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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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