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found out how to use \underbrace without causing unreadably small fonts
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darij grinberg
darij grinberg
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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{\left[ x,y_{\sigma i}\right] }_{\substack{=x\ast y_{\sigma i}-y_{\sigma i}\ast x\\\text{(since the inclusion of }\mathfrak{L}\\\text{into }\operatorname*{Sym} \nolimits^{\ast}\mathfrak{L}\text{ is a morphism}\\\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,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{\left( x\ast y_{\sigma i}-y_{\sigma i}\ast x\right) \ast\cdots\ast y_{\sigma n} }_{\substack{=x\ast y_{\sigma i}\ast\cdots\ast y_{\sigma n}-y_{\sigma i}\ast x\ast\cdots\ast y_{\sigma n}\\\text{(by the induction hypothesis, since }i>1\text{)}}}$$=\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{y_{\sigma1}\ast\cdots\ast x\ast y_{\sigma \left(i+1\right)}\ast\cdots\ast y_{\sigma n}}_{=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{\left( \dfrac{1}{n!}-\dfrac{1}{\left( n+1\right) !}\right) }_{=\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}$$=\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.

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

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

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

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

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

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

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

$\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$

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

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

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.

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.

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darij grinberg
darij grinberg
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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

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

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

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

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

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

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

$\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}n\left[ x,y_{\sigma1}\right] \ast y_{\sigma2}\ast\cdots\ast y_{\sigma n}+\dfrac{1}{\left( n+1\right) !}\sum\limits_{\sigma}ny_{\sigma1}\ast x\ast\cdots\ast y_{\sigma n}$

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

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

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.