A decomposition for the space of games with externalities

Abstract

The main goal of this paper is to present a different perspective than the more ‘traditional’ approaches to study solutions for games with externalities. We provide a direct sum decomposition for the vector space of these games and use the basic representation theory of the symmetric group to study linear symmetric solutions. In our analysis we identify all irreducible subspaces that are relevant to the study of linear symmetric solutions and we then use such decomposition to derive some applications involving characterizations of classes of solutions.

Appendix

A reference for basic representation theory is Fulton and Harris (1991). Nevertheless, we recall all basic facts that we need.

The symmetric group $S_{n}$ acts on G via linear transformations (i.e., G is a representation of $S_{n}$). That is, there is a group homomorphism $\rho :S_{n}\rightarrow GL(G)$, where GL(G) is the group of invertible linear maps in G. This action is given by:

$$\begin{aligned} (\sigma \cdot w)(S,Q):=[\rho (\sigma )(w)](S,Q)=w[\sigma ^{-1}(S,Q)] \end{aligned}$$

for every $\sigma \in S_{n}$, $w\in G$ and $(S,Q)\in EC$.

Definition 9

Let H be an arbitrary group. A representation for H is a homomorphism $\rho :H\rightarrow GL(X)$, where X is a vector space and $GL(X)=\{T:X\rightarrow X\mid T$ linear and invertible$\}$.

In other words, a representation of H is a map assigning to each element $h\in H$ a linear map $\rho (h):X\rightarrow X$ that respects multiplication:

$$\begin{aligned} \rho (h_{1}h_{2})=\rho (h_{1})\rho (h_{2}) \end{aligned}$$

for all $h_{1},h_{2}\in H$.

One usually abuses notation and talks about the representation X without explicitly mentioning the homomorphism $\rho$. Thus, when applying the linear transformation corresponding to $h\in H$ on the element $x\in X$, we write $h\cdot x$ rather than $(\rho (h))(x)$.

The space of payoff vectors, $\mathbb {R} ^{n}$, is also a $S_{n}$-representation:

$$\begin{aligned} \sigma (x_{1},x_{2},\ldots ,x_{n}):=[\rho (\sigma )](x_{1},x_{2},\ldots ,x_{n})=(x_{\sigma (1)},x_{\sigma (2)},\ldots ,x_{\sigma (n)}) \text {.} \end{aligned}$$

Definition 10

Let $X_{1}$ and $X_{2}$ be two representations for the group H.

• A linear map $T:X_{1}\rightarrow X_{2}$ is said to be H-equivariant if $T(h\cdot x)=h\cdot T(x)$, for every $h\in H$ and every $x\in X_{1}$.

• $X_{1}$ and $X_{2}$ are said to be isomorphic H-representations, $X_{1}\simeq X_{2}$, if there exists an H-equivariant isomorphism between them.

Thus, two representations that are isomorphic are, as far as all problems dealing with linear algebra with a group of symmetries, the same. They are vector spaces of the same dimension where the actions are seen to correspond under a linear isomorphism.

Definition 11

A representation X is irreducible if it does not contain a nontrivial invariant subspace. That is, if $Y\subset X$ is also a representation for H (meaning that $h\cdot y\in Y$ $\forall h\in H$), then Y is either $\{0\}$ or all of X.

Proposition 8

For any representation X of a finite group H, there is a decomposition

$$\begin{aligned} X=X_{1}^{\oplus a_{1}}\oplus X_{2}^{\oplus a_{2}}\oplus \cdot \cdot \cdot \oplus X_{j}^{\oplus a_{j}}\text {,} \end{aligned}$$

where the $X_{i}$ are distinct irreducible representations. The decomposition is unique, as are the $X_{i}$ that occur and their multiplicities $a_{i}$.

This property is called “complete reducibility” and the extent to which the decomposition of an arbitrary representation into a direct sum of irreducible ones is unique is one of the consequences of the following:

Theorem 4

(Schur’s Lemma) Let $X_{1}$,$X_{2}$ be irreducible representations of a group H. If $T:X_{1}\rightarrow X_{2}$ is H-equivariant, then $T=0$ or T is an isomorphism.

Moreover, if $X_{1}$ and $X_{2}$ are complex vector spaces, then T is unique up to multiplication by a scalar $\lambda \in \mathbb {C}$.

The previous theorem is one of the reasons why it is worth carrying around the group action when there is one. Its simplicity hides the fact that it is a very powerful tool.

There is a remarkably effective technique for decomposing any given finite dimensional representation into its irreducible components. The secret is character theory.

Definition 12

Let $\rho :H\rightarrow GL(X)$ be a representation. The character of X is the complex-valued function $\chi _{X}:H\rightarrow \mathbb {C}$, defined as:

$$\begin{aligned} \chi _{X}(h)=Tr\left( \rho (h)\right) . \end{aligned}$$

The character of a representation is easy to compute. If H acts on an n -dimensional space X , we write each element h as an $n\times n$ matrix according to its action expressed in some convenient basis, then sum up the diagonal elements of the matrix for h to get $\chi _{X}(h)$. For example, the trace of the identity map of an n-dimensional vector space is the trace of the $n\times n$ identity matrix, or n. In fact, $\chi _{X}(e)=\dim X$ for any finite dimensional representation X of any group.

