Skip to main content
MathOverflow

Return to Answer

Mod Removes Wiki by Stefan Kohl♦
corrected non-rendering mathjax
Source Link
Ricardo Andrade
Ricardo Andrade
  • 6.2k
  • 5
  • 42
  • 69

This should be a comment, but I am not sure the OP reads comments, so...

Sorry, I still don't understand this. How is $\left(A\\#H\right)\\#H^{\ast}$$\left(A\#H\right)\#H^{\ast}$ an $A$-bimodule? The only way $A$ can act on $\left(A\\#H\right)\\#H^{\ast}$$\left(A\#H\right)\#H^{\ast}$ is by acting on the $A$ factor, from both sides. And it leads to a rather boring $A$-$A$-bimodule, canonically isomorphic to $\left(A\otimes H\right)\otimes H^{\ast}$ because multiplication doesn't matter. Of course this has $A$ as a direct summand, since $\left(A\otimes H\right)\otimes H^{\ast}\cong A\otimes \left(H\otimes H^{\ast}\right)$.

What I also fail to understand is how $\left(A\\#H\right)\\#H^{\ast}$$\left(A\#H\right)\#H^{\ast}$ is defined in the first place. In order to define $B\\#H$$B\#H$, I need $B$ to be a left $H$-module algebra. Now even if $A$ is a left $H$-module algebra, how is $A\\#H$$A\#H$ a left $H^{\ast}$-module algebra?

This should be a comment, but I am not sure the OP reads comments, so...

Sorry, I still don't understand this. How is $\left(A\\#H\right)\\#H^{\ast}$ an $A$-bimodule? The only way $A$ can act on $\left(A\\#H\right)\\#H^{\ast}$ is by acting on the $A$ factor, from both sides. And it leads to a rather boring $A$-$A$-bimodule, canonically isomorphic to $\left(A\otimes H\right)\otimes H^{\ast}$ because multiplication doesn't matter. Of course this has $A$ as a direct summand, since $\left(A\otimes H\right)\otimes H^{\ast}\cong A\otimes \left(H\otimes H^{\ast}\right)$.

What I also fail to understand is how $\left(A\\#H\right)\\#H^{\ast}$ is defined in the first place. In order to define $B\\#H$, I need $B$ to be a left $H$-module algebra. Now even if $A$ is a left $H$-module algebra, how is $A\\#H$ a left $H^{\ast}$-module algebra?

This should be a comment, but I am not sure the OP reads comments, so...

Sorry, I still don't understand this. How is $\left(A\#H\right)\#H^{\ast}$ an $A$-bimodule? The only way $A$ can act on $\left(A\#H\right)\#H^{\ast}$ is by acting on the $A$ factor, from both sides. And it leads to a rather boring $A$-$A$-bimodule, canonically isomorphic to $\left(A\otimes H\right)\otimes H^{\ast}$ because multiplication doesn't matter. Of course this has $A$ as a direct summand, since $\left(A\otimes H\right)\otimes H^{\ast}\cong A\otimes \left(H\otimes H^{\ast}\right)$.

What I also fail to understand is how $\left(A\#H\right)\#H^{\ast}$ is defined in the first place. In order to define $B\#H$, I need $B$ to be a left $H$-module algebra. Now even if $A$ is a left $H$-module algebra, how is $A\#H$ a left $H^{\ast}$-module algebra?

Post Made Community Wiki
Source Link
darij grinberg
darij grinberg
  • 33.8k
  • 4
  • 118
  • 253

This should be a comment, but I am not sure the OP reads comments, so...

Sorry, I still don't understand this. How is $\left(A\\#H\right)\\#H^{\ast}$ an $A$-bimodule? The only way $A$ can act on $\left(A\\#H\right)\\#H^{\ast}$ is by acting on the $A$ factor, from both sides. And it leads to a rather boring $A$-$A$-bimodule, canonically isomorphic to $\left(A\otimes H\right)\otimes H^{\ast}$ because multiplication doesn't matter. Of course this has $A$ as a direct summand, since $\left(A\otimes H\right)\otimes H^{\ast}\cong A\otimes \left(H\otimes H^{\ast}\right)$.

What I also fail to understand is how $\left(A\\#H\right)\\#H^{\ast}$ is defined in the first place. In order to define $B\\#H$, I need $B$ to be a left $H$-module algebra. Now even if $A$ is a left $H$-module algebra, how is $A\\#H$ a left $H^{\ast}$-module algebra?