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

arXiv:hep-th/0106122v3 8 May 2002