Notice that in particular, we have $\chi _{X}(h)=\chi _{X}(ghg^{-1})$ for $g,h\in H$. So that $\chi _{X}$ is constant on the conjugacy classes of H; such a function is called a class function.

Definition 13

Let $\mathbb {C} _{class}(H)=\{f:H\rightarrow \mathbb {C} \mid f$ is a class function on $H\}$. If $\chi _{1},\chi _{2}\in \mathbb {C} _{class}(H)$, we define an Hermitian inner product on $\mathbb {C} _{class}(H)$ by

$$\begin{aligned} \left\langle \chi _{1},\chi _{2}\right\rangle =\frac{1}{\left| H\right| }\underset{h\in H}{\sum }\overline{\chi _{1}(h)}\cdot \chi _{2}(h)\text {.} \end{aligned}$$

(6)

Multiplicities of irreducible subspaces in a representation can be calculated via

Proposition 9

If $Z=Z_{1}^{\oplus a_{1}}\oplus Z_{2}^{\oplus a_{2}}\oplus \cdot \cdot \cdot \oplus Z_{j}^{\oplus a_{j}}$, then the multiplicity $Z_{i}$ (irreducible representation) in Z, is:

$$\begin{aligned} a_{i}=\left\langle \chi _{Z},\chi _{Z_{i}}\right\rangle \text {,} \end{aligned}$$

where $\left\langle ,\right\rangle$ is the inner product given by (6).

Proof of Theorem 1

First, $\left\langle \chi _{G_{\lambda }^{k}},\chi _{U}\right\rangle$ and $\left\langle \chi _{G_{\lambda }^{k}},\chi _{V}\right\rangle$ are the number of subspaces isomorphic to the trivial (U) and standard representation (V) within $G_{\lambda }^{k}$, respectively.

We start by computing the number of copies of U in $G_{\lambda }^{k}$:

$$\begin{aligned} \left\langle \chi _{G_{\lambda }^{k}},\chi _{U}\right\rangle =\frac{1}{n!} \underset{\sigma \in S_{n}}{\sum }\chi _{G_{\lambda }^{k}}(\sigma )\chi _{U}(\sigma )=\frac{1}{n!}\underset{\sigma \in S_{n}}{\sum }\chi _{G_{\lambda }^{k}}(\sigma )\text {.} \end{aligned}$$

Notice that, $\chi _{G_{\lambda }^{k}}(\sigma )$ is just the number of pairs $(S,Q)\in EC$ with $\left| S\right| =k$ and $\lambda _{Q}=\lambda$, that are fixed under $\sigma \in S_{n}$.

Define

$$\begin{aligned} \left\{ \sigma \right\} _{(S,Q)}=\left\{ \begin{array}{ll} 1 &{}\quad \hbox { if } \sigma (S,Q)=(S,Q) \\ 0 &{}\quad \hbox {otherwise} \end{array} \right. \text {.} \end{aligned}$$

Then,

$$\begin{aligned} \chi _{G_{\lambda }^{k}}(\sigma )=\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{\underset{(S,Q)\in EC}{\sum }}\left\{ \sigma \right\} _{(S,Q)} \end{aligned}$$

and so,

$$\begin{aligned} \left\langle \chi _{G_{\lambda }^{k}},\chi _{U}\right\rangle =\frac{1}{n!} \underset{\sigma \in S_{n}}{\sum }\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{\underset{(S,Q)\in EC}{\sum }}\left\{ \sigma \right\} _{(S,Q)}=\frac{1}{n!}\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{\underset{(S,Q)\in EC}{\sum }}\underset{\sigma \in S_{n}}{ \sum }\left\{ \sigma \right\} _{(S,Q)}\text {;} \end{aligned}$$

where,

$$\begin{aligned} \underset{\sigma \in S_{n}}{\sum }\left\{ \sigma \right\} _{(S,Q)}=\left| \left\{ \sigma \in S_{n}\mid \sigma (Q)=Q,\text { }\sigma (S)=S\right\} \right| \text {.} \end{aligned}$$

Now, $S_{n}$ acts on the set $Q_{\lambda }$ and take $Q\in Q_{\lambda }$. The orbit of Q under $S_{n}$ is

$$\begin{aligned} S_{n}Q=\left\{ \sigma (Q)\mid \sigma \in S_{n}\right\} =Q_{\lambda }\text {,} \end{aligned}$$

and the isotropy subgroup of Q is

$$\begin{aligned} (S_{n})_{Q}=\left\{ \sigma \in S_{n}\mid \sigma (Q)=Q\right\} \text {.} \end{aligned}$$

By Lagrange theorem, we get:

$$\begin{aligned} \left| S_{n}Q\right| =\frac{\left| S_{n}\right| }{\left| (S_{n})_{Q}\right| }=\left| Q_{\lambda }\right| \quad \Rightarrow \quad \left| (S_{n})_{Q}\right| =\frac{n!}{ \left| Q_{\lambda }\right| }\text {.} \end{aligned}$$

Notice that $H=(S_{n})_{Q}$ acts on Q and take $S\in Q$ such that $\left| S\right| =k$. The orbit of S under H is

$$\begin{aligned} HS=\left\{ hS\mid h\in H\right\} =\{T\in Q\mid \left| T\right| =k\} \text {.} \end{aligned}$$

