The Unapologetic Mathematician

Mathematics for the interested outsider

Galois Connections

I want to mention a topic I thought I’d hit back when we talked about adjoint functors. We know that every poset is a category, with the elements as objects and a single arrow from a to b if a\leq b. Functors between such categories are monotone functions, preserving the order. Contravariant functors are so-called “antitone” functions, which reverse the order, but the same abstract nonsense as usual tells us this is just a monotone function to the “opposite” poset with the order reversed.

So let’s consider an adjoint pair F\dashv G of such functors. This means there is a natural isomorphism between \hom(F(a),b) and \hom(a,G(b)). But each of these hom-sets is either empty (if a\not\leq b) or a singleton (if a\leq b). So the adjunction between F and G means that F(a)\leq b if and only if a\leq G(b). The analogous condition for an antitone adjoint pair is that b\leq F(a) if and only if a\leq G(b).

There are some immediate consequences to having a Galois connection, which are connected to properties of adjoints. First off, we know that a\leq G(F(a)) and F(G(b))\leq b. This essentially expresses the unit and counit of the adjunction. For the antitone version, let’s show the analogous statement more directly: we know that F(a)\leq F(a), so the adjoint condition says that a\leq G(F(a)). Similarly, b\leq F(G(b)). This second condition is backwards because we’re reversing the order on one of the posets.

Using the unit and the counit of an adjunction, we found a certain quasi-inverse relation between some natural transformations on functors. For our purposes, we observe that since a\leq G(F(a)) we have the special case G(b)\leq G(F(G(b))). But F(G(b))\leq b, and G preserves the order. Thus G(F(G(b)))\leq G(b). So G(b)=G(F(G(b))). Similarly, we find that F(G(F(a)))=F(a), which holds for both monotone and antitone Galois connections.

Chasing special cases further, we find that G(F(G(F(a))))=G(F(a)), and that F(G(F(G(b))))=F(G(b)) for either kind of Galois connection. That is, F\circ G and G\circ F are idempotent functions. In general categories, the composition of two adjoint functors gives a monad, and this idempotence is just the analogue in our particular categories. In particular, these functions behave like closure operators, but for the fact that general posets don’t have joins or bottom elements to preserve in the third and fourth Kuratowski axioms.

And so elements left fixed by G\circ F (or F\circ G) are called “closed” elements of the poset. The images of F and G consist of such closed elements

May 18, 2009 Posted by John Armstrong | Algebra, Category theory, Lattices | | 5 Comments

The Category of Representations of a Group

Sorry for missing yesterday. I had this written up but completely forgot to post it while getting prepared for next week’s trip back to a city. Speaking of which, I’ll be heading off for the week, and I’ll just give things here a rest until the beginning of December. Except for the Samples, and maybe an I Made It or so…

Okay, let’s say we have a group G. This gives us a cocommutative Hopf algebra. Thus the category of representations of G is monoidal — symmetric, even — and has duals. Let’s consider these structures a bit more closely.

We start with two representations \rho:G\rightarrow\mathrm{GL}(V) and \sigma:G\rightarrow\mathrm{GL}(W). We use the comultiplication on \mathbb{F}[G] to give us an action on the tensor product V\otimes W. Specifically, we find

\begin{aligned}\left[\left[\rho\otimes\sigma\right](g)\right](v\otimes w)=\left[\rho(g)\otimes\sigma(g)\right](v\otimes w)\\=\left[\rho(g)\right](v)\otimes\left[\sigma(g)\right](w)\end{aligned}

That is, we make two copies of the group element g, use \rho to act on the first tensorand, and use \sigma to act on the second tensorand. If \rho and \sigma came from actions of G on sets, then this is just what you’d expect from linearizing the product of the G-actions.

Symmetry is straightforward. We just use the twist on the underlying vector spaces, and it’s automatically an intertwiner of the actions, so it defines a morphism between the representations.

Duals, though, take a bit of work. Remember that the antipode of \mathbb{F}[G] sends group elements to their inverses. So if we start with a representation \rho:G\rightarrow\mathrm{GL}(V) we calculate its dual representation on V^*:

\begin{aligned}\left[\rho^*(g)\right](\lambda)=\left[\rho(g^{-1})^*\right](\lambda)\\=\lambda\circ\rho(g^{-1})\end{aligned}

Composing linear maps from the right reverses the order of multiplication from that in the group, but taking the inverse of g reverses it again, and so we have a proper action again.

