A New Look at the Study of Solutions for Games in Partition Function Form

Abstract

This chapter studies the structure of games in partition function and according to an axiomatic point of view, we provide a global description of linear symmetric solutions by means of a decomposition of the set of such games (as well as of a decomposition of the space of payoff vectors). The exhibition of relevant subspaces in such decomposition and based on the idea that every permutation of the set of players may be thought of as a linear map, allow for a new look at linear symmetric solutions.

Appendix

The point of view we take in this work depends heavily on a certain decomposition of G (and $\mathbb{R}^{n}$) as a direct sum of subspaces, as well as on the Schur’s Lemma. A reference for this subject is [4]; nevertheless, we recall all the 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 ρ: S _n  → GL(G), where GL(G) is the group of invertible linear maps in G. This action is given by:

$$\displaystyle{ (\theta \cdot w)(S,Q):= [\rho (\theta )(w)](S,Q) = w[\theta ^{-1}(S,Q)] }$$

for every θ ∈ S _n , w ∈ G and (S, Q) ∈ EC.

The space of payoff vectors, $\mathbb{R}^{n}$, is also an S _n -representation:

$$\displaystyle{ \theta (x_{1},x_{2},\ldots,x_{n}):= [\rho (\theta )](x_{1},x_{2},\ldots,x_{n}) = (x_{\theta (1)},x_{\theta (2)},\ldots,x_{\theta (n)}) }$$

In general, for any representation X of a finite group H, there is a decomposition

$$\displaystyle{ X = X_{1}^{\oplus a_{1} } \oplus X_{2}^{\oplus a_{2} } \oplus \cdot \cdot \cdot \oplus X_{j}^{\oplus a_{j} } }$$

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 → 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 5.

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

$$\displaystyle{ \chi _{X}(h) = Tr\left (\rho (h)\right ) }$$

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 × n matrix according to its action expressed in some convenient basis, then sum up the diagonal elements of the matrix for h to get χ _X (h). For example, the trace of the identity map of an n-dimensional vector space is the trace of the n × n identity matrix, or n. In fact, χ _X (e) = dimX for any finite dimensional representation X of any group.

Notice that, in particular, we have χ _X (h) = χ _X (ghg ⁻¹) for g, h ∈ H. So that χ _X is constant on the conjugacy classes of H; such a function is called a class function.

Definition 6.

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

$$\displaystyle{ \left \langle \chi _{1},\chi _{2}\right \rangle = \frac{1} {\left \vert H\right \vert }\mathop{\sum }\limits_{h \in H}\overline{\chi _{1}(h)} \cdot \chi _{2}(h) }$$

(5)

The multiplicities of irreducible subspaces in a representation can be calculated via the previous inner product. Formally, 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 $a_{i} = \left \langle \chi _{Z},\chi _{Z_{i}}\right \rangle$.

Proof (of Theorem 1).

First, $\left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}}\right \rangle$ and $\left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}^{\perp }}\right \rangle$ are the number of subspaces isomorphic to the trivial (Δ _n ) and standard representation (Δ _n ^⊥) within G _λ ^k, respectively.

We start by computing the number of copies of Δ _n in G _λ ^k:

$$\displaystyle{ \left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}}\right \rangle = \frac{1} {n!}\mathop{\sum }\limits_{\theta \in S_{n}}\chi _{G_{\lambda }^{k}}(\theta )\chi _{\varDelta _{n}}(\theta ) = \frac{1} {n!}\mathop{\sum }\limits_{\theta \in S_{n}}\chi _{G_{\lambda }^{k}}(\theta ) }$$

Notice that $\chi _{G_{\lambda }^{k}}(\theta )$ is just the number of pairs (S, Q) ∈ EC with $\left \vert S\right \vert = k$ and λ _Q  = λ, that are fixed under θ ∈ S _n .

