# 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.

## Tuesday, December 18, 2012

### What I learned today: Coadjoint action of group Hopf algebras

This is another post in my psuedo-daily'' series What I learned Today''

Yesterday we talked about actions of a Hopf algebra $H$ on an algebra $A$. Today, let's talk about some examples of this (I didn't cover as much ground as I hoped to today, but we're trying to make these blogs daily!). Every Hopf algebra acts on itself by

$\displaystyle h \triangleright g = \sum_h h_{(1)} g Sh_{(2)}$

This is the adjoint action. It's not hard to prove that it's 1) an action and 2) a Hopf action (that it plays well with the Hopf algebra structure of $H$ and the algebra structure of $A$). As a more concrete example, let's look at the group Hopf algebra $H=kG$ for a finite group $G$. The coproduct is given by $\Delta g = g \otimes g$ and the antipode is group inversion: $Sg = g^-1$. In this case, the adjoint action becomes:

$\displaystyle h \triangleright g = hgh^{-1}$

Which is just group conjugation.

Hopf algebras can act just as well on coalgebras - in this case we require that the action commutes with the coproduct of $A$. Now if $H'$ and $H$ are dually paired, then we also have adjoint action, or rather, a coadjoint action of $H'$ on the coalgebra $H$, given by:

$\displaystyle \phi \triangleright h = \sum_h h_{(2)} \langle \phi, (Sh_{(1)})h_{(3)} \rangle$

We know that the algebra of functions on $G$, $k(G)$ is the Hopf algebra dual of $kG$, where $\langle f, g \rangle = f(g)$, so let's see what the this action becomes for $H' = k(G)$ and $H = kG$:

$\displaystyle f \triangleright g = g \langle f, (Sg)g \rangle = f(1) \cdot g$

Additionally, we also have the notion of a coaction. Generally, if $H$ is a Hopf algebra and $A$ an algebra, we say $\beta : A \rightarrow H \otimes A$ is a coaction of $H$ on $A$ (or that $A$ is a $H$-comodule algebra if $\beta$ is a algebra morphism, $(I \otimes \beta) \circ \beta = (\Delta \otimes I) \circ \beta$ and $I = (\epsilon \otimes I) \circ \beta$, where $I$ is the identity and $\Delta$ and $\epsilon$ are the coalgebra structure maps.

The most interesting case of this is the quantum group $SL_q(2)$ coacting on the quantum plane $\mathbb{A}^2_q$. This is a topic for another post, though.

## Monday, December 17, 2012

### What I learned today: Weird differences between the action of a Hopf algebra and its dual

This is another post in my psuedo-daily'' series What I learned Today''

Let $H$ be a Hopf algebra. Today we're going to talk about its actions (and its coactions, time permitting). A Hopf algebra is just an algebra, which is just a ring, and ring's have actions (equiv. modules), and we're comfortable with that already. For this to make sense on a Hopf algebra, we merely require that the action agrees with all the necessary units, products, coproducts, et cetera.

More precisely, we say that $H$ acts on an algebra $A$ (or that $A$ is an $H$-module algebra) if we have a linear map $\triangleright : H \otimes A \rightarrow A$, written $h \triangleright a$ such that

$\displaystyle h \triangleright (ab) = m(\Delta(h) \triangleright (a \otimes b))$

with $\triangleright$ extended to tensor products and $m$ the product of $A$, and

$\displaystyle h\triangleright 1_A = \epsilon(h) 1_A$

Some weirdness emerges here, for instance, if $H=kG$, the group hopf algebra of a finite group $G$ (the vector space having elements of $G$ as a basis, the product being the group operation, and the coproduct defined by $\Delta(g) = g \otimes g$, antipode, units, et cetera are the obvious choice), then a hopf action collapses to a typical group action:

$\displaystyle g\triangleright(ab) = (g\triangleright a) (g\triangleright b)$

But if we forget the Hopf algebra structure for a second, we know that $kG$ is exactly the same as $k(G)$, that is, as functions defined on $G$ with values in $k$, as any vector'' in $kG$ defines a function and vice versa. Hence, as vector spaces $kG$ and $k(G)$ are naturally isomorphic. This is not the case with the Hopf algebra structure, however (though we have seen how the two are dually paired already), as the coproduct on $k(G)$ is $\Delta(f)(x,y) = f(xy)$, where we identify $k(G \times G)$ with $k(G) \otimes k(G)$ (this is exactly what a tensor product is for, actually.) In this case, the hopf algebra action on an algebra $A$ is the same as a $G$-grading on $A$. Let me explain this (with help from a kind person on math.stackexchange):

Say $A$ is $G$-graded and for each $a \in A$ let $|a|$ denote the element in $G$ such that $a \in A_{|a|}$. We can get a $k(G)$ action from this by $f \triangleright a = f(|a|) a$. To go the other way, start from a Hopf algebra action $\triangleright: k(G) \otimes A \rightarrow A$. Then for each $g \in G$, let $A_g$ be the set of all elements of $A$ on which every $f\in k(G)$ acts via scalar multiplication by $f(g)$. One then has to prove that these sets are nonempty, that they are subspaces, that $A$ decomposes into direct sums of them, and that they obey the axioms of a grading. I've been assured by various sources that it can be done!

## Thursday, December 6, 2012

### What I learned today: Dually Paired Hopf Algebras

To help keep me motivated and mathematically active, I will be blogging about "what I learned today" in my various projects. This is the second post in this "psuedo-daily" series.

