It’s a standard phenomenon that categories of functors inherit many good properties from their targets. In our case, the category of Mackey functors is easily seen to be complete, cocomplete, and even abelian (or stable, in the spectral case). Functor categories also inherit monoidal structures through a general recipe known as Day convolution. I will remind now how this works, at least in our particular case. The monoidal product on Mackey functors is customarily called box product, and denoted by the box symbol $\square$.

Intuitively, if $\underline{M},\underline{N}$ are two Mackey functors, what should their box product $\underline{M}\square\underline{N}$ be? As with classical tensor products, we expect a map out of $\underline{M}\square\underline{N}$ to correspond to a “bilinear” map out of the “pointwise” product. More specifically, I shall first define the x-box product1 by

\[\underline{M}\boxtimes\underline{N} : \mathbf{B}_G\times\mathbf{B}_G \to \mathbf{Ab} : (T_1,T_2) \mapsto \underline{M}(T_1)\otimes\underline{N}(T_2)\]

then define the actual box product as the left Kan extension of the x-box product along the cartesian product functor $\mathbf{B}_G\times\mathbf{B}_G \to \mathbf{B}_G$. Being a left Kan extension simply means being left adjoint to the restriction, so we get the following natural isomorphism (which characterizes $\square$):

\[\operatorname{Hom}(\underline{M}\square\underline{N},\underline{P}) \cong \operatorname{Hom}(\underline{M}\boxtimes\underline{N},\underline{P}(-\times -))\]

As with any Kan extension, one can give an explicit expression for the box product as a coend, but I won’t need that. There’s one more missing detail: every monoidal category should have a unit object, which serves as the unit for the monoidal product. For Day convolutions, this is just the functor represented by the unit object of the original category. In our case $\mathbf{B}_G$ is regarded as monoidal with its usual direct product, so the unit is the one-object $G$-set with the trivial action, and the functor that it represents is precisely what we called the Burnside Mackey functor $\underline{A}_G$.

With a monoidal structure at hand, we can now define in our category monoids and modules over them. A monoid in Mackey functors is called a Green functor, and explicitly it just consists of some object $\underline{R}$ along with a multiplication map $\mu : \underline{R}\square\underline{R}\to\underline{R}$ and a unit map $\eta : \underline{A}_G \to \underline{R}$ satisfying standard compatibilities, for associativity and unitality:

A (left) module over a Green functor $\underline{R}$ is a Mackey functor $\underline{M}$ equipped with an action map $\alpha : \underline{R}\square\underline{M}\to\underline{M}$ which satisfies the expected compatibility:

Recall that a monoid in $\mathbf{Ab}$ (with respect to the tensor product monoidal structure) is just a ring, so we expect Green functors to be sort of ring-analogues in the Mackey world, and likewise for modules over them. In fact this can be made precise: abusively, let $\mu : \underline{R}\boxtimes\underline{R}\to\underline{R}(-\times-)$ be the bilinear map corresponding to the multiplication map (this is actually a natural transformation – keep that in mind). Then for every $T_1,T_2 \in \mathbf{B}_G$, we have the “external product” of the Green functor

\[\mu_{T_1,T_2} : \underline{R}(T_1)\otimes\underline{R}(T_2) \cong (\underline{R}\boxtimes\underline{R})(T_1,T_2) \xrightarrow{\qquad\qquad} \underline{R}(T_1\times T_2)\]

so for every $T \in \mathbf{B}_G$ we have the “internal product” induced by pulling back along the diagonal $\delta_T : T \to T\times T$

\[\mu_T : \underline{R}(T)\otimes\underline{R}(T) \xrightarrow{\mu_{T,T}} \underline{R}(T\times T) \xrightarrow{\delta^*} \underline{R}(T)\]

The monoid’s unit map $\eta_T : \mathbb{Z} \to \underline{R}(T)$ is obtained from $\eta : \underline{A}_G(T) \to \underline{R}(T)$ by treating each integer $n$ as a finite set of cardinality $n$ and trivial $G$-action (and extending additively for negative integers). Associativity and unitality of the Green functor readily implies the same for $\mu_{T,T}$, so $\underline{R}(T)$ is a monoid in abelian groups, aka a ring.

So great, Green functors send each object to an actual ring. Do they also send morphisms in $\mathbf{B}_G$ to ring homomorphisms? Not quite… here is why. Since the Burnside category is a category of spans, we can check separately what happens to right-way maps and wrong-way maps.