Define

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

Then,

$$\displaystyle{ \chi _{G_{\lambda }^{k}}(\theta ) =\mathop{\mathop{\sum }\limits_{ (S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } \left \{\theta \right \}_{(S,Q)} }$$

and so,

$$\displaystyle{ \left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}}\right \rangle = \frac{1} {n!}\mathop{\sum }\limits_{\theta \in S_{n}}\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } \left \{\theta \right \}_{(S,Q)} = \frac{1} {n!}\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda }\mathop{\sum }\limits_{\theta \in S_{n}}\left \{\theta \right \}_{(S,Q)} }$$

where,

$$\displaystyle{ \mathop{\sum }\limits_{\theta \in S_{n}}\left \{\theta \right \}_{(S,Q)} = \left \vert \left \{\theta \in S_{n}:\theta (Q) = Q,\text{ }\theta (S) = S\right \}\right \vert }$$

Now, S _n acts on the set Q _λ and take Q ∈ Q _λ . The orbit of Q under S _n is

$$\displaystyle{ S_{n}Q = \left \{\theta (Q)\mid \theta \in S_{n}\right \} = Q_{\lambda } }$$

and the isotropy subgroup of Q is

$$\displaystyle{ (S_{n})_{Q} = \left \{\theta \in S_{n}\mid \theta (Q) = Q\right \} }$$

By Lagrange theorem, we get

$$\displaystyle{ \left \vert S_{n}Q\right \vert = \frac{\left \vert S_{n}\right \vert } {\left \vert (S_{n})_{Q}\right \vert } = \left \vert Q_{\lambda }\right \vert \ \ \Rightarrow \ \ \left \vert (S_{n})_{Q}\right \vert = \frac{n!} {\left \vert Q_{\lambda }\right \vert } }$$

Notice that H = (S _n ) _Q acts on Q and takes S ∈ Q such that $\left \vert S\right \vert = k$. The orbit of S under H is

$$\displaystyle{ HS = \left \{hS\mid h \in H\right \} =\{ T \in Q\mid \left \vert T\right \vert = k\} }$$

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

$$\displaystyle{ H_{S} = \left \{h \in H\mid h(S) = S\right \} = \left \{\theta \in S_{n}\mid \theta (Q) = Q,\text{ }\theta (S) = S\right \} }$$

Again, by Lagrange theorem, we get

$$\displaystyle{ \left \vert HS\right \vert = \frac{\left \vert H\right \vert } {\left \vert H_{S}\right \vert } = \frac{\left \vert (S_{n})_{Q}\right \vert } {\left \vert H_{S}\right \vert } = m_{k}^{\lambda } \ \ \Rightarrow \ \ \left \vert H_{ S}\right \vert = \frac{\left \vert (S_{n})_{Q}\right \vert } {m_{k}^{\lambda }} = \frac{n!} {\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} }$$

And therefore,

$$\displaystyle{ \left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}}\right \rangle = \frac{1} {n!}\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } \frac{n!} {\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} = \frac{1} {n!}\left \vert Q_{\lambda }\right \vert m_{k} \frac{n!} {\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} = 1 }$$

Now, we compute the multiplicity of Δ _n ^⊥ in G _λ ^k. Since $\mathbb{R}^{n} =\varDelta _{n} \oplus \varDelta _{n}^{\perp }$, then $\chi _{\mathbb{R}^{n}} =\chi _{\varDelta _{n}} +\chi _{\varDelta _{n}^{\perp }}\ \Rightarrow \ \left \langle \chi _{G_{\lambda }^{k}},\chi _{\mathbb{R}^{n}}\right \rangle = \left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}}\right \rangle + \left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}^{\perp }}\right \rangle \ \Rightarrow \left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}^{\perp }}\right \rangle = \left \langle \chi _{G_{\lambda }^{k}},\chi _{\mathbb{R}^{n}}\right \rangle - 1$.

