Can anyone give an explicit basis of the universal (noncommutative) differential calculus over an algebra $A$ with basis ${e_i}$. (The universal calculus over $A$ is the kernel of the multiplication map $m:A \otimes A \to A$.)

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    $\begingroup$ What do you know about A? For a general algebra, there is surely no good answer. $\endgroup$ – David E Speyer Nov 17 '09 at 18:22
  • $\begingroup$ Let's take $SU_q(2)$ or $SUq(3)$. $\endgroup$ – Abtan Massini Nov 17 '09 at 18:35

You can describe $\Omega_A=\ker(m:A\otimes A\to A)$ as the quotient of the free $A$-bimodule generated by symbols $d(a)$, one for each element $a\in A$, by the sub-bimodule generated by the elements of the form $$d(ab)-d(a)\\,b-a\\,d(b), \qquad a,b\in A,$$ together with the elements of the form $$d(\lambda 1), \qquad \lambda\in k$$ with $k$ being the base field. The elements $\{a d(b):a,b\in A\}$, when seen in $\Omega_A$, span $\Omega_A$ over $K$ but are not linearly independent over $k$.

To extract from this a $k$-basis of $\Omega_A$ you need to know more than a basis of $A$. For example, if you know a presentation of $A$ given by generators and relations, you can obtain a basis using essentially Groebner bases.

  • $\begingroup$ Does there exist an obvious basis over $A$? $\endgroup$ – Abtan Massini Nov 17 '09 at 18:28
  • $\begingroup$ If $A$ is projective over $k$ (for example, if $k$ is a field), then the short exact sequence $0\to \Omega_A\to A\otimes_kA\to A\to 0$ splits in the category of left $A$-modules by the map $s:a\in A\mapsto a\otimes 1\in A\ot A$. It follows that $\Omega_A$ is a direct summand of the free left $A$-module $A\ot A$, so it is projective. It is not free, though, in general, I think. $\endgroup$ – Mariano Suárez-Álvarez Nov 17 '09 at 18:37
  • $\begingroup$ What is that "unknown control sequence" \ot supposed to be? Maybe \otimes or maybe \leftarrow? $\endgroup$ – Jon Awbrey Nov 17 '09 at 20:28
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    $\begingroup$ \ot should be \otimes. (Being able to edit comments would be very, very useful!) $\endgroup$ – Mariano Suárez-Álvarez Nov 17 '09 at 21:16

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