November 21, 2008 Posted by John Armstrong | Algebra, Category theory, Group theory, Representation Theory | | 1 Comment

Cocommutativity

One things I don’t think I’ve mentioned is that the category of vector spaces over a field \mathbb{F} is symmetric. Indeed, given vector spaces V and W we can define the “twist” map \tau_{V,W}:V\otimes W\rightarrow W\otimes V by setting \tau_{V,W}(v\otimes w)=w\otimes v and extending linearly.

Now we know that an algebra A is commutative if we can swap the inputs to the multiplication and get the same answer. That is, if m(a,b)=m(b,a)=m\left(\tau_{A,A}(a,b)\right). Or, more succinctly: m=m\circ\tau_{A,A}.

Reflecting this concept, we say that a coalgebra C is cocommutative if we can swap the outputs from the comultiplication. That is, if \tau_{C,C}\circ\Delta=\Delta. Similarly, bialgebras and Hopf algebras can be cocommutative.

The group algebra \mathbb{F}[G] of a group G is a cocommutative Hopf algebra. Indeed, since \Delta(e_g)=e_g\otimes e_g, we can twist this either way and get the same answer.

So what does cocommutativity buy us? It turns out that the category of representations of a cocommutative bialgebra B is not only monoidal, but it’s also symmetric! Indeed, given representations \rho:B\rightarrow\hom_\mathbb{F}(V,V) and \sigma:B\rightarrow\hom_\mathbb{F}(W,W), we have the tensor product representations \rho\otimes\sigma on V\otimes W, and \sigma\otimes\rho on W\otimes V. To twist them we define the natural transformation \tau_{\rho,\sigma} to be the twist of the vector spaces: \tau_{V,W}.

We just need to verify that this actually intertwines the two representations. If we act first and then twist we find

\begin{aligned}\tau_{V,W}\left(\left[\left[\rho\otimes\sigma\right](a)\right](v\otimes w)\right)=\tau_{V,W}\left(\left[\rho\left(a_{(1)}\right)\otimes\sigma\left(a_{(2)}\right)\right](v\otimes w)\right)\\=\tau_{V,W}\left(\left[\rho\left(a_{(1)}\right)\right](v)\otimes\left[\sigma\left(a_{(2)}\right)\right](w)\right)\\=\left[\sigma\left(a_{(2)}\right)\right](w)\otimes\left[\rho\left(a_{(1)}\right)\right](v)\end{aligned}

On the other hand, if we twist first and then act we find

\begin{aligned}\left[\left[\sigma\otimes\rho\right](a)\right]\left(\tau_{V,W}(v\otimes w)\right)=\left[\sigma\left(a_{(1)}\right)\otimes\rho\left(a_{(2)}\right)\right]\left(w\otimes v\right)\\=\left[\sigma\left(a_{(1)}\right)\right](w)\otimes\left[\rho\left(a_{(2)}\right)\right](v)\end{aligned}

It seems there’s a problem. In general this doesn’t work. Ah! but we haven’t used cocommutativity yet! Now we write

a_{(1)}\otimes a_{(2)}=\Delta(a)=\tau_{B,B}\left(\Delta(a)\right)=\tau_{B,B}\left(a_{(1)}\otimes a_{(2)}\right)=a_{(2)}\otimes a_{(1)}

Again, remember that this doesn’t mean that the two tensorands are always equal, but only that the results after (implicitly) summing up are equal. Anyhow, that’s enough for us. It shows that the twist on the underlying vector spaces actually does intertwine the two representations, as we wanted. Thus the category of representations is symmetric.

November 19, 2008 Posted by John Armstrong | Algebra, Category theory, Representation Theory | | 1 Comment

The Category of Representations of a Hopf Algebra

It took us two posts, but we showed that the category of representations of a Hopf algebra H has duals. This is on top of our earlier result that the category of representations of any bialgebra B is monoidal. Let’s look at this a little more conceptually.

Earlier, we said that a bialgebra is a comonoid object in the category of algebras over \mathbb{F}. But let’s consider this category itself. We also said that an algebra is a category enriched over \mathbb{F}, but with only one object. So we should really be thinking about the category of algebras as a full sub-2-category of the 2-category of categories enriched over \mathbb{F}.