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

$$\displaystyle\begin{array}{rcl} \left \langle \chi _{G_{\lambda }^{k}},\chi _{\mathbb{R}^{n}}\right \rangle & =& \left \langle \chi _{G_{\lambda }^{k}},\chi _{G_{[1,1,\ldots,1]}^{1}}\right \rangle = \frac{1} {n!}\mathop{\sum }\limits_{\theta \in S_{n}}\chi _{G_{\lambda }^{k}}(\theta )\chi _{G_{[1,1,\ldots,1]}^{1}}(\theta ) {}\\ & =& \frac{1} {n!}\mathop{\sum }\limits_{\theta \in S_{n}}\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } \left \{\theta \right \}_{(S,Q)}\mathop{\mathop{\mathop{\sum }\limits_{(S^{{\prime}},Q^{{\prime}}) \in EC}}\limits_{\left \vert S^{{\prime}}\right \vert = 1}}\limits_{Q^{{\prime}} = Q_{ [1,1,\ldots,1]}}\left \{\theta \right \}_{(S^{{\prime}},Q^{{\prime}})} {}\\ & =& \frac{1} {n!}\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda }\mathop{\mathop{\sum }\limits_{ (S^{{\prime}},Q^{{\prime}}) \in EC}}\limits_{\left \vert S^{{\prime}}\right \vert = 1,\lambda _{ Q^{{\prime}}} = [1,1,\ldots,1]}\mathop{\sum }\limits_{\theta \in S_{n}}\left \{\theta \right \}_{(S,Q)}\left \{\theta \right \}_{(S^{{\prime}},Q^{{\prime}})}{}\\ \end{array}$$

For x ∈ S:

$$\displaystyle{ \mathop{\sum }\limits_{\theta \in S_{n}}\left \{\theta \right \}_{(S,Q)}\left \{\theta \right \}_{(S^{{\prime}},Q^{{\prime}})} = \left \{\theta \in S_{n}\mid \theta (Q) = Q,\text{ }\theta (S) = S,\text{ }\theta (x) = x\right \} }$$

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

$$\displaystyle{ Mx = \left \{\theta (x)\mid \theta \in M\right \} = S }$$

and the isotropy subgroup of x is

$$\displaystyle{ M_{x} = \left \{\theta \in M\mid \text{ }\theta (x) = x\right \} = \left \{\theta \in S_{n}\mid \theta (Q) = Q,\text{ }\theta (S) = S,\text{ }\theta (x) = x\right \} }$$

By Lagrange theorem,

$$\displaystyle{ \left \vert Mx\right \vert = \frac{\left \vert M\right \vert } {\left \vert M_{x}\right \vert } = \frac{\left \vert H_{S}\right \vert } {\left \vert M_{x}\right \vert } = k \ \ \Rightarrow \ \ \left \vert M_{x}\right \vert = \frac{\left \vert H_{S}\right \vert } {k} = \frac{n!} {k\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} }$$

Thus,

$$\displaystyle\begin{array}{rcl} \left \langle \chi _{G_{\lambda }^{k}},\chi _{\mathbb{R}^{n}}\right \rangle & =& \frac{1} {n!}\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } \left [k \frac{n!} {k\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} + m_{\lambda _{2}}^{\lambda }\lambda _{2} \frac{n!} {m_{\lambda _{2}}^{\lambda }\lambda _{2}\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} + \cdot \cdot \cdot \right. {}\\ & +& \left.m_{\lambda _{l}}^{\lambda }\lambda _{l} \frac{n!} {m_{\lambda _{l}}^{\lambda }\lambda _{l}\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }}\right ] {}\\ & =& \frac{1} {n!}\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } \left \vert \lambda ^{\circ }\right \vert \frac{n!} {\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} = \frac{1} {n!}\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }\left [\left \vert \lambda ^{\circ }\right \vert \frac{n!} {\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }}\right ] {}\\ & =& \left \vert \lambda ^{\circ }\right \vert {}\\ \end{array}$$

