I'm trying to work on the graded Leibniz's law for Shouten bracket, but I've got the wrong sign whenever how hard I tried. Here's the problem$\newcommand{\vt}{\vartheta} \newcommand{\zt}{\zeta} \newcommand{\sig}{\sigma} \newcommand{\R}{\mathbb{R}} \newcommand{\X}{\mathfrak{X}} \newcommand{\we}{\wedge}$:
A Schouten Bracket of $\vt$ and $\zt$ is the multivector field $[\vt,\zt]\in \X^{k+l-1}(M)$(The multivector field on $M$, that is, the exterior product of vector field) with $$[\vt,\zt]=\vt\circ \zt-(-1)^{(k-1)(l-1)}\zt\circ\vt,$$ where we set $$\zt\circ\vt(df_1,\ldots,d f_{k+l-1})=\sum_{\sigma\in S_{k,l-1}}(-1)^\sigma\bar{\zt}(\bar{\vt}(f_{\sig(1)},\ldots,f_{\sig(k)}),f_{\sig(k+1)},\ldots,f_{\sig(k+l-1)}),$$where the summation is taken over $(k,l-1)$-shuffles. Here shuffle means $\sig(1)<\sig(2)<\ldots<\sig(k),\sig(k+1)<\ldots<\sig(k+l-1)$. Now we need to prove $$[\vt,\zt\we\tau]=[\vt,\zt]\we\tau+(-1)^{(k-1)l}\zt\we[\vt,\tau]$$
This seems to be a routine check, but I'm stuck in the sign problem:
\begin{align*}&\vt\circ(\zt\we\tau)(df_1,\ldots,d f_{k+l+m-1})\\&=\sum_{\sig\in S_{l+m,k-1}}(-1)^{\sig}\bar{\vt}((\bar{\zt}\we\bar{\tau})(f_{\sig(1)},\ldots,f_{\sig(l+m)}),f_{\sig(l+m+1)},\ldots,f_{\sig(l+m+k-1)})\\ &=\sum_{\sig\in S_{l,m,k-1}}(-1)^{\sig}\bar{\vt}(\bar{\zt}(f_{\sig(1)},\ldots,f_{\sig(l)})\bar{\tau}(f_{\sig(l+1)},\ldots,f_{\sig(l+m)}),f_{\sig(l+m+1)},\ldots,f_{\sig(l+m+k-1)})\\ &=\sum_{\sig\in S_{l,m,k-1}}(-1)^{\sig}\bar{\vt}(\bar{\zt}(f_{\sig(1)},\ldots,f_{\sig(l)}),f_{\sig(l+m+1)},\ldots,f_{\sig(l+m+k-1)})\bar{\tau}(f_{\sig(l+1)},\ldots,f_{\sig(l+m)})\\&+\sum_{\sig\in S_{l,m,k-1}}(-1)^{\sig}\bar{\vt}(\bar{\tau}(f_{\sig(l+1)},\ldots,f_{\sig(l+m)}),f_{\sig(l+m+1)},\ldots,f_{\sig(l+m+k-1)})\bar{\zt}(f_{\sig(1)},\ldots,f_{\sig(l)})\\ &=(-1)^{k(m-1)}\sum_{\sig\in S_{l,m,k-1}}(-1)^{\sig}\bar{\vt}(\bar{\zt}(f_{\sig(1)},\ldots,f_{\sig(l)}),f_{\sig(l+1)},\ldots,f_{\sig(l+k-1)})\bar{\tau}(f_{\sig(l+k)},\ldots,f_{\sig(l+k+m-1)})\\&+\sum_{\sig\in S_{l,m,k-1}}(-1)^{\sig}\bar{\zt}(f_{\sig(1)},\ldots,f_{\sig(l)})\bar{\vt}(\bar{\tau}(f_{\sig(l+1)},\ldots,f_{\sig(l+m)}),f_{\sig(l+m+1)},\ldots,f_{\sig(l+m+k-1)}) \end{align*} Here $S_{l,m,n}$ is the corresponding $(l,m,n)$-shuffle in $l+m+n$(analougous definition for shuffle).
However, this seems to be the leading term for $(-1)^{k(m-1)}[\vt,\zt]\we\tau+\zt\we[\vt,\tau]$ instead of the required result. So my question is, what's wrong with my calculation? Thanks for your help!
@Golbez, In the definition of the wedge product by means of shuffles you have to be careful: $$\overline{\zeta}\wedge\overline{\tau}(f_{\sigma(1)},...,f_{\sigma(\ell+m)})=\sum_{\sigma'\in S_{\ell,m}}\varepsilon(\sigma')\overline{\zeta}(f_{\sigma'\circ\sigma(1)},...,f_{\sigma'\circ\sigma(\ell)})\overline{\tau}(f_{\sigma'\circ\sigma(\ell+1)},...,f_{\sigma'\circ\sigma(\ell+m)})$$