User david wehlau - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T00:03:34Z http://mathoverflow.net/feeds/user/16684 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/129287/generate-a-higher-degree-symmetric-polynomial-from-an-existing-one/129784#129784 Answer by David Wehlau for Generate a higher degree symmetric polynomial from an existing one David Wehlau 2013-05-05T22:11:04Z 2013-05-05T22:11:04Z <p>Your sum is (up to a scalar) the transfer or trace of $u(x_{n+1})p(x_1,x_2,\dots,x_n)$. Given any polynomial $h(x_1,x_2,\dots,x_n,x_{n+1})$ its transfer or trace (with respect to the symmetric group, $S_{n+1}$ is the symmetric polynomial $\sum_{\sigma\in S_{n+1}} \sigma \cdot h(x_1,x_2,\dots,x_n,x_{n+1})$. This construction works for any finite group.</p> <p>More precisely, you have a relative transfer $\sum_{\sigma\in S_{n+1}/S_n} \sigma \cdot h(x_1,x_2,\dots,x_n,x_{n+1})$ where $h=u(x_{n+1})p(x_1,x_2,\dots,x_n)$ is $S_n$-invariant and the sum is over a (any) set of (left) coset representatives of $S_n$ in $S_{n+1}$. </p> http://mathoverflow.net/questions/125454/cyclically-symmetric-functions/127958#127958 Answer by David Wehlau for Cyclically symmetric functions David Wehlau 2013-04-18T12:49:52Z 2013-04-18T13:00:21Z <p>Consider the action of the cyclic group $G$ of order $n$ acting on an $m$-dimensional vector space $V$. I'll assume you are working over an algebraically closed field $k$ (or at least a field containing the nth roots of unity). Over such a field the elements of the cyclic group are diagonalizable. Thus the group is generated by $g=\diag(\omega^{-a_1},\dots,\omega^{-a_m})$ where $\omega$ is a primitive nth root of unity and the $a_i$ are integers (which may be assumed to lie between 0 and $n-1$. </p> <p>If ${x_1,\dots,x_m}$ is a basis for $V^*$ dual to the diagonal basis for $V$, then $g$ acts via $g\cdot x_i = \omega^{a_i} x_i$. Hence if $t = x_1^{e_1}\cdots x_m^{e_m}$ is a monomial then $g\cdot t = \omega^{a_1e_1 + \cdots a_me_m} t$. Since a function is invariant iff it is a linear combination of invariant monomials, the invariant theory reduces to solving the linear congruence $a_1e_1 + \cdots a_me_m \equiv 0 \pmod{n}$ for $(e_1,\dots,e_m)$. These solutions form a monoid in $N^m$. Finding a set of minimal generators for $k[V]^G$ amounts to finding minimal generators for this monoid. This is not really solved in general but there has been a lot of study, as you might imagine. </p> <p>I suggest the following references.</p> <p>Weidong Gao, Alfred Geroldinger, Zero-sum problems in finite abelian groups: A survey, Expo. Math. 24 (2006) 337 – 369</p> <p>John C. Harris and David L. Wehlau, Non-Negative Integer Linear Congruences, Indagationes Mathematicae {\bf 17} No. 1 (2006) 37-44. arXiv:math/0409489v1</p> <p>Pingzhi Yuan, On the index of minimal zero-sum sequences over finite cyclic groups, Journal of Combinatorial Theory, Series A 114 (2007) 1545–1551</p> <p>Finally, I'll mention that things are much much more complicated (equivalent to the invariant theory of SL(2,C)) when the characteristic of $k$ divides $n$.</p> http://mathoverflow.net/questions/14613/invariant-polynomials-under-a-group-action-hidden-git/71168#71168 Answer by David Wehlau for Invariant Polynomials under a Group Action (hidden GIT) David Wehlau 2011-07-25T00:53:49Z 2013-03-24T22:26:10Z <p>The minimal number of invariants needed to generate ${\mathbb Z}[x_1,...,x_n]^{{\mathbb Z}_n}$ has been considered by a number of authors including Erdos, Dixmier and Kac. (see the references in [John C. Harris and David L. Wehlau Non-Negative Integer Linear Congruences, Indagationes Mathematicae 17 No. 1 (2006) 37-44]. It is easily seen to be bounded below by the number of partitions of n, ${\mathcal P}(n)$. Dixmier produced a number of papers giving the asymptotic behavior of this number as a function of $n$. The results in the above Harris-Wehlau paper are completed by using the main result in [Pingzhi Yuan, On the index of minimal zero-sum sequences over finite cyclic groups, Journal of Combinatorial Theory, Series A 114 (2007) 1545–1551]. These two papers combine to show that the number of homogeneous generators of this ring of invariants of degree $k$ is exactly $\phi(n){\mathcal P}(n-k)$ if $k \geq \lfloor n/2\rfloor + 2$ (here $\phi$ is Euler's totient function). Surprisingly (at least to me) much less is known about the number of generators in lower degrees.