Timeline for On a dual of Kaplansky's $2^{nd}$ conjecture: admissible algebras?
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Aug 29, 2018 at 16:34 | comment | added | tj_ | Hence $(A, \Delta) $ is no Hopf algebra in this case if the characteristic of $k$ is not 2. | |
Aug 29, 2018 at 16:31 | comment | added | tj_ | Matter cleared: In the realm of algebraic topology the product in $A \otimes A$ is defined to be graded, i.e. $(a_1 \otimes b_1)(a_2\otimes b_2)=(-1)^{|b_1||a_2|}a_1a_2\otimes b_1b_2$. For example, if $A=H^\ast(S^1,k)=k[x]/(x^2)$ then multiplication on $S^1$ induces the coproduct $\Delta: A \to A \otimes A, x \mapsto x\otimes 1 + 1 \otimes x$. The graded product then makes $\Delta(x)\cdot \Delta(x)=0$. However, if (as in your proposition) the product in $A \otimes A$ is taken to be ungraded, then $\Delta(x)\cdot \Delta(x)=2\cdot x \otimes x$ which is non-zero if $k$ is not of char. 2. ... | |
Aug 28, 2018 at 21:58 | comment | added | tj_ | About your proposition in Konstantinos Kanakoglou's answer (Prop. 2.7 in arxiv.org/pdf/1601.06687v1.pdf): To my knowledge, the cohomology ring of a connected topological group is a (connected graded) Hopf algebra. The cohomology ring of the 3-torus $T=S^1\times S^1 \times S^1$ is the exterior algebra $H^\ast(T,\mathbb{C})=\bigwedge(x,y,z)$ with degree one generators. In particular, $xy=-yx$, contradicting the proposition. Am I missing something ? | |
Mar 21, 2017 at 2:05 | comment | added | Konstantinos Kanakoglou | In your second paragraph you mention that "algebras whose abelianazation is not a domain are non-admissible". Do such algebras -like $k[x,y]/\langle [x,y] = y^2 \rangle$ for example- fall into the class of semisimple, non-separable algebras mentioned in my answer? | |
Mar 21, 2017 at 1:57 | comment | added | Konstantinos Kanakoglou | thank you for the feedback and the ideas contributed. I didn't yet find time to further study your paper but it seems interesting and relevant. thank you very much for citing it! | |
Mar 13, 2017 at 20:56 | history | answered | Paul Gilmartin | CC BY-SA 3.0 |