# The Unapologetic Mathematician

## Sweedler notation

As we work with coalgebras, we’ll need a nice way to write out the comultiplication of an element. In the group algebra we’ve been using as an example, we just have $\Delta(e_g)=e_g\otimes e_g$, but not all elements are so cleanly sent to two copies of themselves. And other comltiplications in other coalgebras aren’t even defined so nicely on any basis. So we introduce the so-called “Sweedler notation”. If you didn’t like the summation convention, you’re going to hate this.

Okay, first of all, we know that the comultiplication of an element $c\in C$ is an element of the tensor square $C\otimes C$. Thus it can be written as a finite sum

$\displaystyle\Delta(c)=\sum\limits_{i=1}^n(c)a_i\otimes b_i$

Now, this uses two whole new letters, $a$ and $b$, which might be really awkward to come up with in practice. Instead, let’s call them $c_{(1)}$ and $c_{(2)}$, to denote the first and second factors of the comultiplication. We’ll also move the indices to superscripts, just to get them out of the way.

$\displaystyle\Delta(c)=\sum\limits_{i=1}^n(c)c_{(1)}^i\otimes c_{(2)}^i$

The whole index-summing thing is a bit awkward, especially because the number of summands is different for each coalgebra element $c$. Let’s just say we’re adding up all the terms we need to for a given $c$:

$\displaystyle\Delta(c)=\sum\limits_{(c)}c_{(1)}\otimes c_{(2)}$

Then if we’re really pressed for space we can just write $\Delta(c)=c_{(1)}\otimes c_{(2)}$. Since we don’t use a subscript in parentheses for anything else, we remember that this is implicitly a summation.

Let’s check out the counit laws $(1_M\otimes\epsilon)\circ\Delta=1_M=(\epsilon\otimes1_M)\circ\Delta$ in this notation. Now they read $c_{(1)}\epsilon(c_{(2)}=c=\epsilon(c_{(1)})c_{(2)}$. Or, more expansively:

$\displaystyle\sum\limits_{(c)}c_{(1)}\epsilon\left(c_{(2)}\right)=c=\sum\limits_{(c)}\epsilon\left(c_{(1)}\right)c_{(2)}$

Similarly, the coassociativity condition now reads

$\displaystyle\sum\limits_{(c)}\left(\sum\limits_{\left(c_{(1)}\right)}\left(c_{(1)}\right)_{(1)}\otimes\left(c_{(1)}\right)_{(2)}\right)\otimes c_{(2)}=\sum\limits_{(c)}c_{(1)}\otimes\left(\sum\limits_{\left(c_{(2)}\right)}\left(c_{(2)}\right)_{(1)}\otimes\left(c_{(2)}\right)_{(1)}\right)$

In the Sweedler notation we’ll write both of these equal sums as

$\displaystyle\sum\limits_{(c)}c_{(1)}\otimes c_{(2)}\otimes c_{(3)}$

Or more simply as $c_{(1)}\otimes c_{(2)}\otimes c_{(3)}$.

As a bit more practice, let’s write out the condition that a linear map $f:C\rightarrow D$ between coalgebras is a coalgebra morphism. The answer is that $f$ must satisfy

$f\left(c_{(1)}\right)\otimes f\left(c_{(2)}\right)=f(c)_{(1)}\otimes f(c)_{(2)}$

Notice here that there are implied summations here. We are not asserting that all the summands are equal, and definitely not that $f\left(c_{(1)}\right)=f(c)_{(1)}$ (for instance). Sweedler notation hides a lot more than the summation convention ever did, but it’s still possible to expand it back out to a proper summation-heavy format when we need to.

November 10, 2008 Posted by | Algebra | 7 Comments