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

$$\displaystyle\begin{array}{rcl} \left \langle \chi _{G_{\lambda }^{k}},\chi _{\mathbb{R}^{n}}\right \rangle & =& \frac{1} {n!}\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } \left [\begin{array}{c} k \frac{n!} {k\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} + k(m_{k}^{\lambda } - 1) \frac{n!} {k(m_{k}^{\lambda }-1)\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }}+ \\ m_{\lambda _{2}}^{\lambda }\lambda _{ 2} \frac{n!} {m_{\lambda _{2}}^{\lambda }\lambda _{2}\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} + \cdot \cdot \cdot + m_{\lambda _{l}}^{\lambda }\lambda _{ l} \frac{n!} {m_{\lambda _{l}}^{\lambda }\lambda _{l}\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} \end{array} \right ] {}\\ & =& \frac{1} {n!}\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } (\left \vert \lambda ^{\circ }\right \vert + 1) \frac{n!} {\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }} = \frac{1} {n!}\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }\left [(\left \vert \lambda ^{\circ }\right \vert + 1) \frac{n!} {\left \vert Q_{\lambda }\right \vert m_{k}^{\lambda }}\right ] {}\\ & =& \left \vert \lambda ^{\circ }\right \vert + 1 {}\\ \end{array}$$

In summary,

$$\displaystyle{ \left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}}\right \rangle = 1 \ \ \ \text{and }\ \ \left \langle \chi _{G_{\lambda }^{k}},\chi _{\varDelta _{n}^{\perp }}\right \rangle = \left \vert I_{\lambda,k}\right \vert }$$

The next task is to identify such copies of Δ _n and Δ _n ^⊥ inside G _λ ^k. For that end, define the functions $L_{\lambda,k}: \mathbb{R}^{n} \rightarrow G_{\lambda }^{k}$ by L _λ, k (x) = x ^(λ, k). These maps are isomorphisms between U _λ ^k and Δ _n (similarly between $\left \{x_{j}^{\left (\lambda,k\right )}\mid x_{j} \in \varDelta _{n}^{\perp }\right \}$ and Δ _n ^⊥, for each j ∈ I _λ, k ) since they are linear, S _n -equivariant, and 1 − 1. Thus, inside of G _λ ^k, we have the images of these two subspaces: U _λ ^k = L _λ, k (Δ _n ) and V _λ ^k = L _λ, k (Δ _n ^⊥).

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.

Proof (of Proposition 2).

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

Thus, the projection of w onto U is

$$\displaystyle{ \mathop{\sum }\limits_{(\lambda,k) \in A_{n}} \frac{\left \langle w,u_{\lambda }^{k}\right \rangle } {\left \langle u_{\lambda }^{k},u_{\lambda }^{k}\right \rangle }u_{\lambda }^{k} }$$

and so,

$$\displaystyle{ a_{\lambda,k} = \frac{\left \langle w,u_{\lambda }^{k}\right \rangle } {\left \langle u_{\lambda }^{k},u_{\lambda }^{k}\right \rangle } = \frac{\mathop{\sum }\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } w(S,Q)} {\left \vert (S,Q) \in EC: \left \vert S\right \vert = k,\lambda _{Q} =\lambda \right \vert } }$$

Now, for each (λ, k) ∈ A _n , we define $f^{\lambda,k}: G \rightarrow \mathbb{R}^{n}$ as

$$\displaystyle{ f_{i}^{\lambda,k}(w) =\mathop{\mathop{\mathop{ \sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{ Q} =\lambda } w(S,Q) }$$

where each f ^λ, k is S _n -equivariant and observe that f ^[n], n(w) = w(N, {N})(1, …, 1). Let z ∈ Δ _n ^⊥, then f ^λ, k(z _γ, i, j ^(γ, i)) = 0 if λ ≠ γ or i ≠ k, whereas (by Schur’s Lemma)  f ^λ, k(z _λ, k, j ^(λ, k)) = z _λ, k, j  ∈ Δ _n ^⊥.

