I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:22:09Z http://mathoverflow.net/feeds/question/66683 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66683/i-dont-get-a-part-of-bernsteins-deligne-morgans-proof-of-poincare-birkhoff-w I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt darij grinberg 2011-06-01T19:45:27Z 2011-06-10T20:54:49Z <p><strong>Question:</strong> I am talking about the proof given on pages 50-52 of Pierre Deligne, Pavel Etingof, Daniel S. Freed, Lisa C. Jeffrey, David Kazhdan, John W. Morgan, David R. Morrison, and Edward Witten (editors), <em>Quantum Fields and Strings: A Course for Mathematicians</em>, Volume 1, AMS 1999 (on <a href="http://books.google.com/books?id=TQIsyvw1KnsC&amp;pg=PA50&amp;lpg=PA50%20" rel="nofollow">google books</a> and in the usual internet sources).</p> <p>The problem is easily described: In the middle of page 52, the authors say "and (1.3.7.7) gives that [...]". But I don't see how (1.3.7.7) gives the equation that follows.</p> <p><strong>Sidenotes:</strong> The proof was rather readable and well-written up to that point, so I assume the blindness is on my side. If anyone wishes to read the proof (or reprint the book ;) ), here are a few minor mistakes to watch out for:</p> <ul> <li><p>On page 51, $\left[xy\right]$ should be $\left[x,y\right]$ in "while the second term $\frac12\left[xy\right]$ is antisymmetric".</p></li> <li><p>On page 51, in the definition of the map $\left\lbrace x_1,...,x_{n+1}\right\rbrace$, all three terms on the right hand side should end with $x_{n+1}$ rather than $x_n$.</p></li> <li><p>On page 52, in the first formula of this page, the commutators $\left[x\left[y,z\right]\right]$ and $\left[z\left[x,y\right]\right]$ should be $\left[x,\left[y,z\right]\right]$ and $\left[z,\left[x,y\right]\right]$ instead.</p></li> <li><p>On page 52, in the middle of this page, "and the $\left\lbrace x_1,...,x_n\right\rbrace$ vanish" should probably be "and the $\left\lbrace x_1,...,x_{n+1}\right\rbrace$ vanish".</p></li> <li><p>On page 52, in the middle of this page, "1.3.7.4" should be "(1.3.7.4)".</p></li> </ul> http://mathoverflow.net/questions/66683/i-dont-get-a-part-of-bernsteins-deligne-morgans-proof-of-poincare-birkhoff-w/66909#66909 Answer by darij grinberg for I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt darij grinberg 2011-06-04T18:08:40Z 2011-06-10T20:54:49Z <p>Here is an explanation Pavel Etingof has given to me in email. Thanks Pavel!</p> <p>Every $\sigma\in S_{n}$ satisfies</p> <p>$\sum\limits_{i}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\left[ x,y_{\sigma i}\right] \ast\cdots\ast y_{\sigma n}$</p> <p>$=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}$</p> <p>$+\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\left[ x,y_{\sigma i}\right] \ast\cdots\ast y_{\sigma n}$.</p> <p>But since</p> <p><code>$\sum\limits_{i&gt;1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left[ x,y_{\sigma i}\right] }_{\displaystyle\substack{\displaystyle =x\ast y_{\sigma i}-y_{\sigma i}\ast x\\\displaystyle \text{(since the inclusion of }\mathfrak{L}\\\displaystyle \text{into }\operatorname*{Sym} \nolimits^{\ast}\mathfrak{L}\text{ is a morphism}\\\displaystyle \text{of Lie algebras)} }}\ast\cdots\ast y_{\sigma n}$</code></p> <p><code>$=\sum\limits_{i&gt;1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left( x\ast y_{\sigma i}-y_{\sigma i}\ast x\right) \ast\cdots\ast y_{\sigma n} }_{\substack{\displaystyle =x\ast y_{\sigma i}\ast\cdots\ast y_{\sigma n}-y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}\\\displaystyle \text{(by the induction hypothesis, since }i&gt;1\text{)}}}$</code></p> <p>$=\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma i}\ast\cdots\ast y_{\sigma n}$</p> <p>$-\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p><code>$= \sum\limits_{i&gt;0}\left( n-\left( i+1\right) +1\right) \underbrace{\displaystyle