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$H^*(Y_n;\mathbb{Q}) \cong (H^*((S^3)^N;\mathbb{Q}))^{\Sigma_n}$.

Hence with rational coefficients this reduces to analyzing the invariant subspace of the action of $\Sigma_n$ on $H^*((S^3)^N;\mathbb{Q})\cong (\mathbb{Q}^2)^{\otimes N}$.

Let $H^*(S^3;\mathbb{Q})$ be generated by $s^0 \in H^0(S^3)$ and $s^3\in H^3(S^3)$. We have a projection $p_i: (S^3)^N \to S^3$ to the $i$th factor. Then $H^*((S^3)^N)$ has a basis $p_1^*(s^{i_1})\cup \cdots \cup p_N^*(s^{i_N})$, where $i_j\in \{0,3\}$. We may identify $H^*((S^3)^N)$ with $H^*(S^3)^{\otimes N}$ via the Kunneth formula with basis $s^{i_1}\otimes\cdots\otimes s^{i_N}$ (note that this is a graded commutative tensor product of graded commutative algebras).

For a vector space $V$ over $\mathbb{R}$ and a finite group $G$ acting on $V$, we have a retract $av: V\to V^G$ (the fixed points of the $G$-action on $V$) given by $av(v)=\frac{1}{|G|}\sum_{g\in G} g(v)$. Hence for any basis of $V$, the image under $av$ will span $V^G$.

Let’s apply this to the $\Sigma_N$ action on $H^*((S^3)^N)\cong H^3(S^3)^{\otimes N}$. For $\sigma \in \Sigma_N$, suppose that $(i_1,\ldots, i_N)=(i_{\sigma(1)},\ldots,i_{\sigma(N)})$. Consider the subset $J=\{ j | i_j=3\}$, then $\sigma$ restricts to a permutation of this set. Let $\epsilon_J(\sigma)$ be the sign of $\sigma_{|J}$. Then $\sigma (s^{i_1}\otimes \cdots\otimes s^{i_N})= s^{i_{\sigma(1)}}\otimes\cdots\otimes s^{i_{\sigma(N)}} = \epsilon_J(\sigma) s^{i_1}\otimes \cdots\otimes s^{i_N}$ due to the rule for changing signs of permutations of graded commutative tensor products. Thus we see that $av(s^{i_1}\otimes \cdots \otimes s^{i_N})$ (with associated $J=\{j|i_j=3\}$) will $= 0$ if there is $\sigma\in \Sigma_N$ such that $\sigma(J)=J$ and $\epsilon_J(\sigma)=-1$, and otherwise will $= \frac{1}{N!} \sum_{\sigma\in\Sigma_N} s^{i_{\sigma(1)}}\otimes \cdots \otimes s^{i_{\sigma(N)}} \neq 0$.

Now lets apply this to $H^*(Y_n;\mathbb{R})\cong H^*((S^3)^N;\mathbb{R})^{\Sigma_n} \cong (H^*(S^3)^{\otimes N})^{\Sigma_n}$, $N=\binom{n}{2}$.

We get a generating set for $H^*((S^3)^N)$ corresponding to subgraphs of the complete graph $K_n$ on $n$ vertices where if an edge does not connect $i,j$ we choose the generator $s^0$ of $H^0(S^3)$ corresponding to the $ij$ factor and if an edge connects $i,j$, we choose the generator $s^3$ of $H^3(S^3)$. Then $\Sigma_n$ acts on the vertices and hence on the edges. For a generator corresponding to the graph $\Gamma$, and a permuation $\sigma\in \Sigma_n$ preserving $\Gamma$, we may consider the sign of the action of $\Sigma_n$ on $E(\Gamma)$. If this sign is trivial for all $\sigma\in Aut(\Gamma) < \Sigma_n$, then the averaging operator on the corresponding generator will be non-zero, and we get an element of $H^*((S^3)^N)^{\Sigma_n}-\{0\}$. If the sign is non-trivial, then the averaging operator will be zero. One may also see that the elements will be linearly independent for each isomorphism type of graph, since the corresponding tensor factors will be distinct.

Thus the dimension of the fixed subspace, and hence $H^*(Y_n;\mathbb{R})$, will be the number of isomorphism classes of simple graphs $\Gamma$ on $n$ vertices such that the sign of the automorphism group $Aut(\Gamma)$ acting on $E(\Gamma)$ is trivial.

$H^*(Y_n;\mathbb{Q}) \cong (H^*((S^3)^N;\mathbb{Q}))^{\Sigma_n}$.

