Timeline for Inner automorphisms of Hopf algebras
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 14, 2017 at 2:28 | history | edited | Konstantinos Kanakoglou | CC BY-SA 3.0 |
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| Nov 13, 2017 at 5:04 | vote | accept | Adam | ||
| Nov 13, 2017 at 4:36 | comment | added | Adam | Oops, of course! :-) | |
| Nov 13, 2017 at 4:08 | history | edited | Konstantinos Kanakoglou | CC BY-SA 3.0 |
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| Nov 13, 2017 at 4:00 | comment | added | Konstantinos Kanakoglou | In fact: $[a,xy]=a\cdot(xy)=\sum (a_1\cdot x)(a_2\cdot y)=x[a,y]+[a,x]y$. | |
| Nov 13, 2017 at 3:52 | comment | added | Konstantinos Kanakoglou | 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: $[a,xy]=a\cdot(xy)=\sum (a_1\cdot x)(a_2\cdot y)\neq \sum [a_1,x][a_2,y]$. | |
| Nov 13, 2017 at 0:07 | comment | added | Adam | 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? | |
| Nov 11, 2017 at 20:53 | history | edited | Konstantinos Kanakoglou | CC BY-SA 3.0 |
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| Nov 11, 2017 at 17:00 | history | edited | Konstantinos Kanakoglou | CC BY-SA 3.0 |
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| Nov 10, 2017 at 3:00 | history | edited | Konstantinos Kanakoglou | CC BY-SA 3.0 |
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| Nov 10, 2017 at 2:42 | history | edited | Konstantinos Kanakoglou | CC BY-SA 3.0 |
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| Nov 10, 2017 at 0:15 | history | edited | Konstantinos Kanakoglou | CC BY-SA 3.0 |
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| Nov 9, 2017 at 20:34 | history | edited | Konstantinos Kanakoglou | CC BY-SA 3.0 |
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| Nov 9, 2017 at 20:12 | history | answered | Konstantinos Kanakoglou | CC BY-SA 3.0 |