-nilpotent-lie-algebras/72396#72396 Answer by Pasha Zusmanovich for System of weights for nilpotent Lie algebras Pasha Zusmanovich 2011-08-08T20:51:22Z 2011-08-08T22:16:35Z <p>Merely some further references, doesn't fit into the comment field: </p> <p>G.F. Leger and E.M. Luks, Cohomology and weight systems for nilpotent Lie algebras, Bull. Amer. Math. Soc. 80 (1974), 77-80 <a href="http://projecteuclid.org/euclid.bams/1183535294" rel="nofollow">http://projecteuclid.org/euclid.bams/1183535294</a></p> <p>L.J. Santharoubane, Kac-Moody Lie algebras and the classification of nilpotent Lie algebras of maximal rank, Canad. J. Math. 34 (1982), 1215-1239 DOI:10.4153/CJM-1982-084-5</p> <p>L.J. Santharoubane, Kac-Moody Lie algebras and the universal element for the category of nilpotent Lie algebras, Math. Ann. 263 (1983), 365-370 <a href="http://resolver.sub.uni-goettingen.de/purl?GDZPPN002323583" rel="nofollow">http://resolver.sub.uni-goettingen.de/purl?GDZPPN002323583</a></p> <p>L. Magnin, Remarks on weight systems on cohomology of nilpotent Lie algebras, Algebras Groups Geom. 9 (1992), No.2, 111-135</p> http://mathoverflow.net/questions/72233/restricted-lie-algebras-of-low-dimension/72400#72400 Answer by Pasha Zusmanovich for Restricted Lie algebras of low dimension Pasha Zusmanovich 2011-08-08T21:12:13Z 2011-08-08T21:12:13Z <p>H. Strade, Lie algebras of small dimension, Contemp. Math. 442 (2007), 232--265; arXiv:math/0601413 - classifies nonsolvable Lie algebras of dimension &lt;7 over a finite field. I think there is a corresponding GAP package.</p> <p>Another seemingly relevant reference: M. Vaughan-Lee, Simple Lie algebras of low dimension over GF(2), LMS J. Comput. Math. 9 (2006), 174-192.</p> http://mathoverflow.net/questions/71144/algebraic-axiomatization-for-abbat-operation-on-matrices/72381#72381 Answer by Pasha Zusmanovich for Algebraic axiomatization for AB+BA^T operation on matrices Pasha Zusmanovich 2011-08-08T18:18:50Z 2011-08-08T18:18:50Z <p>Let me reformulate the question a bit and then provide an empirical argument in favor of the negative answer.</p> <p>For an associative algebra $A$ with involution $J$, define a new binary operation on $A$ as $a$ $\square$ $b = ab + ba^J$. The question is whether this operation defines a "nice" class of algebras.</p> <p>This seems to be a correct generalization of the initial question, as by analogy with Lie ($[a,b] = ab - ba$) and Jordan ($a \circ b = ab + ba$) algebras constructed from associative algebras in a similar way, "interesting" and "natural" examples of such algebras do not confined to matrix algebras $A$.</p> <p>What is a "nice" class of algebras? Again, by analogy with Lie and Jordan situations we might require that it satisfies some nontrivial identity (like Jacobi or Jordan identity), i.e., form a variety of algebras. Then the answer is "no": the class of all associative algebras with involution under the opertion $\square$ do not satisfy any nontrivial identity. For, suppose such identity exist. Than each associative algebra $A$ with involution satisfies some idenity of the form $f(x_1, \dots, x_n, x_1^J, \dots, x_n^J) = 0$ with nontrivial occurences of involutions. Then by theorem of Amitsur, $A$ satisfies a power $n$ of a standard identity of degree $d$, both $n$ and $d$ are determined by $f$ (see, for example, I.N. Herstein, Rings with Involution, The Univ. of Chicago Press, 1976, $\S$ 5). As, obviously, not all algebras with involution would satisfy the latter identity, we get a contradiction.</p> <p>Even if we confine ourselves with matrix algebars with transposition, as in the initial question, we can choose a matrix algebra of sufficiently large degree not satisfying a given power of a standard identity of a given degree, so even that narrower class of algebras do not form a variety.