symmetric monoidal (∞,1)-category of spectra
Given an algebraic theory $V$, a $V$-algebra is a model of $V$ in the category $Set$. A Tall–Wraith $V$-monoid is the kind of thing that acts on $V$-algebras.
Let $V$ be an algebraic theory and let $V Alg$ be the category of models of this theory in $Set$. Then a Tall–Wraith $V$-monoid is a monoid object in the category of co-$V$-objects in $V Alg$.
To see why these are what acts on $V$-algebras one needs to understand what a co-$V$-object in $V Alg$ actually is. A co-$V$-object in some category $D$ is a representable covariant functor from $D$ to $V Alg$. To give a particular $D$-object, $d$, the structure of a co-$V$-object is to give a lift of the $Set$-valued $Hom$-functor $D(d,-)$ to $V Alg$. Thus a co-$V$-object in $V Alg$ is a representable covariant functor from $V Alg$ to itself.
One can therefore consider composition of such representable covariant functors. The main result of this can be simply stated:
The composition of representable covariant functors $V Alg \to V Alg$ is again representable.
This is a basic result in general algebra, and is not stated here in its full generality.
An almost corollary of this is that the category of representable covariant functors from $V Alg$ to itself is monoidal (the “almost” refers to the fact that you have to show that the identity functor is representable, but this is not hard).
Thus for two co-$V$-algebra objects in $V$, say $R_1$ and $R_2$, there is a product $R_1 \odot R_2$ and a natural isomorphism
for any $V$-algebra, $A$.
A Tall–Wraith $V$-monoid is thus a triple $(P,\mu,\eta)$ with $\mu : P \odot P \to P$ and $\eta : I \to P$ (where $I$ is the free $V$-algebra on one element — this represents the identity functor), satisfying the obvious coherence diagrams. An action of $P$ on a $V$-algebra, say $A$, is then a morphism $\rho : P \odot A \to A$ again satisfying certain coherence diagrams.
Ah, but I have not told you what $P \odot A$ is! At the moment, one can take the “product” of two co-$V$-algebra objects in $V Alg$ but now I want to take the product of a co-$V$-algebra object with a $V$-algebra. How do I do this? I do this by observing that a $V$-algebra is a co-$Set$-algebra object in $V Alg$! That’s a complicated way of saying that $V$ represents a covariant functor $V Alg \to Set$. Precomposing this with the functor represented by $P$ yields again a covariant functor $V Alg \to Set$. This is again representable and we write its representing object $P \odot A$.
As an aside, we note a consequence. As we’ve seen, the category of co-$V$-algebra objects in $V$ is a monoidal category, with the tensor product $\odot$. Now we’re seeing this monoidal category acts on the category of $V$-algebras. Indeed, it acts on the categories of $V$-algebra and co-$V$-algebra objects in a reasonably arbitrary base category.
One postscript to this is that although the category of co-$V$-algebra objects in $V Alg$ is not a variety of algebras, for a specific Tall–Wraith $V$-monoid $P$, the category of $P$-modules is a variety of algebras.
If $V$ is the theory of commutative unital rings, a $V$-algebra is a commutative unital ring, a co-$V$-algebra object in $V$ is a biring and the corresponding sort of Tall–Wraith $V$-monoid is called, in Tall and Wraith’s original paper, a biring triple.
If $V$ is the theory of commutative associative algebras over a field $k$, then a $V$-algebra is a commutative associative algebra over $k$, and the corresponding sort of Tall–Wraith $V$-monoid is called a plethory.
If $V$ is the theory of abelian groups, than a $V$-algebra is an abelian group, and the corresponding sort of Tall–Wraith $V$-monoid is a ring.
To understand the last example, we need to think about co-abelian group objects in the category of abelian groups. Abstractly, such a thing is an abelian group object internal to $AbGp^{op}$ (though this picture gets the morphisms the wrong way around; in full abstraction then the category of co-abelian group objects in $AbGrp$ is the opposite category of the category of abelian group objects in $AbGrp^{op}$). Concretely, such a thing is an abelian group $A$ together with group homomorphisms
where $I$ is the initial object in the category of abelian groups. These homomorphisms must satisfy certain laws: just the abelian group axioms written out diagrammatically, with all the arrows turned around.
In fact, $I = \{0\}$. Thus $\epsilon$ is forced to be the map that sends everything to $0$: we have no choice here.
We also have that $A \coprod A = A \oplus A$. That means that for $a \in A$, $\mu(a) = (a_1,a_2)$ for some $a_1, a_2 \in A$. Now, one of the laws says that $\epsilon$ is a counit for $\mu$. This means that $(\epsilon \oplus 1) \mu = 1$ and similarly for $1 \oplus \epsilon$. Thus $a_1 = a_2 = a$ and $\mu$ is the diagonal map. So, we have no choice here either.
