**Question:** 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), *Quantum Fields and Strings: A Course for Mathematicians*, Volume 1, AMS 1999 (on [google books](http://books.google.com/books?id=TQIsyvw1KnsC&pg=PA50&lpg=PA50%20) and in the usual internet sources).


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.


**Sidenotes:** 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:


* On page 51, $\left[xy\right]$ should be $\left[x,y\right]$ in "while the second term $\frac12\left[xy\right]$ is antisymmetric".
* 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 $x\_n$.
* 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.
* 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".
* On page 52, in the middle of this page, "1.3.7.4" should be "(1.3.7.4)".



I don't get a part of Bernstein's / Deligne-Morgan's proof of Poincaré-Birkhoff-Witt
Here is an explanation Pavel Etingof has given to me in email. Thanks Pavel!


Every $\sigma\in S\_{n}$ satisfies


$\sum\limits\_{i}\left( n-i+1\right) y\_{\sigma1}\ast\cdots\ast\left[ x,y\_{\sigma i}\right] \ast\cdots\ast y\_{\sigma n}$


$=n\left[ x,y\_{\sigma1}\right] \ast y\_{\sigma2}\ast\cdots\ast y\_{\sigma n}$


$+\sum\limits\_{i>1}\left( n-i+1\right) y\_{\sigma1}\ast\cdots\ast\left[ x,y\_{\sigma i}\right] \ast\cdots\ast y\_{\sigma n}$.


But since


$\sum\limits\_{i>1}\left( n-i+1\right) y\_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left[
x,y\_{\sigma i}\right] }\_{\displaystyle\substack{\displaystyle =x\ast y\_{\sigma i}-y\_{\sigma i}\ast
x\\\displaystyle \text{(since the inclusion of }\mathfrak{L}\\\displaystyle \text{into }\operatorname\*{Sym}
\nolimits^{\ast}\mathfrak{L}\text{ is a morphism}\\\displaystyle \text{of Lie algebras)}
}}\ast\cdots\ast y\_{\sigma n}$


$=\sum\limits\_{i>1}\left( n-i+1\right) y\_{\sigma1}\ast\cdots\ast\underbrace{\displaystyle \left(
x\ast y\_{\sigma i}-y\_{\sigma i}\ast x\right) \ast\cdots\ast y\_{\sigma n}
}\_{\substack{\displaystyle =x\ast y\_{\sigma i}\ast\cdots\ast y\_{\sigma n}-y\_{\sigma i}\ast
x\ast\cdots\ast y\_{\sigma n}\\\displaystyle \text{(by the induction hypothesis, since }i>1\text{)}}}$


$=\sum\limits\_{i>1}\left( n-i+1\right) y\_{\sigma1}\ast\cdots\ast x\ast y\_{\sigma
i}\ast\cdots\ast y\_{\sigma n}$


$-\sum\limits\_{i>1}\left( n-i+1\right) y\_{\sigma1}\ast\cdots\ast y\_{\sigma i}\ast
x\ast\cdots\ast y\_{\sigma n}$


$= \sum\limits\_{i>0}\left( n-\left( i+1\right) +1\right) \underbrace{\displaystyle y\_{\sigma1}\ast\cdots\ast x\ast y\_{\sigma
\left(i+1\right)}\ast\cdots\ast y\_{\sigma n}}\_{\displaystyle =y\_{\sigma1}\ast\cdots\ast
y\_{\sigma i}\ast x\ast\cdots\ast y\_{\sigma n}}$


$-\sum\limits\_{i>1}\left( n-i+1\right) y\_{\sigma1}\ast\cdots\ast y\_{\sigma i}\ast
x\ast\cdots\ast y\_{\sigma n}$ (here we substituted $i+1$ for $i$ in the first sum)


$=\sum\limits\_{i>0}\left( n-\left( i+1\right) +1\right) y\_{\sigma1}\ast\cdots\ast
y\_{\sigma i}\ast x\ast\cdots\ast y\_{\sigma n}$


$-\sum\limits\_{i>1}\left( n-i+1\right) y\_{\sigma1}\ast\cdots\ast y\_{\sigma i}\ast
x\ast\cdots\ast y\_{\sigma n}$


$=\left( n-1\right) y\_{\sigma1}\ast x\ast\cdots\ast y\_{\sigma n}-\sum\limits
\_{i>1}y\_{\sigma1}\ast\cdots\ast y\_{\sigma i}\ast x\ast\cdots\ast y\_{\sigma n}$,


this becomes


$\sum\limits\_{i}\left( n-i+1\right) y\_{\sigma1}\ast\cdots\ast\left[ x,y\_{\sigma
i}\right] \ast\cdots\ast y\_{\sigma n}$


$=n\left[ x,y\_{\sigma1}\right] \ast y\_{\sigma2}\ast\cdots\ast y\_{\sigma
n}+\left( n-1\right) y\_{\sigma1}\ast x\ast\cdots\ast y\_{\sigma n}$