Observe that $\left| HS\right| =m_{k}^{\lambda }$ and the isotropy subgroup of S is

$$\begin{aligned} H_{S}=\left\{ h\in H\mid h(S)=S\right\} =\left\{ \sigma \in S_{n}\mid \sigma (Q)=Q,\text { }\sigma (S)=S\right\} \text {.} \end{aligned}$$

Again, by Lagrange theorem, we get

$$\begin{aligned} \left| HS\right| =\frac{\left| H\right| }{\left| H_{S}\right| }=\frac{\left| (S_{n})_{Q}\right| }{\left| H_{S}\right| }=m_{k}^{\lambda }\quad \Rightarrow \quad \left| H_{S}\right| =\frac{\left| (S_{n})_{Q}\right| }{ m_{k}^{\lambda }}=\frac{n!}{\left| Q_{\lambda }\right| m_{k}^{\lambda }}\text {.} \end{aligned}$$

And therefore,

$$\begin{aligned} \left\langle \chi _{G_{\lambda }^{k}},\chi _{U}\right\rangle =\frac{1}{n!} \underset{\left| S\right| =k,\lambda _{Q}=\lambda }{\underset{ (S,Q)\in EC}{\sum }}\frac{n!}{\left| Q_{\lambda }\right| m_{k}^{\lambda }}=\frac{1}{n!}\left| Q_{\lambda }\right| m_{k}\frac{ n!}{\left| Q_{\lambda }\right| m_{k}^{\lambda }}=1\text {.} \end{aligned}$$

Now, we compute the multiplicity of V in $G_{\lambda }^{k}$. Since $\mathbb {R} ^{n}=U\oplus V$ , then $\chi _{ \mathbb {R} ^{n}}=\chi _{U}+\chi _{V}$ $\Rightarrow$ $\left\langle \chi _{G_{\lambda }^{k}},\chi _{ \mathbb {R} ^{n}}\right\rangle =\left\langle \chi _{G_{\lambda }^{k}},\chi _{U}\right\rangle +\left\langle \chi _{G_{\lambda }^{k}},\chi _{V}\right\rangle$ $\Rightarrow$ $\left\langle \chi _{G_{\lambda }^{k}},\chi _{V}\right\rangle =\left\langle \chi _{G_{\lambda }^{k}},\chi _{ \mathbb {R} ^{n}}\right\rangle -1$.

Notice that $G_{[1,1,\ldots ,1]}^{1}\simeq \mathbb {R} ^{n}$ (as a representation for $S_{n}$). Let us compute

$$\begin{aligned} \left\langle \chi _{G_{\lambda }^{k}},\chi _{ \mathbb {R} ^{n}}\right\rangle= & {} \left\langle \chi _{G_{\lambda }^{k}},\chi _{G_{[1,1,\ldots ,1]}^{1}}\right\rangle =\frac{1}{n!}\underset{\sigma \in S_{n}}{ \sum }\chi _{G_{\lambda }^{k}}(\sigma )\chi _{G_{[1,1,\ldots ,1]}^{1}}(\sigma ) \\= & {} \frac{1}{n!}\underset{\sigma \in S_{n}}{\sum }\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{\underset{(S,Q)\in EC}{\sum }}\left\{ \sigma \right\} _{(S,Q)}\underset{Q^{\prime }=Q_{[1,1,\ldots ,1]}}{\underset{ \left| S^{\prime }\right| =1}{\underset{(S^{\prime },Q^{\prime })\in EC}{\sum }}}\left\{ \sigma \right\} _{(S^{\prime },Q^{\prime })} \\= & {} \frac{1}{n!}\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{ \underset{(S,Q)\in EC}{\sum }}\underset{\left| S^{\prime }\right| =1,\lambda _{Q^{\prime }}=[1,1,\ldots ,1]}{\underset{(S^{\prime },Q^{\prime })\in EC}{\sum }}\underset{\sigma \in S_{n}}{\sum }\left\{ \sigma \right\} _{(S,Q)}\left\{ \sigma \right\} _{(S^{\prime },Q^{\prime })}\text {.} \end{aligned}$$

For $x\in S$:

$$\begin{aligned} \underset{\sigma \in S_{n}}{\sum }\left\{ \sigma \right\} _{(S,Q)}\left\{ \sigma \right\} _{(S^{\prime },Q^{\prime })}=\left\{ \sigma \in S_{n}\mid \sigma (Q)=Q,\text { }\sigma (S)=S,\text { }\sigma (x)=x\right\} \end{aligned}$$

Without loss of generality, suppose $\left| S\right| =k=\lambda _{1}$ and take the case $m_{k}^{\lambda }=m_{\lambda _{1}}^{\lambda }=1$. Here, $M=H_{S}=\left\{ \sigma \in S_{n}\mid \sigma (Q)=Q,\text { }\sigma (S)=S\right\}$ acts on $S\in Q\in Q_{\lambda }$ and take $x\in S$. The orbit of x under M is

$$\begin{aligned} Mx=\left\{ \sigma (x)\mid \sigma \in M\right\} =S\text {,} \end{aligned}$$

and the isotropy subgroup of x is

$$\begin{aligned} M_{x}=\left\{ \sigma \in M\mid \text { }\sigma (x)=x\right\} =\left\{ \sigma \in S_{n}\mid \sigma (Q)=Q,\text { }\sigma (S)=S,\text { }\sigma (x)=x\right\} \text {.} \end{aligned}$$

