Capelli determinant = Duflo ( determinant) - was it known ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:07:57Z http://mathoverflow.net/feeds/question/92348 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/92348/capelli-determinant-duflo-determinant-was-it-known Capelli determinant = Duflo ( determinant) - was it known ? Alexander Chervov 2012-03-27T06:59:16Z 2012-04-04T21:05:16Z <p>Question briefly. Was this fact known: Capelli determinant = Duflo (determinant) ? (This is an equality of the two central elements in universal enveloping of Lie algebra \$gl_n\$).</p> <p>I googled a lot and ask some colleagues - seems was not stated explicitly in literature (although proof is simple), am I wrong ? So I wrote this observation as "seems to be new" in <a href="http://arxiv.org/abs/1203.5759" rel="nofollow">arXiv:1203.5759</a> page 12 proposition 1.</p> <hr> <h2>More details.</h2> <p>Capelli determinant is certain element in the center of universal enveloping of \$U(gl_n)\$ (<a href="http://en.wikipedia.org/wiki/Capelli%27s_identity" rel="nofollow">Wikipedia on Capelli</a>). The Duflo map allows to construct elements in the center of universal enveloping \$U(g)\$ of any Lie algebra \$g\$ from the elements in Poisson center of symmetric algebra \$S(g)\$. </p> <p>Capelli elements in the centers are down-to-earth, simple-to-work, but somewhat ad hoc, elements, on the other hand Duflo map gives general and conceptual way to construct such elements, but it is difficult to write down them explicitly in some PBW-basis. The equality between the two elements gives a pleasant conclusion that we have elements which are both easy-to-work and conceptually clear.</p> <p>Consider \$gl_n\$ denote by \$E_{ij}\$ its standard basis - the matrices with zeros everywhere except 1 at position \$(ij)\$. Form the following Capelli matrix \$C_{ij}\$ = \$E_{ij}\$ i.e. \$(ij)\$-th element of matrix \$C\$ is element \$E_{ij} \$ of the Lie algebra \$gl_n\$.</p> <p>Rather clearly the element \$det(C)\$ considered as element of commutative algebra \$S(gl_n)\$ gives \$gl_n\$-invariant element, so it means that it belongs to the Poisson center of \$S(gl_n)\$. And so by Duflo's theorem \$Duflo(det(C))\$ belongs to the center of the \$U(gl_n)\$.</p> <p>On the other hand we can consider Capelli determinant. \$det^{column} (C + diag(n-1,n-2,...,1,0) - (t)Id) \$ where calculations are performed in non-commutative algebra \$U(gl_n)\$ and determinant is defined by the convention that elements from the first column come first, second column come second and so on. It is known that this element is also in the center of the \$U(gl_n)\$ for any \$t\$.</p> <p><strong>The claim is that</strong>: \$Duflo(det(C)) = det^{column} (C + diag(n-1,n-2,...,1,0) - (n-1)/2 Id) \$.</p> <p>This seems can be extended to characteristic polynoms and other simple Lie algebras...</p> <hr> <hr> <p>The Duflo map is discussed e.g. here: </p> <p>D. Calaque, C. Rossi "Lectures on Duflo isomorphisms in Lie algebras and complex geometry"</p> <p><a href="http://people.mpim-bonn.mpg.de/crossi/LectETHbook.pdf" rel="nofollow">http://people.mpim-bonn.mpg.de/crossi/LectETHbook.pdf</a></p> <p>See <a href="http://mathoverflow.net/questions/80025/is-the-duflo-map-for-lie-algs-unique" rel="nofollow">http://mathoverflow.net/questions/80025/is-the-duflo-map-for-lie-algs-unique</a> for some info on Duflo map.</p> <p>Remark.</p> <p>In his seminal paper on quantization M. Kontsevich generalized the Duflo map to the quantization of an arbitrary Poisson manifolds - he was able to construct isomorphism of commutative algebras Center(Quantization(M)) = PoissonCenter(M). He showed that the classical Duflo map is the particulat case for \$M= g\$ , here \$g\$ is Lie algebra with its natural Poisson bracket.</p>