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1. • CommentRowNumber1.
   • CommentAuthordomenico_fiorenza
   • CommentTimeSep 24th 2012
   • (edited Sep 24th 2012)
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItexLet $A$ be an unital algbra over a field $k$, and let $A^o$ be its opposite algebra. Since $A$ is naturally an $A$-$A$-bimodule, $A$ is naturally an $A\otimes_k A^o$ left module, and similarly $A^o$ is an $A\otimes_k A^o$ right module. So one can consider the $(A^o\otimes_k A)$-$(A^o\otimes_k A)$-bimoule $A\otimes_k A^o$. On the other hand, since both $A$ and $A^o$ are algebras, so is $A\otimes_k A^o$, which therefore has a natural $(A^o\otimes_k A)$-$(A^o\otimes_k A)$-bimoule structure. My guess is that these two $(A^o\otimes_k A)$-$(A^o\otimes_k A)$-bimoule structures on $A\otimes_k A^o$ do not coincide (in the first one a $A\otimes{k}A^o$-scalar cannot "travel from the left to the right", or at least so it seems to me), but I'd like to see this clearly, and also to have a better notation to distinguish the two bimodule structures (if they are indeed different). Another guess that the two bimodule structures should be different is the following: let $B=A^o\otimes_{A^o\otimes_k A}A$. Then, if the two bimodule structures coincide one has $$ B\otimes_k B = (A^o\otimes_{A^o\otimes_k A}A)\otimes_k (A^o\otimes_{A^o\otimes_k A}A) = A^o\otimes_{A^o\otimes_k A}(A\otimes_k A^o)\otimes_{A^o\otimes_k A}A=A^o\otimes_{A^o\otimes_k A}A=B $$ which would imply $\dim_k B=0$ or $\dim_k B=1$, and this seems to me not to be the case in general.

   Let $A$ be an unital algbra over a field $k$, and let $A^o$ be its opposite algebra. Since $A$ is naturally an $A$-$A$-bimodule, $A$ is naturally an $A\otimes_k A^o$ left module, and similarly $A^o$ is an $A\otimes_k A^o$ right module. So one can consider the $(A^o\otimes_k A)$-$(A^o\otimes_k A)$-bimoule $A\otimes_k A^o$.

   On the other hand, since both $A$ and $A^o$ are algebras, so is $A\otimes_k A^o$, which therefore has a natural $(A^o\otimes_k A)$-$(A^o\otimes_k A)$-bimoule structure.

   My guess is that these two $(A^o\otimes_k A)$-$(A^o\otimes_k A)$-bimoule structures on $A\otimes_k A^o$ do not coincide (in the first one a $A\otimes{k}A^o$-scalar cannot “travel from the left to the right”, or at least so it seems to me), but I’d like to see this clearly, and also to have a better notation to distinguish the two bimodule structures (if they are indeed different).

   Another guess that the two bimodule structures should be different is the following: let $B=A^o\otimes_{A^o\otimes_k A}A$. Then, if the two bimodule structures coincide one has

   $$B\otimes_k B = (A^o\otimes_{A^o\otimes_k A}A)\otimes_k (A^o\otimes_{A^o\otimes_k A}A) = A^o\otimes_{A^o\otimes_k A}(A\otimes_k A^o)\otimes_{A^o\otimes_k A}A=A^o\otimes_{A^o\otimes_k A}A=B$$

   which would imply $\dim_k B=0$ or $\dim_k B=1$, and this seems to me not to be the case in general.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeSep 24th 2012
   • (edited Sep 24th 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItex> do not coincide (in the first one a [...] scalar cannot “travel from the left to the right”, Yes, only that maybe "scalar" is misleading. Not sure, but of course the elements in $k \hookrightarrow A \hookrightarrow A \otimes_K A^o$ do "travel from left to right". And back. > I’d like to see this clearly, Here is a suggestion: draw multiplication in [[string diagram]] notation. Then you see that the two module structure differ by a permutation/braiding of strings. But maybe I am not properly addressing your question. Let me know.

   do not coincide (in the first one a […] scalar cannot “travel from the left to the right”,

   Yes, only that maybe “scalar” is misleading. Not sure, but of course the elements in $k \hookrightarrow A \hookrightarrow A \otimes_K A^o$ do “travel from left to right”. And back.

   I’d like to see this clearly,

   Here is a suggestion: draw multiplication in string diagram notation. Then you see that the two module structure differ by a permutation/braiding of strings.

   But maybe I am not properly addressing your question. Let me know.

3. • CommentRowNumber3.
   • CommentAuthordomenico_fiorenza
   • CommentTimeSep 24th 2012
   • PermaLink
   Author: domenico_fiorenza
   Format: MarkdownItex> Here is a suggestion: draw multiplication in string diagram notation. Yes, that is where I started from! :) ok, now I'm reassured I wasn't wrong. Thanks a lot!

   Here is a suggestion: draw multiplication in string diagram notation.

   Yes, that is where I started from! :)

   ok, now I’m reassured I wasn’t wrong. Thanks a lot!