</p> <p>In fact, this argument is valid for any binary operation defined in terms of addition, multiplication, and involution in an associative algebra, provided involution occurs non-trivailly, not necessary $\square$. </p> http://mathoverflow.net/questions/65963/deceptively-short-proof-of-regevs-a-otimes-b-theorem Deceptively short proof of Regev's $A \otimes B$ theorem Pasha Zusmanovich 2011-05-25T13:33:33Z 2011-08-08T10:21:38Z <p>The following theorem, due to Regev, is one of the cornerstones of the theory of PI algebras (i.e., associative algebras satisfying a nontrivial polynomial identity): </p> <p>Let $A$, $B$ be two PI algebras over a field $K$. Then their tensor product $A \otimes_K B$ is PI.</p> <p>Consider the following "proof" of this theorem. Since $A$ and $B$ are PI, their Jacobson radicals $J(A)$ and $J(B)$ are nilpotent, and $A/J(A)$ and $B/J(B)$ are semisimple PI algebras which are known to be embedded into matrix algebras over a commutative ring, say $M_n(C)$ and $M_m(D)$. Now, $J(A) \otimes B + A \otimes J(B)$ is a nilpotent ideal of $A \otimes B$, quotient by which is isomorphic to $A/J(A) \otimes B/J(B)$ and hence is embedded into $M_n(C) \otimes M_m(D)$, which, in its turn, is embedded into $M_{nm} (C \otimes D)$. Therefore, $A \otimes B$ contains a nilpotent ideal quotient by which is embedded into a matrix algebra over a commutative ring, and hence is PI.</p> <p>Regev's theorem is a relatively difficult result, first conjectured by Jacobson, and having resisted attempts by a few mathematicians. Thus it hardly admits such a short simple proof. Where is the catch? </p> <p>The only weak spot a can think of, is that nilpotence of the Jacobson radical of a PI algebra is a relatively new (at least proved long after Regev's theorem in 1970) complicated result whose proof probably involves appeal to Regev's theorem. Is it true? Or am I missing something else?</p> <p><b>Edit May 26, 28 2011</b>: As was pointed out by Bugs Bunny, we should require that $A$ and $B$ are finitely generated, as this is the hypothesis of the Razmyslov-Kemer-Brown theorem about nilpotence of the Jacobson radical of a PI algebra (and the theorem does not hold for infinitely-generated algebras). <strike>But, the general statement of Regev's theorem obviously reduces to this case.</strike></p> http://mathoverflow.net/questions/65963/deceptively-short-proof-of-regevs-a-otimes-b-theorem/72346#72346 Answer by Pasha Zusmanovich for Deceptively short proof of Regev's $A \otimes B$ theorem Pasha Zusmanovich 2011-08-08T10:21:38Z 2011-08-08T10:21:38Z <p>I spoke about this with Louis Rowen (who, among other, wrote a few books on the subject) and here is what I got from this conversation:</p> <ol> <li>There is no circular dependency in this argument.</li> <li>The Razmyslov-Kemer-Brown theorem is more difficult and complicated result than Regev's theorem, so there is no much point to infer the latter from the former.</li> </ol> http://mathoverflow.net/questions/67717/lie-algebras-with-trivial-second-cohomology-group/68198#68198 Answer by Pasha Zusmanovich for Lie algebras with trivial second cohomology group Pasha Zusmanovich 2011-06-19T05:58:36Z 2011-06-19T05:58:36Z <p>This is merely an additional comment, too long for a comment field. To see that classification of finite-dimensional Lie algebras $L$ with $H^2(L,L) = 0$ is unfeasible, consider, for example, Lie algebras of the form $g \otimes A$, for a classical simple $g$ and finite-dimensional associative commutative algebra $A$. By (cohomological interpretation of) the result of J.