So what’s a comonoid object in this 2-category? When we defined comonoid objects we used a model category \mathrm{Th}(\mathbf{CoMon}). Now let’s augment it to a 2-category in the easiest way possible: just add an identity 2-morphism to every morphism!

But the 2-category language gives us a bit more flexibility. Instead of demanding that the morphism \Delta:C\rightarrow C\otimes C satisfy the associative law on the nose, we can add a “coassociator” 2-morphism \gamma:(\Delta\otimes1)\circ\Delta\rightarrow(1\otimes\Delta)\circ\Delta to our model 2-category. Similarly, we dispense with the left and right counit laws and add left and right counit 2-morphisms. Then we insist that these 2-morphisms satisfy pentagon and triangle identities dual to those we defined when we talked about monoidal categories.

What we’ve built up here is a model 2-category for weak comonoid objects in a 2-category. Then any weak comonoid object is given by a 2-functor from this 2-category to the appropriate target 2-category. Similarly we can define a weak monoid object as a 2-functor from the opposite model 2-category to an appropriate target 2-category.

So, getting a little closer to Earth, we have in hand a comonoid object in the 2-category of categories enriched over \mathbb{F} — our algebra B. But remember that a 2-category is just a category enriched over categories. That is, between H (considered as a category) and \mathbf{Vect}(\mathbb{F}) we have a hom-category \hom(H,\mathbf{Vect}(\mathbb{F})). The entry in the first slot H is described by a 2-functor from the model category of weak comonoid objects to the 2-category of categories enriched over \mathbb{F}. This hom-functor is contravariant in the first slot (like all hom-functors), and so the result is described by a 2-functor from the opposite of our model 2-category. That is, it’s a weak monoid object in the 2-category of all categories. And this is just a monoidal category!

This is yet another example of the way that hom objects inherit structure from their second variable, and inherit opposite structure from their first variable. I’ll leave it to you to verify that a monoidal category with duals is similarly a weak group object in the 2-category of categories, and that this is why a Hopf algebra — a (weak) cogroup object in the 2-category of categories enriched over \mathbb{F} has dual representations.

November 18, 2008 Posted by John Armstrong | Algebra, Category theory, Representation Theory | | No Comments Yet

Representations of Hopf Algebras II

Now that we have a coevaluation for vector spaces, let’s make sure that it intertwines the actions of a Hopf algebra. Then we can finish showing that the category of representations of a Hopf algebra has duals.

Take a representation \rho:H\rightarrow\hom_\mathbb{F}(V,V), and pick a basis \left\{e_i\right\} of V and the dual basis \left\{\epsilon^i\right\} of V^*. We define the map \eta_\rho:\mathbf{1}\rightarrow V^*\otimes V by \eta_\rho(1)=\epsilon^i\otimes e_i. Now \left[\rho(a)\right](1)=\epsilon(a), so if we use the action of H on \mathbf{1} before transferring to V^*\otimes V, we get \epsilon(a)\epsilon^i\otimes e_i. Be careful not to confuse the counit \epsilon with the basis elements \epsilon^i.

On the other hand, if we transfer first, we must calculate

\begin{aligned}\left[\left[\rho^*\otimes\rho\right](a)\right](\epsilon^i\otimes e_i)=\left[\rho^*\left(a_{(1)}\right)\otimes\rho\left(a_{(2)}\right)\right](\epsilon^i\otimes e_i)\\=\left[\rho\left(S\left(a_{(1)}\right)\right)^*\otimes\rho\left(a_{(2)}\right)\right](\epsilon^i\otimes e_i)\\=\left[\rho\left(S\left(a_{(1)}\right)\right)^*\right](\epsilon^i)\otimes\left[\rho\left(a_{(2)}\right)\right](e_i)\end{aligned}

Now let’s use the fact that we’ve got this basis sitting around to expand out both \rho\left(S\left(a_{(1)}\right)\right) and \rho\left(a_{(2)}\right) as matrices. We’ll just take on matrix indices on the right for our notation. Then we continue the calculation above:

\begin{aligned}\left[\rho\left(S\left(a_{(1)}\right)\right)^*\right](\epsilon^i)\otimes\left[\rho\left(a_{(2)}\right)\right](e_i)=\epsilon^j\rho\left(S\left(a_{(1)}\right)\right)_j^i\otimes\rho\left(a_{(2)}\right)_i^ke_k\\=\epsilon^j\otimes\rho\left(S\left(a_{(1)}\right)\right)_j^i\rho\left(a_{(2)}\right)_i^ke_k\\=\epsilon^j\otimes\left[\rho\left(\mu\left(\left[S\otimes1_H\right]\left(\Delta(a)\right)\right)\right)\right](e_j)\\=\epsilon^j\otimes\left[\rho\left(\iota\left(\epsilon(a)\right)\right)\right](e_j)=\epsilon^j\otimes\epsilon(a)e_j\end{aligned}