Let $p: \mathbb{R}^{n} \rightarrow \varDelta _{n}^{\perp }$ be the projection of $\mathbb{R}^{n}$ onto Δ _n ^⊥ given by

$$\displaystyle{ p_{i}(x) = x_{i} - \frac{1} {n}\mathop{\mathop{\sum }\limits^{n}}\limits_{j = 1}x_{j} }$$

This projection is equivariant, sends Δ _n to zero and it is the identity on Δ _n ^⊥.

Next, define L ^λ, k: G → Δ _n ^⊥ as L ^λ, k = p ∘ f ^λ, k. Observe that

$$\displaystyle{ L^{\lambda,k}(w) =\mathop{\sum }\limits_{ j \in I_{\lambda,k}}z_{\lambda,k,j} }$$

since by equivariance, $f^{\lambda,k}\left (U\right ) \subset \varDelta _{n}$ and $f^{\lambda,k}\left (W\right ) = 0$. Moreover, f ^λ, k(z _γ, i, j ^(γ, i)) = 0 if λ ≠ γ or i ≠ k. Then,

$$\displaystyle\begin{array}{rcl} L_{i}^{\lambda,k}(w)& =& p_{ i}(f^{\lambda,k}(w)) =\mathop{\mathop{\mathop{ \sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{ Q} =\lambda } w(S,Q) - \frac{1} {n}\mathop{\sum }\limits_{l \in N}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni l,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda } w(S,Q) {}\\ & =& \frac{n - k} {n} \mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda } w(S,Q) -\frac{k} {n}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S\not\ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda } w(S,Q) {}\\ \end{array}$$

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

$$\displaystyle{ \mathop{\sum }\limits_{j \in I_{\lambda,k}}\left [\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda }\mathop{\mathop{\sum }\limits_{ T \in Q\setminus \{S\}}}\limits_{\left \vert T\right \vert = j}\frac{j} {n}w(S,Q) -\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S\not\ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda,\left \vert Q^{i}\right \vert = j}\frac{k} {n}w(S,Q)\right ] }$$

Proof (of Proposition 3).

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

$$\displaystyle{ w =\mathop{\sum }\limits_{ (\lambda,k) \in A_{n}}a_{\lambda,k}u_{\lambda }^{k} +\mathop{\sum }\limits_{ (\lambda,k,j) \in B_{ n}}z_{\lambda,k,j}^{(\lambda,k)} + r }$$

where by linearity,

$$\displaystyle{ \varphi _{i}(w) =\mathop{\sum }\limits_{ (\lambda,k) \in A_{n}}a_{\lambda,k}\varphi _{i}\left (u_{\lambda }^{k}\right ) +\mathop{\sum }\limits_{ (\lambda,k,j) \in B_{ n}}\varphi _{i}\left (z_{\lambda,k,j}^{(\lambda,k)}\right ) +\varphi _{ i}\left (r\right ) }$$

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

$$\displaystyle{ \varphi _{i}(w) =\mathop{\sum }\limits_{ (\lambda,k) \in A_{n}}a_{\lambda,k}\alpha \left (\lambda,k\right ) +\mathop{\sum }\limits_{ (\lambda,k,j) \in B_{n}}\beta \left (\lambda,k,j\right ) \cdot \left (z_{\lambda,k,j}\right )_{i} }$$

for some constants {α(λ, k)∣(λ, k) ∈ A _n } ∪{β(λ, k, j)∣(λ, k, j) ∈ B _n }.