-L. Cathelineau, Homologie de degre trois d'algebres de Lie simple deployees etendues a une algebre commutative, Enseign. Math. 33 (1987), 159-173 <a href="http://dx.doi.org/10.5169/seals-87889" rel="nofollow">http://dx.doi.org/10.5169/seals-87889</a> , $H^2(g\otimes A, g\otimes A)$ is isomorphic either to Hochschild of dihedral cohomology of $A$, depending on the type of $g$. So, the purported classification would include, as a very particular case, classification of associative commutative algebras with the vanishing second Hochschild cohomology. By modification of this tensor product construction, one possible to get other variants of this classification problem which seem to be out of reach.</p> http://mathoverflow.net/questions/66745/on-engel-anticommutative-algebras/67430#67430 Answer by Pasha Zusmanovich for On Engel-anticommutative algebras Pasha Zusmanovich 2011-06-10T12:06:49Z 2011-06-11T06:55:20Z <ol> <li><p>To rephrase José Figueroa-O'Farrill's comment above - a 7-dimensional simple exceptional Malcev algebra (which is a quotient of octonions under the commutator by 1-dimensional center). But, really, I find the question is formulated badly: just skew-commutativity is a very mild restriction to say something meaningful about an algebra in general, so it's almost the same as to ask for "famous" examples of a (nonassociative) algebra.</p></li> <li><p>No, this is not true. I was able to construct counterexamples on computer, as finite-dimensional quotients of free algebras in respective "Engel varieties", but all they are large and cumbersome. Shorther examples can be found in the following paper by Koreshkov and Kharitonov, which, apparently, was published twice:</p> <p>About nilpotency of Engel algebras, Formal Power Series and Algebraic Combinatorics (ed. D. Krob et al.), Springer, 2000, 461-467; ZBL: 0983.17003 [available on google books and amazon]</p> <p>Nilpotency of the Engel algebras, Russ. Math. (Izv. VUZ) 45 (2001), No. 11, 15-18; ZBL: 1103.17300</p></li> </ol> <p>They prove also that this is true for algebras of dimension $\le 4$. It is also easy to see (and is recorded, for example, in: V.T. Filippov, Binary Lie algebras satisfying the third Engel condition, Siber. Math. J. 49 (2008), N4, 744-748; DOI: 10.1007/s11202-008-0071-3) that the second Engel condition implies nilpotency of degree $4$.</p> <p>On the other hand, Engel(-like) theorems were established for many particular ("famous"?) classes of anticommutative algebras considered in the literature: binary-Lie, Malcev, and some other generalizations of Lie algebras, and it is an open question, as far as I know, for Sagle algebras.</p> http://mathoverflow.net/questions/66162/appendices-to-papers-authored-by-others Appendices to papers authored by others Pasha Zusmanovich 2011-05-27T08:00:00Z 2011-05-28T11:35:32Z <p>Every now and then, one sees in mathematical papers appendices, authored by person(s) different from the authors of the main body of the paper. </p> <ol> <li><p>What is the rationale behind such appendices? In most of the cases I have seen, it can be a bona fide short separate paper (granted, tight closely to the main paper in question, but many papers are tight closely to each other). </p></li> <li><p>What is the complete list of authors of a paper with an appendix? I belong to the school of thought claiming that references should be given according to the first letters of authors names. Each time, when citing a paper with appendix, I am agonizing over whether the authors of the appendix should be included or not in this abbreviation. More seriously, if a person is a (co)author of such an appendix (but not of the main body of the paper), how it is reflected in his list of publications?</p></li> <li><p>My impression is that the number of such appendices has been increased drastically in the recent years. Is it true? If yes, what may be reason(s) for that?