alasdairforsythe/text-english-code-fiction-nonfiction

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The transition to the classical function algebra on the other hand is
achieved by setting $q=1$.
This may be depicted as follows:
\[\begin{array}{c @{} c @{} c @{} c}
& \ensuremath{U_q(\lalg{b_+})} \cong \ensuremath{C_q(B_+)} && \\
& \diagup \hspace{\stretch{1}} \diagdown && \\
\begin{array}{l} q=e^{-t} \\ g=e^{tH} \end{array} \Big\| _{t\to 0}
&& q=1 &\\
\swarrow &&& \searrow \\
\ensuremath{U(\lalg{b_+})} & <\cdots\textrm{dual}\cdots> && \ensuremath{C(B_+)}
\end{array}\]
The self-duality of \ensuremath{U_q(\lalg{b_+})}{} is expressed as a pairing
$\ensuremath{U_q(\lalg{b_+})}\times\ensuremath{U_q(\lalg{b_+})}\to\k$
with
itself:
\[\langle X^n g^m, X^r g^s\rangle =
\delta_{n,r} [n]_q!\, q^{-n(n-1)/2} q^{-ms}
\qquad\forall n,r\in\mathbb{N}_0\: m,s\in\mathbb{Z}\]
In the classical limit this becomes the pairing $\ensuremath{U(\lalg{b_+})}\times\ensuremath{C(B_+)}\to\k$
\begin{equation}
\langle X^n H^m, X^r g^s\rangle =
\delta_{n,r} n!\, s^m\qquad \forall n,m,r\in\mathbb{N}_0\: s\in\mathbb{Z}
\label{eq:pair_class}
\end{equation}



\subsection{Differential Calculi and Quantum Tangent Spaces}

In this section we recall some facts about differential calculi
along the lines of Majid's treatment in \cite{Majid_calculi}.

Following Woronowicz \cite{Wor_calculi}, first order bicovariant differential
calculi on a quantum group $A$ (of
function algebra type) are in one-to-one correspondence to submodules
$M$ of $\ker\cou\subset A$ in the category $^A_A\cal{M}$ of (say) left
crossed modules of $A$ via left multiplication and left adjoint
coaction:
\[
a\triangleright v = av \qquad \mathrm{Ad_L}(v)
=v_{(1)}\antip v_{(3)}\otimes v_{(2)}
\qquad \forall a\in A, v\in A
\]
More precisely, given a crossed submodule $M$, the corresponding
calculus is given by $\Gamma=\ker\cou/M\otimes A$ with $\diff a =
\pi(\cop a - 1\otimes a)$ ($\pi$ the canonical projection).
The right action and coaction on $\Gamma$ are given by
the right multiplication and coproduct on $A$, the left action and
coaction by the tensor product ones with $\ker\cou/M$ as a left
crossed module. In all of what follows, ``differential calculus'' will
mean ``bicovariant first order differential calculus''.

Alternatively \cite{Majid_calculi}, given in addition a quantum group $H$
dually paired with $A$
(which we might think of as being of enveloping algebra type), we can
express the coaction of $A$ on
itself as an action of $H^{op}$ using the pairing:
\[
h\triangleright v = \langle h, v_{(1)} \antip v_{(3)}\rangle v_{(2)}
\qquad \forall h\in H^{op}, v\in A
\]
Thereby we change from the category of (left) crossed $A$-modules to
the category of left modules of the quantum double $A\!\bowtie\! H^{op}$.

In this picture the pairing between $A$ and $H$ descends to a pairing
between $A/\k 1$ (which we may identify with $\ker\cou\subset A$) and
$\ker\cou\subset H$. Further quotienting $A/\k 1$ by $M$ (viewed in
$A/\k 1$) leads to a pairing with the subspace $L\subset\ker\cou H$
that annihilates $M$. $L$ is called a ``quantum tangent space''
and is dual to the differential calculus $\Gamma$ generated by $M$ in
the sense that $\Gamma\cong \Lin(L,A)$ via
\begin{equation}
A/(\k 1+M)\otimes A \to \Lin(L,A)\qquad
v\otimes a \mapsto \langle \cdot, v\rangle a
\label{eq:eval}
\end{equation}
if the pairing between $A/(\k 1+M)$ and $L$ is non-degenerate.

The quantum tangent spaces are obtained directly by dualising the
(left) action of the quantum double on $A$ to a (right) action on
$H$. Explicitly, this is the adjoint action and the coregular action
\[
h \triangleright x = h_{(1)} x \antip h_{(2)} \qquad
a \triangleright x = \langle x_{(1)}, a \rangle x_{(2)}\qquad
\forall h\in H, a\in A^{op},x\in A
\]
where we have converted the right action to a left action by going
from \mbox{$A\!\bowtie\! H^{op}$}-modules to \mbox{$H\!\bowtie\! A^{op}$}-modules.
Quantum tangent spaces are subspaces of $\ker\cou\subset H$ invariant
under the projection of this action to $\ker\cou$ via \mbox{$x\mapsto
x-\cou(x) 1$}. Alternatively, the left action of $A^{op}$ can be
converted to a left coaction of $H$ being the comultiplication (with
subsequent projection onto $H\otimes\ker\cou$).

We can use the evaluation map (\ref{eq:eval})
to define a ``braided derivation'' on elements of the quantum tangent
space via
\[\partial_x:A\to A\qquad \partial_x(a)={\diff a}(x)=\langle
x,a_{(1)}\rangle a_{(2)}\qquad\forall x\in L, a\in A\]
This obeys the braided derivation rule

The transition to the classical function algebra on the other hand is

achieved by setting $q=1$.

