Is there a reasonable notion of an inner automorphism of a Hopf algebra $H$ which in the case of a group ring $H=\mathbb KG$ for a group $G$ reduces to a conjugation by $g\in G$?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
I am not sure if the following is the kind of answer you are expecting, but take the (left) adjoint action $(ad_l h)\triangleright k=\sum h_1 kS(h_2)$ of a hopf algebra $H$ on itself.
(It is known that the left adjoint action of $H$ on itself is always inner (in the sense of the def. 6.1.1, p. 87, in Montgomery's book "Hopf algebras and their action on rings") and turns $H$ into a $H$-module algebra).
If you set $H=kG$, then, the left adjoint action for $g\in G$ reduces to action by conjugation: $(ad_l g)\triangleright k=gkg^{-1}$.
More generally, for any inner action $\ \triangleright : H\otimes A\rightarrow A$, it is easy to show that any grouplike element $g\in G(H)$, acts as an inner automorphism of $A$. In case $H=A=kG$ and $\triangleright$ is the left adjoint action then this inner automorphism is the action by conjugation (of the corresponding group element).
Concluding, i would say that the notion of an "inner action" (p.674-675) may be the reasonable notion you are looking for.
answered Nov 9, 2017 at 20:12
-
$\begingroup$ Many thanks! That is indeed on the spot. I was however hoping to generalize inner automorphisms beyond conjugation by grouplike elements and I got confused. By Montgomery's book, for $a,x\in H,$ $a\cdot x=\sum a_1xS(a_2)$ is a measuring and, hence, $a\cdot (xy)=\sum (a_1\cdot x)(a_2\cdot y)$. If $a$ is primitive then $a\cdot x=[a,x]=ax-xa$ (Eg. 6.1.6) implying $[a, xy]=\sum [a_1,x][a_2,y].$ But this is not true, since this sum is $[a,x] [1,y]+[1,x][a,y]=0.$ Where do I make a mistake? $\endgroup$– AdamCommented Nov 13, 2017 at 0:07
-
$\begingroup$ Your claim is correct in that, if $a$ is primitive then $a\cdot x=[a,x]$. However, if $a$ is primitive then $a_1$ and $a_2$ cannot both be primitives. One of them is equal to $1$. Unlike the primitive case: $1\cdot x=x \neq [1,x]=0$. Consequently, your mistake lies at the following: $[a,xy]=a\cdot(xy)=\sum (a_1\cdot x)(a_2\cdot y)\neq \sum [a_1,x][a_2,y]$. $\endgroup$ Commented Nov 13, 2017 at 3:52
-
$\begingroup$ In fact: $[a,xy]=a\cdot(xy)=\sum (a_1\cdot x)(a_2\cdot y)=x[a,y]+[a,x]y$. $\endgroup$ Commented Nov 13, 2017 at 4:00
-
1