</p></li> </ol> <p><b>Edit May 27, 2011</b>: DamienC made an interesting (and looking plausible to me) suggestion that the purported recent increase of separately-authored appendices is a way to cope with no longer fully adequate convention of authors ordering. If so, it would be interesting to compare with how the things evolved, say, in life sciences, from the situation where coauthors were not acceptable - a situation, as far as I understand, common to all scientific papers at the beginning of XX century - till today ugly state of affairs with convoluted and self-contradictory rules who goes first, who goes last, how the authors should be clustered, etc. This may suggest that in mathematics things evolve in a different direction, which, though not perfect, is sort of reassuring. </p> <p><b>2nd edit May 27, 28 2011</b>. My question is not about appendices in general, including differences in various subjects and subcultures (for example, in statistics it is common to defer all proofs to appendices) -- this is an <strike>entirely</strike> different topic. My question is about the sutuation when the set of the authors of an appenix is different from the set of the authors of the main paper -- and, bibliographic (and bibliometric, and social, if you wish) difficulties (or at least what I see as bibliographic, etc. difficulties) arising from such situations. Sorry if that was not clear enough from the first place.</p> http://mathoverflow.net/questions/65115/nilpotent-lie-algebras-of-vector-fields Nilpotent Lie algebras of vector fields Pasha Zusmanovich 2011-05-16T07:35:25Z 2011-05-16T15:58:27Z <p>Let $L$ be a finite-dimensional nilpotent subalgebra of the Lie algebra $W_n$ of all vector fields in $n$ variables (I am interested both in polynomial and formal vector fields). Does there exist a bound in terms of $n$ on the index of nilpotency of $L$?</p> <p>For $n=1$ the answer is trivially "yes": every finite dimensional subalgebra is of dimension $\le 3$, concentrated in degrees $-1$, $0$, $1$. For $n>1$, I don't know.</p> <p><b>Motivation</b>: the question looks for me interesting in its own right, but also arises in control theory. There, one has a criterion for nonlinear systems to be so-called finitely discretizable (roughly, to admit a polynomial solution of some sort) in terms of nilpotency of the Lie algebra generated by the corresponding vector fields. So, when one applies this criterion on practice, one wants to be sure that it is enough to check the vanishing of the iterated Lie brackets up to a given degree. For the control theory application, the base field is $\mathbb R$, but the question does not seem to be dependent on the base field (as long as it is of characteristic zero at least).</p> http://mathoverflow.net/questions/104091/universal-enveloping-ringsymmetric-algebra-isomorphism-for-lie-rings/104112#104112 Comment by Pasha Zusmanovich Pasha Zusmanovich 2013-05-10T15:38:40Z 2013-05-10T15:38:40Z Paper by Cohn: <a href="http://justpasha.org/math/links/files/c/cohn/197.pdf" rel="nofollow">justpasha.org/math/links/files/c/cohn/197.pdf</a> http://mathoverflow.net/questions/98997/determinants-and-polynomials-in-matrices Comment by Pasha Zusmanovich Pasha Zusmanovich 2013-05-10T14:51:52Z 2013-05-10T14:51:52Z MacDuffee's book should be available here: <a href="http://libgen.info/view.php?id=4344" rel="nofollow">libgen.info/view.php?id=4344</a> http://mathoverflow.net/questions/112017/generalization-of-traces Comment by Pasha Zusmanovich Pasha Zusmanovich 2012-11-14T14:50:27Z 2012-11-14T14:50:27Z No idea whether it is related to your question, but here is a reference: H. Rohrl, A categorical setting for determinants and traces, Nagoya Math. J. 34 (1969), 35-76 <a href="http://projecteuclid.org/euclid.nmj/1118797662" rel="nofollow">projecteuclid.org/euclid.nmj/1118797662</a> http://mathoverflow.net/questions/111400/good-computer-package-for-calculating-inverse-of-a-formal-power-series Comment by Pasha Zusmanovich Pasha Zusmanovich 2012-11-06T16:47:48Z 2012-11-06T16:47:48Z Pari/GP's serreverse(). http://mathoverflow.net/questions/100265/not-especially-famous-long-open-problems-which-anyone-can-understand/100432#100432 