By Lagrange theorem,

$$\begin{aligned} \left| Mx\right| =\frac{\left| M\right| }{\left| M_{x}\right| }=\frac{\left| H_{S}\right| }{\left| M_{x}\right| }=k\quad \Rightarrow \quad \left| M_{x}\right| =\frac{\left| H_{S}\right| }{k}=\frac{n!}{ k\left| Q_{\lambda }\right| m_{k}^{\lambda }}\text {.} \end{aligned}$$

Thus,

$$\begin{aligned} \left\langle \chi _{G_{\lambda }^{k}},\chi _{ \mathbb {R} ^{n}}\right\rangle= & {} \frac{1}{n!}\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{\underset{(S,Q)\in EC}{\sum }}\left[ k\frac{n!}{ k\left| Q_{\lambda }\right| m_{k}^{\lambda }}+m_{\lambda _{2}}^{\lambda }\lambda _{2}\frac{n!}{m_{\lambda _{2}}^{\lambda }\lambda _{2}\left| Q_{\lambda }\right| m_{k}^{\lambda }}\right. \nonumber \\&\left. +\cdot \cdot \cdot +m_{\lambda _{l}}^{\lambda }\lambda _{l}\frac{n!}{m_{\lambda _{l}}^{\lambda }\lambda _{l}\left| Q_{\lambda }\right| m_{k}^{\lambda }}\right] \\= & {} \frac{1}{n!}\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{ \underset{(S,Q)\in EC}{\sum }}\left| \{\lambda \}\right| \frac{n!}{ \left| Q_{\lambda }\right| m_{k}^{\lambda }}=\frac{1}{n!}\left| Q_{\lambda }\right| m_{k}^{\lambda }\left[ \left| \{\lambda \}\right| \frac{n!}{\left| Q_{\lambda }\right| m_{k}^{\lambda }} \right] \\= & {} \left| \{\lambda \}\right| \text {.} \end{aligned}$$

Finally, without loss of generality, suppose $\left| S\right| =k=\lambda _{1}$ and take the case $m_{k}^{\lambda }>1$. Following the same line of reasoning as above, we obtain

$$\begin{aligned} \left\langle \chi _{G_{\lambda }^{k}},\chi _{ \mathbb {R} ^{n}}\right\rangle= & {} \frac{1}{n!}\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{\underset{(S,Q)\in EC}{\sum }}\left[ \begin{array}{c} k\frac{n!}{k\left| Q_{\lambda }\right| m_{k}^{\lambda }} +k(m_{k}^{\lambda }-1)\frac{n!}{k(m_{k}^{\lambda }-1)\left| Q_{\lambda }\right| m_{k}^{\lambda }}+ \\ m_{\lambda _{2}}^{\lambda }\lambda _{2}\frac{n!}{m_{\lambda _{2}}^{\lambda }\lambda _{2}\left| Q_{\lambda }\right| m_{k}^{\lambda }}+\cdot \cdot \cdot +m_{\lambda _{l}}^{\lambda }\lambda _{l}\frac{n!}{m_{\lambda _{l}}^{\lambda }\lambda _{l}\left| Q_{\lambda }\right| m_{k}^{\lambda }} \end{array} \right] \\= & {} \frac{1}{n!}\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{ \underset{(S,Q)\in EC}{\sum }}(\left| \{\lambda \}\right| +1)\frac{n! }{\left| Q_{\lambda }\right| m_{k}^{\lambda }}=\frac{1}{n!} \left| Q_{\lambda }\right| m_{k}^{\lambda }\left[ (\left| \{\lambda \}\right| +1)\frac{n!}{\left| Q_{\lambda }\right| m_{k}^{\lambda }}\right] \\= & {} \left| \{\lambda \}\right| +1\text {.} \end{aligned}$$

In summary,

$$\begin{aligned} \left\langle \chi _{G_{\lambda }^{k}},\chi _{U}\right\rangle =1\text { and }\left\langle \chi _{G_{\lambda }^{k}},\chi _{V}\right\rangle =\left| I_{\lambda ,k}\right| \text {.} \end{aligned}$$

The next task is to identify such copies of U and V inside $G_{\lambda }^{k}$. For that end, define the functions $L_{\lambda ,k}: \mathbb {R} ^{n}\rightarrow G_{\lambda }^{k}$ by $L_{\lambda ,k}(x)=x^{(\lambda ,k)}$. These maps are isomorphisms between $U_{\lambda }^{k}$ and U (similarly between $\left\{ x_{j}^{\left( \lambda ,k\right) }\mid x_{j}\in V\right\}$ and V, for each $j\in I_{\lambda ,k}$) since they are linear, $S_{n}$ -equivariant and $1-1$. Thus, inside of $G_{\lambda }^{k}$, we have the images of these two subspaces: $U_{\lambda }^{k}=L_{\lambda ,k}(U)$ and $V_{\lambda }^{k}=L_{\lambda ,k}(V)$.

Finally, the invariant inner product $\left\langle ,\right\rangle$ gives an equivariant isomorphism, in particular must preserve the decomposition. This implies orthogonality of the decomposition. $\square$

Proof of Proposition 2

We start by computing the orthogonal projection of w onto $U_{G}$. Notice that $\{u_{\lambda }^{k}\}$ is an orthogonal basis for $U_{G}$, and that $\left\| u_{\lambda }^{k}\right\| ^{2}=m_{k}^{\lambda }\left| Q_{\lambda }\right|$.

