Dual of a Basis for a Hopf Algebra Conatined in all Dually Paired Hopf Algebras - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:24:04Z http://mathoverflow.net/feeds/question/56679 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56679/dual-of-a-basis-for-a-hopf-algebra-conatined-in-all-dually-paired-hopf-algebras Dual of a Basis for a Hopf Algebra Conatined in all Dually Paired Hopf Algebras Janos Erdmann 2011-02-25T20:45:02Z 2011-02-26T00:49:01Z <p>For an infinite dimensional Hopf algebra $H$, a non-degenerate dually pairing Hopf algebra $H'$, and a choice of basis $e_i$ of $H$, is the dual basis $e^i$ (defined of course by $e^i(e_j) = \delta_{ij}$) contained in $H'$?</p> <p>I am interested in the specific case of $SU_q(N)$ and the dually paired Hopf algebra $\mathfrak{sl}_N$.</p> http://mathoverflow.net/questions/56679/dual-of-a-basis-for-a-hopf-algebra-conatined-in-all-dually-paired-hopf-algebras/56692#56692 Answer by Theo Johnson-Freyd for Dual of a Basis for a Hopf Algebra Conatined in all Dually Paired Hopf Algebras Theo Johnson-Freyd 2011-02-26T00:49:01Z 2011-02-26T00:49:01Z <p>No. Let $k$ be a field of characteristic $0$. Consider the symmetric algebra on one generator $k[x]$, with comultiplication $x \mapsto x\otimes 1 + 1\otimes x$. It has a Hopf pairing with itself, given by <code>$\langle x^m,x^n\rangle = n! \hspace{.5ex} \delta_{m=n}$</code>. Then consider the bases of $k[x]$ given by expansion around $1$, i.e. $e_n = (x-1)^n$. The dual basis, if it exists, includes $e^0$ such that <code>$\langle e^0, (x-1)^n\rangle = \delta_{0,n}$</code>. So suppose that <code>$e^0 = \sum a_n x^n$</code>; then: <code>\begin{aligned} 1 &amp; = a_0 \\ 0 &amp; = a_1 - a_0 \\ 0 &amp; = a_2 - 2a_1 + a_0 \\ \dots &amp; \phantom= \dots \\ 0 &amp; = \sum_{k=0}^n (-1)^{n-k} \hspace{.5ex} n!\hspace{.5ex} a_n \hspace{.5ex} \binom{n}{k} \\ \dots &amp; \phantom= \dots \\ \end{aligned}</code> The solution is that all $a_n = 1/n!$, i.e. $e^0 = \sum x^n/n! = \exp(x)$. But this is not a polynomial.</p> <p>In the case you ask about, you will similarly have some bases with dual bases, and some without.</p>