# The Unapologetic Mathematician

## Group objects

Just like we have monoid objects, we can construct a category called $\mathrm{Th}(\mathbf{Grp})$, which encodes the notion of a “group object”.

Groups are a lot like monoids, but we’ll need to be able to do a few more things to describe groups than we needed for monoids. So let’s start with all the same setup as for monoid objects, but let’s make the monoidal structure in our toy category be an actual cartesian structure. That is, we start with an object $G$ and we build up all the “powers”, but now we insist that they be built from categorical products rather than just some arbitrary monoidal structure. Then $G^{\times n}$ is the categorical product of $n$ copies of $G$. Not only does this work like a monoidal structure, but it come equipped with a bunch of extra arrows. For example, there are arrows to “project out” a copy of $G$ from a product, and every object $X$ has a unique arrow $t_X$ from $X$ to the terminal object — the product of zero copies of $G$.

More importantly it has an arrow $\Delta:G\rightarrow G\times G$ called the “diagonal” that is defined as the unique arrow satisfying $\pi_1\circ\Delta=1_G=\pi_2\circ\Delta$. That is, it makes an “identical copy” of $G$. For instance, in the context of sets this is the function $\Delta:S\rightarrow S\times S$ defined by $\Delta(s)=(s,s)$.

Now we do everything like we did for monoid objects. There’s a morphism $m:G\times G\rightarrow G$, and one $e:G^{\otimes0}\rightarrow G$, and these satisfy the identity and associativity relations. Now we also throw in an arrow $i:G\rightarrow G$ satisfying $m\circ(i\times1_G)\circ\Delta=e\circ t_G=m\circ(1_G\times i)\circ\Delta$. That is, we can start with an “element” of $G$ and split it into two copies. Then we can either copy with $i$ and leave the other alone. Then we can multiply together the copies. Either choice of which one to hit with $i$ will give us the exact same result as if we’d just “forgotten” the original element of $G$ by passing to the terminal object, and then created a copy of the identity element with $e$.

Wow, that looks complicated. Well, let’s take a functor from this category to $\mathbf{Set}$ that preserves products. Then what does the equation say in terms of elements of sets? We read off $m(x,i(x))=e=m(i(x),x)$. That is, the product of $x$ and $i(x)$ on either side is just the single element in the image of the arrow described by $e$ — the identity element of the monoid. But this is the condition that $i(x)$ be the inverse of $x$. So we’re just saying that (when we read the condition in sets) every element of our monoid has an inverse, which makes it into a group! Now a group object in any other category $\mathcal{C}$ is a product-preserving functor from $\mathrm{Th}(\mathbf{Grp})$ to $\mathcal{C}$.

We can do even better. Since the monoidal structure in $\mathrm{Th}(\mathbf{Grp})$ is cartesian, it comes with a symmetry. The twist $\tau_{A,B}:A\times B\rightarrow B\times A$ is defined as the unique arrow satisfying $\pi_1\circ\tau_{A,B}=\pi_2$ and $\pi_2\circ\tau_{A,B}=\pi_1$. In sets, this means $\tau_{A,B}(a,b)=(b,a)$. Now we can add the relation $m\circ\tau_{G,G}=m$ to our category. In sets this reads that $m(x,y)=m(y,x)$, which says that the multiplication is commutative. The resulting category is $\mathrm{Th}(\mathbf{Ab})$ — the “theory of abelian groups” — and an “abelian group object” in a category $\mathcal{C}$ is a product-preserving functor in $\mathcal{C}^{\mathrm{Th}(\mathbf{Ab})}$.

August 4, 2007 - Posted by | Category theory

1. So, if I ask you now what a group-object looks like in some of the more familiar categories, you’ll just shush me and tell me to wait till tomorrow – or possibly till Thursday. Right?

Comment by Mikael Johansson | August 4, 2007 | Reply

2. No, but the examples I could reasonably give at this point are all but trivial. If you’re up for it, you can show for yourself that a group object in $k$-vector spaces is a Hopf algebra over $k$. It’ll make a nice little project for the trip on your return home.

