Quantum group
Can group be quantized?
The answer is quantum group.
Let's deformation quantize the Poisson Lie-Group.
Deformation quantization of (co)-Poisson structures
Definition: Let (A_0, m_0, \eta_0, \{.,.\}) be a Poisson algebra over the complex numbers \mathbf{C}. A quantization of A_0 is a non-commutative algebra (A,m,\eta) over the ring \mathbf{C}[[\hbar]], where \hbar is a formal parameter (interested as Planck's constant), such that
1. A/\hbar A \simeq A_0 (put \hbar=0, classical limit)
2. m_0 \circ (\pi \otimes \pi) = \pi \circ m
3. \pi \circ \eta = \eta_0
4. \{ \pi (a), \pi (b) \} = \pi (\frac{[a,b]}{\hbar}) for all a, b\in A
where \pi denotes the canonical quotient map
\pi : A \rightarrow A /\hbar A \simeq A_0
and [.,.] = m - m\circ \tau denotes the ordinart commutator (with respect to the multiplication m).
Non-commutative product on A is Moyal product, \star, defined by m(a\otimes b) \equiv a\star b.
It can express like a\star b = \sum^\infty_{n=0} f_n(a, b) \hbar^n where f_n:A\times A\rightarrow A.
Definition: Let(A_0, m_0, \Delta_0, \eta_0, \epsilon_0;\delta) be a co-Poisson bi-algebra where \delta denotes the co-Poisson structure. A quantization of A_0 is a non co-commutative bi-algebra (A, m, \Delta, \eta, \epsilon) over the ring \mathbf{C}[[\hbar]] such that
1. A/\hbar A \simeq A_0
2. (\pi \otimes \pi)\circ \Delta = \Delta_0 \circ \pi
3. m_0 \circ (\pi \otimes \pi) = \pi \circ m
4. \pi\circ \eta = \eta_0
5. \epsilon \circ \pi = \epsilon_0
6. \delta(\pi(a)) = \pi (\frac{1}{\hbar} (\Delta(a) - \tau\circ\Delta(a)) for all a in A
where \pi again denotes the canonical quotient map \pi : A\rightarrow A_0.
Also co-multiplication have to defined.
\Delta_0 (a_0.b_0) = \Delta_0 (a_0) .\Delta(b_0) and \Delta (a \star b) = \Delta (a) \star \Delta(b).
The quantization of \mathcal{U} (sl_2)
co-Poisson sturcture of \mathcal{U}(sl_2) is given by the extension to the entire universal enveloping algebra of the map
\delta(H) = 0
\delta(E) = \frac{1}{2} E \wedge H
\delta(F) = \frac{1}{2} F \wedge H
Now let's find co-product \Delta satisfy these:
1. \Delta must be co-associative (to be Hopf algebra)
2. \delta(\pi(a)) = \pi(\frac{1}{\hbar}(\Delta(a) - \tau\circ\Delta(a)))
3. In the classical limit (\hbar\rightarrow 0) the coproduct \Delta must reduce to the ordinary coproduct on \mathcal{U}(sl_2).
The comultiplication \Delta has the general form \Delta = \sum^\infty_{n=0} \frac{\hbar^n}{n!} \Delta_{(n)}.
The third requirement fixes \Delta_{(0)}:
\Delta_{(0)}(H) = H\otimes 1 + 1 \otimes H
\Delta_{(0)}(E) = E\otimes 1 + 1 \otimes E
\Delta_{(0)}(F) = F\otimes 1 + 1 \otimes F
The second requirement gives:
\delta(H) = 0 = \Delta_{(1)}(H) - \tau \circ \Delta_{(1)}(H)
\delta(E) = \frac{1}{2}E\wedge H = \Delta_{(1)}(E) - \tau \circ \Delta_{(1)}(E)
\delta(F) = \frac{1}{2}F\wedge H = \Delta_{(1)}(F) - \tau \circ \Delta_{(1)}(F)
So solution is
$\Delta_{(1)}(H) = 0!
\Delta_{(1)}(E) = \frac{1}{4} E\wedge H
\Delta_{(1)}(F) = \frac{1}{4} F \wedge H
To get higher order, use first requirement
\sum^n_{k=0} {n \choose k} (\Delta_{(k)} \otimes 1 - 1\otimes \Delta_{(k)})\Delta_{(n-k)} = 0
Solve,
\Delta_{(n)}(H) = 0
\Delta_{(n)}(E) = \frac{1}{4^n}(E\otimes H^n + (-1)^n H^n\otimes E)
\Delta_{(n)}(F) = \frac{1}{4^n}(F\otimes H^n + (-1)^n H^n\otimes F)
where H^n = H \star H \star \cdots \star H.
Complete solution is
\Delta(H) = H\otimes 1 + 1 \otimes H
\Delta(E) = E \otimes q^H + q^{-H} \otimes E
\Delta(F) = F\otimes q^H + q^{-H}\otimes F
where q = e^{\hbar/4}.
To check Hopf algebra still satisfy, see multiplication map
\Delta(a\star b) = \Delta(a) \star \Delta (b)
Which means
\Delta([H,E]) = [\Delta(H),\Delta(E)]
\Delta([H,F]) = [\Delta(H),\Delta(F)]
\Delta([E,F]) = [\Delta(E),\Delta(F)]
Since [H,E]=2E and [H,F]=-2F, first two relations are saristied.
But third is not. \Delta([E,F]) = [E,F]\otimes q^{2H} + q^{-2H}\otimes[E,F], [E,F]=\frac{q^{2H}-q^{-2H}}{q-q^{-1}}.
Reference
T.Tjin - An introduction to quantized Lie groups and algebras
A Guide to Quantum Groups