Bi, Hopf algebras

\(\newcommand{\rar}{\rightarrow} \newcommand{\mb}{\mathbb} \newcommand{\mf}{\mathfrak}\)Last time we talked about coalgebras as duals to algebras. Now we can consider what we get when we have a vector space with both an algebra and a coalgebra structure.
Let's consider a group algebra, \(\mb kG\). Recall that for a finite group \(G\), this means a vector space with a basis \(\{\hat g\}\) corresponding to elements \(g \in G\), and multiplication given by composition in \(G\).
We defined a comultiplication on this vector space by setting
$$\Delta(\hat g) = \hat g \otimes \hat g$$This is one reason why we use the letter \(\Delta\) for the comultiplication: in some cases \(\Delta\) is the diagonal map \(x \mapsto (x, x)\) whose image, when applied to \(\mb R\), is a diagonal line in \(\mb R^2\). This comultiplication is coassociative and cocommutative.
We also defined the counit \(\epsilon\) by
$$\epsilon(\hat g) = 1$$How does this interact with the group algebra structure? Well, first we note that \(\mb kG \otimes \mb kG\) has a multiplication on it, via
$$(a \otimes b)(c \otimes d) = ac \otimes bd$$In other words, we have a multiplication \(m_2\) on \(\mb kG\) defined by
$$m_2 = (m \otimes m) \circ (id \otimes \sigma_{\mb kG\mb kG} \otimes id)$$where the \(\sigma\) moves the \(c\) to the left of the \(b\).
The next thing we note is that \(\Delta\) turns group multiplication in \(\mb kG\) into multiplication in \(\mb kG \otimes \mb kG\):
$$\Delta(\widehat{gh}) = \widehat{gh} \otimes \widehat{gh} = \hat g\hat h \otimes \hat g \hat h = (\hat g \otimes \hat g)(\hat h \otimes \hat h) = \Delta(\hat g)\Delta(\hat h)$$Also that \(\Delta(\hat e) = \hat e \otimes \hat e\), where \(e\) is the identity element of \(G\) and hence \(\hat e = \eta(1)\) is the multiplicative identity of \(\mb kG\). Hence \(\hat e \otimes \hat e\) is the multiplicative identity of \(\mb kG \otimes \mb kG\), so \(\Delta\) and \(\eta\) also play well together. Dually, we have \(\epsilon(\hat f)\epsilon(\hat g) = \epsilon(\hat f \hat g)\), so the multiplication \(m\) and \(\epsilon\) play well together too.
These compatibility rules give us what we call a bialgebra.

We can then write down the rules for a general bialgebra \(B\): \(B\) is a vector space with an associative, unital algebra structure \(m, \eta\) and a coassociative counital coalgebra structure \(\Delta, \epsilon\) that are compatible.
Written out as tensors, we get that the \(m, \Delta\) compatibility rule as:
$$m_{ij}^h\Delta_h^{kl} = \Delta_i^{pq}\Delta_j^{rs}m_{pr}^km_{qs}^l$$Written as maps, we get
$$\Delta \circ m = (m \otimes m)\circ (id \otimes \sigma_{BB} \otimes id) \circ (\Delta \otimes \Delta)$$For the \(\eta, \Delta\) compatibility, it looks like
$$\Delta_i^{jk} \eta^i = \eta^j \eta^k \text{ i.e. }\Delta \circ \eta = (\eta \otimes \eta) \circ \Delta_{\mb k}$$where \(\Delta_{\mb k}\) is the comultiplication on \(\mb k\) as a 1-dimensional coalgebra over itself; \(\Delta_{\mb k}(c) = c \otimes 1\) for \(c \in \mb k\).
The \(m, \epsilon\) compatibility then becomes
$$m_{ij}^k \epsilon_k = \epsilon_i\epsilon_j\text{ i.e. }\epsilon \circ m = m_{\mb k}\circ (\epsilon \otimes \epsilon)$$Note that we usually don't even think about \(m_{\mb k}\) and just identify the tensor product of two scalars as the product of them.
The best thing about bialgebras is that the dual of a finite-dimensional bialgebra is also a bialgebra. The multiplication in \(B\) becomes a comultiplication in \(B^\vee\), and the comultiplication in \(B\) becomes a multiplication in \(B^\vee\); the unit becomes a counit, the counit becomes a unit, and the compatibility conditions for \(B\) dualize to the compatibility conditions for \(B^\vee\).

