How do you define the composition of two $\mathbb{S}$-modules?

Question

I am reading the book "Algebraic Operads" by J. L. Loday and B. Vallete. I am stuck with the definition of composition of $\mathbb{S}$-modules in Sec-5.1.6, pg. 99. Below I have written down all the required definitions and have posted my question at the end.

An $\mathbb{S}$-module over $\mathbb{K}$ is a family $M = (M(0), M(1), M(2), \ldots, M(N), \ldots)$ of right $\mathbb{K}[\mathbb{S}_n]$-modules $M(n)$.

Tensor product of two $\mathbb{S}$-module:

The tensor product of two $\mathbb{S}$-module is defined as follows (cf. Section 5.1.4): Let $M$ and $N$ be two $\mathbb{S}$-modules. Then their tensor product is the $\mathbb{S}$-module $M \otimes N$ defined by $$M \otimes N (n) := \bigoplus_{i+j=n} \mathrm{Ind}^{\mathbb{S}_n}_{\mathbb{S}_i \times \mathbb{S}_j} M(i) \otimes M(j) \cong \bigoplus_{i+j=n} M(i) \otimes M(j) \otimes \mathbb{K}[Sh(i,j)]$$ where $\mathrm{Ind}^{\mathbb{S}_n}_{\mathbb{S}_i \times \mathbb{S}_j} M(i) \otimes M(j)$ is the induced representation of $\mathbb{S}_n$ defined in Appendix A.1.3, pg 458. It says the following: Let $G$ be a group and $H$ be a subgroup of $G$. If $M$ is a right $H$-module, then the induced representation is the following representation of $G$ (i.e., a left $K[G]$-module) $$\mathrm{Ind}^G_H M := M \otimes_H \mathbb{K}[G].$$

Question 1:

I am a bit confused with the definition of the tensor product of two $\mathbb{S}$-module $M$ and $N$. The definition of $\mathbb{S}$-module says it is a family of right $\mathbb{K}[\mathbb{S}_n]$-modules. Then by definition $M \otimes N (n)$ should be a right $\mathbb{K}[\mathbb{S}_n]$-module. But the induced representation $\mathrm{Ind}^{\mathbb{S}_n}_{\mathbb{S}_i \times \mathbb{S}_j} M(i) \otimes M(j)$ is a representation of $\mathbb{S}_n$, hence is a left $\mathbb{K}[\mathbb{S}_n]$-module and not a right $\mathbb{K}[\mathbb{S}_n]$-module. This has confused me, as I am unable to understand how to make $\mathrm{Ind}^{\mathbb{S}_n}_{\mathbb{S}_i \times \mathbb{S}_j} M(i) \otimes M(j)$ a right $\mathbb{K}[\mathbb{S}_n]$-module? Therefore, assuming the definition of tensor product to be $$M \otimes N(n) = \bigoplus_{i+j=n} M(i) \otimes M(j) \otimes \mathbb{K}[Sh(i,j)]$$ I have proceeded further.

Question 2:

Let $N$ be an $\mathbb{S}$-module. Then can we tell that the following equality holds? If yes, how should I proceed to prove it? I also could not prove it for the case $k=2$.
$$N^{\otimes k} (n) = \bigoplus_{i_1 + \cdots + i_k =n} \mathrm{Ind}^{\mathbb{S}_n}_{\mathbb{S}_{i_1} \times \cdots \times \mathbb{S}_{i_k}} \big(N(i_1) \otimes \cdots \otimes N(i_k)\big) \cong \bigoplus_{i_1 + \cdots + i_k=n} N(i_1) \otimes \cdots \otimes N(i_k) \otimes \mathbb{K}[Sh(i_1,\ldots,i_k)]$$

Composition of two $\mathbb{S}$-module:

Given two $\mathbb{S}$-module $M$ and $N$ their composite is the $\mathbb{S}$-module $$M \circ N(n) := \bigoplus_{k \ge 0} M(k) \otimes_{\mathbb{S}_k} N^{\otimes k}(n) \cong \bigoplus_{k \ge 0} \Big(M(k) ~~\otimes_{\mathbb{S}_k} \bigoplus_{i_1 + \cdots + i_k =n} \mathrm{Ind}^{\mathbb{S}_n}_{\mathbb{S}_{i_1} \times \cdots \times \mathbb{S}_{i_k}} \big(N(i_1) \otimes \cdots \otimes N(i_k)\big)\Big)$$ Assuming the equality mentioned in "Question 2" holds, we get $$M \circ N(n) = \bigoplus_{k \ge 0} P(k) = \bigoplus_{k \ge 0} \Big(M(k) ~~\otimes_{\mathbb{S}_k} \bigoplus_{i_1 + \cdots + i_k =n} N(i_1) \otimes \cdots \otimes N(i_k) \otimes \mathbb{K}[Sh(i_1,\ldots,i_k)]\Big)$$

