In the book [Algebras of Functions on Quantum Groups: Part I](http://bookstore.ams.org/surv-56), Remark 3.1.4, we have the following result. 


Let $A$ be a Poisson Hopf algebra. That is, $A$ is both a Hopf algebra and a Poisson algebra and $\Delta: A \to A \otimes A$ is a Poisson algebra homomorphism. Then the antipode $S: A \to A$ is a Poisson algebra anti-automorphism ($S\{a,b\}=-\{S(a), S(b)\}$, $a, b \in A$) and the counit $\epsilon: A \to \mathbb{C}$ is a Poisson algebra homomorphism. 


If we do not require the Jacobian condition in a Poisson algebra. Then we obtained a bracked algebra as defined in [R-Matrix Poisson Algebras and Their Deformations](http://arxiv.org/abs/0706.0351). 


An algebra $A$ is bracked if their is a bilinear map $\{,\}: A \otimes A \to A$ such that $\{,\}$ is skew-symmetric and satisfies the Leibniz rule (but not necessarily the Jacobian identity). 


Now suppose that $A$ is a bracked Hopf algebra. That is, $A$ is both a Hopf algebra and a bracked algebra and $\Delta: A \to A \otimes A$ is a bracked algebra homomorphism. Do we also have the following result: the antipode is anti-bracked ($S\{a,b\}=-\{S(a), S(b)\}$, $a, b \in A$)? Thank you very much. 


