Coalgebras

\( \newcommand{\ncmd}{\newcommand} \ncmd{\rar}{\rightarrow} \ncmd{\mb}{\mathbb} \ncmd{\mf}{\mathfrak} \) At some point there's going to be a plan. Not at the moment. But I'm introducing coalgebras because it's easier to do representation theory of Lie algebras as infinitesimal group stuff if I have comultiplication available. Plus duality is the best thing ever.

Last time we saw an algebra \(A\) as a vector space with a multiplication operation \(m: A \otimes A \rar A\). Maybe it obeyed some laws, like associativity or commutativity, and maybe it had a multiplicative identity, given by the unit map \(\eta: \mb k\rar A\).
This time we're going to flip things around.

The motivation here is duality. Recall the dual of a vector space \(V\): it's the set \(V^\vee\) of linear maps from \(V\) to \(\mb k\), and happens to form another vector space.
Given a linear map \(T\) from a vector space \(W\) to \(V\), and an element \(f\) in \(V^\vee\), we can look at the composition \(f\circ T\), which takes a vector from \(W\) and spits out numbers in \(\mb k\). Since both \(T\) and \(f\) are linear, we get that \(f\circ T\) is in \(W^\vee\). Hence the map \(T\) from \(W\) to \(V\) gives us a way to get maps from \(V^\vee\) to \(W^\vee\). The technical term is that the operation of taking the dual is "contravariant", with maps becoming maps in the other direction.
So our multiplication \(m: A\otimes A\rar A\) dualizes to a map from \(A^\vee\) to \((A\otimes A)^\vee\). If \(A\) is finite-dimensional, then \((A\otimes A)^\vee = A^\vee \otimes A^\vee\), so we want to look at a vector space \(A^\vee\) equipped with a map from \(A^\vee\) to \(A^\vee \otimes A^\vee\). The multiplication map goes from the tensor product to the vector space, this dualized "comultiplication" map goes from the vector space to the tensor product.

So, ignoring the original vector space, a coalgebra \(C\) is a vector space with a comultiplication map \(\Delta: C \rightarrow C \otimes C\). Everything is tensors, so we can write \(\Delta\) with tensor indices. Letting \(\{e_i\}\) be a basis for \(C\), we write
$$\Delta(e_i) = \Delta_i^{jk} e_j\otimes e_k$$ Compare to the multiplication map, which when written as a tensor has two lower indices and an upper index.

Let's put in some properties. Recall that associativity looked like
$$m_{ab}^e m_{ec}^d = m_{ae}^d m_{bc}^e$$ $$m \circ (m \otimes id) = m \circ (id \otimes m)$$ Flipping that around gives the notion of coassociativity, which looks like
$$\Delta_h^{ij} \Delta_l^{hk} = \Delta_l^{ih} \Delta_h^{jk}$$ $$(\Delta \otimes id) \circ \Delta = (id \otimes \Delta) \circ \Delta$$ We could write this out in terms of the behavior on vectors, but that would require instating some specific notation for \(\Delta(v)\). The usual route is to use Sweedler notation, writing \(\Delta(v) = \sum_i v_{(i1)} \otimes v_{(i2)}\) and then abbreviating the sum as \(v_{(1)} \otimes v_{(2)}\), but I'm going to hold off on using that for now.

We can also talk about commutativity for algebras, using a swapping map \(\sigma_{AA}\) that sends \(u \otimes v\) to \(v\otimes u\). Commutativity looks like
$$m_{ab}^c = m_{ba}^c$$ $$m = m \circ \sigma_{AA}$$ Flipping this around we get the rule for cocommutativity
$$\Delta_i^{jk} = \Delta_i^{kj}$$ $$\Delta = \sigma_{CC} \circ \Delta$$
Finally we can talk about the counit. The unit map \(\eta\) goes from \(\mb k\) to \(A\), so the counit map \(\epsilon\) goes from \(C\) to the dual of \(\mb k\), which is just \(\mb k\) again.
The law for units in tensor indices is
$$m_{ij}^k \eta^j = m_{ji}^k \eta^j = I_i^k$$ So the counit looks like
$$\Delta_i^{jk}\epsilon_j = \Delta_i^{kj} \epsilon_j = I_i^k$$ In terms of maps, for the unit we write
$$m \circ (\eta \otimes id) = lm$$ $$m \circ (id \otimes \eta) = rm$$ where \(lm\) is the scalar multiplication map \(c \otimes v \mapsto cv\), and \(rm\) is scalar multiplication with the scalar on the right.
So we also need to dualize \(lm\) and \(rm\). We define \(ld\) to be the map that sends \(v\) in \(C\) to \(1 \otimes v\) in \(\mathbb{k}\otimes C\), and similarly \(rd\) to be the map that sends \(v\) to \(v\otimes 1\). Then our counit map acts as
$$(\epsilon \otimes id) \circ \Delta = ld$$ $$(id \otimes \epsilon) \circ \Delta = rd$$
So now we can talk about coassociative, cocommutative, counital coalgebras.

