most recent 30 from http://mathoverflow.net 2013-05-24T17:01:23Z http://mathoverflow.net/feeds/question/102855 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/102855/about-the-term-tangential-derivation-on-a-free-lie-algebra Zhaoting Wei 2012-07-22T04:43:24Z 2012-07-22T18:05:15Z

<p>Let $\mathcal{lie}_n$ be the free Lie algebra generated by $n$ elements $x_1,\ldots, x_n$. A derivation $u\in \text{Der}(\mathcal{lie}_n)$ is called tangential if there exist $a_i\in \mathcal{lie}_n, i=1\ldots n$ such that $u(x_i)=[x_i,a_i]$. The definition can be found in section 3 of Alekseev and Torossian's paper "The Kashiwara-Vergne conjecture and Drinfeld's associators" arXiv:0802.4300v1.</p> <p>My question is: why it has the name "tangential derivation", is there any historical reason or geometric intuition of it?</p>

http://mathoverflow.net/questions/102855/about-the-term-tangential-derivation-on-a-free-lie-algebra/102885#102885 Adrien 2012-07-22T18:05:15Z 2012-07-22T18:05:15Z

<p>You may be interested in these <a href="http://katlas.math.toronto.edu/drorbn/dbnvp/wClips-120530.php" rel="nofollow">Bar Natan's lecture</a>.</p> <p>The geometric intuition is that the $n$th Lie algebra of tangential derivations can be realized into the algebra of tangential differential operators (not derivations !) on $\mathfrak g^n$ for any (say finite dimensional) Lie algebra $\mathfrak g$. If $G$ is a Lie group with Lie algebra $\mathfrak g$, then it acts on $\mathfrak g$ by the adjoint action, and the action of a tangential differential operators is a differentiation in a direction which is tangential to the orbits of the adjoint action, hence the name.</p>