Example:

The authors have provided an example of the composition of two $\mathbb{S}$-modules in 5.1.9, pg 100: Let $M$ and $N$ be two $\mathbb{S}$-modules with $M(0)=0=N(0)$ and $M(1) = \mathbb{K} = N(1)$ then $$M \circ N (2) = M(2) \oplus N(2)$$ $$M \circ N(3) = M(3) \oplus \big(M(2) \otimes \mathrm{Ind}^{\mathbb{S}^3}_{\mathbb{S}_2} N(2)\big) \oplus N(3)$$

Question 3:

Assuming the equality in "Question 2" I did the computation for $M \circ N(3) = \bigoplus_{k \ge 0} P(k) = P(1) \oplus P(2) \oplus P(3)$ (note that since $M(0)=0 \implies P(0)=0$ and for $k > 3$ we have $P(k) = 0$ since $i_1 + \cdots + i_k = 3$ implies one of $i_1,\ldots,i_k$ say $i_j$ is $0$, which implies $N(i_j)=0$). Now computing $P(1)$, $P(2)$, and $P(3)$ we get: $$P(1) = M(1) \otimes_{\mathbb{S}_1} \big(N(3) \otimes \mathbb{K}[Sh(3)]\big) \cong M(1)~ \otimes_{\mathbb{S}_1} N(3) \cong N(3)$$ $$P(2) = M(2) \otimes_{\mathbb{S}_2} \Big( \big(N(1) \otimes N(2) \otimes \mathbb{K}[Sh(1,2)]\big) \oplus \big(N(2) \otimes N(1) \otimes \mathbb{K}[Sh(2,1)]\big)\Big)$$ $$P(3) = M(3) \otimes_{\mathbb{S}_3} \big(N(1) \otimes N(1) \otimes N(1) \otimes \mathbb{K}[Sh(1,1,1)]\big) \cong M(3) \otimes_{\mathbb{S}_3} \mathbb{K}[\mathbb{S}_3] \cong M(3)$$ My question is how to show $P(2) \cong M(2) \otimes\mathrm{Ind}^{\mathbb{S}_3}_{\mathbb{S}_2}N(2)$? The authors have mentioned, "Since $\mathbb{S}_2$ is exchanging the two summands we get the expected result". I could not decipher the meaning of this.

Answer 1

(These multi-part questions are always awkward to handle, because different people may answer different parts, and then how to award credit?)

For question 1, I find that the shuffle notation doesn't make it easier to keep track of what's going on, which is that the tensor product is defined by

$$(M \otimes N)(n) = \bigoplus_{i+j=n} (M(i) \otimes N(j)) \otimes_{S_i \times S_j} \mathbb{K}S_n.$$

Here $S_i \times S_j$ acts on the right on $M(i) \otimes N(j)$ and on the left on $\mathbb{K}S_n$ (because any subgroup inclusion $G \hookrightarrow H$ or for that matter any homomorphism $\phi: G \to H$ gives a left action of $G$ on $\mathbb{K}H$, via $g \cdot h = \phi(g)h$). So you take the tensor product of a right $(S_i \times S_j)$-module with a left $(S_i \times S_j)$-module. But at the same time, $\mathbb{K}S_n$ is not merely a left $(S_i \times S_j)$-module; it's a left $(S_i \times S_j)$- right $S_n$-module, in other words a bimodule, and the right $S_n$-module structure on $\mathbb{K}S_n$ induces a right $S_n$-module structure on

$$\bigoplus_{i+j=n} (M(i) \otimes N(j)) \otimes_{S_i \times S_j} \mathbb{K}S_n$$

which answers the first question. (Where you say $M \otimes_H \mathbb{K}[G]$ is a left $G$-module, I say it's a right $G$-module.)

I'll come back to the other questions later when I have more time, unless someone else gets there first.

