Sunday, December 30, 2012

The Quantum Group $SL_q(2)$ and its coaction on the Quantum Plane

The algebra $SL(2)$ and its coproduct.

Recall our earlier discussion about a universal group structure on algebras. In particular, consider

$\displaystyle \text{Hom}_\text{Alg}(k[a,b,c,d], A) \cong A^4 \cong M_2(A)$

as a vector space. Let $M(2)$ denote the polynomial algebra $k[a,b,c,d]$. Last time we pulled back addition on $A$ to a map $\Delta k[x] \rightarrow k[x] \otimes k[x]$. This time, we're going to follow the same pattern to take matrix multiplication (a map $ m : M_4(A) \otimes M_2(A) \rightarrow M_2(A)$) back to $\Delta M(2) \rightarrow M(2) \otimes M(2)$.

In particular, we're looking for a $\Delta$ so that when we have $\alpha \in \text{Hom}_\text{Alg}(M_2(A) \otimes M_2(A), A)$, $\alpha \circ \Delta = m (\alpha)$

The above notation is slightly confusing, let me try to explain more clearly: We have that the matrix algebra $M_2(A)$ is isomorphic as a vector space to $\text{Hom}_\text{Alg}(M(2),A)$ by the map taking

$\displaystyle f \mapsto \begin{pmatrix} f(a) & f(b) \\ f(c) & f(d) \end{pmatrix} = \hat{f}$

We also have that matrix multiplication is a map $m: M_2(A) \otimes M_2(A) \rightarrow M_2(A)$, and that $M_2(A) \otimes M_2(A)$ is isomorphic as a vector space to $\text{Hom}_\text{Alg}(M(2) \otimes M(2),k)$ by the map

$\displaystyle \alpha \mapsto \begin{pmatrix} \alpha(a) & \alpha(b) \\ \alpha(c) & \alpha(d) \end{pmatrix} \otimes \begin{pmatrix} \alpha(a') & \alpha(b') \\ \alpha(c') & \alpha(d') \end{pmatrix} = \tilde{\alpha}$

I write $a'$, et al, just so it's clear that $\alpha$ varies on different elements of the tensor product basis $a \otimes a$, et al. So for a map to implement multiplication on the $M(2)$ side, it must be $\Delta: M(2) \rightarrow M(2) \otimes M(2)$ and we want $m(\tilde{\alpha}) = \alpha \circ \Delta$.

From this requirement it's obvious what $\Delta$ needs to be; $\Delta(M)$ for $M\in M(2)$ needs to ensure that the generators $a,b,c,d$, et cetera, get mapped to the items that will correspond to matrix multiplication after they are acted upon by $\alpha$. So $\Delta(a) = a \otimes a + b \otimes c$, et cetera. In matrix notation, we can write this

$\displaystyle \Delta \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \otimes \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

This looks group-like! But it's not! The matrix isn't actually an element of $M(2)$, it's just a convenient way for us to write the action of $\Delta$. The action elements are $k$-linear combinations of $a,b,c$, and $d$.

Quantizing things

I bet the reader is guessing that $\Delta$ is a coproduct, making $M(2)$ into a bialgebra. Such a reader would be correct. But before we discuss this further, let's take a quotient of $M(2)$ by the relation $ad -bc =1$. Call this new bialgebra $SL(2)$. We can make it into a Hopf algebra by introducing the antipode:

$\displaystyle S\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} d & -b \\ -c & a\end{pmatrix}$

We're going to take a ``quantum deformation'' of this new Hopf algebra. It's actually simple to do, we're going to modify the commutativity of $a$, et al, by an element $q \in k^\ast$. In particular, let $ca = qac$, $ba =qab$, $db = qbd$, $dc = qcd$, $bc = cb$, $da -ad = (q-q^{-1})bc$, and the ``q-determinant'' relation $ad -q^{-1}bc = 1$. The coproduct remains the same.

The Quantum Plane and the coaction

The classical group we're mimicking, $SL_2(\mathbb{R})$, acts on the affine plane $\mathbb{R}^2$ by transforming it in a way that preserves orientation and area of all geometric shapes on the plane. Since $SL(2)$'s role is as the base of a set of homomorphisms to the algebra $A$, we expect any equivelant ``action'' to have the arrows reversed, we'll discuss this later. But first, we're going to need a notion of an affine plane in our polynomial algebra-geometry language:

This isn't so bad, as $\text{Hom}_\text{Alg}(k[x,y],A) \cong A^2$, hence we call $k[x,y]$ the affine plane. Quantizing it is easy, too: we define the quantum plane $\mathbb{A}^2_q$ to be the free algebra $k\langle x,y\rangle$ quotiented by the relation $yx = qxy$, id est, it's $k[x,y]$ but with a multiplication deformed by the element $q \in k$.

Now back to are ``arrow reversed'' version of an action, or a coaction. One can arrive at this definition by reversing the arrows in the commutative diagram that captures the axioms of an algebra acting on a vector space. In particular, we say that a Hopf algebra $H$ coacts on an algebra $A$ by an algebra morphism $\beta: A \rightarrow H \otimes A$ such that $(I \otimes \beta) \circ \beta = (\Delta \otimes I)\circ \beta$, and $I = (\epsilon \otimes I) \circ \beta)$, where $I$ is the identity map and $\Delta$ and $\epsilon$ are the coproduct structure maps for $H$.

We can define the coaction $\beta$ on the generators $x$ and $y$ and extend it as an algebra morphism should, so the reader can check that $\beta(x) = x\otimes a + y\otimes c$ and $\beta(y) = x\otimes b + y\otimes d$ defines a coaction. In matrix notation, we have:

$\displaystyle \beta(x,y) = (x,y) \otimes \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

I do not have a geometric interpretation for this.

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