A field \(\mb k\) is a coalgebra as a 1-dimensional vector space over itself, with the comultiplication:
$$\Delta(c) = c \otimes 1 = 1 \otimes c$$ and the counit being the identity. Compare with the structure of \(\mb k\) as an algebra.

The coalgebra \(\mb C^\vee\) as a 2-dimensional vector space over \(\mb R\) has as a basis \(x\) and \(y\), where \(x\) returns the real part of a complex number and \(y\) the imaginary part. Note that \(x\) and \(y\) are linear functions.
The comultiplication in \(\mb C^\vee\) comes from the multiplication in \(\mb C\), i.e. the following rule: \((a + bi)(c + di) = (ac - bd) + (ad + bc)i\). So our coalgebra looks like:
$$\Delta(x) = x \otimes x - y \otimes y, \Delta(y) = x \otimes y + y \otimes x$$ This is the general idea for comultiplication in the dual of a finite-dimensional algebra: if we have \(A\) and \(A^\vee\), then for \(a\) and \(b\) in \(A\) and \(f\) in \(A^\vee\), we have
$$f(m(a\otimes b)) = \Delta(f)(a \otimes b)$$ Since \(\mb C\) has a unit, sending the real number \(a\) to the complex number \(a + 0i\), \(\mathbb{C}\) has a counit, which is just evaluating a function at the number \(1 + 0i\). So \(\epsilon(x) = 1, \epsilon(y) = 0\). This is the general idea for counits in the dual of a finite-dimensional algebra: evaluate at the multiplicative identity.
Note that \(\Delta\) in this case is coassociative, cocommutative, and counital; this is the general case for duals; if the algebra is ___, then the dual coalgebra is co___, and vice-versa.

Consider the set of \(n \times n\) matrices, \(M_n = Mat_n(\mathbb{k})\). It's a unital, associative algebra. The dual space \(M_n^\vee\) is an \(n^2\)-dimensional counital, coassociative coalgebra. If we write the basis of \(M_n\) as \(e_i^j\), then we get a basis of \(M_n^\vee\) as \(f_p^q\), where \(f_p^q(e_i^j)\) is \(1\) if \(p = l\) and \(q = j\) and is \(0\) if either of those conditions doesn't hold.
Multiplication in \(M_n\) is that \(e_i^j e_k^l\) equals \(e_i^l\) if \(j = k\) and is \(0\) otherwise. So comultiplication in \(M_n^\vee\) is
$$\Delta(f_p^q) = f_r^q \otimes f_p^r$$ Again, since \(M_n\) is unital, \(M_n^\vee\) has a counit which sends \(f_p^q\) to 1 if \(p = q\) and to 0 otherwise, i.e. evaluating on the identity matrix.

One example that will be important for us is the group algebra, \(\mb kG\). The comultiplication sends the basis elements \(\hat g\) to \(\hat g \otimes \hat g\) and the counit sends \(\hat g\) to 1. We can even discuss a comultiplication for \(G\), which sends \(g\) to \(g \otimes g\). This is not a coalgebra structure because \(G\) does not have a vector-space structure that this respects, but it is still a structure on \(G\). The counit just sends everything in \(G\) to 1. This comultiplication indicates why we use the symbol \(\Delta\): as a stand-in for the diagonal map \(g \rar (g, g)\). In particular, if \(G\) is the real line then the image of \(\Delta\) in \(\mb R^2\) would be a diagonal line.

You can build similar setups for the other finite-dimensional algebras we mentioned. For infinite dimensional algebras we have a bit of an issue taking duals, from a number of standpoints. The duals of infinite dimensional vector spaces are themselves a bit tricky to deal with, because they're generally "bigger" than the original vector spaces. Also, dualizing the multiplication in an infinite dimensional algebra to get a comultiplication in the dual often leads to needing to take infinite sums of tensor products, which is generally a bad thing if you don't want to keep careful track of convergence issues. There are infinite-dimensional coalgebras, but duality takes a little bit of doing.
In particular, we remove the "dual space" aspect and simply ask for a duality map: an algebra \(A\) is dual to a coalgebra \(C\) if there is a map \(ev\) from \(C \otimes A\) to \(\mb k\) such that, viewed as maps from \(C \otimes A \otimes A\) to \(\mb k\), the following holds:
$$ev \circ (id \otimes m) = (ev \otimes ev) \circ (id \otimes \sigma_{CA} \otimes id)\circ (\Delta \otimes id \otimes id)$$ where we use \(\sigma_{CA}\) to move an element of \(a\) to the left of an element of \(C\).
In terms of individual elements \(f\) in \(C\) and \(a\) and \(b\) in \(A\), we have
$$(ev(f \otimes m(a \otimes b)) = (ev(f_{(1)} \otimes a))(ev(f_{(2)} \otimes b))$$ where we're using the Sweedler notation for the comultiplication.