Hence with rational coefficients this reduces to analyzing the invariant subspace of the action of $\Sigma_n$ on $H^*((S^3)^N;\mathbb{Q})\cong (\mathbb{Q}^2)^{\otimes N}$.

Let $H^*(S^3;\mathbb{Q})$ be generated by $s^0 \in H^0(S^3)$ and $s^3\in H^3(S^3)$. We have a projection $p_i: (S^3)^N \to S^3$ to the $i$th factor. Then $H^*((S^3)^N)$ has a basis $p_1^*(s^{i_1})\cup \cdots \cup p_N^*(s^{i_N})$, where $i_j\in \{0,3\}$. We may identify $H^*((S^3)^N)$ with $H^*(S^3)^{\otimes N}$ via the Kunneth formula with basis $s^{i_1}\otimes\cdots\otimes s^{i_N}$ (note that this is a graded commutative tensor product of graded commutative algebras).

For a vector space $V$ over $\mathbb{Q}$ and a finite group $G$ acting on $V$, we have a retract $av: V\to V^G$ (the fixed points of the $G$-action on $V$) given by $av(v)=\frac{1}{|G|}\sum_{g\in G} g(v)$. Hence for any basis of $V$, the image under $av$ will span $V^G$.

Let’s apply this to the $\Sigma_N$ action on $H^*((S^3)^N)\cong H^3(S^3)^{\otimes N}$. For $\sigma \in \Sigma_N$, suppose that $(i_1,\ldots, i_N)=(i_{\sigma(1)},\ldots,i_{\sigma(N)})$. Consider the subset $J=\{ j | i_j=3\}$, then $\sigma$ restricts to a permutation of this set. Let $\epsilon_J(\sigma)$ be the sign of $\sigma_{|J}$. Then $\sigma (s^{i_1}\otimes \cdots\otimes s^{i_N})= s^{i_{\sigma(1)}}\otimes\cdots\otimes s^{i_{\sigma(N)}} = \epsilon_J(\sigma) s^{i_1}\otimes \cdots\otimes s^{i_N}$ due to the rule for changing signs of permutations of graded commutative tensor products. Thus we see that $av(s^{i_1}\otimes \cdots \otimes s^{i_N})$ (with associated $J=\{j|i_j=3\}$) will $= 0$ if there is $\sigma\in \Sigma_N$ such that $\sigma(J)=J$ and $\epsilon_J(\sigma)=-1$, and otherwise will $= \frac{1}{N!} \sum_{\sigma\in\Sigma_N} s^{i_{\sigma(1)}}\otimes \cdots \otimes s^{i_{\sigma(N)}} \neq 0$.

Now lets apply this to $H^*(Y_n;\mathbb{Q})\cong H^*((S^3)^N;\mathbb{Q})^{\Sigma_n} \cong (H^*(S^3)^{\otimes N})^{\Sigma_n}$, $N=\binom{n}{2}$.

We get a generating set for $H^*((S^3)^N)$ corresponding to subgraphs of the complete graph $K_n$ on $n$ vertices where if an edge does not connect $i,j$ we choose the generator $s^0$ of $H^0(S^3)$ corresponding to the $ij$ factor and if an edge connects $i,j$, we choose the generator $s^3$ of $H^3(S^3)$. Then $\Sigma_n$ acts on the vertices and hence on the edges. For a generator corresponding to the graph $\Gamma$, and a permuation $\sigma\in \Sigma_n$ preserving $\Gamma$, we may consider the sign of the action of $\Sigma_n$ on $E(\Gamma)$. If this sign is trivial for all $\sigma\in Aut(\Gamma) < \Sigma_n$, then the averaging operator on the corresponding generator will be non-zero, and we get an element of $H^*((S^3)^N)^{\Sigma_n}-\{0\}$. If the sign is non-trivial, then the averaging operator will be zero. One may also see that the elements will be linearly independent for each isomorphism type of graph, since the corresponding tensor factors will be distinct.

Thus the dimension of the fixed subspace, and hence $H^*(Y_n;\mathbb{Q})$, will be the number of isomorphism classes of simple graphs $\Gamma$ on $n$ vertices such that the sign of the automorphism group $Aut(\Gamma)$ acting on $E(\Gamma)$ is trivial.

$H^*(Y_n;\mathbb{Q}) \cong (H^*((S^3)^N;\mathbb{Q}))^{\Sigma_n}$.