</p> http://mathoverflow.net/questions/111894/modules-of-invariants/117183#117183 Answer by David Wehlau for Modules of invariants? David Wehlau 2012-12-25T04:25:16Z 2012-12-25T04:38:54Z <p>The polynomial ring Sym(V) is naturally graded: $Sym(V) = \oplus Sym(V)_d$ Suppose you have can compute the isotypic decomposition of these graded components </p> <p>Sym(V$)_d$ =</p> <p>${\oplus_{\chi \in A_d} U_\chi}$</p> <p>where the $U_\chi$ are irreducible representations of $G$. Then $M = \oplus_d \oplus_{\chi \in A_d} (U_\chi \otimes W)^G$. Now $(U_\chi \otimes W)^G = 0$ unless $W \cong U_\chi^*$ in which case $(U_\chi \otimes W)^G$ is one dimensional. Thus your problem really amounts to decomposing $Sym(V)$ as a $G$-representation. So far none of these depends on $G$ being finite (but it should be reductive).</p> <p>$M$ is called the module of $W$-covariants of $V$. The Hilbert Series of $M$ may be computed using Molien's Theorem. A minimal generating set for $M$ contains only elements of degree less than or equal to the order of $G$. Lots of other things are known but I suggest you read about it. One good reference for this is {\it Invariants of Finite Groups and Their Applications to Combinatorics} R. Stanley Bull. A.M.S. 1979.</p> http://mathoverflow.net/questions/106965/is-kxg-integral-closed-in-kx/106972#106972 Answer by David Wehlau for Is $k[X]^G$ integral closed in $k[X]$. David Wehlau 2012-09-12T03:10:27Z 2012-09-12T03:47:02Z <p>whoops - made a silly mistake. read the question as asking whether k[X]^G is integrally closed in its field of fractions.</p> http://mathoverflow.net/questions/104223/justifying-explaining-math-research-in-a-public-address Justifying/Explaining math research in a public address David Wehlau 2012-08-07T19:17:23Z 2012-08-10T01:48:28Z <p>I have been chosen by my university to give a 1 hour public research lecture. Every year a researcher is chosen for this honour. Traditionally people explain their own research about designing improved airplanes or the life of a peasant in the middle ages or whatever. I don't feel that I can explain my research in an hour to the general public in any sort of intellectually honest way. </p> <p>I have decided that it would be more interesting and more useful for me to try to say some things about math research in general. I hope to explain what math research is like, why it is important to society and how it differs from research in other scientific fields. </p> <p>I think 3 or 4 really good examples of advances arising from recent mathematical research would go a long way to making my points in an interesting manner. I am thinking of examples like the page rank algorithm used by Google. Or perhaps modern cryptography techniques such as the RSA algorithm. </p> <p><em>Question:</em> What are some other examples that help show the public why math research is important? </p> <p>I think it is important that these be relatively recent examples and that they relate directly to things people experience themselves.</p> <p>I would be grateful for any suggestions which might improve my talk. </p> http://mathoverflow.net/questions/69471/what-is-the-ring-of-invariants-of-gl-acting-on-quaternary-cubic-forms/82525#82525 Answer by David Wehlau for What is the ring of invariants of GL acting on quaternary cubic forms? David Wehlau 2011-12-03T00:17:00Z 2011-12-03T00:17:00Z <p>Don't be misled into thinking that the answer over $\Bbb C$ tells you very much about the answer over your finite field $K$. The space of cubic forms in 4 variables is 20 dimensional. The group $GL(4,K)$ is a finite group and so the ring of invariants you seek has Krull dimension 