The quantum general linear group

The passage from quantum to classical mechanics is quite well defined by taking the limit as Planck's constant $\hbar$ tends to 0. The passage in the other direction is not so clear cut, and may not be uniquely determined. On the algebraic side, ``quantization'' involves deforming commutative algebras to non-commutative ones:

$$\mbox{e.g.} \qquad x\,y \;=\; y\,x \qquad \mbox{becomes} \qquad x\, y \;=\; e^\hbar\,y\, x\;.$$

Usually we deal with $q = e^{\hbar}$ rather than $\hbar$, so classical results correspond to the case q = 1. Quantum spaces correspond to more general $\arbK$-algebras, not necessarily commutative.

Let $\arbK$ be a fixed field and fix $q \in \arbK$ with $q \neq 0$. Write $\arbK \langle x_1^{}\,, \ldots\,, x_n^{}\, \rangle$ for the $\arbK$-algebra of polynomials in non-commuting indeterminates $x_1^{}\,, \ldots\,, x_n^{}$. As a vector space over $\arbK$, a basis is given by those elements

$$x_{\xi(1)}^{m_1^{}}\, x_{\xi(2)}^{m_2^{}}\, \cdots\, x_{\xi(r)}^{m_r^{}}$$

for which $r\in \N$ and $m_1^{}, \ldots, m_r^{} \in \Z^{+}$ and $\map \xi \from \{ 1, \ldots ,r \} \to {\{1, \ldots , n\}}$ is any function. Notice that

$$\arbK[\,x\,,y\,] = \arbK \langle x\,,y\, \rangle /(\,x\,y = y\,x\,) \;.$$

The coordinate algebra of the space of quantum 2 x 2 matrices is defined by

$$\Mq(2) \;=\; \arbK \langle a\,,b\,,c\,,d\, \rangle /R$$

where R is the system of equations

$$\displaylines{ a b \;=\; q^\inv b a \;\;,\;\; a c \;=\; q^\... ...\urdiag \rto & d \enddiagram\hfill\hphantom{(mnemonic)}}}\cr}$$

The monomials a^m₁ b^m₂ c^m₃ d^m₄ form a basis for the algebra, as a vector space over $\arbK$.

$$\Alg_\arbK \big(\Mq(2)\,, A\big) \;\isom\; \Big\{ \Big(\matr... ...r}\Big) \in \Mat(2\,,A) \;\vert\; R \,\hbox{ holds } \Big \}$$

Theorem 4.1  

Let $\Big(\matrix{a & b \cr c & d \cr}\Big)$ and $\Big(\matrix{a' & b' \cr c' & d' \cr}\Big)$ be two A-points of $\Mq(2)$ such that each entry of the first commutes with each entry of the second.

(i) The product $\Big( \matrix{a & b \cr c & d \cr} \Big) \Big( \matrix{a' & b' \cr c' & d' \cr} \Big)$ (as matrices) is an A-point of $\Mq(2)$.

(ii) The ``q-determinant'' $\;\detq \Big( \matrix {a & b \cr c & d \cr} \Big) \;=\; (a d - q^\inv b c)\;$ commutes with each of a ,b ,c ,d and

$$\detq \left(\Big( \matrix {a & b \cr c & d \cr} \Big) \Big... ...times \detq \Big( \matrix {a' & b' \cr c' & d' \cr} \Big)\;.$$

(iii) If $\detq \Big( \matrix {a & b \cr c & d \cr} \Big)$ is invertible in A then

$$\Big(\matrix {a & b \cr c & d \cr} \Big) ^\inv \;=\; \left( ... ...ht)^\inv \Big( \matrix {d & -qb \cr -q^\inv c & a \cr} \Big)$$

is an A-point of $M_{\!q^\inv}(2)$.

The above result can be proved by direct calculation, but this gives little insight into the special nature of the relations R. Examples such as this arose in work of L.D. Faddeev [FRT87] and his school on the quantum inverse scattering transform (QIST) method. The version I present here comes from some lectures of Yu Manin [Man88] given at Université de Montréal in June 1988. The following ``explanation" of this theorem is due to Yu Kobyzev (Moscow, winter 1986-87).

Introduce the quantum plane, as defined by the $\arbK$-algebra

$$\Affq20 \;=\; \arbK \langle x\,,y \,\rangle / (xy = q^\inv\, yx) \;.$$

The monomials x^m yⁿ with $m\,,n \in \N$ form a basis for this as a vector space. We also need to consider a quantized version of a Grassmannian algebra in two variables:

$$\Affq02 \;=\; \arbK \langle \xi\, ,\eta \,\rangle / (\xi^2 = \eta^2 = 0 = \xi \eta + q\,\eta \xi) \;.$$

The monomials $\xi^m\, \eta^n$ with $m\,,n \in \{ 0,1 \}$ form a basis for this algebra. The reason for the funny superscripts ${\scriptstyle 2\vert}$ and ${\scriptstyle 0\vert 2}$ comes from ``supergeometry'' where dimensions are represented by pairs d | d' of numbers. This $\Affq02$ is a quantum superplane.

An A-point of B is called generic when the algebra morphism $\map \from B \to A$ is injective.

Theorem 4.2  

Suppose (x,y) is a generic A-point of $\Affq20$ and $(\xi\,,\eta)$ is a generic A-point of $\Affq02$. Suppose $a\,,b\,,c\,,d\in A$ all commute with $x\,,y\,, \xi\,, \eta$. Put

$$\Big( \matrix{x' \cr y'\cr} \Big) = \Big( \matrix {a & b \cr... ...cr c & d \cr} \Big) \Big( \matrix{\xi \cr \eta \cr}\Big) \;.$$

If $q^2 \ne -1$, the following conditions are equivalent :

(i) (x', y') and (x''!, y'') are points of $\Affq20$;

(ii) (x', y') is a point of $\Affq20$ and $(\xi', \eta')$ is a point of $\Affq02$;

(iii) $\Big( \matrix {a & c \cr b & d \cr} \Big)$ is a point of M_q(2).

[For q² = -1 we only have (iii)$\implies$(i) & (ii).]

In other words, the relations R are precisely what is needed for $\Big(\matrix{a & b \cr c & d \cr}\Big)$ and its transpose to both transform the quantum plane into itself; or for $\Big(\matrix{a & b \cr c & d \cr}\Big)$ to transform both the quantum plane and superplane into themselves.

The quantum general linear group is defined from 2 x 2 matrices by inverting the determinant:

$$\GL{}_q(2) \;=\; \Mq(2) [t]/(t\,a = a\,t \;, \; t\,b = b\,t \;, \; t\,c = c\,t \;, \; d\,t = t\,d \;, \; t\detq = 1) \;.$$

Similarly, the quantum special linear group is defined by requiring that the determinant be equal to 1:

$$\SL{}_q(2) \;=\; \Mq(2) / (\detq = 1) \;.$$

The 2nd theorem describes the representations of these ``groups'' on quantum spaces $\Affq20$ and $\Affq02$.