And so the coevaluation map does indeed intertwine the two actions of H. Together with the evaluation map, it provides the duality on the category of representations of a Hopf algebra H that we were looking for.

November 14, 2008 Posted by John Armstrong | Algebra, Category theory, Representation Theory | | 3 Comments

Representations of Hopf Algebras I

We’ve seen that the category of representations of a bialgebra is monoidal. What do we get for Hopf algebras? What does an antipode buy us? Duals! At least when we restrict to finite-dimensional representations.

Again, we base things on the underlying category of vector spaces. Given a representation \rho:H\rightarrow\hom_\mathbb{F}(V,V), we want to find a representation \rho^*:H\rightarrow\hom_\mathbb{F}(V^*,V^*). And it should commute with the natural transformations which make up the dual structure.

Easy enough! We just take the dual of each map to find \rho(h)^*:V^*\rightarrow V^*. But no, this can’t work. Duality reverses the order of composition. We need an antiautomorphism S to reverse the multiplication on H. Then we can define \rho^*(h)=\rho(S(h))^*.

The antiautomorphism we’ll use will be the antipode. Now to make these representations actual duals, we’ll need natural transformations \eta_\rho:\mathbf{1}\rightarrow\rho^*\otimes\rho and \epsilon_\rho:\rho\otimes\rho^*\rightarrow\mathbf{1}. This natural transformation \epsilon is not to be confused with the counit of the Hopf algebra. Given a representation \rho on the finite-dimensional vector space V, we’ll just use the \eta_V and \epsilon_V that come from the duality on the category of finite-dimensional vector spaces.

Thus we find that \epsilon_\rho is the pairing v\otimes\lambda\mapsto\lambda(v). Does this commute with the actions of H? On the one side, we calculate

\begin{aligned}\left[\left[\rho\otimes\rho^*\right](h)\right](v\otimes\lambda)=\left[\rho\left(h_{(1)}\right)\otimes\rho^*\left(h_{(2)}\right)\right](v\otimes\lambda)\\=\left[\rho\left(h_{(1)}\right)\otimes\rho\left(S\left((h_{(2)}\right)\right)^*\right](v\otimes\lambda)\\=\left[\rho\left(h_{(1)}\right)\right](v)\otimes\left[\rho\left(S\left(h_{(2)}\right)\right)^*\right](\lambda)\end{aligned}

Then we apply the evaluation to find

\begin{aligned}\left[\left[\rho\left(S\left(h_{(2)}\right)\right)^*\right](\lambda)\right]\left(\left[\rho\left(h_{(1)}\right)\right](v)\right)=\lambda\left(\left[\rho\left(S\left(h_{(2)}\right)\right)\right]\left(\left[\rho\left(h_{(1)}\right)\right](v)\right)\right)\\=\lambda\left(\left[\rho\left(h_{(1)}S\left(h_{(2)}\right)\right)\right](v)\right)=\lambda\left(\left[\rho\left(\mu\left(h_{(1)}\otimes S\left(h_{(2)}\right)\right)\right)\right](v)\right)\\=\lambda\left(\left[\rho\left(\iota\left(\epsilon(h)\right)\right)\right](v)\right)=\epsilon(h)\lambda(v)\end{aligned}

Which is the same as the result we’d get by applying the “unit” action after evaluating. Notice how we used the definition of the dual map, the fact that \rho is a representation, and the antipodal property in obtaining this result.

This much works whether or not V is a finite-dimensional vector space. The other direction, though, needs more work, especially since I waved my hands at it when I used \mathbf{FinVect} as the motivating example of a category with duals. Tomorrow I’ll define this map.

November 12, 2008 Posted by John Armstrong | Algebra, Category theory, Representation Theory | | 3 Comments

Representations of Bialgebras

What’s so great about bialgebras? Their categories of representations are monoidal!