Thus, the projection of w onto $U_{G}$ is

and so,

$$\begin{aligned} a_{\lambda ,k}=\frac{\left\langle w,u_{\lambda }^{k}\right\rangle }{ \left\langle u_{\lambda }^{k},u_{\lambda }^{k}\right\rangle }=\frac{\underset{\left| S\right| =k,\lambda _{Q}=\lambda }{\sum }w(S,Q)}{\left| (S,Q)\in EC\mid \left| S\right| =k,\lambda _{Q}=\lambda \right| } \text {.} \end{aligned}$$

Now, for each $(\lambda ,k)\in C_{n}$, we define $f^{\lambda ,k}:G\rightarrow \mathbb {R} ^{n}$ as

$$\begin{aligned} f_{i}^{\lambda ,k}(w)=\underset{\lambda _{Q}=\lambda }{\underset{S\ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}}}w(S,Q)\text {;} \end{aligned}$$

where each $f^{\lambda ,k}$ is $S_{n}$-equivariant and observe that $f^{[n],n}(w)=w(N,\{N\})(1,\ldots ,1)$. Let $z\in V$, then $f^{\lambda ,k}(z_{\gamma ,i,j}^{(\gamma ,i)})=0$ if $\lambda \ne \gamma$ or $i\ne k$ , whereas (by Schur’s Lemma) $f^{\lambda ,k}(z_{\lambda ,k,j}^{(\lambda ,k)})=z_{\lambda ,k,j}\in V$.

Let $p: \mathbb {R} ^{n}\rightarrow V$ be the projection of $\mathbb {R} ^{n}$ onto V given by

$$\begin{aligned} p_{i}(x)=x_{i}-\frac{1}{n}\underset{j=1}{\overset{n}{\sum }}x_{j}\text {.} \end{aligned}$$

This projection is equivariant, sends U to zero and it is the identity on V.

Next, define $L^{\lambda ,k}:G\rightarrow V$ as $L^{\lambda ,k}=p\circ f^{\lambda ,k}$. Observe that

$$\begin{aligned} L^{\lambda ,k}(w)=\underset{j\in I_{\lambda ,k}}{\sum }z_{\lambda ,k,j}\text { ,} \end{aligned}$$

since by equivariance, $f^{\lambda ,k}\left( U_{G}\right) \subset U$ and $f^{\lambda ,k}\left( W_{G}\right) =0$. Moreover, $f^{\lambda ,k}(z_{\gamma ,i,j}^{(\gamma ,i)})=0$ if $\lambda \ne \gamma$ or $i\ne k$. Then,

$$\begin{aligned} L_{i}^{\lambda ,k}(w)= & {} p_{i}(f^{\lambda ,k}(w))=\underset{\lambda _{Q}=\lambda }{\underset{S\ni i,\left| S\right| =k}{\underset{ (S,Q)\in EC}{{\sum }}}}w(S,Q)-\frac{1}{n}\underset{l\in N}{\sum }\underset{ \lambda _{Q}=\lambda }{\underset{S\ni l,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}}}w(S,Q) \\= & {} \frac{n-k}{n}\underset{\lambda _{Q}=\lambda }{\underset{S\ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}}}w(S,Q)-\frac{k}{n}\underset{\lambda _{Q}=\lambda }{\underset{S\not \ni i,\left| S\right| =k}{ \underset{(S,Q)\in EC}{{\sum }}}}w(S,Q)\text {.} \end{aligned}$$

The value of the component $\left( z_{\lambda ,k,j}\right) _{i}$ follows from the fact that the last expression can be written as

$$\begin{aligned} \underset{j\in I_{\lambda ,k}}{\sum }\left[ \underset{\lambda _{Q}=\lambda }{ \underset{S\ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}} }\underset{\left| T\right| =j}{\underset{T\in Q\backslash \{S\}}{ \sum }}\frac{j}{n}w(S,Q)-\underset{\lambda _{Q}=\lambda ,\left| Q^{i}\right| =j}{\underset{S\not \ni i,\left| S\right| =k}{ \underset{(S,Q)\in EC}{{\sum }}}}\frac{k}{n}w(S,Q)\right] \text {.} \end{aligned}$$

$\square$

Proof of Proposition 3

Let $\varphi :G\rightarrow \mathbb {R} ^{n}$ be a linear symmetric solution. From Proposition 2, $w\in G$ decomposes as

$$\begin{aligned} w=\underset{(\lambda ,k)\in C_{n}}{\sum }a_{\lambda ,k}u_{\lambda }^{k}+ \underset{(\lambda ,k,j)\in D_{n}}{\sum }z_{\lambda ,k,j}^{(\lambda ,k)}+r \text {,} \end{aligned}$$

where by linearity,

$$\begin{aligned} \varphi _{i}(w)=\underset{(\lambda ,k)\in C_{n}}{\sum }a_{\lambda ,k}\varphi _{i}\left( u_{\lambda }^{k}\right) +\underset{(\lambda ,k,j)\in D_{n}}{\sum } \varphi _{i}\left( z_{\lambda ,k,j}^{(\lambda ,k)}\right) +\varphi _{i}\left( r\right) \text {.} \end{aligned}$$

Now, $\varphi _{i}\left( r\right) =0$ by Corollary 1 and Schur’s Lemma implies

$$\begin{aligned} \varphi _{i}(w)=\underset{(\lambda ,k)\in C_{n}}{\sum }a_{\lambda ,k}\alpha \left( \lambda ,k\right) +\underset{(\lambda ,k,j)\in D_{n}}{\sum }\beta \left( \lambda ,k,j\right) \cdot \left( z_{\lambda ,k,j}\right) _{i} \end{aligned}$$

for some constants $\{\alpha (\lambda ,k)\mid (\lambda ,k)\in C_{n}\}\cup \{\beta (\lambda ,k,j)\mid (\lambda ,k,j)\in D_{n}\}$.