Edit: the following is an example of bracketed algebra. We take the following bracket on $\mathbb{C}[GL\_3]$ (the indices starts from $0$). 
\begin{align}
& \{ c\_{0 0}, c\_{0 1} \} = 2 c\_{0 0} c\_{0 1}, \\ 
& \{ c\_{0 0}, c\_{0 2} \} = 4 {c\_{0 1}}^2 \\ 
& \{ c\_{0 0}, c\_{1 0} \} = 2 c\_{0 0} c\_{1 0}, \\ 
& \{ c\_{0 0}, c\_{1 1} \} = 4 c\_{0 1} c\_{1 0}, \\ 
& \{ c\_{0 0}, c\_{1 2} \} = 4 c\_{0 1} c\_{1 1} + 2 c\_{1 0} c\_{0 2}, \\ 
& \{ c\_{0 0}, c\_{2 0} \} = 4 {c\_{1 0}}^2 \\ 
& \{ c\_{0 0}, c\_{2 1} \} = 2 c\_{0 1} c\_{2 0} + 4 c\_{1 0} c\_{1 1}, \\ 
& \{ c\_{0 0}, c\_{2 2} \} = 4 c\_{0 1} c\_{2 1} + 4 c\_{1 0} c\_{1 2}, \\ 
& \{ c\_{0 1}, c\_{0 0} \} = - 2 c\_{0 0} c\_{0 1}, \\ 
& \{ c\_{0 1}, c\_{0 2} \} = 2 c\_{0 1} c\_{0 2}, \\ 
& \{ c\_{0 1}, c\_{1 0} \} = 0 \\ 
& \{ c\_{0 1}, c\_{1 1} \} = 2 c\_{0 1} c\_{1 1} - c\_{0 0} c\_{1 2} + c\_{1 0} c\_{0 2}, \\ 
& \{ c\_{0 1}, c\_{1 2} \} = 4 c\_{0 2} c\_{1 1}, \\ 
& \{ c\_{0 1}, c\_{2 0} \} = 4 c\_{1 0} c\_{1 1} - 2 c\_{0 0} c\_{2 1}, \\ 
& \{ c\_{0 1}, c\_{2 1} \} = 4 {c\_{1 1}}^2 - c\_{0 0} c\_{2 2} + c\_{0 2} c\_{2 0}, \\ 
& \{ c\_{0 1}, c\_{2 2} \} = 2 c\_{0 2} c\_{2 1} + 4 c\_{1 1} c\_{1 2}, \\ 
& \{ c\_{0 2}, c\_{0 0} \} = - 4 {c\_{0 1}}^2 \\ 
& \{ c\_{0 2}, c\_{0 1} \} = - 2 c\_{0 1} c\_{0 2}, \\ 
& \{ c\_{0 2}, c\_{1 0} \} = 2 c\_{0 0} c\_{1 2} - 4 c\_{0 1} c\_{1 1}, \\ 
& \{ c\_{0 2}, c\_{1 1} \} = 0 \\ 
& \{ c\_{0 2}, c\_{1 2} \} = 2 c\_{0 2} c\_{1 2}, \\ 
& \{ c\_{0 2}, c\_{2 0} \} = 4 c\_{1 0} c\_{1 2} - 4 c\_{0 1} c\_{2 1}, \\ 
& \{ c\_{0 2}, c\_{2 1} \} = 4 c\_{1 1} c\_{1 2} - 2 c\_{0 1} c\_{2 2}, \\ 
& \{ c\_{0 2}, c\_{2 2} \} = 4 {c\_{1 2}}^2 \\ 
& \{ c\_{1 0}, c\_{0 0} \} = - 2 c\_{0 0} c\_{1 0}, \\ 
& \{ c\_{1 0}, c\_{0 1} \} = 0 \\ 
& \{ c\_{1 0}, c\_{0 2} \} = 4 c\_{0 1} c\_{1 1} - 2 c\_{0 0} c\_{1 2}, \\ 
& \{ c\_{1 0}, c\_{1 1} \} = c\_{0 1} c\_{2 0} - c\_{0 0} c\_{2 1} + 2 c\_{1 0} c\_{1 1}, \\ 
& \{ c\_{1 0}, c\_{1 2} \} = 4 {c\_{1 1}}^2 - c\_{0 0} c\_{2 2} + c\_{0 2} c\_{2 0}, \\ 
& \{ c\_{1 0}, c\_{2 0} \} = 2 c\_{1 0} c\_{2 0}, \\ 
& \{ c\_{1 0}, c\_{2 1} \} = 4 c\_{1 1} c\_{2 0}, \\ 
& \{ c\_{1 0}, c\_{2 2} \} = 4 c\_{1 1} c\_{2 1} + 2 c\_{2 0} c\_{1 2}, \\ 
& \{ c\_{1 1}, c\_{0 0} \} = - 4 c\_{0 1} c\_{1 0}, \\ 
& \{ c\_{1 1}, c\_{0 1} \} = c\_{0 0} c\_{1 2} - 2 c\_{0 1} c\_{1 1} - c\_{1 0} c\_{0 2}, \\ 
& \{ c\_{1 1}, c\_{0 2} \} = 0 \\ 
& \{ c\_{1 1}, c\_{1 0} \} = c\_{0 0} c\_{2 1} - c\_{0 1} c\_{2 0} - 2 c\_{1 0} c\_{1 1}, \\ 
& \{ c\_{1 1}, c\_{1 2} \} = c\_{0 2} c\_{2 1} - c\_{0 1} c\_{2 2} + 2 c\_{1 1} c\_{1 2}, \\ 
& \{ c\_{1 1}, c\_{2 0} \} = 0 \\ 
& \{ c\_{1 1}, c\_{2 1} \} = 2 c\_{1 1} c\_{2 1} - c\_{1 0} c\_{2 2} + c\_{2 0} c\_{1 2}, \\ 
& \{ c\_{1 1}, c\_{2 2} \} = 4 c\_{1 2} c\_{2 1}, \\ 
& \{ c\_{1 2}, c\_{0 0} \} = - 4 c\_{0 1} c\_{1 1} - 2 c\_{1 0} c\_{0 2}, \\ 
& \{ c\_{1 2}, c\_{0 1} \} = - 4 c\_{0 2} c\_{1 1}, \\ 