Let’s say we have two algebra representations \rho:A\rightarrow\hom_\mathbb{F}(V,V) and \sigma:A\rightarrow\hom_\mathbb{F}(W,W). These are morphisms in the category of \mathbb{F}-algebras, and so of course we can take their tensor product \rho\otimes\sigma. But this is not a representation of the same algebra. It’s a representation of the tensor square of the algebra:

\rho\otimes\sigma:A\otimes A\rightarrow\hom_\mathbb{F}(V,V)\otimes\hom_\mathbb{F}(W,W)\cong\hom_\mathbb{F}(V\otimes W,V\otimes W)

Ah, but if we have a way to send A to A\otimes A (an algebra homomorphism, that is), then we can compose it with this tensor product to get a representation of A. And that’s exactly what the comultiplication \Delta does for us. We abuse notation slightly and write:

\rho\otimes\sigma:A\rightarrow\hom_\mathbb{F}(V\otimes W,V\otimes W)

where the homomorphism of this representation is the comultiplication \Delta followed by the tensor product of the two homomorphisms, followed by the equivalence of \hom algebras.

Notice here that the underlying vector space of the tensor product of two representations \rho\otimes\sigma is the tensor product of their underlying vector spaces V\otimes W. That is, if we think (as many approaches to representation theory do) of the vector space as fundamental and the homomorphism as extra structure, then this is saying we can put the structure of a representation on the tensor product of the vector spaces.

Which leads us to the next consideration. For the tensor product to be a monoidal structure we need an associator. And the underlying linear map on vector spaces must clearly be the old associator for \mathbf{Vect}(\mathbb{F}). We just need to verify that it commutes with the action of A.

So let’s consider three representations \rho:A\rightarrow\hom_\mathbb{F}(U,U), \sigma\rightarrow\hom_\mathbb{F}(V,V), and \tau:A\rightarrow\hom_\mathbb{F}(W,W). Given an algebra element a and vectors u, v, and w, we have the action

\begin{aligned}\left[\left[(\rho\otimes\sigma)\otimes\tau\right](a)\right]((u\otimes v)\otimes w)=\\\left(\left[\rho\left(\left(a_{(1)}\right)_{(1)}\right)\right](u)\otimes\left[\sigma\left(\left(a_{(1)}\right)_{(2)}\right)\right](v)\right)\otimes\left[\tau\left(a_{(2)}\right)\right](v)\end{aligned}

On the other hand, if we associate the other way we have the action

\begin{aligned}\left[\left[\rho\otimes(\sigma\otimes\tau)\right](a)\right](u\otimes(v\otimes w))=\\\left[\rho\left(a_{(1)}\right)\right](u)\otimes\left(\left[\sigma\left(\left(a_{(2)}\right)_{(1)}\right)\right](v)\otimes\left[\tau\left(\left(a_{(2)}\right)_{(2)}\right)\right](v)\right)\end{aligned}

Where we have used the Sweedler notation to write out the comultiplications of a. But now we can use the coassociativity of the comultiplication — along with the fact that, as algebra homomorphisms, the representations are linear maps — to show that the associator on \mathbf{Vect}(\mathbb{F}) intertwines these actions, and thus acts as an associator for the category of representations of A as well.

We also need a unit object, and similar considerations to those above tell us it should be based on the vector space unit object. That is, we need a homomorphism A\rightarrow\hom_\mathbb{F}(\mathbb{F},\mathbb{F}). But linear maps from the base field to itself (considered as a one-dimensional vector space) are just multiplications by field elements! That is, the \hom algebra is just the field \mathbb{F} itself, and we need a homomorphism A\rightarrow\mathbb{F}. This is precisely what the counit \epsilon provides! I’ll leave it to you to verify that the left and right unit maps from vector spaces intertwine the relevant representations.

November 11, 2008 Posted by John Armstrong | Algebra, Category theory, Representation Theory | | 5 Comments

Bialgebras

In yesterday’s post I used the group algebra \mathbb{F}[G] of a group G as an example of a coalgebra. In fact, more is true.

A bialgebra is a vector space A equipped with both the structure of an algebra and the structure of a coalgebra, and that these two structures are “compatible” in a certain sense. The traditional definitions usually consist in laying out the algebra maps and relations, then the coalgebra maps and relations. Then they state that the algebra structure preserves the coalgebra structure, and that the coalgebra structure preserves the algebra structure, and they note that really you only need to require one of these last two conditions because they turn out to be equivalent.

