TITLE: Dimension formula for Cartan-type abelian.group Nichols algebra?
QUESTION [6 upvotes]: Existence of a root system has been established for Nichols algebras $B(V)$ of a Yetter-Drinfel'd-module $V$ (resp. braided vectorspaces $V$) over abelian groups (resp. with diagonal braiding $x_i\otimes x_j\mapsto q_{ij}x_j\otimes x_i$) already in 2000 by Kharchenko. In 2008 Schneider and Heckenberger established root systems over nonabelian groups, under the condition that the root system / Nichols algebra be finite (-dimensional)!  
Having a root system means among others that one has a set of irreducible sub-Yetter-modules $X_1\ldots X_n\subset B(V)$, with multiplying as below a bijection (no algebra morphism!):
$$B(V)\cong\bigotimes_{i=1}^nB(X_i)$$
Dynkin-diagrams can be drawn with Cartan Matrix $$q_{ij}q_{ji}=q_{ii}^{-A_{ij}}$$ but there are sporadic cases not corresponding to finite semisimple Lie algebras (triangle, exotic edge...) for low prime exponents. In these cases the classification of pointed Hopf algebras (Schneider /
 Andruskiewitsch) fails; the Weyl group is replaced by a -groupoid between different Dynkin diagrams.   
But in Cartan-type it's the same as in the semisimple case - a root system $\Phi$!

Is is generally true that the "roots" in both formulas coicide?
$$|\{X_1,X_2,\ldots X_n\}|=n=|\Phi^+|\qquad or=\ldots?$$
So can I calculate the dimension of  $B(X)$ like that? 

$$dim(B(X))=\prod_{\alpha\in\Phi^+}dim(X_i)$$
...clearly $B(X_i)\cong k[x]/(x^{ord(q)})$ with dimension $ord(q_{ii})$) or infinite for $q=1$ (the bosonic $k[x]$). 

So I just multiply the orders of respective the $q_{ii}$ for all roots?

$$dim(B(X))=\prod_{\alpha\in\Phi^+}ord(q_{\alpha\alpha})$$

So especially for $G=\mathbb{Z}_2^k$ the dimension is the following?

$$dim(B(X))=2^{|\Phi^+|}$$

REPLY [2 votes]: Yes, careful application of the theory of rootsystems verifies indeed these formulae.....farily streight-forward!