There's one more piece of group structure that we haven't even touched: inversion. Every group element has an inverse.
So we are going to introduce a new map, \(S\), which is short for "antipode", that goes from \(\mb kG\) to \(\mb kG\) and sends \(\hat g\) to \(\widehat{g^{-1}}\).
A few observations about \(S\): \(S(\hat e) = \hat e\), \(S(\widehat{gh}) = S(\hat h)S(\hat g)\), \(\epsilon(S(\hat g)) = 1 = \epsilon(\hat g)\), and \(\Delta(S(\hat g)) = S(\hat g)\otimes S(\hat g)\); there's a swap there that we can't see because \(\Delta\) here is cocommutative, but we can detect the swap by duality with the multiplication rule. Finally, the most important rule,
$$m(S(\hat g)\otimes \hat g) = \hat e = m(\hat g \otimes S(\hat g))$$which is kind of why we care about \(S\) at all.
Again, generalizing, we say that a Hopf algebra \(H\) is a bialgebra with an antipode map \(S\) such that
$$S_i^j \eta^i = \eta^j, S_i^j m^i_{kl} = m_{ba}^j S_k^a S_l^b,$$$$S_i^j \epsilon_j = \epsilon_i, S_i^j \Delta_j^{kl} = \Delta_i^{ba} S_a^k S_b^l,$$ $$m_{hi}^j S_k^h \Delta_l^{ki} = I_l^j = m_{hi}^j S_k^i \Delta_l^{hk}$$ $$S \circ \eta = \eta, S \circ m = m \circ \sigma_{HH} \circ (S\otimes S),$$ $$ \epsilon \circ S = \epsilon, \Delta \circ S = (S \otimes S) \circ \sigma_{HH} \circ \Delta.$$ $$m \circ (S \otimes id) \circ \Delta = id = m \circ (id \otimes S)\circ \Delta$$Like with biaglebras, the duals of finite-dimensional Hopf algebras are Hopf algebras. The dual of \(\mb kG\) is the space \(Fun(G)\) of functions from \(G\) to \(\mb k\). It's clearly an algebra, since you can multiply pointwise: \((st)(g) = s(g)t(g)\), and thus, treating \(s\) and \(t\) as linear functions on \(\mb kG\), we get
$$m(s \otimes t)(\hat g) = (s \otimes t)(\Delta(\hat g))$$and the unit is the function that always returns 1.
We can take as a basis of \(Fun(G)\) the set of functions \(d_g\) where \(d_g(\hat g) = 1\) and \(d_g(\hat h) = 0\) if \(h \neq g\), so that the unit then becomes \(\eta(c) = c\sum_{g \in G}d_g\)
The comultiplication is dual to the multiplication in \(\mb kG\):
$$\Delta(s)(\hat g \otimes \hat h) = s(\hat g\hat h)$$and more explicitly,
$$\Delta(d_g) = \sum_{h \in G} d_{gh^{-1}} \otimes d_h$$and the counit is evaluation at \(\hat e\).
The antipode is then \(S(d_g) = d_{g^{-1}}\), dual to the antipode of \(\mb kG\).
\(\mb kG\) is always cocommutative, but commutative only if \(G\) is commutative; \(Fun(G)\) is commutative, but cocommutative only if \(G\) is commutative.

The tensor algebra \(\bigotimes^* V) of a vector space \(V\) is itself a Hopf algebra. Its multiplication and unit we already know as tensor product \(\otimes\) and \(1 \in \mb k\), but to make things easier we'll write said multiplication without the \(\otimes\) symbol.
The comultiplication \(\Delta\) sends \(v \in V\) to \(v \otimes 1 + 1 \otimes v\), treated as an element of \(\bigotimes^* V \otimes \bigotimes^* V\). We extend this to the rest of \(\bigotimes^* V\) by the usual multiplication rules: \(Delta(uv) =  \Delta(u)\Delta(v) = uv \otimes 1 + u\otimes v + v\otimes u + 1\otimes uv\) and so on. The general rule is a little complicated.
In contrast the counit is easy: \(v\) gets sent to 0, and so does any multiple of \(v\). So only things in \(\mb k\) have nonzero counit.
Finally, the antipode sends \(v \in V\) to \(-v\), and again we extend by multiplication. So \(S(uv) = (-v)(-u) = vu\), and \(S(uvw) = -wvu\) and so on.
And you can check that this does give us a Hopf algebra.

One last weird example before we end:
The Sweedler Hopf algebra \(H\) is four dimensional. We'll denote \(\eta(1)\) as just \(1\) here to match conventional notation. A basis for \(H\) is given as \(\{1, c, x, y\}\), with the following properties:
$$c^2 = 1, cx = y = -xc, x^2 = xy = yx = y^2 = 0$$ $$\Delta(c) = c \otimes c, \Delta(x) = x \otimes 1 + c \otimes x,\Delta(y) = y \otimes c + 1 \otimes y$$ $$\epsilon(c) = 1, \epsilon(x) = 0,\epsilon(y) = 0$$ $$S(c) = c, S(x) = -y, S(y) = x$$This is a noncommutative, noncocommutative Hopf algebra and, along with its dual, is the lowest-dimensional example of such. Compute its dual, which looks even weirder.