Hence with rational coefficients this reduces to analyzing the invariant subspace of the action of $\Sigma_n$ on $H^*((S^3)^N;\mathbb{Q})\cong (\mathbb{Q}^2)^{\otimes N}$.

To analyze the invariant subspace, let’s switch to $\mathbb{R}$ coefficients and de Rham cohomology. Let $H^*(S^3;\mathbb{R})$ be generated by a $0$-form the constant function $s^0 \in \Omega^0(S^3) = C^{\infty}(S^3)$ and a volume form $s^3\in \Omega^3(S^3)$. We have a projection $p_i: (S^3)^N \to S^3$ to the $i$th factor. Then $H^*((S^3)^N)$ has a basis $p_1^*(s^{i_1})\wedge \cdots \wedge p_N^*(s^{i_N})$, where $i_j\in \{0,3\}$. We may identify $H^*((S^3)^N)$ with $H^*(S^3)^{\otimes N}$ via the Kunneth formula with basis $s^{i_1}\otimes\cdots\otimes s^{i_N}$ (note that this is a graded commutative tensor product of graded commutative algebras).

For a vector space $V$ over $\mathbb{R}$ and a finite group $G$ acting on $V$, we have a retract $av: V\to V^G$ (the fixed points of the $G$-action on $V$) given by $av(v)=\frac{1}{|G|}\sum_{g\in G} g(v)$. Hence for any basis of $V$, the image under $av$ will span $V^G$.

Let’s apply this to the $\Sigma_N$ action on $H^*((S^3)^N)\cong H^3(S^3)^{\otimes N}$. For $\sigma \in \Sigma_N$, suppose that $(i_1,\ldots, i_N)=(i_{\sigma(1)},\ldots,i_{\sigma(N)})$. Consider the subset $J=\{ j | i_j=3\}$, then $\sigma$ restricts to a permutation of this set. Let $\epsilon_J(\sigma)$ be the sign of $\sigma_{|J}$. Then $\sigma (s^{i_1}\otimes \cdots\otimes s^{i_N})= s^{i_{\sigma(1)}}\otimes\cdots\otimes s^{i_{\sigma(N)}} = \epsilon_J(\sigma) s^{i_1}\otimes \cdots\otimes s^{i_N}$ due to the rule for changing signs of permutations of graded commutative tensor products. Thus we see that $av(s^{i_1}\otimes \cdots \otimes s^{i_N})$ (with associated $J=\{j|i_j=3\}$) will $= 0$ if there is $\sigma\in \Sigma_N$ such that $\sigma(J)=J$ and $\epsilon_J(\sigma)=-1$, and otherwise will $= \frac{1}{N!} \sum_{\sigma\in\Sigma_N} s^{i_{\sigma(1)}}\otimes \cdots \otimes s^{i_{\sigma(N)}} \neq 0$.

Now lets apply this to $H^*(Y_n;\mathbb{R})\cong H^*((S^3)^N;\mathbb{R})^{\Sigma_n} \cong (H^*(S^3)^{\otimes N})^{\Sigma_n}$, $N=\binom{n}{2}$.

We get a generating set for $H^*((S^3)^N)$ corresponding to subgraphs of the complete graph $K_n$ on $n$ vertices where if an edge does not connect $i,j$ we choose the generator $s^0$ of $H^0(S^3)$ corresponding to the $ij$ factor and if an edge connects $i,j$, we choose the generator $s^3$ of $H^3(S^3)$. Then $\Sigma_n$ acts on the vertices and hence on the edges. For a generator corresponding to the graph $\Gamma$, and a permuation $\sigma\in \Sigma_n$ fixing $\Gamma$, we may consider the sign of the action of $\Sigma_n$ on $E(\Gamma)$. If this sign is trivial for all $\sigma\in Aut(\Gamma) < \Sigma_n$, then the averaging operator on the corresponding generator will be non-zero, and we get an element of $H^*((S^3)^N)^{\Sigma_n}-\{0\}$. If the sign is non-trivial, then the averaging operator will be zero. One may also see that the elements will be linearly independent for each isomorphism type of graph, since the corresponding tensor factors will be distinct.

Thus the dimension of the fixed subspace, and hence $H^*(Y_n;\mathbb{R})$, will be the number of isomorphism classes of simple graphs $\Gamma$ on $n$ vertices such that the sign of the automorphism group $Aut(\Gamma)$ acting on $E(\Gamma)$ is trivial.