In fact, our perspective allows this equivalence to come to the fore. The algebra structure makes the bialgebra a monoid object in the category of vector space over \mathbb{F}. Then a compatible coalgebra structure makes it a comonoid object in the category of algebras over \mathbb{F}. Or in the other order, we have a monoid object in the category of comonoid objects in the category of vector spaces over \mathbb{F}. And these describe essentially the same things because internalizations commute!

Okay, let’s be explicit about what we mean by “compatibility”. This just means that — on the one side — the coalgebra maps are not just linear maps between the underlying vector spaces, but actually are algebra homomorphisms. On the other side, it means that the algebra maps are actually coalgebra homomorphisms.

Multiplication and comultiplication being compatible actually mean the same thing. Take two algebra elements and multiply them, then comultiply the result. Alternatively, comultiply each of them, and the multiply corresponding factors of the result. We should get the same answer whether we multiply or comultiply first. That is: \Delta\circ\mu=(\mu\otimes\mu)\circ(1_A\otimes\tau_{A,A}\otimes1_A)\circ(\Delta\otimes\Delta), where \tau is the twist map, exchanging two factors.

Let’s check this condition for the group algebra \mathbb{F}[G]:

\begin{aligned}\left[\mu\otimes\mu\right]\left(\left[1_A\otimes\tau_{A,A}\otimes1_A\right]\left(\left[\Delta\otimes\Delta\right](e_g\otimes e_h)\right)\right)=\\\left[\mu\otimes\mu\right]\left(\left[1_A\otimes\tau_{A,A}\otimes1_A\right](e_g\otimes e_g\otimes e_h\otimes e_h)\right)=\\\left[\mu\otimes\mu\right](e_g\otimes e_h\otimes e_g\otimes e_h)=e_{gh}\otimes e_{gh}=\\\Delta(e_{gh})=\Delta\left(\mu(e_g\otimes e_h)\right)\end{aligned}

Similarly, if we multiply two algebra elements and then take the counit, it should be the same as the product (in \mathbb{F}) of the counits of the elements. Dually, the product of two copies of the algebra unit should be the algebra unit again, and the counit of the algebra unit should be the unit in \mathbb{F}. It’s straightforward to verify that these hold for \mathbb{F}[G].

November 6, 2008 Posted by John Armstrong | Algebra, Category theory | | 4 Comments

Coalgebras

Okay, back to business. We’re about to need a little more algebraic structure floating around. This is something that’s always present, but many approaches don’t explicitly mention it until much later. Since I’m taking a categorical view of things, it’s easier to show what’s really going on right away.

Remember that an \mathbb{F}-algebra is a monoid object in the category of vector spaces over \mathbb{F}. Dually, an \mathbb{F}-coalgebra is a comonoid object in the category of vector spaces over \mathbb{F}. That’s all well and good, but what’s a comonoid object? We’ve mentioned them before, but let’s be more explicit this time around.

Remember that a monoid object was a functor from a certain category we cooked up to mirror the axioms of a monoid. We gave the category objects M^{\otimes n} corresponding to the natural numbers, corresponding to lists of monoid elements. We have a map \mu:M\otimes M\rightarrow M corresponding to multiplication, and a map \iota:\mathbf{1}\rightarrow M picking out the unit in the monoid.

So a comonoid object will be a functor from the dual of this category! That is, we’ve still got all the same objects, but now we have a “comultiplication” arrow \Delta:C\rightarrow C\otimes C, and a “counit” arrow \epsilon:C\rightarrow\mathbf{1}.

Now, the model category describing monoid objects isn’t just objects and arrows. We also have the relations that make a monoid a monoid: the associative law \mu\circ(\mu\otimes1_M)=\mu\circ(1_m\otimes\mu), and the left and right unit laws \mu\circ(1_M\otimes\iota)=1_M=\mu\circ(\iota\otimes1_M).

Dually, we must have dual relations for comonoid objects. We have a coassociative law (\Delta\otimes1_M)\circ\Delta=(1_M\otimes\Delta)\circ\Delta, and left and right counit laws (1_M\otimes\epsilon)\circ\Delta=1_M=(\epsilon\otimes1_M)\circ\Delta.

We could write these down in terms of commuting diagrams, but it’s even more instructive to look at “string diagrams” like we did before. This makes the sense of what’s going on all the clearer.