Comment by Pasha Zusmanovich Pasha Zusmanovich 2012-06-23T15:04:39Z 2012-06-23T15:04:39Z That's interesting. Do you have some further pointers? http://mathoverflow.net/questions/88850/twisted-affine-lie-algebras Comment by Pasha Zusmanovich Pasha Zusmanovich 2012-04-06T08:40:47Z 2012-04-06T08:40:47Z Silly remark: isn't it possible to write the whole thing in the first order language of the theory of fields? If so, then it will be automatically true in any algebraically closed field of characteristic 0. http://mathoverflow.net/questions/92972/homomorphic-images-of-a-cartesian-product-of-finite-groups Comment by Pasha Zusmanovich Pasha Zusmanovich 2012-04-04T02:17:31Z 2012-04-04T02:17:31Z @Mark and @Emil: my understanding is that there is a discrepancy in terminology: in group theory it is more (but not exclusively) like Mark is saying, and in universal algebra (and linear algebras, including Lie algebras for that matter) it seems the common usage of &quot;direct product&quot; corresponds to &quot;Cartesian product&quot; in group theory. http://mathoverflow.net/questions/92972/homomorphic-images-of-a-cartesian-product-of-finite-groups/92973#92973 Comment by Pasha Zusmanovich Pasha Zusmanovich 2012-04-03T13:27:33Z 2012-04-03T13:27:33Z Thanks! (amended the question) http://mathoverflow.net/questions/92972/homomorphic-images-of-a-cartesian-product-of-finite-groups Comment by Pasha Zusmanovich Pasha Zusmanovich 2012-04-03T07:44:08Z 2012-04-03T07:44:08Z @Mark: &quot;same first order theory&quot;: thanks, of course, corrected it. &quot;Cartesian product&quot;: it seems that I am confused about the terminology. Aren't &quot;Cartesian product&quot; and &quot;direct product&quot; the same, as opposed to &quot;direct sum&quot;? http://mathoverflow.net/questions/79528/embedding-of-relatively-free-groups-of-bigger-rank-into-ones-of-smaller-rank Comment by Pasha Zusmanovich Pasha Zusmanovich 2011-10-31T07:43:46Z 2011-10-31T07:43:46Z @Arturo Magidin: It is, in the context of universal algebra. Thanks! Adding &quot;universal algebra&quot; tag. http://mathoverflow.net/questions/71144/algebraic-axiomatization-for-abbat-operation-on-matrices/79530#79530 Comment by Pasha Zusmanovich Pasha Zusmanovich 2011-10-31T07:32:46Z 2011-10-31T07:32:46Z Ok then. This is a commonly exploited situation in the context of (associative) rings with involutions. http://mathoverflow.net/questions/79528/embedding-of-relatively-free-groups-of-bigger-rank-into-ones-of-smaller-rank Comment by Pasha Zusmanovich Pasha Zusmanovich 2011-10-30T19:59:39Z 2011-10-30T19:59:39Z @Benjamin Steinberg: Yes. I will edit the question accordingly. http://mathoverflow.net/questions/78456/koethe-conjecture-on-non-associative-rings Comment by Pasha Zusmanovich Pasha Zusmanovich 2011-10-30T17:55:32Z 2011-10-30T17:55:32Z You should definitely state what you are after. For example, what is &quot;nil&quot; for nonassociative rings? http://mathoverflow.net/questions/74682/twisted-affine-algebras Comment by Pasha Zusmanovich Pasha Zusmanovich 2011-10-22T06:44:43Z 2011-10-22T06:44:43Z @Chris: Sorry, no. This is a sheer speculation, like I said. http://mathoverflow.net/questions/77431/can-i-express-variable-occurence-ranges-in-logic Comment by Pasha Zusmanovich Pasha Zusmanovich 2011-10-22T06:38:48Z 2011-10-22T06:38:48Z @Countably Infinite: &quot;I deleting on the principle that a certain communication has been accomplished. I don't judge my comments so important for curation.&quot; : awfully wrong, IMHO. First, you should not to consider yourself a sole proprietor of a specific page with a specific discussion, even if you have initiated it and technically you can edit it as you please. Once it appears publicly, it is more like a public property. Second, what have been said, have been said, and do have the courage to stand for it, even if you regret it afterwards.