$H^*(Y_n;\mathbb{Q}) \cong (H^*((S^3)^N;\mathbb{Q}))^{\Sigma_n}$.

Hence with rational coefficients this reduces to analyzing the invariant subspace of the action of $\Sigma_n$ on $H^*((S^3)^N;\mathbb{Q})\cong (\mathbb{Q}^2)^{\otimes N}$.

Let $H^*(S^3;\mathbb{Q})$ be generated by $s^0 \in H^0(S^3)$ and $s^3\in H^3(S^3)$. We have a projection $p_i: (S^3)^N \to S^3$ to the $i$th factor. Then $H^*((S^3)^N)$ has a basis $p_1^*(s^{i_1})\cup \cdots \cup p_N^*(s^{i_N})$, where $i_j\in \{0,3\}$. We may identify $H^*((S^3)^N)$ with $H^*(S^3)^{\otimes N}$ via the Kunneth formula with basis $s^{i_1}\otimes\cdots\otimes s^{i_N}$ (note that this is a graded commutative tensor product of graded commutative algebras).

For a vector space $V$ over $\mathbb{R}$ and a finite group $G$ acting on $V$, we have a retract $av: V\to V^G$ (the fixed points of the $G$-action on $V$) given by $av(v)=\frac{1}{|G|}\sum_{g\in G} g(v)$. Hence for any basis of $V$, the image under $av$ will span $V^G$.

Let’s apply this to the $\Sigma_N$ action on $H^*((S^3)^N)\cong H^3(S^3)^{\otimes N}$. For $\sigma \in \Sigma_N$, suppose that $(i_1,\ldots, i_N)=(i_{\sigma(1)},\ldots,i_{\sigma(N)})$. Consider the subset $J=\{ j | i_j=3\}$, then $\sigma$ restricts to a permutation of this set. Let $\epsilon_J(\sigma)$ be the sign of $\sigma_{|J}$. Then $\sigma (s^{i_1}\otimes \cdots\otimes s^{i_N})= s^{i_{\sigma(1)}}\otimes\cdots\otimes s^{i_{\sigma(N)}} = \epsilon_J(\sigma) s^{i_1}\otimes \cdots\otimes s^{i_N}$ due to the rule for changing signs of permutations of graded commutative tensor products. Thus we see that $av(s^{i_1}\otimes \cdots \otimes s^{i_N})$ (with associated $J=\{j|i_j=3\}$) will $= 0$ if there is $\sigma\in \Sigma_N$ such that $\sigma(J)=J$ and $\epsilon_J(\sigma)=-1$, and otherwise will $= \frac{1}{N!} \sum_{\sigma\in\Sigma_N} s^{i_{\sigma(1)}}\otimes \cdots \otimes s^{i_{\sigma(N)}} \neq 0$.

Now lets apply this to $H^*(Y_n;\mathbb{R})\cong H^*((S^3)^N;\mathbb{R})^{\Sigma_n} \cong (H^*(S^3)^{\otimes N})^{\Sigma_n}$, $N=\binom{n}{2}$.

We get a generating set for $H^*((S^3)^N)$ corresponding to subgraphs of the complete graph $K_n$ on $n$ vertices where if an edge does not connect $i,j$ we choose the generator $s^0$ of $H^0(S^3)$ corresponding to the $ij$ factor and if an edge connects $i,j$, we choose the generator $s^3$ of $H^3(S^3)$. Then $\Sigma_n$ acts on the vertices and hence on the edges. For a generator corresponding to the graph $\Gamma$, and a permuation $\sigma\in \Sigma_n$ preserving $\Gamma$, we may consider the sign of the action of $\Sigma_n$ on $E(\Gamma)$. If this sign is trivial for all $\sigma\in Aut(\Gamma) < \Sigma_n$, then the averaging operator on the corresponding generator will be non-zero, and we get an element of $H^*((S^3)^N)^{\Sigma_n}-\{0\}$. If the sign is non-trivial, then the averaging operator will be zero. One may also see that the elements will be linearly independent for each isomorphism type of graph, since the corresponding tensor factors will be distinct.

Thus the dimension of the fixed subspace, and hence $H^*(Y_n;\mathbb{R})$, will be the number of isomorphism classes of simple graphs $\Gamma$ on $n$ vertices such that the sign of the automorphism group $Aut(\Gamma)$ acting on $E(\Gamma)$ is trivial.

$H^*(Y_n;\mathbb{Q}) \cong (H^*((S^3)^N;\mathbb{Q}))^{\Sigma_n}$.

