Well, I was out of town for a weekend, and then had a miserable cold that went away but only after sleeping about 4 extra hours per day for a few days. So it’s been a while since I continued the story here.

To recap: I first explained how to turn a span of sets into a linear operator between the free vector spaces on those sets. Then I described the “free” 2-vector space on a groupoid $X$ – namely, the category of functors from $X$ to $\mathbf{Vect}$. So now the problem is to describe how to turn a span of groupoids into a 2-linear map. Here’s a span of groupoids:

Here we have a span $Y \stackrel{s}{\leftarrow} X \stackrel{t}{\rightarrow} Z$, of groupoids. In fact, they’re skeletal groupoids: there’s only one object in each isomorphism class, so they’re completely described, up to isomorphism, by the automorphism groups of each object. The object $y_2 \in Y$, for instance, has automorphism group $H_2$, and the object $x_1 \in X$ has automorphism group $G_1$. This diagram shows the object maps of the “source” and “target” functors $s$ and $t$ explicitly, but note that with each arrow indicated in the diagram, there is a group homomorphism. So, since the object map for $s$ sends $x_1$ to $y_2$, that strand must be labelled with a group homomorphism $s_1 : G_1 \rightarrow H_2$. (We’re leaving these out of the diagram for clarity).

So, we want to know how to transport a $\mathbf{Vect}$-valued functor $F : Y \rightarrow \mathbf{Vect}$ – along this span. We know that such a functor attaches to each $y_i \in Y$ a representation of $H_i$ on some vector space $F(y_i)$. As with spans of sets, the first stage is easy: we have the composable pair of functors $X \stackrel{s}{\longrightarrow} Y \stackrel{F}{\longrightarrow} \mathbf{Vect}$, so “pulling back” $F$ to $X$ gives $s^{\ast}F = F \circ s : X \rightarrow \mathbf{Vect}$.

What about the other leg of the span? Remember back in Part 1 what happened when we pushed down a function (not a functor) along the second leg of a span. To find the value of the pushed-forward function on an element $z$, we took a sum of the complex values on every element of the preimage $t^{-1}(z)$. For vector-space-valued functors, we expect to use a direct sum of some terms. Since we’re dealing with functors, things are a little more complex than before, but there should still be a contribution from each object in the preimage (or, if we’re not talking about skeletal groupoids, the essential preimage) of the object $z$ we look at.

However, we have to deal with the fact that there are morphisms. Instead of adding scalars, we have to combine vector spaces using the fact that they are given as representation spaces for some particular groups.

To see what needs to be done, consider the situation of groupoids with just one object, so the only important information is the homomorphism of groups. These can be seen as one-object groupoids, which we can just call $G$ and $H$. A functor between them is given by the single group homomorphism $h : G \rightarrow H$.

Now suppose we have a representation $R$ of the group $G$ on $V$ (so that $R(g) \in GL(V)$ and $R(gg') = R(g)R(g')$). Then somehow we need to get a representation of $H$ which is “induced” by the homomorphism $h$, $Ind(R)$:

This diagram shows “the answer” – but how does it work? Essentially, we use the fact that there’s a nice, convenient representation of any group $G$, namely the regular representation of $G$ on the group algebra $\mathbb{C}[G]$. Elements of $\mathbb{C}[G]$ are just complex linear combinations of elemenst of $G$, which are acted on by $G$ by left multiplication. The group $H$ also has regular representation, on $\mathbb{C}[H]$. These are the most easily available building blocks with which to build the “push-forward” of $R$ onto $H$.

To see how, we use the fact that $\mathbb{C}[H]$ has a right-action of $G$, and hence $\mathbb{C}[G]$, by way of $h$. An element $g \in G$ acts on $\mathbb{C}[H]$ by right-multiplication by $h(g)$ – and this extends linearly to $\mathbb{C}[G]$. So we can combine this with the left action of $\mathbb{C}[G]$ on $V$ (also extended linearly from $G$) by taking a tensor product of $\mathbb{C}[H]$ with $V$ over $\mathbb{C}[G]$. This lets us “mod out” by the actions of $G$ which are not detected in $\mathbb{C}[H]$. The result, called the induced representation $Ind(R)$ of $H$, in turn gives us back a left-action of $H$ on $\mathbb{C}[H] \otimes_{\mathbb{C}[G]} V$. I’ll call this $h_{\ast} R$.

(Note that usually this name refers to the situation where $G$ is a subgroup of $H$, but in fact this can be defined for any homomorphism.)

This tells us what to do for single-object groupoids. As we remarked earlier, if more than one object is sent to the same $z \in Z$, we should get a direct sum of all their contributions. So I want to describe the 2-linear map, which I’ll now call $V(X) : V(Y) \rightarrow V(Z)$ which we get from the span above, thought of as $X : Y \rightarrow Z$ in $Span(\mathbf{Grpd})$. Here $V(X) = hom(X,\mathbf{Vect})$ and $V(Y) = hom(Y,\mathbf{Vect})$ (where I’m now being more explicit that this whole process is a functor in some reasonable sense).

I have to say what $V(X)$ does to a given 2-vector (what it does to morphisms between 2-vectors is straightforward to work out, since every operation we do is a tensor product or direct sum). Suppose we have $F : Y \rightarrow \mathbf{Vect}$ is one. Then $V(X)(F) = t_{\ast} s^{ast} F= t_{\ast} (F \circ s) : Z \rightarrow \mathbf{Vect}$. We can now say what this works out to. At some object $z \in Z$, we get (still assuming everything is skeletal for simplicity):

$V(X)(F) = \bigoplus_{t(x)=z} \mathbb{C}[Aut(z)] \otimes_{\mathbb{C}[Aut(x)]} F(s(x))$

And this is a direct sum of a bunch of such expressions where $F$ is a basis 2-vector – i.e. assigns an irreducible representation to some one object, and the trivial rep on the zero vector space to every other. That allows this to be written as a matrix with vector-space components, just like any 2-linear map.

So the 2-linear map $V(X)$ has a matrix representation. The indices of the matrix are the simple objects in $hom(Y,\mathbf{Vect}$ and $hom(Z,\mathbf{Vect}$, which consist of a choice of (a) object in $Y$ or $Z$ (which we assume are skeletal – otherwise it’s a choice of isomorphism class), and (b) irreducible representation of the automorphism group of that object. Given a choice of index on each side, the corresponding coefficient in the matrix is a vector space. Namely the direct sum, over all the objects $x \in X$ that restrict down to our chosen pair, of a bunch of terms like $\mathbb{C}[Aut(z)] \otimes_{\mathbb{C}[Aut(x)]} \mathbb{C}$. This is just a quotient space of the one group algebra by the image of the other.

Next up: a quick finisher about what happens at the 2-morphism level, then back to TQFT and gravity!