This should be a math.stack question, but I am posting it on math.overflow so that someone who knows the theory of operads might provide some helpful comments and answers. 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 $\mathbb{S}$-module 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.

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.