So a coalgebra is a comonoid object in the category of vector spaces over \mathbb{F}. That is, it’s an \mathbb{F} vector space C, equipped with a linear comultiplication \Delta and a linear counit \epsilon, which satisfy the coassociative and counit laws. I’ll admit that this seems an extremely quirky structure to discuss, so an example is in order. The one we care most about right now is the group algebra. Yes, it turns out to also be a coalgebra!

To really wrap our heads around it, let’s start with a finite group G. Then we get a finite-dimensional vector space \mathbb{F}[G], with a basis e_g indexed by elements of G. Let’s forget, for the moment, that we have a multiplication and a unit. Instead, we define the comultiplication by \Delta(e_g)=e_g\otimes e_g for each basis element. We also define the counit by \epsilon(e_g)=1 for each element g\in G. Both of these maps extend by linearity.

Now, let’s check the coassociative property. It suffices to check it on basis elements, because the extensions by linearity have to agree. In this case we have

\begin{aligned}\left[\Delta\otimes1_{\mathbb{F}[G]}\right]\left(\Delta(e_g)\right)=\left[\Delta\otimes1_{\mathbb{F}[G]}\right](e_g\otimes e_g)=e_g\otimes e_g\otimes e_g\\=\left[1_{\mathbb{F}[G]}\otimes\Delta\right](e_g\otimes e_g)=\left[1_{\mathbb{F}[G]}\otimes\Delta\right]\left(\Delta(e_g)\right)\end{aligned}

Similarly, we can check the right counit law:

\begin{aligned}\left[1_{\mathbb{F}[G]}\otimes\epsilon\right]\left(\Delta(e_g)\right)=\left[1_{\mathbb{F}[G]}\otimes\epsilon\right](e_g\otimes e_g)=e_g\\=\left[\epsilon\otimes1_{\mathbb{F}[G]}\right](e_g\otimes e_g)=\left[\epsilon\otimes1_{\mathbb{F}[G]}\right]\left(\Delta(e_g)\right)\end{aligned}

and the left counit law is similar. Thus these maps do indeed describe the structure of a coalgebra.

November 5, 2008 Posted by John Armstrong | Algebra, Category theory | | 3 Comments

The Category of Representations

Now let’s narrow back in to representations of algebras, and the special case of representations of groups, but with an eye to the categorical interpretation. So, representations are functors. And this immediately leads us to the category of such functors. The objects, recall, are functors, while the morphisms are natural transformations. Now let’s consider what, exactly, a natural transformation consists of in this case.

Let’s say we have representations \rho:A:\rightarrow\hom_\mathbb{F}(V,V) and \sigma:A\rightarrow\hom_\mathbb{F}(W,W). That is, we have functors \rho and \sigma with \rho(*)=V, \sigma(*)=W — where * is the single object of A, when it’s considered as a category — and the given actions on morphisms. We want to consider a natural transformation \phi:\rho\rightarrow\sigma.

Such a natural transformation consists of a list of morphisms indexed by the objects of the category A. But A has only one object: *. Thus we only have one morphism, \phi_*, which we will just call \phi.

Now we must impose the naturality condition. For each arrow a:*\rightarrow * in A we ask that the diagram

\displaystyle\begin{matrix}V&\xrightarrow{\phi}&W\\\downarrow^{\rho(a)}&&\downarrow^{\sigma(a)}\\V&\xrightarrow{\phi}&W\end{matrix}

commute. That is, we want \phi\circ\rho(a)=\sigma(a)\circ\phi for every algebra element a. We call such a transformation an “intertwiner” of the representations. These intertwiners are the morphisms in the category of \mathbf{Rep}(A) of representations of A. If we want to be more particular about the base field, we might also write \mathbf{Rep}_\mathbb{F}(A).

Here’s another way of putting it. Think of \phi as a “translation” from V to W. If \phi is an isomorphism of vector spaces, for instance, it could be a change of basis. We want to take a transformation from the algebra A and apply it, and we also want to translate. We could first apply the transformation in V, using the representation \rho, and then translate to W. Or we could first translate from V to W and then apply the transformation, now using the representation \sigma. Our condition is that either order gives the same result, no matter which element of A we’re considering.

October 28, 2008 Posted by John Armstrong | Category theory, Group theory, Representation Theory, Ring theory | | 7 Comments

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