The diagram for $\iota$ (representing the inverse map) is a little more complicated. As $I$ is the initial object in $AbGrp$, there is a unique morphism $I \to A$ (inclusion of the zero). Composing this with $\epsilon$ yields a morphism $A \to A$ which maps every element to the zero in $A$. Using $\mu$ and $\iota$ we can construct another morphism $A \to A$ as
where $\Delta^c$ is the co-diagonal. The relations for abelian groups say that this morphism must be the same as the zero morphism $A \to A$. Using the fact that $A \coprod A \cong A \oplus A$ and that $\mu$ is the diagonal, this says that $a + \iota(a) = 0$. Hence, by the uniqueness of inverses for abelian groups, $\iota(a) = -a$.
Thus if $(A, \mu, \iota, \epsilon)$ is a co-abelian group object in $AbGrp$ then $\mu$ is the diagonal, $\iota$ the inverse from abelian groups, and $\epsilon$ the zero morphism.
However, that is still not quite the same as saying that $(A, \mu, \iota, \epsilon)$ is a co-abelian group object in $AbGrp$. Certainly, $(A, \mu, \epsilon)$ is a co-commutative co-monoid object in $AbGrp$ since $\mu$ is the diagonal, which is automatically co-commutative and co-associative, and $\epsilon$ the zero map, which is the co-unit for the diagonal. What remains is to fit $\iota$ into the structure.
The first issue is that $\iota$ is not automatically a morphism in $AbGrp$. That is to say, when defining an algebraic theory then the operations are defined on the underlying objects. It is a consequence of the relations of abelian groups that the operations lift to morphisms of abelian groups (algebraic theories where this happens for all operations are sometimes called commutative). Thus $\iota$ is a morphism of abelian groups and so $(A, \mu, \iota, \epsilon)$ is a co-commutative co-monoid with an extra unary co-operation. In fact, it is an involution from the relations for abelian groups.
The final relation is that $\iota$ is the inverse for $\mu$. The relation that $\iota$ is the inverse for addition (let us write it as, say, $\alpha$) is that
is the zero map $A \to I \to A$. This is precisely the relation that $\iota$ is the inverse for $\mu$ since we have the following identifications: $\mu = \Delta$, $A \coprod A = A \times A$, and $\Delta^c = \alpha$. Also, $\epsilon = 0$ and $\eta : I \to A$ is the initial morphism in $AbGrp$.
Thus the fact that $\iota$ is the inverse for the diagonal+zero co-monoidal structure is due to the fact that $\iota$ is the inverse for $(\alpha,\eta)$ and $\alpha : A \oplus A \to A$ is the co-diagonal in $AbGrp$ and $\eta : I \to A$ is the unit.
It is part of the general theory that the category of co-$V$-objects in $V$ is monoidal (though not, in general, symmetric). For details on this see The Hunting of the Hopf Ring, referred to belelow. This monoidal structure for abelian groups turns out to be the tensor product.
Thus a Tall–Wraith monoid for abelian groups is actually an ordinary monoid in the category of abelian groups: in other words, a ring!
We now recapitulate the discussion above in a slightly more general context.
For now our context is that of monads $T$ on $Set$. The category of $T$-algebras is denoted $Set^T$, with forgetful functor $U: Set^T \to Set$ and free functor $F: Set \to Set^T$, whose composite is the monad $T = U F$, and whose counit is denoted $\epsilon: F U \to 1_{Set^T}$.
For each $T$-algebra $R$, there is an adjoint pair of functors
with associated monad $\hom(R, - \cdot R)$. We define a $T$-bialgebra to be a $T$-algebra $R$ equipped with a morphism of monads $\phi: T \to \hom(R, -\cdot R)$. The datum $\phi$ is equivalent to a left $T$-algebra structure
on $\hom(R, -): Set^T \to Set$, or to a right $T$-algebra (aka right $T$-module) structure
where $W = W_R \coloneqq - \cdot R: Set \to Set^T$. A $T$-bialgebra map is a $T$-algebra map $f: R \to S$ such that the induced map $W_f: W_R \to W_S$ is a morphism of right $T$-modules.
