In the article, a Nichols algebra is defined as follows. Let ${\displaystyle V\in {}_{H}^{H}{\mathcal {YD}}}$. There exists a largest ideal ${\displaystyle {\mathfrak {I}}\subset TV} $ with the following properties: \begin{align} & {\displaystyle {\mathfrak {I}}\subset \bigoplus _{n=2}^{\infty }T^{n}V,} \\ & {\displaystyle \Delta ({\mathfrak {I}})\subset {\mathfrak {I}}\otimes TV+TV\otimes {\mathfrak {I}}} \quad (\text{this is automatic}) \end{align} The Nichols algebra is \begin{align} {\displaystyle {\mathfrak {B}}(V):=TV/{\mathfrak {I}}}. \end{align} Why in the definition of a Nichols algebra we require that $V$ is a Yetter-Drinfeld module? If we take $V$ to be any vector space and define $\mathfrak{B}(V)$ using the same formulas as above, is $\mathfrak{B}(V)$ also some interesting algebra (or coalgebra)? Thank you very much.
$\begingroup$
$\endgroup$
2
asked Mar 2, 2017 at 15:38
-
$\begingroup$ By the way, the standard definition is to let $I$ be the maximal coideal in $TV$ contained in that subspace, and then show that it is an ideal. The braiding of $V$ appears in the comultiplication of $TV$, so the condition does involve it. $\endgroup$– Mariano Suárez-ÁlvarezCommented Mar 2, 2017 at 17:54
-
2$\begingroup$ You can define a Nichols algebra for every braided vectior space. See en.wikipedia.org/wiki/Nichols_algebra?wprov=sfla1 $\endgroup$– Leandro VendraminCommented Mar 3, 2017 at 8:07
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
If you start with a vector space $V$ (which amounts to letting $H$ be the trivial algebra $k$), then the Nichols algebra is just the symmetric algebra on $V$.
answered Mar 2, 2017 at 17:51