Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm trying to understand the composition product (Tall-Wraith monoid) for $R$-rings (aka commutative $R$-algebras). Multivariable polynomials over $R$ are examples $R$-rings. I think that $R[X_1,...,X_n] ⊙_R R[Y_1,...,Y_m]$ is (isomorphic to) $R[Z_{1,1},...,Z_{n,m}]$, and that the composition product of $p \in R[X_1,...,X_n]$ and $q \in R[Y_1,...,Y_m]$ is

$p⊙q := p(q(Z_{1,1},...,Z_{1,m}),...,q(Z_{n,1},...,Z_{n,m}))$.

My questions are

  1. Is this correct?
  2. What do I need to do to verify that some construction (such as above) is in fact a composition product?
share|improve this question

1 Answer 1

up vote 4 down vote accepted

Yes, this is correct, except that the plethystic monoidal product refers not to a monoidal product on the category of $R$-rings, but to a monoidal product on the category of co-$R$-ring objects in the category of $R$-rings. (What one might call an $R$-biring.) Or, it could also refer to a right action

$$\odot \colon R\text{-Ring} \times R\text{-Biring} \to R\text{-Ring}$$

of this monoidal category, acting on the category of $R$-rings (we describe this below).

A good way to think about this is that an $R$-biring is an $R$-ring $S$ that carries enough extra structure so as to endow each hom-set $\hom(S, A)$ with an $R$-ring structure, in a way that is natural in $R$-rings $A$. Thus, an $R$-biring is essentially the same thing as an endofunctor

$$R\text{-Ring} \to R\text{-Ring}$$

of the form $\hom(S, -)$. Yes, the hom-functor $\hom(S, -)$ goes from $R\text{-Ring}$ to $Set$, but the idea is that we also have natural maps representing addition

$$a: \hom(S, A) \times \hom(S, A) \to \hom(S, A),$$

multiplication, etc. on each hom-set, to make each such hom-set an $R$-ring. Using the fact that we have a natural isomorphism

$$\hom(S, -) \times \hom(S, -) \cong \hom(S \otimes_R S, -)$$

we see that the natural map $a$ must (by an application of the Yoneda lemma) be represented by an $R$-ring map of the form $\alpha: S \to S \otimes_R S$. Similarly, the natural map $m$ for multiplication is represented by another $R$-ring map $\mu: S \to S \otimes_R S$, and so on for each defining operation in the theory of $R$-rings.

Then, if $S$ and $T$ are two birings, the plethystic product $S \odot T$ is defined to be the biring that represents the composite

$$R\text{-Ring} \stackrel{\hom(T, -)}{\to} R\text{-Ring} \stackrel{\hom(S, -)}{\to} R\text{-Ring}.$$

(Yes, any two such endofunctors represented by birings compose to give another such endofunctor.)

How does this help us? Let's take the polynomial algebra $S = R[x_1, x_2, \ldots, x_n]$., and see it as a biring. (Here $\alpha \colon S \to S \otimes_R S$ is identified with the $R$-ring homomorphism

$$R[x_1, \ldots x_n] \to R[x_1, \ldots, x_n, y_1, \ldots, y_n]$$

taking $x_i$ to $x_i + y_i$. The other co-operations are similar.) We have $\hom(S, A) \cong A^n$, so clearly the hom-set here does carry natural $R$-ring structure, namely the $R$-ring $A^n$. Thus, the biring $S$ represents the functor

$$R\text{-Ring} \stackrel{n^{th}\text{-power}}{\to} R\text{-Ring}$$

and the composite of two such functors is by multiplying powers. Thus we indeed have a natural identification of birings

$$R[X] \odot R[Y] \cong R[X \times Y]$$

where $X$ and $Y$ are finite sets with $m$ and $n$ elements. This is completely general, and $m$ and $n$ do not need to be finite.


How does one construct the plethystic product explicitly from this point of view? The general point is that endofunctors of the form $\hom(S, -)$ are continuous (preserve limits) and have left adjoints. The left adjoint is denoted $- \odot S: R\text{-Ring} \to R\text{-Ring}$ (this gives the right monoidal category action mentioned above). To calculate it, use the fact that left adjoints preserve colimits, and hence any presentation of an $R$-ring $A$ as a coequalizer of the form

$$R[Y] \stackrel{\to}{\to} R[X] \to A.$$

Thus, applying $- \odot S$ to this sequence, we obtain a coequalizer diagram

$$S^{\otimes Y} \stackrel{\to}{\to} S^{\otimes X} \to A \odot S.$$

Here $S^{\otimes X}$ is simply the $X$-fold coproduct of copies of $S$ in the category of $R$-rings. This gives an explicit construction of the ring $A \odot S$. If $A$ is a biring, this construction lifts to a biring structure on $A \odot S$; the details are easy, but I will omit them.

The monoidal unit for $\odot$ is the polynomial algebra in one generator, since this represents the identity endofunctor. Now suppose we have (following your notation) an $R$-ring map $p \colon R[x] \to A$ representing an element of $A$, and an $R$-biring map $q: R[x] \to S$. Then we can indeed form

$$R[x] \cong R[x] \odot R[x] \stackrel{p \odot q}{\to} A \odot S$$

and this represents an element of $A \odot S$, which by abuse of notation we denote by $p \odot q$. Taking the case $A = R[x_1, \ldots, x_n]$ and following the definitions above, one easily calculates $p \odot q$ to be the element represented by

$$R[x] \stackrel{p}{\to} R[x_1, \ldots, x_n] \cong R[x]^{\otimes n} \stackrel{q^{\otimes n}}{\to} S^{\otimes n} \cong A \odot S$$

and from here it is not difficult to assure yourself that your formula for $p \odot q$ is indeed correct.

share|improve this answer
    
A measly +1 (from me) for an answer that deserves much, much more. –  Jacques Carette Nov 27 '11 at 23:06
    
Thanks, Jacques. It benefited me to work through some details here. –  Todd Trimble Nov 27 '11 at 23:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.