Set $\left| (S,Q)_{\lambda ,k}\right| =\left| (S,Q)\in EC\mid \left| S\right| =k,\lambda _{Q}=\lambda \right|$, then

$$\begin{aligned} \varphi _{i}(w)= & {} \underset{(\lambda ,k)\in C_{n}}{\sum }\frac{\alpha \left( \lambda ,k\right) }{\left| (S,Q)_{\lambda ,k}\right| } \underset{\left| S\right| =k,\lambda _{Q}=\lambda }{\underset{ (S,Q)\in EC}{\sum }}w(S,Q) \\&+\underset{(\lambda ,k,j)\in D_{n}}{\sum }\beta \left( \lambda ,k,j\right) \left[ \underset{\lambda _{Q}=\lambda }{\underset{S\ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}}}\underset{\left| T\right| =j}{\underset{T\in Q\backslash \{S\}}{\sum }}jw(S,Q)-\underset{ \lambda _{Q}=\lambda ,\left| Q^{i}\right| =j}{\underset{S\not \ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}}}kw(S,Q)\right] \\= & {} \underset{(\lambda ,k)\in C_{n}}{\sum }\underset{\lambda _{Q}=\lambda }{ \underset{S\ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{\sum }}} \frac{\alpha \left( \lambda ,k\right) }{\left| (S,Q)_{\lambda ,k}\right| }w(S,Q)+\underset{(\lambda ,k)\in C_{n}}{\sum }\underset{ \lambda _{Q}=\lambda }{\underset{S\not \ni i,\left| S\right| =k}{ \underset{(S,Q)\in EC}{\sum }}}\frac{\alpha \left( \lambda ,k\right) }{ \left| (S,Q)_{\lambda ,k}\right| }w(S,Q) \\&+\underset{(\lambda ,k,j)\in D_{n}}{\sum }\underset{\lambda _{Q}=\lambda }{ \underset{S\ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}} }\underset{\left| T\right| =j}{\underset{T\in Q\backslash \{S\}}{ \sum }}\frac{j}{n}\beta \left( \lambda ,k,j\right) w(S,Q)\\&-\underset{(\lambda ,k,j)\in D_{n}}{\sum }\underset{\lambda _{Q}=\lambda ,\left| Q^{i}\right| =j}{\underset{S\not \ni i,\left| S\right| =k}{ \underset{(S,Q)\in EC}{{\sum }}}}\frac{k}{n}\beta \left( \lambda ,k,j\right) w(S,Q)\text {.} \end{aligned}$$

Observe that

$$\begin{aligned} \underset{(\lambda ,k)\in C_{n}}{\sum }\underset{\lambda _{Q}=\lambda }{ \underset{S\not \ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{\sum }}}w(S,Q)=\underset{(\lambda ,k,j)\in D_{n}}{\sum }\underset{\lambda _{Q}=\lambda ,\left| Q^{i}\right| =j}{\underset{S\not \ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}}}w(S,Q)\text {,} \end{aligned}$$

and

$$\begin{aligned} \underset{(\lambda ,k,j)\in D_{n}}{\sum }\underset{\lambda _{Q}=\lambda }{ \underset{S\ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}} }\underset{\left| T\right| =j}{\underset{T\in Q\backslash \{S\}}{ \sum }}w(S,Q)=\underset{(\lambda ,k)\in C_{n}}{\sum }\underset{\lambda _{Q}=\lambda }{\underset{S\ni i,\left| S\right| =k}{\underset{ (S,Q)\in EC}{{\sum }}}}\underset{j\in I_{\lambda ,k}}{\sum }m_{j}^{\lambda - \left[ k\right] }w(S,Q)\text {.} \end{aligned}$$

Thus

$$\begin{aligned} \varphi _{i}(w)= & {} \underset{(\lambda ,k)\in C_{n}}{\sum }\underset{\lambda _{Q}=\lambda }{\underset{S\ni i,\left| S\right| =k}{\underset{ (S,Q)\in EC}{\sum }}}\frac{\alpha \left( \lambda ,k\right) }{\left| (S,Q)_{\lambda ,k}\right| }w(S,Q)\\&+\underset{(\lambda ,k,j)\in D_{n}}{ \sum }\underset{\lambda _{Q}=\lambda ,\left| Q^{i}\right| =j}{ \underset{S\not \ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{ \sum }}}}\frac{\alpha \left( \lambda ,k\right) }{\left| (S,Q)_{\lambda ,k}\right| }w(S,Q) \\&+\underset{(\lambda ,k)\in C_{n}}{\sum }\underset{\lambda _{Q}=\lambda }{ \underset{S\ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}} }\underset{j\in I_{\lambda ,k}}{\sum }\frac{j}{n}m_{j}^{\lambda -\left[ k \right] }\beta \left( \lambda ,k,j\right) w(S,Q)\\&-\underset{(\lambda ,k,j)\in D_{n}}{\sum }\underset{\lambda _{Q}=\lambda ,\left| Q^{i}\right| =j}{ \underset{S\not \ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{ \sum }}}}\frac{k}{n}\beta \left( \lambda ,k,j\right) w(S,Q)\text {.} \end{aligned}$$