Hence with rational coefficients this reduces to analyzing the invariant subspace of the action of $\Sigma_n$ on $H^*((S^3)^N;\mathbb{Q})\cong (\mathbb{Q}^2)^{\otimes N}$.

To analyze the invariant subspace, let’s switch to $\mathbb{R}$ coefficients and de Rham cohomology. Let $H^*(S^3;\mathbb{R})$ be generated by a $0$-form the constant function $s^0 \in \Omega^0(S^3) = C^{\infty}(S^3)$ and a volume form $s^3\in \Omega^3(S^3)$. We have a projection $p_i: (S^3)^N \to S^3$ to the $i$th factor. Then $H^*((S^3)^N)$ has a basis $p_1^*(s^{i_1})\wedge \cdots \wedge p_N^*(s^{i_N})$, where $i_j\in \{0,3\}$. We may identify $H^*((S^3)^N)$ with $H^*(S^3)^{\wedge N}$ via the Kunneth formula with basis $s^{i_1}\wedge\cdots\wedge s^{i_N}$.

For a vector space $V$ over $\mathbb{R}$ and a finite group $G$ acting on $V$, we have a retract $av: V\to V^G$ (the fixed points of the $G$-action on $V$) given by $av(v)=\frac{1}{|G|}\sum_{g\in G} g(v)$. Hence for any basis of $V$, the image under $av$ will span $V^G$.

Let’s apply this to the $\Sigma_N$ action on $H^*((S^3)^N)\cong H^3(S^3)^{\wedge N}$. For $\sigma \in \Sigma_N$, suppose that $(i_1,\ldots, i_N)=(i_{\sigma(1)},\ldots,i_{\sigma(N)})$. Consider the subset $J=\{ j | i_j=3\}$, then $\sigma$ restricts to a permutation of this set. Let $\epsilon_J(\sigma)$ be the sign of $\sigma_{|J}$. Then $\sigma (s^{i_1}\wedge \cdots\wedge s^{i_N})= s^{i_{\sigma(1)}}\wedge\cdots\wedge s^{i_{\sigma(N)}} = \epsilon_J(\sigma) s^{i_1}\wedge \cdots\wedge s^{i_N}$ due to the rule for changing signs of permutations of differential forms. Thus we see that $av(s^{i_1}\wedge \cdots \wedge s^{i_N})$ (with associated $J=\{j|i_j=3\}$) will $= 0$ if there is $\sigma\in \Sigma_N$ such that $\sigma(J)=J$ and $\epsilon_J(\sigma)=-1$, and otherwise will $= \frac{1}{N!} \sum_{\sigma\in\Sigma_N} s^{i_{\sigma(1)}}\wedge \cdots \wedge s^{i_{\sigma(N)}} \neq 0$.

Now lets apply this to $H^*(Y_n;\mathbb{R})\cong H^*((S^3)^N;\mathbb{R})^{\Sigma_n} \cong (H^*(S^3)^{\wedge N})^{\Sigma_n}$, $N=\binom{n}{2}$.

We get a generating set for $H^*((S^3)^N)$ corresponding to subgraphs of the complete graph $K_n$ on $n$ vertices where if an edge does not connect $i,j$ we choose the generator $s^0$ of $H^0(S^3)$ corresponding to the $ij$ factor and if an edge connects $i,j$, we choose the generator $s^3$ of $H^3(S^3)$. Then $\Sigma_n$ acts on the vertices and hence on the edges. For a generator corresponding to the graph $\Gamma$, and a permuation $\sigma\in \Sigma_n$ fixing $\Gamma$, we may consider the sign of the action of $\Sigma_n$ on $E(\Gamma)$. If this sign is trivial for all $\sigma\in Aut(\Gamma) < \Sigma_n$, then the averaging operator on the corresponding generator will be non-zero, and we get an element of $H^*((S^3)^N)^{\Sigma_n}-\{0\}$. If the sign is non-trivial, then the averaging operator will be zero. One may also see that the elements will be linearly independent for each isomorphism type of graph, since the corresponding tensor factors will be distinct.

Thus the dimension of the fixed subspace, and hence $H^*(Y_n;\mathbb{R})$, will be the number of isomorphism classes of simple graphs $\Gamma$ on $n$ vertices such that the sign of the automorphism group $Aut(\Gamma)$ acting on $E(\Gamma)$ is trivial.

$H^*(Y_n;\mathbb{Q}) \cong (H^*((S^3)^N;\mathbb{Q}))^{\Sigma_n}$.