$-\sum\limits\_{i>1}y\_{\sigma1}\ast\cdots\ast y\_{\sigma i}\ast x\ast\cdots\ast
y\_{\sigma n}$


$=n\left[ x,y\_{\sigma1}\right] \ast y\_{\sigma2}\ast\cdots\ast y\_{\sigma
n}+ny\_{\sigma1}\ast x\ast\cdots\ast y\_{\sigma n}$


$-y\_{\sigma1}\ast x\ast\cdots\ast y\_{\sigma n}-\sum\limits\_{i>1}y\_{\sigma1}\ast
\cdots\ast y\_{\sigma i}\ast x\ast\cdots\ast y\_{\sigma n}$


$=n\left[ x,y\_{\sigma1}\right] \ast y\_{\sigma2}\ast\cdots\ast y\_{\sigma
n}+ny\_{\sigma1}\ast x\ast\cdots\ast y\_{\sigma n}$


$-\sum\limits\_{i>0}y\_{\sigma1}\ast\cdots\ast y\_{\sigma i}\ast x\ast\cdots\ast y\_{\sigma
n}$.


Thus, (1.3.7.7) rewrites as


$\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) $


$+\dfrac{1}{\left( n+1\right) !}\sum\limits\_{\sigma}n\left[ x,y\_{\sigma1}\right]
\ast y\_{\sigma2}\ast\cdots\ast y\_{\sigma n}+\dfrac{1}{\left( n+1\right)
!}\sum\limits\_{\sigma}ny\_{\sigma1}\ast x\ast\cdots\ast y\_{\sigma n}$


$-\dfrac{1}{\left( n+1\right) !}\sum\limits\_{\sigma}\sum\limits\_{i>0}y\_{\sigma1}\ast\cdots\ast y\_{\sigma
i}\ast x\ast\cdots\ast y\_{\sigma n}$.


Since


$\left( \text{symmetrized product of }x,y\_{1},...,y\_{n}\right) $


$=\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) $,


this simplifies to


$\dfrac{1}{n!}x\ast\sum\limits\_{\sigma}y\_{\sigma1}\ast\cdots\ast y\_{\sigma n}
=\dfrac{1}{\left( n+1\right) !}\sum\limits\_{\sigma}x\ast y\_{\sigma1}\ast\cdots\ast
y\_{\sigma n}$


$+\dfrac{1}{\left( n+1\right) !}\sum\limits\_{\sigma}n\left[ x,y\_{\sigma1}\right]
\ast y\_{\sigma2}\ast\cdots\ast y\_{\sigma n}+\dfrac{1}{\left( n+1\right)
!}\sum\limits\_{\sigma}ny\_{\sigma1}\ast x\ast\cdots\ast y\_{\sigma n}$.


Thus


$\dfrac{1}{\left( n+1\right) !}\sum\limits\_{\sigma}n\left[ x,y\_{\sigma1}\right]
\ast y\_{\sigma2}\ast\cdots\ast y\_{\sigma n}+\dfrac{1}{\left( n+1\right)
!}\sum\limits\_{\sigma}ny\_{\sigma1}\ast x\ast\cdots\ast y\_{\sigma n}$


$=\dfrac{1}{n!}x\ast\sum\limits\_{\sigma}y\_{\sigma1}\ast\cdots\ast y\_{\sigma n}
-\dfrac{1}{\left( n+1\right) !}\sum\limits\_{\sigma}x\ast y\_{\sigma1}\ast\cdots\ast
y\_{\sigma n}$


$=\underbrace{\displaystyle \left( \dfrac{1}{n!}-\dfrac{1}{\left( n+1\right) !}\right)
}\_{\displaystyle =\dfrac{n}{\left( n+1\right) !}}\sum\limits\_{\sigma}x\ast y\_{\sigma1}\ast
\cdots\ast y\_{\sigma n}=\dfrac{n}{\left( n+1\right) !}\sum\limits\_{\sigma}x\ast
y\_{\sigma1}\ast\cdots\ast y\_{\sigma n}$.


Divide this by $\dfrac{n}{\left( n+1\right) !}$ to obtain


$\sum\limits\_{\sigma}\left[ x,y\_{\sigma1}\right] \ast y\_{\sigma2}\ast\cdots\ast
y\_{\sigma n}+\sum\limits\_{\sigma}y\_{\sigma1}\ast x\ast\cdots\ast y\_{\sigma n}$


$=\sum\limits\_{\sigma}x\ast y\_{\sigma1}\ast\cdots\ast y\_{\sigma n}$.


In other words,


$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) $


$=\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 $


(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 $i=\sigma1$).


Thus, $\sum\limits\_{i}\left\lbrace x,y\_{i},y\_{1},...,\widehat{y\_{i}},...,y\_{n}\right\rbrace
=0$, qed.