The result follows by setting $\gamma \left( \lambda ,k\right) =\frac{\alpha \left( \lambda ,k\right) }{\left| (S,Q)_{\lambda ,k}\right| }+ \underset{j\in I_{\lambda ,k}}{\sum }\frac{j}{n}m_{j}^{\lambda -\left[ k \right] }\beta \left( \lambda ,k,j\right)$ and $\delta \left( \lambda ,k,j\right) =\frac{\alpha \left( \lambda ,k\right) }{\left| (S,Q)_{\lambda ,k}\right| }-\frac{k}{n}\beta \left( \lambda ,k,j\right)$ . $\square$

Proof of Proposition 6

First, it is clear that the collection of games $\{e_{(S,Q)}\mid (S,Q)\in EC,S\ne \varnothing \}$ defined by

$$\begin{aligned} e_{(S,Q)}(T,P)=\left\{ \begin{array}{ll} 1 &{} \text {if }(T,P)=(S,Q) \\ 0 &{} \text {otherwise} \end{array} \right. \end{aligned}$$

constitutes a basis of G. Now, from Proposition 3,

$$\begin{aligned} \varphi _{i}(w)=\underset{(\lambda ,k)\in C_{n}}{\sum }\underset{\lambda _{Q}=\lambda }{\underset{S\ni i,\left| S\right| =k}{\underset{ (S,Q)\in EC}{\sum }}}\gamma \left( \lambda ,k\right) \cdot w(S,Q)+\underset{ (\lambda ,k,j)\in D_{n}}{\sum }\underset{\lambda _{Q}=\lambda ,\left| Q^{i}\right| =j}{\underset{S\not \ni i,\left| S\right| =k}{ \underset{(S,Q)\in EC}{\sum }}}\delta \left( \lambda ,k,j\right) \cdot w(S,Q) \end{aligned}$$

is a linear and symmetric solution for arbitrary constants $\{\gamma (\lambda ,k)\mid (\lambda ,k)\in C_{n}\}\cup \{\delta (\lambda ,k,j)\mid (\lambda ,k,j)\in D_{n}\}$. Suppose $i\in N$ is a null player in $e_{(S,Q)}$ for every $(S,Q)\in EC$; that is, $e_{(S,Q)}(R,P)=e_{(S,Q)}\left( R_{-i},\left\{ R_{-i},\{i\}\right\} \cup P_{-R}\right)$ for each (R, P) such that $i\in R$.

Nullity implies:

• 1.
  $$\begin{aligned} 0=\varphi _{i}(e_{(S,Q)})=\gamma (\lambda _{Q},s)+\delta (\lambda _{Q_{S,i}},s-1,1) \end{aligned}$$

  for every $(S,Q)\in EC$ such that $i\in S$ and $s>1$.

• 2.
  $$\begin{aligned} 0=\varphi _{i}(e_{(S,Q)})=\delta \left( \lambda _{Q},s,\left| Q^{i}\right| \right) \end{aligned}$$

  for every pair (S, Q) such that $i\notin S$ and $\left| Q^{i}\right| >1$.

Thus,

$$\begin{aligned} \delta (\lambda ,k,j)=\left\{ \begin{array}{ll} -\gamma (\left( \lambda -[k,1]\right) +[k+1],k+1) &{} \hbox { if } j=1 \\ 0 &{} \hbox { if } j\ne 1 \end{array} \right. \text {.} \end{aligned}$$

Therefore,

$$\begin{aligned} \varphi _{i}(w)= & {} \underset{(\lambda ,k)\in C_{n}}{\sum }\underset{\lambda _{Q}=\lambda }{\underset{S\ni i,\left| S\right| =k}{\underset{ (S,Q)\in EC}{\sum }}}\gamma \left( \lambda ,k\right) \cdot w(S,Q)\nonumber \\&-\underset{ (\lambda ,k,1)\in D_{n}}{\sum }\underset{\lambda _{Q}=\lambda ,\left| Q^{i}\right| =1}{\underset{S\not \ni i,\left| S\right| =k}{ \underset{(S,Q)\in EC}{\sum }}}\gamma (\left( \lambda -[k,1]\right) +[k+1],k+1)\cdot w(S,Q)\text {.} \end{aligned}$$

(7)