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For question 2: first of all, the shuffle description doesn't seem right to me. I'll come back to this in a moment. (It seems Bruno Vallete is not a member of the MO community; I wonder how he would respond.) Anyway, it would be better to describe $N^{\otimes k}$ in terms of induced representations:

$$N^{\otimes k} (n) = \bigoplus_{i_1 + \cdots + i_k =n} \mathrm{Ind}^{\mathbb{S}_n}_{\mathbb{S}_{i_1} \times \cdots \times \mathbb{S}_{i_k}} \big(N(i_1) \otimes \cdots \otimes N(i_k)\big)$$

or in other words

$$N^{\otimes k} (n) = \bigoplus_{i_1 + \cdots + i_k =n} \big(N(i_1) \otimes \cdots \otimes N(i_k)\big) \otimes_{S_{i_1} \times \ldots \times S_{i_k}} \mathbb{K}S_n.$$

To see this is correct, it is clarifying to understand the tensor product of $\mathbb{S}$-modules in terms of universal properties (a useful catchphrase here is "Day convolution"). Consider $\mathbb{S}$ as equivalent to the symmetric monoidal groupoid of finite sets and bijections, with tensor product obtained by restricting the coproduct = cocartesian monoidal product $+$ on finite sets and all functions between them, to finite sets and bijections between them. Let

$$y: \mathbb{S} \to \mathrm{Mod}_K(\mathbb{S})$$

be the functor that sends the $n$-element set $[n]$ to $\mathbb{K}[S_n]$. Then the assignment $(M, N) \mapsto M \otimes N$, called variously the Cauchy product or the Day convolution, is up to unique isomorphism the unique functor that extends the functor

$$\mathbb{S} \times \mathbb{S} \overset{+}{\to} \mathbb{S} \overset{y}{\to} \mathrm{Mod}_K(\mathbb{S})$$

along $y \times y$ to a functor

$$\mathrm{Mod}_K(\mathbb{S}) \times \mathrm{Mod}_K(\mathbb{S}) \to \mathrm{Mod}_K(\mathbb{S})$$

that is $\mathbb{K}$-linear and cocontinuous (colimit-preserving) separately in the first and second arguments $M, N$. Once this universal property is understood as the main point of the construction, then it is relatively straightforward to see that symmetric monoidal product $+$ induces a symmetric monoidal structure for the Cauchy product $\otimes$. This implies that iterated tensor products $M_1 \otimes \ldots \otimes M_k$ are given by the formula

$$\bigoplus_{i_1 + \cdots + i_k =n} \big(M_1(i_1) \otimes \cdots \otimes M_k(i_k)\big) \otimes_{S_{i_1} \times \ldots \times S_{i_k}} \mathbb{K}S_n$$

precisely because this construction is also $\mathbb{K}$-linear and cocontinuous in each of its separate arguments $M_1, \ldots, M_k$. This answers question 2.

The reason the shuffle formula looks wrong to me can be illustrated very simply. Take $M$ so that $M(i) = 0$ for $i \neq 1$ and $M(1) = \mathbb{K}[S_1]$; take $N$ so that $N(i) = 0$ for $i \neq 1$ and $N(2) = \mathbb{K}[S_2]$. As a module over $S_1 \times S_2 \cong S_2$, the only nontrivial $M(i) \otimes N(j)$ is isomorphic to the right $S_2$-module $\mathbb{K}[S_2]$. The induced representation formula gives

$$(M(1) \otimes N(2)) \otimes_{S_1 \times S_2} \mathbb{K}[S_3] \cong \mathbb{K}[S_2] \otimes_{S_2} \mathbb{K}[S_3] \cong \mathbb{K}[S_3]$$

as the only nontrivial $(M \otimes N)(n)$; here $n = 3$. But if we try the shuffle formula, we get

$$\mathbb{K}[S_2] \otimes \mathbb{K}[S_2\backslash S_3]$$

which is a very different right $S_3$-module. Indeed, if we identify $S_2$ with the subgroup of $S_3$ generated by the transposition $(1\; 2)$, then $(1\; 2)$ acts trivially on the construction that uses the shuffle product (because it acts trivially on the right on $\mathbb{K}[S_2\backslash S_3]$), but it acts nontrivially on $\mathbb{K}[S_3]$.

(Will get back to question 3 later.)