Hence with rational coefficients this reduces to analyzing the invariant subspace of the action of $\Sigma_n$ on $H^*((S^3)^N;\mathbb{Q})\cong (\mathbb{Q}^2)^{\otimes N}$.

To analyze the invariant subspace, let’s switch to $\mathbb{R}$ coefficients and de Rham cohomology. Let $H^*(S^3;\mathbb{R})$ be generated by a $0$-form the constant function $s^0 \in \Omega^0(S^3) = C^{\infty}(S^3)$ and a volume form $s^3\in \Omega^3(S^3)$. We have a projection $p_i: (S^3)^N \to S^3$ to the $i$th factor. Then $H^*((S^3)^N)$ has a basis $p_1^*(s^{i_1})\wedge \cdots \wedge p_N^*(s^{i_N})$, where $i_j\in \{0,3\}$. We may identify $H^*((S^3)^N)$ with $H^*(S^3)^{\otimes N}$ via the Kunneth formula with basis $s^{i_1}\otimes\cdots\otimes s^{i_N}$ (note that this is a graded commutative tensor product of graded commutative algebras).

For a vector space $V$ over $\mathbb{R}$ and a finite group $G$ acting on $V$, we have a retract $av: V\to V^G$ (the fixed points of the $G$-action on $V$) given by $av(v)=\frac{1}{|G|}\sum_{g\in G} g(v)$. Hence for any basis of $V$, the image under $av$ will span $V^G$.

Let’s apply this to the $\Sigma_N$ action on $H^*((S^3)^N)\cong H^3(S^3)^{\otimes N}$. For $\sigma \in \Sigma_N$, suppose that $(i_1,\ldots, i_N)=(i_{\sigma(1)},\ldots,i_{\sigma(N)})$. Consider the subset $J=\{ j | i_j=3\}$, then $\sigma$ restricts to a permutation of this set. Let $\epsilon_J(\sigma)$ be the sign of $\sigma_{|J}$. Then $\sigma (s^{i_1}\otimes \cdots\otimes s^{i_N})= s^{i_{\sigma(1)}}\otimes\cdots\otimes s^{i_{\sigma(N)}} = \epsilon_J(\sigma) s^{i_1}\otimes \cdots\otimes s^{i_N}$ due to the rule for changing signs of permutations of graded commutative tensor products. Thus we see that $av(s^{i_1}\otimes \cdots \otimes s^{i_N})$ (with associated $J=\{j|i_j=3\}$) will $= 0$ if there is $\sigma\in \Sigma_N$ such that $\sigma(J)=J$ and $\epsilon_J(\sigma)=-1$, and otherwise will $= \frac{1}{N!} \sum_{\sigma\in\Sigma_N} s^{i_{\sigma(1)}}\otimes \cdots \otimes s^{i_{\sigma(N)}} \neq 0$.

Now lets apply this to $H^*(Y_n;\mathbb{R})\cong H^*((S^3)^N;\mathbb{R})^{\Sigma_n} \cong (H^*(S^3)^{\otimes N})^{\Sigma_n}$, $N=\binom{n}{2}$.

We get a generating set for $H^*((S^3)^N)$ corresponding to subgraphs of the complete graph $K_n$ on $n$ vertices where if an edge does not connect $i,j$ we choose the generator $s^0$ of $H^0(S^3)$ corresponding to the $ij$ factor and if an edge connects $i,j$, we choose the generator $s^3$ of $H^3(S^3)$. Then $\Sigma_n$ acts on the vertices and hence on the edges. For a generator corresponding to the graph $\Gamma$, and a permuation $\sigma\in \Sigma_n$ fixing $\Gamma$, we may consider the sign of the action of $\Sigma_n$ on $E(\Gamma)$. If this sign is trivial for all $\sigma\in Aut(\Gamma) < \Sigma_n$, then the averaging operator on the corresponding generator will be non-zero, and we get an element of $H^*((S^3)^N)^{\Sigma_n}-\{0\}$. If the sign is non-trivial, then the averaging operator will be zero. One may also see that the elements will be linearly independent for each isomorphism type of graph, since the corresponding tensor factors will be distinct.

Thus the dimension of the fixed subspace, and hence $H^*(Y_n;\mathbb{R})$, will be the number of isomorphism classes of simple graphs $\Gamma$ on $n$ vertices such that the sign of the automorphism group $Aut(\Gamma)$ acting on $E(\Gamma)$ is trivial.