Here is an answer to question (1). I recommend that you split of question (2) as a separate question.

Define the *quantum plane* to be the "spectrum" of the noncommutative ring $\mathbb K\langle x,y\rangle / (xy = qyx)$, where $\mathbb K$ is some ground commutative ring in which $q$ is invertible (e.g. $\mathbb K = \mathbb C(q)$). So a "point" in this "plane" is determined by the "values" of the two "coordinates" $x$ and $y$. Note that these "values" don't commute: they are "valued" in some space of "noncommutative numbers".

Now, recall that $2\times 2$ matrices act on the usual plane. Are there "$2 \times 2$ matrices" that act on the quantum plane?

Well, in the usual case, a matrix $\bigl( \begin{smallmatrix} a & b \\ c & d\end{smallmatrix}\bigr)$ takes the vector $\bigl( \begin{smallmatrix} x \\ y \end{smallmatrix}\bigr)$ to:
$$ \begin{pmatrix} ax + by \\ cx + dy \end{pmatrix}.$$
Let's try to keep the same rule. For this to work, the coordinates of the new point need to satisfy the same "$q$-mmutation" (as opposed to "commutation") law as $\{x,y\}$, which is to say:

$$ (ax + by)(cx+ dy) = q(cx+dy)(ax+by). $$

If we suppose that variables $a,b,c,d$ commute with variables $x,y$, then we get:

$$ ac = qca \quad bd = qdb \quad ad + q^{-1}bc = qcb + da. \quad\quad (\star)$$

What about, say, how $a$ and $b$ commute? Well, I should have mentioned that the quantum plane has a "dual quantum plane". Recall that in classical linear algebra, the dual to the space of column vectors is the space of row vectors. Let's, then, take the dual quantum plane to consist of row vectors $(v,w)$ whose coordinates satisfy $vw = qwv$. Then by letting quantum matrices act from the right on quantum row vectors, you can compute:

$$ ab = qba \quad cd = qdc \quad ad + q^{-1}cb = qbc + da. \quad\quad (\star\star)$$

Note that the last lines in each are slightly different. Together, the rules $(\star)$ and $(\star\star)$ define the space of "quantum $2\times 2$ matrices". Let me simplify the last two rules:

$$ bc = cb \quad ad - da = (q-q^{-1})bc $$

Now here's an **exercise:** suppose that $\bigl( \begin{smallmatrix} a_1 & b_1 \\ c_1 & d_1\end{smallmatrix}\bigr)$ and $\bigl( \begin{smallmatrix} a_2 & b_2 \\ c_2 & d_2\end{smallmatrix}\bigr)$ are each quantum $2\times 2$ matrices, by which I mean that their coordinates each independently satisfy the rules $(\star,\star\star)$, and suppose that the $r_1$s commute with the $r_2$s. Then their product
$$ \begin{pmatrix} a_1a_2 + b_1 c_2 & a_1 b_2 + b_1 d_2 \\ c_1 a_2 + d_1 c_2 & c_1 b_2 + d_1 d_2 \end{pmatrix}$$
is again a quantum $2\times 2$ matrix.

Thus our "quantum space" of quantum $2\times 2$ matrix (defined by $(\star,\star\star)$, which just determines if at some space in which each point is determined by the values of 4 noncommuting coordinates) is in face a "ring", or at least a "monoid". Oh, also you should check that (**exercise**) $\bigl( \begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix} \bigr)$ is a quantum matrix, and is the unit. So now always do the coordinates have to evaluate to noncommuting numbers.

Now, an important property of matrices is the "determinant". **Exercise:** the number $\Delta = ad - qbc$ commutes with all coordinates.

**Exericse:** If $X = \bigl( \begin{smallmatrix} a & b \\ c & d\end{smallmatrix}\bigr)$ is a quantum $2\times 2$ matrix in which $\Delta^{-1}$ exists, then $X$ is invertible, with inverse $\Delta^{-1}\bigl( \begin{smallmatrix} d & -b \\ -qc & a\end{smallmatrix}\bigr)$.

Anyway, by definition, the group *quantum $SL(2)$* is the subgroup of the quantum $2\times 2$ matrices consisting of those matrices for which $\Delta = 1$. By definition, the group *quantum $GL(2)$* is the subgroup for which $\Delta^{-1}$ exists (thus you could say that a point in quantum $GL(2)$ has five coordinates, $a,b,c,d,\Delta^{-1}$, satisfying the equations $(\star,\star\star)$ and $\Delta^{-1}(ad - qbc) = 1$).

Similar constructions give other quantum groups as well, by starting with higher-dimensional quantum spaces and/or quantum versions of symmetric or skew-symmetric pairings.

Finally, you might have seen "groups" like $U_q \mathfrak{sl}(2)$, which really are the universal enveloping algebras of "lie algebras" of these groups.

This post imported from StackExchange MathOverflow at 2015-04-28 14:42 (UTC), posted by SE-user Theo Johnson-Freyd