This may be depicted as follows:

\[\begin{array}{c @{} c @{} c @{} c}

& \ensuremath{U_q(\lalg{b_+})} \cong \ensuremath{C_q(B_+)} && \\

& \diagup \hspace{\stretch{1}} \diagdown && \\

\begin{array}{l} q=e^{-t} \\ g=e^{tH} \end{array} \Big| _{t\to 0}

&& q=1 &\\

\swarrow &&& \searrow \\

\ensuremath{U(\lalg{b_+})} & <\cdots\textrm{dual}\cdots> && \ensuremath{C(B_+)}

\end{array}\]

The self-duality of \ensuremath{U_q(\lalg{b_+})}{} is expressed as a pairing

$\ensuremath{U_q(\lalg{b_+})}\times\ensuremath{U_q(\lalg{b_+})}\to\k$

with

itself:

\[\langle X^n g^m, X^r g^s\rangle =

\delta_{n,r} [n]_q!\, q^{-n(n-1)/2} q^{-ms}

\qquad\forall n,r\in\mathbb{N}_0\: m,s\in\mathbb{Z}\]

In the classical limit this becomes the pairing $\ensuremath{U(\lalg{b_+})}\times\ensuremath{C(B_+)}\to\k$

\begin{equation}

\langle X^n H^m, X^r g^s\rangle =

\delta_{n,r} n!\, s^m\qquad \forall n,m,r\in\mathbb{N}_0\: s\in\mathbb{Z}

\label{eq:pair_class}

\end{equation}

\subsection{Differential Calculi and Quantum Tangent Spaces}

In this section we recall some facts about differential calculi

along the lines of Majid's treatment in \cite{Majid_calculi}.

Following Woronowicz \cite{Wor_calculi}, first order bicovariant differential

calculi on a quantum group $A$ (of

function algebra type) are in one-to-one correspondence to submodules

$M$ of $\ker\cou\subset A$ in the category $^A_A\cal{M}$ of (say) left

crossed modules of $A$ via left multiplication and left adjoint

coaction:

\[

a\triangleright v = av \qquad \mathrm{Ad_L}(v)

=v_{(1)}\antip v_{(3)}\otimes v_{(2)}

\qquad \forall a\in A, v\in A

\]

More precisely, given a crossed submodule $M$, the corresponding

calculus is given by $\Gamma=\ker\cou/M\otimes A$ with $\diff a =

\pi(\cop a - 1\otimes a)$ ($\pi$ the canonical projection).

The right action and coaction on $\Gamma$ are given by

the right multiplication and coproduct on $A$, the left action and

coaction by the tensor product ones with $\ker\cou/M$ as a left

crossed module. In all of what follows, ``differential calculus'' will

mean ``bicovariant first order differential calculus''.

Alternatively \cite{Majid_calculi}, given in addition a quantum group $H$

dually paired with $A$

(which we might think of as being of enveloping algebra type), we can

express the coaction of $A$ on

itself as an action of $H^{op}$ using the pairing:

\[

h\triangleright v = \langle h, v_{(1)} \antip v_{(3)}\rangle v_{(2)}

\qquad \forall h\in H^{op}, v\in A

\]

Thereby we change from the category of (left) crossed $A$-modules to

the category of left modules of the quantum double $A\!\bowtie\! H^{op}$.

In this picture the pairing between $A$ and $H$ descends to a pairing

between $A/\k 1$ (which we may identify with $\ker\cou\subset A$) and

$\ker\cou\subset H$. Further quotienting $A/\k 1$ by $M$ (viewed in

$A/\k 1$) leads to a pairing with the subspace $L\subset\ker\cou H$

that annihilates $M$. $L$ is called a ``quantum tangent space''

and is dual to the differential calculus $\Gamma$ generated by $M$ in

the sense that $\Gamma\cong \Lin(L,A)$ via

\begin{equation}

A/(\k 1+M)\otimes A \to \Lin(L,A)\qquad

v\otimes a \mapsto \langle \cdot, v\rangle a

\label{eq:eval}

\end{equation}

if the pairing between $A/(\k 1+M)$ and $L$ is non-degenerate.

The quantum tangent spaces are obtained directly by dualising the

(left) action of the quantum double on $A$ to a (right) action on

$H$. Explicitly, this is the adjoint action and the coregular action

\[

h \triangleright x = h_{(1)} x \antip h_{(2)} \qquad

a \triangleright x = \langle x_{(1)}, a \rangle x_{(2)}\qquad

\forall h\in H, a\in A^{op},x\in A

\]

where we have converted the right action to a left action by going

from \mbox{$A\!\bowtie\! H^{op}$}-modules to \mbox{$H\!\bowtie\! A^{op}$}-modules.

Quantum tangent spaces are subspaces of $\ker\cou\subset H$ invariant

under the projection of this action to $\ker\cou$ via \mbox{$x\mapsto

x-\cou(x) 1$}. Alternatively, the left action of $A^{op}$ can be

converted to a left coaction of $H$ being the comultiplication (with

subsequent projection onto $H\otimes\ker\cou$).

We can use the evaluation map (\ref{eq:eval})

to define a ``braided derivation'' on elements of the quantum tangent

space via

\[\partial_x:A\to A\qquad \partial_x(a)={\diff a}(x)=\langle

x,a_{(1)}\rangle a_{(2)}\qquad\forall x\in L, a\in A\]

This obeys the braided derivation rule