If we set $\lambda =\left( \widetilde{\lambda }-[\widetilde{k}]\right) +[ \widetilde{k}-1,1]$, $k=\widetilde{k}-1$, $S=\widetilde{S}_{-i}$ and $Q=\left\{ \widetilde{S}_{-i},\{i\}\right\} \cup \widetilde{Q}_{-\widetilde{S} }$, then the second part of the above expression reduces to

$$\begin{aligned}&\underset{(\lambda ,k,1)\in D_{n}}{\sum }\underset{\lambda _{Q}=\lambda ,\left| Q^{i}\right| =1}{\underset{S\not \ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{\sum }}}\gamma (\left( \lambda -[k,1]\right) +[k+1],k+1)\cdot w(S,Q) \\&\quad \quad =\underset{\widetilde{k}\ne 1}{\underset{(\widetilde{\lambda },\widetilde{ k})\in C_{n}}{\sum }}\underset{\widetilde{\lambda }_{\widetilde{Q}}= \widetilde{\lambda }}{\underset{\widetilde{S}\ni i,\left| \widetilde{S} \right| =\widetilde{k}}{\underset{(\widetilde{S},\widetilde{Q})\in EC}{ \sum }}}\gamma \left( \widetilde{\lambda },\widetilde{k}\right) \cdot w\left( \widetilde{S}_{-i},\left\{ \widetilde{S}_{-i},\{i\}\right\} \cup \widetilde{Q}_{-\widetilde{S}}\right) \text {.} \nonumber \end{aligned}$$

(8)

The result follows from substitution of the equallity (8) in the expression given by (7). $\square$

Proof of Proposition 7

( $\Rightarrow$ ) From Theorem 3,

$$\begin{aligned} \varphi _{i}(w)&=\frac{w(N,\{N\})}{n}+\underset{(\lambda ,k,j)\in D_{n}}{\sum }\delta (\lambda ,k,j)\\&\quad \times \left[ \underset{\lambda _{Q}=\lambda }{\underset{S\ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}}}\underset{ \left| T\right| =j}{\underset{T\in Q\backslash \{S\}}{\sum }}jw(S,Q)- \underset{\lambda _{Q}=\lambda ,\left| Q^{i}\right| =j}{\underset{ S\not \ni i,\left| S\right| =k}{\underset{(S,Q)\in EC}{{\sum }}}} kw(S,Q)\right] \end{aligned}$$

is a linear, symmetric and efficient solution for arbitrary constants $\{\delta (\lambda ,k,j)\mid (\lambda ,k,j)\in D_{n}\}$.

For $(S,Q)\in EC$, $i\in S$ and $T\in Q_{-S}$, denote $Q_{i,S,T}=\left\{ S_{-i},T_{+i}\right\} \cup Q_{-S,-T}$. Now, suppose $i\in N$ is a null player in $e_{(S,Q)}$ in the strong sense, for every $(S,Q)\in EC$; that is, $e_{(S,Q)}(R,P)=u_{(S,Q)}\left( R_{-i},P_{i,S,T}\right)$ for each (R, P) such that $i\in R$ and each $T\in P_{-R}$.

Weak nullity implies:

• 1.
  $$\begin{aligned} 0=\varphi _{i}(e_{(N,\{N\})})=\frac{1}{n}-(n-1)\delta ([n-1,1],n-1,1) \end{aligned}$$

  Hence,

  $$\begin{aligned} \delta ([n-1,1],n-1,1)=\frac{1}{n(n-1)} \end{aligned}$$
• 2.
  $$\begin{aligned} 0=\varphi _{i}(u_{(S,Q)})=\underset{T\in Q_{-S}}{\sum }\left[ t\delta (\lambda _{Q},s,t)-\left( s-1\right) \delta (\lambda _{Q_{i,S,T}},s-1,t+1) \right] \end{aligned}$$

  (9)

  for every pair (S, Q) such that $s\notin \{1,n\}$. Notice that the above relation yields many repeated equations. In particular, relation (9) provides the same equation for (S, Q) and $(S^{\prime },Q^{\prime })$, if $s=s^{\prime }$ and $\lambda _{Q}=\lambda _{Q^{\prime }}$.

  Now, for a fixed (S, Q) such that $s\notin \{1,n\}$, it holds:

  $$\begin{aligned} \underset{T\in Q_{-S}}{\sum }t\delta (\lambda _{Q},s,t)=\left\{ \begin{array}{ll} \left( m_{s}^{\lambda _{Q}}-1\right) s\delta (\lambda _{Q},s,s) &{} \hbox { if } t=s \\ \underset{z\ne s}{\underset{z\in \{\lambda _{Q}\}\cup \{0\}}{\sum }} zm_{z}^{\lambda _{Q}}\delta (\lambda _{Q},s,z) &{} \hbox { if } t\ne s \end{array} \right. \end{aligned}$$

  (10)

  and

  $$\begin{aligned} \underset{T\in Q_{-S}}{\sum }\delta (\lambda _{Q_{i,S,T}},s-1,t+1)=\left\{ \begin{array}{ll} \left( m_{s}^{\lambda _{Q}}-1\right) \delta \left( \left( \lambda _{Q}\right) _{s}^{s},s-1,s+1\right) &{}\quad \hbox {if}\,\, t=s \\ \underset{z\ne s}{\underset{z\in \{\lambda _{Q}\}\cup \{0\}}{\sum }} m_{z}^{\lambda _{Q}}\delta \left( \left( \lambda _{Q}\right) _{z}^{s},s-1,z+1\right) &{}\quad \hbox {if}\,\, t\ne s \end{array} \right. \end{aligned}$$

  (11)

  The system given in b) follows from the substitution of equalities (10) and (11) in relation (9).

( $\Leftarrow$ ) The converse is a straightforward computation in view of the equalities in part ($\Rightarrow$). $\square$