20. In particular, it has at least 20 generators. Probably, however, it has a few thousand generators if not many more. </p> <p>There are computer algebra packages (e.g. Magma) that include routines that can compute generators for your ring of invariants in principle, but in practice you will find that they run out of memory long before getting very far on a problem of this size. They will be able to compute vector space bases for the invariants in low degrees. On the other hand, it is doubtful that having a list of thousands of generators (or even just their degrees) for the ring of invariants will help you very much. On the other hand if you want to know something about the properties of the ring of invariants as a ring, then it's possible that may be known.</p> http://mathoverflow.net/questions/20106/generalized-symmetric-algebras-and-dickson-algebras-over-mathbb-f-p/71460#71460 Answer by David Wehlau for Generalized symmetric algebras and Dickson algebras over ${\mathbb F}_p$. David Wehlau 2011-07-28T03:27:37Z 2011-07-28T03:27:37Z <p>The paper [P. FLEISCHMANN, A NEW DEGREE BOUND FOR VECTOR INVARIANTS OF SYMMETRIC GROUPS, TRANS. AMS Volume 350, Number 4, April 1998, Pages 1703-1712] shows that this ring is generated by homogeneous invariants whose degree does not exceed max{n, k(n − 1)} (where i runs over an index set of size k). Also this bound is sharp if $n=p^s$ for some prime $p$ and either $R=\mathbb Z$ or $R$ has characteristic $p$. </p> <p>Some work has been done on the Dickson invariants version as well. I think that is considered in the article [Steinberg, Robert, On Dickson's theorem on invariants. J. Fac. Sci. Univ. Tokyo Sect. IA Math. 34 (1987), no. 3, 699–707.]</p> http://mathoverflow.net/questions/18860/a-ring-of-invariants-in-characteristic-2/71161#71161 Answer by David Wehlau for A ring of invariants in characteristic 2 David Wehlau 2011-07-24T23:13:40Z 2011-07-24T23:13:40Z <p>The paper [H.E.A. Campbell, J. Harris and D.L. Wehlau, Internal duality for resolutions of rings, J. of Algebra, 215 (1999) 1--33.] considers this question (or a closely related one) when n=3. Also note that since the group acts via permutations, the answer is (essentially) the same for all fields of characteristic 2 so it suffices to work over ${\mathbb F}_2$.</p> http://mathoverflow.net/questions/24537/weyl-group-invariants/71098#71098 Answer by David Wehlau for Weyl group Invariants David Wehlau 2011-07-24T03:01:10Z 2011-07-24T03:01:10Z <p>I don't think the answer is known. The paper [Hunziker, Classical invariant theory for finite reflection groups. Transform. Groups 2 (1997), no. 2, 147–163] is relevant. The author conjectures an answer and shows his answer is correct for $F_4$.</p> http://mathoverflow.net/questions/32450/chevalleyshephardtodd-theorem/71097#71097 Answer by David Wehlau for Chevalley–Shephard–Todd theorem David Wehlau 2011-07-24T02:37:28Z 2011-07-24T02:37:28Z <p>Chevalley was interested in the action of (real) Weyl groups and so a reflection to him had determinant -1 and so was a real reflection, i.e. order 2. My understanding is that Serre had seen the paper by Shepard and Todd and so he knew that pseudo-reflections were relevant. He pointed out that Chevalley's proof was valid for pseudo-reflections.</p> http://mathoverflow.net/questions/59248/the-ring-of-sl-2-invariants-in-sums-of-conjugation-and-tautological-modules/71095#71095 Answer by David Wehlau for The ring of SL_2 invariants in sums of conjugation and tautological modules David Wehlau 2011-07-24T02:09:33Z 2011-07-24T02:09:33Z <p>I don't have the literature with me, but yes the answer to your question was well known to classical invariant theorists. My recollection is that it is one of the examples (sections) in Grace and Young (J.H. Grace and A. Young, The algebra of invariants, Cambridge Univ. Press, Cambridge, 1903). They give an answer without using polarization. All the generating invariants are the ones you describe or else invariants of degree 3 which are quadratic in (one or 2) of the