For a co-quasi-triangular Hopf algebra $H$, with universal $r$-form $r$, there exists an important map $Q$ defined by $$ Q:H \otimes H \to k, ~~~~~~h \otimes g \mapsto r(g_{(1)}\otimes h_{(1)})r(h_{(2)}\otimes g_{(2)}). $$ The map is usually called the quantum Killing form.

In some papers I have read, it seems that the authors have tacitly assumed that the kernel of $Q$ is a right ideal. Is this true? If so, why?

  • $\begingroup$ I'm hearing about co-quasi-triangular Hopf algebras for the first time, but if they're just the dual notion of quasi-triangular Hopf algebras, then are you sure about your $Q$? The $R$ in a quasi-triangular Hopf algebra is an element of $H\otimes H$, so I assume the $r$ will be a map $H\otimes H\to k$, and thus your $Q$ should go to $k$ as well? $\endgroup$ – darij grinberg Aug 20 '10 at 20:44
  • $\begingroup$ Yes of course. Sorry for the typos. $\endgroup$ – John McCarthy Aug 20 '10 at 21:06
  • $\begingroup$ I am a bit surprised here. Wouldn't that mean that $Q$ is a $k$-algebra homomorphism (because whenever $Q:A\to k$ is a $k$-linear map from a $k$-algebra $A$ to the base field $k$ satisfying $Q\left(1\right)=1$, then $\mathrm{Ker}Q$ is an ideal if and only if $Q$ is a $k$-algebra homomorphism)? Which would then more or less yield that any arbitrary bicharacter of an abelian group is a product of characters? (More or less because we can conclude this only for squares of bicharacters, but I don't think this is much of a difference.) $\endgroup$ – darij grinberg Aug 24 '10 at 12:01
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    $\begingroup$ Actually, a bicharacter of an abelian group is almost never a product of characters, and neither is the square (example: group = $\mathbb Z\diagup 3\mathbb Z$, and the bicharacter sends $\left(i,j\right)$ to $\zeta^{ij}$ with $\zeta$ being a primitive $3$-rd root of unity). $\endgroup$ – darij grinberg Aug 24 '10 at 12:08
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    $\begingroup$ John, can you please edit your question accordingly to my remarks? (Not 100% mine, Mark Pedron helped me construct the counterexample.) $\endgroup$ – darij grinberg Aug 26 '10 at 15:53

I doubt this can be true. I claim that:

Lemma. Let $k$ be a commutative ring, $A$ be a $k$-algebra, and $Q:A\to k$ be a $k$-linear map such that $Q\left(1\right)=1$. Then, the following four assertions (1), (2), (3), (4) are pairwise equivalent:

(1) The kernel $\mathrm{Ker} Q$ is a two-sided ideal of $A$.

(2) The kernel $\mathrm{Ker} Q$ is a right ideal of $A$.

(3) The kernel $\mathrm{Ker} Q$ is a left ideal of $A$.

(4) The map $Q$ is a $k$-algebra homomorphism.

Proof of Lemma. Clearly, (4) $\Longrightarrow$ (1) $\Longrightarrow$ (2). Now let us prove that (2) $\Longrightarrow$ (4): Assume that (2) holds. That is, we assume that $\mathrm{Ker} Q$ is a right ideal of $A$. Clearly, every $a\in A$ satisfies $Q\left(a\right)\cdot 1-a\in\mathrm{Ker} Q$ (since the $k$-linearity of $Q$ yields $Q\left(Q\left(a\right)\cdot 1-a\right)=Q\left(a\right)\cdot \underbrace{Q\left(1\right)}_{=1}-Q\left(a\right)=0$). Thus, every $a\in A$ and $b\in A$ satisfy $\underbrace{\left(Q\left(a\right)\cdot 1-a\right)} _ {\in\mathrm{Ker} Q} b \in\mathrm{Ker} Q$ (since $\mathrm{Ker} Q$ is a right ideal), so that

$0=Q\left(\left(Q\left(a\right)\cdot 1-a\right)b\right)=Q\left(Q\left(a\right)b-ab\right)=Q\left(a\right)Q\left(b\right)-Q\left(ab\right)$

(by the $k$-linearity of $Q$), so that $Q\left(a\right)Q\left(b\right)=Q\left(ab\right)$. Together with the $k$-linearity of $Q$ and $Q\left(1\right)=1$, this yields that $Q$ is a $k$-algebra homomorphism, so that assertion (4) holds. Thus we have shown that (2) $\Longrightarrow$ (4), which completes the (4) $\Longrightarrow$ (1) $\Longrightarrow$ (2) $\Longrightarrow$ (4) circle. Thus, (4) $\Longleftrightarrow$ (1) $\Longleftrightarrow$ (2). Similarly (4) $\Longleftrightarrow$ (1) $\Longleftrightarrow$ (3). This proves that all four assertions (1), (2), (3), (4) are pairwise equivalent, and the lemma is proven.

The Lemma shows that as long as you want the kernel of $r$ to be an ideal (one- or two-sided), $r$ will be forced to be a $k$-algebra homomorphism. Considering the main example of co-quasi-triangular Hopf algebras, namely the group algebra with a bicharacter, the counterexample we gave in the comments above will hold.

Maybe the "right ideal" that your references claimed refered to a different algebra structure? One of my main sources of confusion in the advanced Hopf algebra theory has always been the presence of many conflicting multiplications, comultiplications, actions etc. on one and the same set.

  • $\begingroup$ Ok that looks pretty convincing. I've gone back rechecked the paper and it looks like what is claimed is actually more complicated, it involves fixing one of the entries and restricting the action to the augmentation ideal $\ker(\epsilon)$. It wouldn't really be fair to rework the question so much at this stage so I'm going to leave it as it is. $\endgroup$ – John McCarthy Aug 27 '10 at 20:18

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