Comment by John Armstrong | August 4, 2007 | Reply

3. Ummmm, it is a vector space with a multiplication, a comultiplication, and the Hopf PROP diagram fulfilled. I’m not complete convinced I see where the inversion appears, but the rest is easy enough.

Comment by Mikael Johansson | August 5, 2007 | Reply

4. Having a multiplication and a (compatible) comultiplication makes a bialgebra. Keep looking.

Comment by John Armstrong | August 5, 2007 | Reply

5. Ok.

I realized I was missing something. Opened the relevant Wikipedia page. And discovered that the inversion is called “antipode”, and that the formal definition of a Hopf algebra, according to Wikipedia, is that it is a group object in k-Mod.

With the right definition, this statement is trivial to the point of almost being ridiculous.

Comment by Mikael Johansson | August 5, 2007 | Reply

6. Be careful. A group object in that category is a Hopf algebra, but not all Hopf algebras are so. What hidden relation does the setup of a group object satisfy that might not be satisfied in a general Hopf algebra?

Comment by John Armstrong | August 5, 2007 | Reply

7. [...] sets the stage. It defines a group as a group object in Set, but without the diagonal map. It produces a minimal definition – the left identity and inverse [...]

Pingback by Michi’s blog » Blog Archive » Coq and simple group theory | August 5, 2007 | Reply

8. [...] at Weighted Graphs and the Minimum Spanning Tree (by Mark C. Chu-Carroll of Good Math, Bad Math), Group objects (at The Unapologetic Mathematician—one of an ongoing series on Category Theory), and Questions [...]

Pingback by Carnival of Mathematics XIV « Vlorbik on Math Ed | August 8, 2007 | Reply

9. John Armstrong commented:

“If you’re up for it, you can show for yourself that a group object in k-vector spaces is a Hopf algebra over k.”

Careful! If you’re using cartesian products in k-Vect, then a monoid object (V, m, e) in k-Vect is really nothing more than V itself, i.e., where m: V x V –> V is addition and e: 1 –> V is the zero element. This is a manifestation of the Eckmann-Hilton lemma, which says that a monoid object (M, m, e) in the category of monoids is the same as an abelian monoid (with m the same as the multiplication of the underlying monoid M, and e the identity of the underlying monoid).

Obviously, it doesn’t help just to use (k-Vect, \otimes) in place of (k-Vect, x), because you need to use cartesian products to define group objects, and \otimes isn’t the cartesian product!

To get something in the ballpark of where you’re aiming, consider instead group objects in the category of *cocommutative coalgebras over k* (i.e., cocommutative comonoids in the symmetric monoidal category k-Vect, with tensor product as monoidal product.) Here, the cartesian product of two cocommutative coalgebras is given by the tensor product \otimes on the underlying vector spaces, and all is well again. (NB: I almost forgot to say “cocommutative”, but without that condition the cartesian product isn’t tensor; it’s something more complicated than that.)

Modulo this fix, John was right: not all cocommutative Hopf algebras are group objects in the category of cocommutative coalgebras, since the antipode in the definition of Hopf algebra is not required to be a coalgebra map.

Comment by Todd Trimble | August 19, 2007 | Reply

10. [...] what I did for group objects can be extended to cover relations as well as functions. [...]

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11. Note: as I mention in the new post today, I was wrong in the earlier comments. Todd is right here.

Comment by John Armstrong | November 7, 2008 | Reply

12. [...] thing that we should point out: this is not a group object in the category of vector spaces over . A group object needs the diagonal we get from the finite [...]

Pingback by Hopf Algebras « The Unapologetic Mathematician | November 7, 2008 | Reply

13. [...] and groups in a compatible way. The fancy way to say it is, of course, that a Lie group is a group object in the category of smooth [...]

Pingback by Lie Groups « The Unapologetic Mathematician | June 6, 2011 | Reply