I'm still trying to catch up on Quantum Groups. Any reader who is familiar with the field will be able to tell from these posts that I'm a long way away from doing research. But we're slowly getting somewhere!
Let $H = kG$ be the group Hopf algebra for a finite group $G$. In particular, the product map $m: H \otimes H \rightarrow H$ is just the group operation $g \otimes h \rightarrow gh$ for $g, h\in G$, the coproduct is $g \mapsto g\otimes g$, the antipode is group inversion $x \mapsto x^{-1}$, the group idenity is the unit and $g \mapsto 1$ is the counit. We want to talk about its dual $H^\ast = \text{Hom}(H,k)$.

Kassel's Quantum Groups text gives us several propositions to prove that the dual of a finite dimensional Hopf algebra is also a Hopf Algebra. Since $G$ is a finite group, $H$ is finite dimensional, and we'll follow Kassel in proving that $H^\ast$ in particular is a Hopf Algebra.

Now, $H^\ast$ is the space of linear functions from $kG$ to $k$. Any such linear functional will be determined by basis elements that themselves are determined by their action on $G$, hence $H^\ast$ is the space $k(G)$ of functions on $G$ with values in $k$. Clearly this space has pointwise multiplication to make it into an algebra - but we want to derive this from the dual of the coproduct map $\Delta: H \rightarrow H \otimes H$.

So let's talk about $\Delta^\ast: (H\otimes H)^\ast \rightarrow H^\ast$. This is, by definition, the map that takes each $\alpha\otimes\beta \in (H\otimes H)^\ast$ to another linear functional $\gamma = \Delta^\ast(\alpha\otimes\beta)$ on $H$ such that $\gamma(h) = \alpha\otimes\beta(\Delta(h))$. But before we can work out how this becomes a product map, we need to work out some technicalities:

Can we say that $(H\otimes H)^\ast \cong H^\ast \otimes H^\ast$ so that $\Delta^\ast$ is actually a product? Yes! this is provided by a theorem in Kassel's book which says that the map $\lambda: \text{Hom}(U, U') \otimes \text{Hom}(V,V') \rightarrow \text{Hom}(V \otimes U, U' \otimes V')$ by $(f\otimes g)(v \otimes u) = f(u)\otimes g(v)$ is an isomorphism when one of the pairs $(U, U')$, $(V, V')$, or $(U,V)$ consists of only finite dimensional spaces. We won't discuss the proof here, but I do want to point that the theorem does require finite dimensionality .

Back to the dual of comultiplication on $H$: We have that $\Delta(h) = h \otimes h$. So $\alpha \otimes \beta (h\otimes h) = \alpha(h) \beta(h) = \gamma(h)$. Hence $\Delta^\ast(\alpha \otimes \beta)(h) = \alpha(h)\beta(h)$ for $\alpha, \beta \in H^\ast$ and $h\in H$. Hence the dual of comultiplication of a group Hopf algebra is just pointwise multiplication, as we expected.

Now what about the dual of multiplication? Same definition as above, $m^\ast: H^\ast \rightarrow H^\ast \otimes H^\ast$ must be the map taking $\alpha \in H^\ast$ to an $m^\ast(\alpha)=\beta \otimes \gamma$ such that $\beta\otimes\gamma(h_1\otimes h_2) = \alpha(m(h_1,h_2)) = \alpha(h_1 h_2)$. Hence we let $m^\ast$ be the function $m^\ast(\alpha)(x\otimes y) = \alpha(xy)$

Continuing in this fashion, you'll see that the antipode is the map $S(\alpha(x)) = \alpha(x^{-1})$, the unit is the identity function, and the counit the map $\alpha \mapsto \alpha(1)$.

Hence we have a Hopf algebra $H^\ast = k(G)$ that is dual to $H = kG$. If we let $\langle, \rangle: H^\ast \otimes H$ be the evaluation map $\langle \alpha, x \rangle = \alpha(x)$, we can see write down the behavior of this map and see if we can come up with a more general situation to more varied sets of Hopf algebras. For instance, extending the map to tensor products pairwise, note that

$\displaystyle \langle \alpha\beta, h \rangle = \langle \alpha\otimes\beta, \Delta(h) \rangle$
$\displaystyle \langle \Delta (\alpha), h\otimes g \rangle = \langle \alpha, hg\rangle$
$\displaystyle \langle S(\alpha), h \rangle = \langle \alpha, S(h) \rangle$
$\displaystyle \langle 1, h \rangle = 1$
$\displaystyle \langle \alpha, 1 \rangle = \alpha(1)$
Replacing those last two conditions with the same thing expressed in units and counits, we have
$\displaystyle \langle 1, h \rangle = \epsilon(h)$
$\displaystyle \langle \alpha, 1 \rangle = \epsilon(\alpha)$

Thus now we have a definition: Let $H_1, H_2$ be Hopf algebras. We say that they are dually paired if there is a linear map $\langle, \rangle : H_2 \otimes H_1 \rightarrow k$ that satisfies the above 5 conditions.

To make things explicit, $k(G)$ and $kG$ are dually paired by the evaluation map. Majid's book gives a more exotic example of a quantum group'' that is paired with itself:

Let $q \in k$ be nonzero and let $U_q(b+)$ be the $k$-algebra generated by elements $g, g^{-1}$, and $X$ with the relation $gX = qgX$. Majid's book assures me we'll see how to find this quantum group'' in the wild later on, but for now he gives it a Hopf algebra structure with the coproduct $\Delta X = X\otimes 1 + g\otimes X$, $\Delta g = g \otimes g$, $\epsilon X = 0$, $\epsilon g = 1$, $SX = -g^{-1} X$, and $Sg = g^{-1}$

This Hopf algebra is dually paired with itself by the map $\langle g, g \rangle = q$, $\langle X, X \rangle = 1$, and $\langle X, g \rangle = \langle g, X \rangle = 0$