Wrong-way maps actually do give ring homomorphisms, essentially because the internal product is defined using the wrong-way of the diagonal. Being a homomorphisms amounts to commutativity of two diagrams; the first is for multiplicativity:

where the left-hand square commutes by naturality and the right-hand square simply by functoriality. The second diagram is for preserving the ring’s unit element, which is expressed rigorously by the next diagram:

In this one, the right-hand square commutes by naturality. The left-hand square is a bit more subtle, since it relies on how the Burnside Mackey functor actually acts on the trivial $G$-set. I remind that the wrong-way maps of $\underline{A}_G$ are just restriction, so they keep the trivial $G$-set as itself, and that’s exactly what makes the left-hand square commute.

We have concluded that $f^* : \underline{R}(T’) \to \underline{R}(T)$ makes $\underline{R}(T)$ into an algebra, hence a module, over $\underline{R}(T’)$. The action map of this module is the composite

\[\alpha : \underline{R}(T')\otimes\underline{R}(T) \xrightarrow{f^*\otimes\mathrm{id}} \underline{R}(T)\otimes\underline{R}(T) \xrightarrow{\mu_T} \underline{R}(T)\]

You can now consider the same two diagrams as before, but replace $f^*$ with $f_*$ (and flip the arrows’ directions). You will see that those fail to commute in general, so $f_* : \underline{R}(T) \to \underline{R}(T’)$ is not a ring homomorphism. But observe that it’s still a map between $\underline{R}(T’)$-modules, so perhaps it could actually be an $\underline{R}(T’)$-module homomorphism? This would amount to commutativity of the following square

I encourage the reader to prove this commutativity to ensure they are following along. Here is another way to write the fact that $f_*$ is a homomorphism of $\underline{R}(T’)$-modules:

Corollary: For every $x \in \underline{R}(T)$ and $y \in \underline{R}(T’)$, we have the projection formula

\[f_*(x)\cdot y = f_*(x\cdot f^*(y))\]

On the left-hand side, the multiplication means $\mu_{T’}$, and on the right-hand side it is the multiplication $\mu_{T}$. In this whole discussion we treated $\underline{R}(T)$ as a left $\underline{R}(T’)$-module, but you can also do everything from the right and get a symmetric version of this formula. One may also define the notion of a commutative Green functor, in which case the symmetric version would follow from this one.

Fun fact: Projection formulas are really cool, and they show up in lots of different places in math.

  • In set theory: $f(A \cap f^{-1}(B)) = f(A)\cap B$.
  • In commutative algebra: $f_*(M\otimes f^*(N)) \cong f_*(M)\otimes N$.
  • In representation theory: $\mathrm{Ind}^G_H(V\otimes\mathrm{Res}^G_H(W)) \cong \mathrm{Ind}^G_H(V)\otimes W$. In fact we can rigorously obtain this in our context by showing that the representation Mackey functor is Green.
  • Even the adjoint functors $f^* \dashv f_*$ we defined in the pervious part satisfy their own projection formula! This suggests that the similarity in notations might not be a mere coincidence…

I’d rather not go through the whole details of modules over Green functors, so I’ll just say you can follow very similar arguments, and end up with

Corollary: Let $\underline{R}$ be a Green functor and $\underline{M}$ a Mackey $\underline{R}$-module. Then $\underline{M}(T)$ is a module over $\underline{R}(T)$, and the action maps $\alpha_T$ of these modules satisfy their own projection formula: for every $x \in \underline{R}(T)$ and $y \in \underline{M}(T’)$, we have

\[f_*(x)\cdot y = f_*(x\cdot f^*(y))\]

where the multiplication on the left-hand side is $\alpha_{T’}$ and on the right-hand side $\alpha_T$.

Finally, let’s specialize this to the case $f : G/K \to G/H$ for some subgroups $K\leq H\leq G$. Then in arrow notation, the projection formula will be

\[\boxed{\uparrow_K^H(x)\cdot y = \uparrow_K^H(x\cdot\downarrow^H_K(y))}\]

Remember from the last part, where I said the composite I care about is $\uparrow_e^G\circ\downarrow^G_e$? Now the projection formula seems pretty useful! If we let $x = 1_K$ be the identity element of the ring $R(K)$, we see that $\uparrow_K^H\circ\downarrow^H_K$ is equal to multiplication by the scalar $\uparrow_K^H(1_T)$. In the next and final part of this lengthy-out-of-control blogpost, I will use this projection formula to finally get an answer (and hopefully a satisfying one) for the question I posed in the beginning, which now falls down to the following, much more approachable question:

In the Burnside Mackey ring there are two interesting elements: the transfer-of-unit $\uparrow_K^H(1_K)$, and the integer $[H:K]$. On which Mackey functors do these elements act the same way?

Footnotes

  1. Mom can I get an x-box? We already have an x-box at home. The x-box at home: ↩