A good case to keep in mind is that of birings, which are $T$-bialgebras for the theory $T$ of commutative rings. The monad morphism $T \to \hom(R, -\cdot R)$ has components $T X \to \hom(R, X \cdot R)$ for each set $X$. Here $X \cdot R$ is an $X$-indexed coproduct of copies of $R$, where coproduct in the category of commutative rings $Set^T$ is given by tensor product. Thus, for example, $2 \cdot R$ is the ring $R \otimes R$. The component $T(2) \to \hom(R, 2\cdot R)$ therefore “interprets” each element $\theta \in T(2)$, i.e., each binary operation in the Lawvere theory, as a binary co-operation $R \to R \otimes R$. This applies in particular to the elements $m, a \in T(2)$ which abstractly represent multiplication and addition (seen as natural operations on the category of commutative rings).
Let $Ladj(Set^T, Set^T)$ ($Radj(Set^T, Set^T)$) be the category of left (right) adjoint functors $\Psi: Set^T \to Set^T$. The functor $T$-$BiAlg \to Ladj(Set^T, Set^T)$ that takes $(R, \phi)$ to the right $T$-module $(W_R, \beta)$ is an equivalence. Or, what is the same, the functor $T$-$BiAlg \to Radj(Set^T, Set^T)^{op}$, taing $(R, \phi)$ to the left $T$-algebra $\hom(R, -), \alpha)$, is an equivalence.
The main thing to check is that the functor $R \mapsto \hom(R, -)$ to $Radj(Set^T, Set^T)$ is essentially bijective. The essential point is that $\Phi$ has a left adjoint iff $U \Phi$ has a left adjoint iff $U \Phi: Set^T \to Set$ is representable: $U \Phi \cong \hom(R, -)$ for some $T$-algebra $R$ (in which case the lift $\Phi$ of $\hom(R, -)$ through $U$ is tantamount to a $T$-algebra structure on $\hom(R, -)$). The only (mildly) tricky part is that $\Phi$ has a left adjoint if $U\Phi$ has a left adjoint $W = W_R$. To define the left adjoint $\Psi$ of $\Phi$ objectwise, we take any $T$-algebra $S$ with its canonical presentation
which is a coequalizer diagram. A left adjoint $\Psi$ must preserve this coequalizer, and we must have $\Psi F \cong W$ since both sides are left adjoint to $U \Phi$. Thus we define $\Psi (S)$ to be a coequalizer
where $\beta: W T \to T$ is the $T$-module structure coming from the monad morphism $\phi: T \to \hom(R, -\cdot R)$. This objectwise definition of $\Psi$ easily extends to morphisms by universality and provides a left adjoint to $\Phi$. Remaining details are left to the reader.
The import of this proposition is that left adjoint endofunctors on $Set^T$ compose, i.e., endofunctor composition gives a monoidal structure on $Ladj(Set^T, Set^T)$, and this monoidal structure transports across the categorical equivalence of the proposition to give a monoidal structure on $T$-$BiAlg$. The resultant monoidal product on $T$-bialgebras is denoted $\odot$.
A direct construction of $\odot$ can be extracted by following the proof of the proposition. If $R, S$ are $T$-bialgebras, then the underlying $T$-algebra of $S \odot R$ (corresponding to composition of $\hom(S, -) \circ \hom(R, -)$ of right adjoints $Set^T \to Set^T$) is computed as a reflexive coequalizer in $Set^T$:
Here $\beta X: T X \cdot R \to X \cdot R$ is a component of the $T$-module structure $W_R T \to W_R$; it is mated by the $- \cdot R \dashv \hom(R, -)$ adjunction to the component of the coalgebra structure $\phi X: T X \to \hom(R, X \cdot R)$.
To extract the $T$-coalgebra structure on $S \odot R$, let us observe generally that if $F: C \to D$ is a left adjoint, then for any category $E$ there is an induced left adjoint $[1_E, F]: [E, C] \to [E, D]$ and similarly an induced left adjoint $Ladj(E, C) \to Ladj(E, D)$. Applying this to the case where $C = D = E = Set^T$ and where $F$ is the left adjoint to the lift $\hom(R, -): Set^T \to Set^T$, we find that
is a left adjoint, and in particular preserves $Y$-indexed copowers $Y \cdot S$. In other words, for each $X$ we have canonical isomorphisms
so that the desired right $T$-module structure is given by a composite
D. Tall, G. Wraith, Representable functors and operations on rings, Proc. London Math. Soc. (3), 1970, 619–643, MR265348, doi
J. Borger, B. Wieland, Plethystic algebra, Adv. Math. 194 (2005), no. 2, 246–283, doi, pdf, MR2006i:13044
A. Stacey and S. Whitehouse, The Hunting of the Hopf Ring, Homology, Homotopy and Applications 11(2), 2009, 75–132, online, arXiv/0711.3722.
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