Basis for Universal Calculus - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:12:42Z http://mathoverflow.net/feeds/question/5845 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5845/basis-for-universal-calculus Basis for Universal Calculus Abtan Massini 2009-11-17T18:10:36Z 2009-11-17T18:28:24Z <p>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 $m:A \otimes A \to A$.)</p> http://mathoverflow.net/questions/5845/basis-for-universal-calculus/5849#5849 Answer by Mariano Suárez-Alvarez for Basis for Universal Calculus Mariano Suárez-Alvarez 2009-11-17T18:20:27Z 2009-11-17T18:28:24Z <p>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 $k$.</p> <p>To extract from this a $k$-<em>basis</em> 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.</p>