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In the book Algebras of Functions on Quantum Groups: Part I, 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.

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}

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  • $\begingroup$ What would be a typical bracketed non-Poisson algebra? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 22, 2016 at 12:30
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    $\begingroup$ @მამუკაჯიბლაძე, I added an example. $\endgroup$
    – Jianrong Li
    Commented Apr 22, 2016 at 12:47
  • $\begingroup$ WOW :D And this is not Poisson? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Apr 22, 2016 at 13:07
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    $\begingroup$ @მამუკაჯიბლაძე, this is not Poisson. For example, $Jacobi(c_{00}, c_{01}, c_{12}) = 4\, c_{0,0}\, c_{0,1}\, c_{1,2} - 8\, c_{0,0}\, c_{0,2}\, c_{1,1} + 4\, c_{0,1}\, c_{1,0}\, c_{0,2}$. Here $Jacobi(a,b,c)=\{a, \{b, c\}\} + \{b, \{c, a\}\} + \{c, \{a,b\}\}$. $\endgroup$
    – Jianrong Li
    Commented Apr 22, 2016 at 14:00

2 Answers 2

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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.

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  • $\begingroup$ thank you very much. What is the antipode of $A[\hbar]/\hbar^2$? $\endgroup$
    – Jianrong Li
    Commented Apr 23, 2016 at 10:56
  • $\begingroup$ Is the antipode of $A[\hbar]/\hbar^2$ obtained by extending the antipode of $A$ linearly to $A[\hbar]/\hbar^2$ ($S(a + h b) = S(a) + h S(b)$, $a, b \in A$)? $\endgroup$
    – Jianrong Li
    Commented Apr 23, 2016 at 11:12
  • $\begingroup$ what are the unit and counit of $A[\hbar]/\hbar^2$? $\endgroup$
    – Jianrong Li
    Commented Apr 23, 2016 at 11:15
  • $\begingroup$ The unit is $1$. The counit if $\epsilon$. These do not deform. (Note that I am thinking of deformed algebra $A[\hbar]/\hbar^2$ as a Hopf algebra over the dual numbers $\mathbb C[\hbar]/\hbar^2$, so $\epsilon(a+b\hbar) = \epsilon(a) + \epsilon(b)\hbar$. The point about "it's an open condition" is that you can solve for $S$. $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 24, 2016 at 1:35
  • $\begingroup$ Let's see. The rule for the antipode is usually $x_{(1)} \cdot S(x_{(2)}) = x$, where in Sweedler notation $\Delta(x) = x_{(1)} \otimes x_{(2)}$, and I don't write the $\sum$ sign (but it is a nontrivial sum). So now let $S_\hbar$ denote the sought-after deformed antipode; I want to solve $x_{(1)} \cdot_\hbar S_\hbar(x_{(2)}) = x$, and it suffices to assume $x\in A \subseteq A[\hbar]/\hbar^2$ by $\mathbb C[\hbar]/\hbar^2$-linearity. By sending $\hbar\to 0$, you see that $S_\hbar(x) = S(x) + S'(x)\hbar$ for $S$ the undeformed antipode. Then the nontrivial part of the equation is ... $\endgroup$
    – Theo Johnson-Freyd
    Commented Apr 24, 2016 at 1:51
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This is also proved in Lemma 2.6 of the paper: CO-POISSON COALGEBRA AND CO-POISSON HOPF ALGEBRA STRUCTURES ON $k[x_1, x_2, \ldots, x_d]$ .

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