>> BTW, I was a bit puzzled when I saw the formula for "plethysm". >> >> sum(g[k]*F(z^k), k) >> = sum(sum(g[d]*f[n/d], d)*z^n, n)) >> = sum(f[k]*G(z^k), k) >> >> That somehow looks as if >> >> f pleth g = g pleth f > > That's right, in the univariate case at least (I have to think twice > about the multivariate case).
As I understand f and g are power sums. There is no restriction on http://www.sciface.com/support/doc/40/en/combinat/countingFunctions.html so let me take f = x + x^2 + x^3 + ... = 1/(1-x) -1 and g=1 g(f) = 1 and f(g) = \infty You will probably tell me that (f pleth g) is something else. Would be nice if the documentation of plethysm also says something about the bounds of the sum. Otherwise the only reasonable bounds are 0..\infty. >> But then f(g) is usually not g(f) so how can this be connected with >> substitution? > > Note that in Kerber's definition, plethysm is not just substitution; > there is some relabeling. You probably know better. But where do you see a relabelling? Just because the z_i go away in favour of the y_j? No. Ralf ------------------------------------------------------------------------- Take Surveys. Earn Cash. Influence the Future of IT Join SourceForge.net's Techsay panel and you'll get the chance to share your opinions on IT & business topics through brief surveys-and earn cash http://www.techsay.com/default.php?page=join.php&p=sourceforge&CID=DEVDEV _______________________________________________ Aldor-combinat-devel mailing list Aldor-combinat-devel@lists.sourceforge.net https://lists.sourceforge.net/lists/listinfo/aldor-combinat-devel