& \{ c\_{1 2}, c\_{0 2} \} = - 2 c\_{0 2} c\_{1 2}, \\ 
& \{ c\_{1 2}, c\_{1 0} \} = - 4 {c\_{1 1}}^2 + c\_{0 0} c\_{2 2} - c\_{0 2} c\_{2 0}, \\ 
& \{ c\_{1 2}, c\_{1 1} \} = c\_{0 1} c\_{2 2} - c\_{0 2} c\_{2 1} - 2 c\_{1 1} c\_{1 2}, \\ 
& \{ c\_{1 2}, c\_{2 0} \} = 2 c\_{1 0} c\_{2 2} - 4 c\_{1 1} c\_{2 1}, \\ 
& \{ c\_{1 2}, c\_{2 1} \} = 0 \\ 
& \{ c\_{1 2}, c\_{2 2} \} = 2 c\_{1 2} c\_{2 2}, \\ 
& \{ c\_{2 0}, c\_{0 0} \} = - 4 {c\_{1 0}}^2 \\ 
& \{ c\_{2 0}, c\_{0 1} \} = 2 c\_{0 0} c\_{2 1} - 4 c\_{1 0} c\_{1 1}, \\ 
& \{ c\_{2 0}, c\_{0 2} \} = 4 c\_{0 1} c\_{2 1} - 4 c\_{1 0} c\_{1 2}, \\ 
& \{ c\_{2 0}, c\_{1 0} \} = - 2 c\_{1 0} c\_{2 0}, \\ 
& \{ c\_{2 0}, c\_{1 1} \} = 0 \\ 
& \{ c\_{2 0}, c\_{1 2} \} = 4 c\_{1 1} c\_{2 1} - 2 c\_{1 0} c\_{2 2}, \\ 
& \{ c\_{2 0}, c\_{2 1} \} = 2 c\_{2 0} c\_{2 1}, \\ 
& \{ c\_{2 0}, c\_{2 2} \} = 4 {c\_{2 1}}^2 \\ 
& \{ c\_{2 1}, c\_{0 0} \} = - 2 c\_{0 1} c\_{2 0} - 4 c\_{1 0} c\_{1 1}, \\ 
& \{ c\_{2 1}, c\_{0 1} \} = - 4 {c\_{1 1}}^2 + c\_{0 0} c\_{2 2} - c\_{0 2} c\_{2 0}, \\ 
& \{ c\_{2 1}, c\_{0 2} \} = 2 c\_{0 1} c\_{2 2} - 4 c\_{1 1} c\_{1 2}, \\ 
& \{ c\_{2 1}, c\_{1 0} \} = - 4 c\_{1 1} c\_{2 0}, \\ 
& \{ c\_{2 1}, c\_{1 1} \} = c\_{1 0} c\_{2 2} - 2 c\_{1 1} c\_{2 1} - c\_{2 0} c\_{1 2}, \\ 
& \{ c\_{2 1}, c\_{1 2} \} = 0 \\ 
& \{ c\_{2 1}, c\_{2 0} \} = - 2 c\_{2 0} c\_{2 1}, \\ 
& \{ c\_{2 1}, c\_{2 2} \} = 2 c\_{2 1} c\_{2 2}, \\ 
& \{ c\_{2 2}, c\_{0 0} \} = - 4 c\_{0 1} c\_{2 1} - 4 c\_{1 0} c\_{1 2}, \\ 
& \{ c\_{2 2}, c\_{0 1} \} = - 2 c\_{0 2} c\_{2 1} - 4 c\_{1 1} c\_{1 2}, \\ 
& \{ c\_{2 2}, c\_{0 2} \} = - 4 {c\_{1 2}}^2 \\ 
& \{ c\_{2 2}, c\_{1 0} \} = - 4 c\_{1 1} c\_{2 1} - 2 c\_{2 0} c\_{1 2}, \\ 
& \{ c\_{2 2}, c\_{1 1} \} = - 4 c\_{1 2} c\_{2 1}, \\ 
& \{ c\_{2 2}, c\_{1 2} \} = - 2 c\_{1 2} c\_{2 2}, \\ 
& \{ c\_{2 2}, c\_{2 0} \} = - 4 {c\_{2 1}}^2 \\ 
& \{ c\_{2 2}, c\_{2 1} \} = - 2 c\_{2 1} c\_{2 2} 
\end{align}



Is the antipode anti-bracketed?
Yes. Suppose $A$ is a bracked Hopf algebra. Equip $A [\hbar]/\hbar^2$ with the multiplication $a \cdot\_\hbar b = ab + \hbar \{a,b\}$. That this is associative follows from the Leibniz rule --- you do not need Jacobi. Note that with undeformed $\Delta$, this is Hopf. (Indeed, it is a bialgebra by inspection. Now suppose $B,\cdot,\Delta$ is a bialgebra, and for $f,g : B \to B$ linear maps, define their *convolution* to be $f\star g = \cdot \circ (f \otimes g) \circ \Delta$. Then $\star$ is an associative multiplication on $\hom(B,B)$ with unit $1 \circ \epsilon$, and the antipode, if it exists, is the $\star$-inverse to $\mathrm{id}:B \to B$. But using that $\cdot\_\hbar$ is a deformation of $\cdot$, you can see that if $A$ is Hopf, then so is $A [\hbar]/\hbar^2$.) But the antipode is always an algebra anti-morphism, and so in particular looking at the order-$\hbar$ part is a bracket anti-morphism.