tautological reps and linear in a conjugation rep. Describing the relations is harder but was understood classically. I am not sure of a reference though. </p> http://mathoverflow.net/questions/121715/invariant-polynomials-for-a-product-of-algebraic-groups Comment by David Wehlau David Wehlau 2013-03-02T20:21:30Z 2013-03-02T20:21:30Z Your more general question, i.e. not just for $SO_n$ see the paper Littelmann, Peter Koregul&#228;re und &#228;quidimensionale Darstellungen. (German) [Coregular and equidimensional representations] J. Algebra 123 (1989), no. 1, 193–222. http://mathoverflow.net/questions/111068/morphism-between-affine-spaces-of-polynomials-by-setting-equal-variables Comment by David Wehlau David Wehlau 2012-12-25T04:42:48Z 2012-12-25T04:42:48Z I think you should read about &quot;restitution&quot;. http://mathoverflow.net/questions/104223/justifying-explaining-math-research-in-a-public-address Comment by David Wehlau David Wehlau 2012-08-07T21:49:20Z 2012-08-07T21:49:20Z @David White: I am not too concerned about trying to draw a line between old and modern. I am just concerned that the older the application the more impressive it has to be in order to use as a modern justification. As you say, many people believe math ended in the 1600s. It won't serve my purpose very well if I convince them it was still alive in 1900. http://mathoverflow.net/questions/104223/justifying-explaining-math-research-in-a-public-address Comment by David Wehlau David Wehlau 2012-08-07T21:44:11Z 2012-08-07T21:44:11Z Although I tried to ask a precise question, I am happy to get any suggestions which would improve the talk. My main goal is to entertain ~200 members of the public while I talk for an hour about math research. A secondary goal is to try to justify to the public why society should invest in math research. A tertiary goal (which may be omitted) is to educate some of my colleagues in other departments about why CV might look different from theirs (# of papers, # of authors, # of students, etc.) Not all of this can be covered well in an hour but that is okay, I don't need to do everything. http://mathoverflow.net/questions/85664/structure-of-s-mathfrakg-otimes-s-mathfrakgg-for-semisimple-mathfr/85750#85750 Comment by David Wehlau David Wehlau 2012-01-28T16:27:56Z 2012-01-28T16:27:56Z For $G=SL(n,C)$ the papers by Teranishi are I believe the best published results. I think this is where you can find $N(n) = n(n+1)/2$ for $n=3,4$. Teranishi, Yasuo The ring of invariants of matrices. Nagoya Math. J. 104 (1986), 149–161. The following paper shows that the ring of invariants is polynomial only for G=SL(2,C) and r=2 (for r&gt;1). Teranishi, Yasuo A theorem on invariants of semi-simple Lie algebras. Perspectives in ring theory (Antwerp, 1987), 37–40, NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., 233, Kluwer Acad. Publ., Dordrecht, 1988. http://mathoverflow.net/questions/20106/generalized-symmetric-algebras-and-dickson-algebras-over-mathbb-f-p/71460#71460 Comment by David Wehlau David Wehlau 2011-12-03T15:38:19Z 2011-12-03T15:38:19Z @Dev: You are welcome. Looking back at my answer I realize I misread question 2. Let i run over an index set of size k. Some things may be known for k=2 but I don't think anything is known for larger values of k except there is a complete answer for general k for $GL(2,F)$. See the paper [Vector invariants for the two dimensional Modular representation of a cyclic group of prime order, Campbell, Shank, Wehlau, Adv. in Math., 225(2) 1069-1094]. http://mathoverflow.net/questions/82181/how-to-interpret-conflicting-formal-proofs-about-a-mod-0/82182#82182 Comment by David Wehlau David Wehlau 2011-11-30T02:39:47Z 2011-11-30T02:39:47Z You don't really need to divide by 0. a and b are congruent mod n means b-a is a multiple of n. For n=0 this just means a and b are equal as Andreas has pointed out. This <i>is</i> is the generally accepted meaning of mod 0 and it is the meaning used in any ring. It makes no sense to interpret a mod 0 as 0 (unless a=0).