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Atrium > Mathematics, Physics & Philosophy: Bimodule structures on the tensor product of an algebra with its opposite

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- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi></mrow><annotation encoding="application/x-tex">A</annotation></semantics></math>$ be an unital algbra over a field $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>k</mi></mrow><annotation encoding="application/x-tex">k</annotation></semantics></math>$ , and let $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">A^o</annotation></semantics></math>$ be its opposite algebra. Since $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi></mrow><annotation encoding="application/x-tex">A</annotation></semantics></math>$ is naturally an $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi></mrow><annotation encoding="application/x-tex">A</annotation></semantics></math>$ - $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi></mrow><annotation encoding="application/x-tex">A</annotation></semantics></math>$ -bimodule, $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi></mrow><annotation encoding="application/x-tex">A</annotation></semantics></math>$ is naturally an $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi><msub><mo>⊗</mo> <mi>k</mi></msub><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">A\otimes_k A^o</annotation></semantics></math>$ left module, and similarly $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">A^o</annotation></semantics></math>$ is an $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi><msub><mo>⊗</mo> <mi>k</mi></msub><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">A\otimes_k A^o</annotation></semantics></math>$ right module. So one can consider the $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">(</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">(A^o\otimes_k A)</annotation></semantics></math>$ - $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">(</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">(A^o\otimes_k A)</annotation></semantics></math>$ -bimoule $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi><msub><mo>⊗</mo> <mi>k</mi></msub><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">A\otimes_k A^o</annotation></semantics></math>$ .

On the other hand, since both $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi></mrow><annotation encoding="application/x-tex">A</annotation></semantics></math>$ and $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">A^o</annotation></semantics></math>$ are algebras, so is $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi><msub><mo>⊗</mo> <mi>k</mi></msub><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">A\otimes_k A^o</annotation></semantics></math>$ , which therefore has a natural $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">(</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">(A^o\otimes_k A)</annotation></semantics></math>$ - $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">(</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">(A^o\otimes_k A)</annotation></semantics></math>$ -bimoule structure.

My guess is that these two $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">(</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">(A^o\otimes_k A)</annotation></semantics></math>$ - $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mo stretchy="false">(</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi><mo stretchy="false">)</mo></mrow><annotation encoding="application/x-tex">(A^o\otimes_k A)</annotation></semantics></math>$ -bimoule structures on $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi><msub><mo>⊗</mo> <mi>k</mi></msub><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">A\otimes_k A^o</annotation></semantics></math>$ do not coincide (in the first one a $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>A</mi><mo>⊗</mo><mi>k</mi><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">A\otimes{k}A^o</annotation></semantics></math>$ -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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>B</mi><mo>=</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mrow><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi></mrow></msub><mi>A</mi></mrow><annotation encoding="application/x-tex">B=A^o\otimes_{A^o\otimes_k A}A</annotation></semantics></math>$ . Then, if the two bimodule structures coincide one has
$<math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>B</mi><msub><mo>⊗</mo> <mi>k</mi></msub><mi>B</mi><mo>=</mo><mo stretchy="false">(</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mrow><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi></mrow></msub><mi>A</mi><mo stretchy="false">)</mo><msub><mo>⊗</mo> <mi>k</mi></msub><mo stretchy="false">(</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mrow><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi></mrow></msub><mi>A</mi><mo stretchy="false">)</mo><mo>=</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mrow><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi></mrow></msub><mo stretchy="false">(</mo><mi>A</mi><msub><mo>⊗</mo> <mi>k</mi></msub><msup><mi>A</mi> <mi>o</mi></msup><mo stretchy="false">)</mo><msub><mo>⊗</mo> <mrow><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi></mrow></msub><mi>A</mi><mo>=</mo><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mrow><msup><mi>A</mi> <mi>o</mi></msup><msub><mo>⊗</mo> <mi>k</mi></msub><mi>A</mi></mrow></msub><mi>A</mi><mo>=</mo><mi>B</mi></mrow><annotation encoding="application/x-tex"> B\otimes_k B = (A^o\otimes_{A^o\otimes_k A}A)\otimes_k (A^o\otimes_{A^o\otimes_k A}A) = A^o\otimes_{A^o\otimes_k A}(A\otimes_k A^o)\otimes_{A^o\otimes_k A}A=A^o\otimes_{A^o\otimes_k A}A=B </annotation></semantics></math>$
which would imply $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>dim</mi> <mi>k</mi></msub><mi>B</mi><mo>=</mo><mn>0</mn></mrow><annotation encoding="application/x-tex">\dim_k B=0</annotation></semantics></math>$ or $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><msub><mi>dim</mi> <mi>k</mi></msub><mi>B</mi><mo>=</mo><mn>1</mn></mrow><annotation encoding="application/x-tex">\dim_k B=1</annotation></semantics></math>$ , and this seems to me not to be the case in general.
- 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 $<math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><semantics><mrow><mi>k</mi><mo>↪</mo><mi>A</mi><mo>↪</mo><mi>A</mi><msub><mo>⊗</mo> <mi>K</mi></msub><msup><mi>A</mi> <mi>o</mi></msup></mrow><annotation encoding="application/x-tex">k \hookrightarrow A \hookrightarrow A \otimes_K A^o</annotation></semantics></math>$ 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.
- 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!

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Atrium > Mathematics, Physics & Philosophy: Bimodule structures on the tensor product of an algebra with its opposite