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\[\partial_x(a b)=(\partial_x a) b
+ a_{(2)} \partial_{a_{(1)}\triangleright x}b\qquad\forall x\in L, a\in A\]
Given a right invariant basis $\{\eta_i\}_{i\in I}$ of $\Gamma$ with a
dual basis $\{\phi_i\}_{i\in I}$ of $L$ we have
\[{\diff a}=\sum_{i\in I} \eta_i\cdot \partial_i(a)\qquad\forall a\in A\]
where we denote $\partial_i=\partial_{\phi_i}$. (This can be easily
seen to hold by evaluation against $\phi_i\ \forall i$.)
\section{Classification on \ensuremath{C_q(B_+)}{} and \ensuremath{U_q(\lalg{b_+})}{}}
\label{sec:q}
In this section we completely classify differential calculi on \ensuremath{C_q(B_+)}{}
and, dually, quantum tangent spaces on \ensuremath{U_q(\lalg{b_+})}{}. We start by
classifying the relevant crossed modules and then proceed to a
detailed description of the calculi.
\begin{lem}
\label{lem:cqbp_class}
(a) Left crossed \ensuremath{C_q(B_+)}-submodules $M\subseteq\ensuremath{C_q(B_+)}$ by left
multiplication and left
adjoint coaction are in one-to-one correspondence to
pairs $(P,I)$
where $P\in\k(q)[g]$ is a polynomial with $P(0)=1$ and $I\subset\mathbb{N}$ is
finite.
$\codim M<\infty$ iff $P=1$. In particular $\codim M=\sum_{n\in I}n$
if $P=1$.
(b) The finite codimensional maximal $M$
correspond to the pairs $(1,\{n\})$ with $n$ the
codimension. The infinite codimensional maximal $M$ are characterised by
$(P,\emptyset)$ with $P$ irreducible and $P(g)\neq 1-q^{-k}g$ for any
$k\in\mathbb{N}_0$.
(c) Crossed submodules $M$ of finite
codimension are intersections of maximal ones.
In particular $M=\bigcap_{n\in I} M^n$, with $M^n$ corresponding to
$(1,\{n\})$.
\end{lem}
\begin{proof}
(a) Let $M\subseteq\ensuremath{C_q(B_+)}$ be a crossed \ensuremath{C_q(B_+)}-submodule by left
multiplication and left adjoint coaction and let
$\sum_n X^n P_n(g) \in M$, where $P_n$ are polynomials in $g,g^{-1}$
(every element of \ensuremath{C_q(B_+)}{} can be expressed in
this form). From the formula for the coaction ((\ref{eq:adl}), see appendix)
we observe that for all $n$ and for all $t\le n$ the element
\[X^t P_n(g) \prod_{s=1}^{n-t} (1-q^{s-n}g)\]
lies in $M$.
In particular
this is true for $t=n$, meaning that elements of constant degree in $X$
lie separately in $M$. It is therefore enough to consider such
elements.
Let now $X^n P(g) \in M$.
By left multiplication $X^n P(g)$ generates any element of the form
$X^k P(g) Q(g)$, where $k\ge n$ and $Q$ is any polynomial in
$g,g^{-1}$. (Note that $Q(q^kg) X^k=X^k Q(g)$.)
We see that $M$ contains the following elements:
\[\begin{array}{ll}
\vdots & \\
X^{n+2} & P(g) \\
X^{n+1} & P(g) \\
X^n & P(g) \\
X^{n-1} & P(g) (1-q^{1-n}g) \\
X^{n-2} & P(g) (1-q^{1-n}g) (1-q^{2-n}g) \\
\vdots & \\
X & P(g) (1-q^{1-n}g) (1-q^{2-n}g) \ldots (1-q^{-1}g) \\
& P(g) (1-q^{1-n}g) (1-q^{2-n}g) \ldots (1-q^{-1}g)(1-g)
\end{array}
\]
Moreover, if $M$ is generated by $X^n P(g)$ as a module
then these elements generate a basis for $M$ as a vector
space by left
multiplication with polynomials in $g,g^{-1}$. (Observe that the
application of the coaction to any of the elements shown does not
generate elements of new type.)
Now, let $M$ be a given crossed submodule. We pick, among the
elements in $M$ of the form $X^n P(g)$ with $P$ of minimal
length,
one
with lowest degree in $X$. Then certainly the elements listed above are
in $M$. Furthermore for any element of the form $X^k Q(g)$, $Q$ must
contain $P$ as a factor and for $k<n$, $Q$ must contain $P(g) (1-q^{1-n}g)$
as a factor. We continue by picking the smallest $n_2$, so that
$X^{n_2} P(g) (1-q^{1-n}g) \in M$. Certainly $n_2<n$. Again, for any
element of $X^l Q(g)$ in $M$ with $l<n_2$, we have that
$P(g) (1-q^{1-n}g) (1-q^{1-n_2}g)$ divides Q(g). We proceed by
induction, until we arrive at degree zero in $X$.
We obtain the following elements generating a basis for $M$ by left
multiplication with polynomials in $g,g^{-1}$ (rename $n_1=n$):
\[ \begin{array}{ll}
\vdots & \\
X^{n_1+1} & P(g) \\
X^{n_1} & P(g) \\
X^{n_1-1} & P(g) (1-q^{1-{n_1}}g) \\
\vdots & \\