y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma \left(i+1\right)}\ast\cdots\ast y_{\sigma n}}_{\displaystyle =y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}}$</code></p> <p>$-\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$ (here we substituted $i+1$ for $i$ in the first sum)</p> <p>$=\sum\limits_{i>0}\left( n-\left( i+1\right) +1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p>$-\sum\limits_{i>1}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p><code>$=\left( n-1\right) y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits _{i&gt;1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$</code>,</p> <p>this becomes</p> <p>$\sum\limits_{i}\left( n-i+1\right) y_{\sigma1}\ast\cdots\ast\left[ x,y_{\sigma i}\right] \ast\cdots\ast y_{\sigma n}$</p> <p>$=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\left( n-1\right) y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p>$-\sum\limits_{i>1}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p>$=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p>$-y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\sum\limits_{i>1}y_{\sigma1}\ast \cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p>$=n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p>$-\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$.</p> <p>Thus, (1.3.7.7) rewrites as</p> <p>$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n}=\left( \text{symmetrized product of }x,y_{1},...,y_{n}\right)$</p> <p>$+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p>$-\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}\sum\limits_{i>0}y_{\sigma1}\ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}$.</p> <p>Since</p> <p>$\left( \text{symmetrized product of }x,y_{1},...,y_{n}\right)$</p> <p>$=\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}\left( \sum\limits_{i>0}y_{\sigma1} \ast\cdots\ast y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}+x\ast y_{\sigma 1}\ast\cdots\ast y_{\sigma n}\right)$,</p> <p>this simplifies to</p> <p>$\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n} =\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$</p> <p>$+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$.</p> <p>Thus</p> <p>$\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p>$=\dfrac{1}{n!}x\ast\sum\limits_{\sigma}y_{\sigma1}\ast\cdots\ast y_{\sigma n} -\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$</p> <p><code>$=\underbrace{\displaystyle \left( \dfrac{1}{n!}-\dfrac{1}{\left( n+1\right) !}\right) }_{\displaystyle =\dfrac{n}{\left( n+1\right) !}}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast \cdots\ast y_{\sigma n}=\dfrac{n}{\left( n+1\right) !}\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$</code>.</p> <p>Divide this by $\dfrac{n}{\left( n+1\right) !}$ to obtain</p> <p>$\sum\limits_{\sigma}\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\sum\limits_{\sigma}y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$</p> <p>$=\sum\limits_{\sigma}x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n}$.</p> <p>In other words,</p> <p>$0=\sum\limits_{\sigma}\left( x\ast y_{\sigma1}\ast\cdots\ast y_{\sigma n} -y_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}-\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}\right)$</p> <p>$=\sum\limits_{\sigma}\left\lbrace x,y_{\sigma1},...,y_{\sigma n}\right\rbrace =\left( n-1\right) !\sum\limits_{i}\left\lbrace x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n} \right\rbrace$</p> <p>(here we used that $\left\lbrace x_{1},...,x_{n+1}\right\rbrace$ is symmetric in the last $n-1$ variables, so that every $\sigma\in S_{n}$ satisfies $\left\lbrace x,y_{\sigma1},...,y_{\sigma n}\right\rbrace =\left\lbrace x,y_{i},y_{1} ,...,\widehat{y_{i}},...,y_{n}\right\rbrace$ for $i=\sigma1$).</p> <p>Thus, $\sum\limits_{i}\left\lbrace x,y_{i},y_{1},...,\widehat{y_{i}},...,y_{n}\right\rbrace =0$, qed.</p>