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

$$\displaystyle\begin{array}{rcl} \varphi _{i}(w)& =& \mathop{\sum }\limits_{(\lambda,k) \in A_{n}} \frac{\alpha \left (\lambda,k\right )} {\left \vert (S,Q)_{\lambda,k}\right \vert }\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{\left \vert S\right \vert = k,\lambda _{Q} =\lambda } w(S,Q) {}\\ & & +\mathop{\sum }\limits_{(\lambda,k,j) \in B_{n}}\beta \left (\lambda,k,j\right )\left [\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda }\mathop{\mathop{\sum }\limits_{ T \in Q\setminus \{S\}}}\limits_{\left \vert T\right \vert = j}jw(S,Q) -\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S\not\ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda,\left \vert Q^{i}\right \vert = j}kw(S,Q)\right ] {}\\ & =& \mathop{\sum }\limits_{(\lambda,k) \in A_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda } \frac{\alpha \left (\lambda,k\right )} {\left \vert (S,Q)_{\lambda,k}\right \vert }w(S,Q) +\mathop{\sum }\limits_{ (\lambda,k) \in A_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S\not\ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda } \frac{\alpha \left (\lambda,k\right )} {\left \vert (S,Q)_{\lambda,k}\right \vert }w(S,Q) {}\\ & & +\mathop{\sum }\limits_{(\lambda,k,j) \in B_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda }\mathop{\mathop{\sum }\limits_{ T \in Q\setminus \{S\}}}\limits_{\left \vert T\right \vert = j}\frac{j} {n}\beta \left (\lambda,k,j\right )w(S,Q) {}\\ & & -\mathop{\sum }\limits_{(\lambda,k,j) \in B_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S\not\ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda,\left \vert Q^{i}\right \vert = j}\frac{k} {n}\beta \left (\lambda,k,j\right )w(S,Q) {}\\ \end{array}$$

Observe that

$$\displaystyle{ \mathop{\sum }\limits_{(\lambda,k) \in A_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S\not\ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda } w(S,Q) =\mathop{\sum }\limits_{ (\lambda,k,j) \in B_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S\not\ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda,\left \vert Q^{i}\right \vert = j}w(S,Q) }$$

and

$$\displaystyle{ \mathop{\sum }\limits_{(\lambda,k,j) \in B_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda }\mathop{\mathop{\sum }\limits_{ T \in Q\setminus \{S\}}}\limits_{\left \vert T\right \vert = j}w(S,Q) =\mathop{\sum }\limits_{ (\lambda,k) \in A_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda }\mathop{\sum }\limits_{ j \in I_{\lambda,k}}m_{j}^{\lambda -\left [\left \vert S\right \vert \right ]}w(S,Q) }$$

Thus

$$\displaystyle\begin{array}{rcl} \varphi _{i}(w)& =& \mathop{\sum }\limits_{(\lambda,k) \in A_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda } \frac{\alpha \left (\lambda,k\right )} {\left \vert (S,Q)_{\lambda,k}\right \vert }w(S,Q) +\mathop{\sum }\limits_{ (\lambda,k,j) \in B_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S\not\ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda,\left \vert Q^{i}\right \vert = j} \frac{\alpha \left (\lambda,k\right )} {\left \vert (S,Q)_{\lambda,k}\right \vert }w(S,Q) {}\\ & & +\mathop{\sum }\limits_{(\lambda,k) \in A_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S \ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda }\mathop{\sum }\limits_{ j \in I_{\lambda,k}} \frac{j} {n}m_{j}^{\lambda -\left [\left \vert S\right \vert \right ]}\beta \left (\lambda,k,j\right )w(S,Q) {}\\ & & -\mathop{\sum }\limits_{(\lambda,k,j) \in B_{n}}\mathop{\mathop{\mathop{\sum }\limits_{(S,Q) \in EC}}\limits_{S\not\ni i,\left \vert S\right \vert = k}}\limits_{\lambda _{Q} =\lambda,\left \vert Q^{i}\right \vert = j}\frac{k} {n}\beta \left (\lambda,k,j\right )w(S,Q) {}\\ \end{array}$$

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