I am investigating a particular map from a product of three-spheres to the moduli space of (non-negative, real edge weight) networks. The map in question is

$$f: \smash{\left( \mathbb{S}^3 \right)}^N \rightarrow \mathbb{R}_{\ge 0}^N$$

where $N=\binom{n}{2}$ is the number of edges of a complete graph. We are placing a qubit on each edge of the underlying graph, given by $g: \mathbb{R}_{\ge 0}^N \rightarrow \{0,1\}^N$, a semi-ring morphism.

A qubit lives in the Bloch sphere, $\mathbb{S}^3$. Define $e_{ij}:=e_{ij}^0 \lvert0\rangle + e_{ij}^1 \rvert1\rangle$, with $\lvert e_{ij}^0\rvert^2 + \lvert e_{ij}^1\rvert^2 = 1$, and $e_{ij}^k \in \mathbb{C}$. $f$ is given by applying $e_{ij} \rightarrow \lvert e_{ij}^1\rvert^2$ for $1 \le i < j \le n$.

To get unlabeled networks, quotient by the permutation group $\Sigma_n$ where $\sigma \cdot e_{ij} = e_{\sigma(i)\sigma(j)}$.

EDIT 1: As pointed out in the comments, this map has fibers which are tori over a generic point, and some of those tori are replaced by circles when the coordinates are 0 (or perhaps 1).

Thus, $f$ is a singular toric fibration. Is it possible to apply a modified Serre spectral sequence to compute the cohomology?

It has been shown by myself and Stephen Rosenberg that the cohomology of the base space is trivial here with code on GitHub.

EDIT 2: Let $Y_n=(\mathbb{S}^3)^N / \Sigma_n$ and $X_N=(\mathbb{S}^3)^N/\Sigma_N$. Claim: $H^*(Y_n)$ is non-trivial. $H^*(Y_n)$ sits in $H^*(X_N)$ as an invariant subspace, since $X_N \rightarrow Y_n$ as $\Sigma_n \subset \Sigma_N$ and $H^*$ is a contravariant functor. Note that $X_N$ is a symmetric space. $H^*(\mathbb{S}^3)$ is non-trivial, so we expect $H^*(X_N)$ to be nontrivial.

EDIT 3: Note that $(P^1)^N/\Sigma_N \cong P^N$ by mapping inhomogeneous coordinates $(Z_1,...,Z_N)$ to coefficients of the symmetric polynomial $(x-Z_1)...(x-Z_N)$. $(S^3)^N/\Sigma_N$ has a $T^N$ fibration over $(P^1)^N/\Sigma_N$ by Hopf $S^3 \rightarrow P^1$. Then there is a spectral sequence on this fiber bundle. The $E_2$ page is the tensor product between $H^*(\text{base}=P^N)$ and $H^*(\text{fiber}=T^N)$. So $H^*((S^3)^N/\Sigma_N)$ has dim. not greater than dim. $$H^*(\text{base}=P^N) \otimes H^*(\text{fiber}=T^N)$$

Acknowledgements: Thank you to Siu-Cheong Lau for helpful comments.

I am investigating a particular map from a product of three-spheres to the moduli space of (non-negative, real edge weight) networks. The map in question is

$$f: \smash{\left( \mathbb{S}^3 \right)}^N \rightarrow \mathbb{R}_{\ge 0}^N$$

where $N=\binom{n}{2}$ is the number of edges of a complete graph. We are placing a qubit on each edge of the underlying graph, given by $g: \mathbb{R}_{\ge 0}^N \rightarrow \{0,1\}^N$, a semi-ring morphism.

A qubit lives in the Bloch sphere, $\mathbb{S}^3$. Define $e_{ij}:=e_{ij}^0 \lvert0\rangle + e_{ij}^1 \rvert1\rangle$, with $\lvert e_{ij}^0\rvert^2 + \lvert e_{ij}^1\rvert^2 = 1$, and $e_{ij}^k \in \mathbb{C}$. $f$ is given by applying $e_{ij} \rightarrow \lvert e_{ij}^1\rvert^2$ for $1 \le i < j \le n$.

To get unlabeled networks, quotient by the permutation group $\Sigma_n$ where $\sigma \cdot e_{ij} = e_{\sigma(i)\sigma(j)}$.

Let $Y_n=(\mathbb{S}^3)^N / \Sigma_n$ and $X_N=(\mathbb{S}^3)^N/\Sigma_N$. What is $H^*(Y_n;\mathbb{Z})$?

I am investigating a particular map from a product of three-spheres to the moduli space of (non-negative, real edge weight) networks. The map in question is

$$f: \smash{\left( \mathbb{S}^3 \right)}^N \rightarrow \mathbb{R}_{\ge 0}^N$$

where $N=\binom{n}{2}$ is the number of edges of a complete graph. We are placing a qubit on each edge of the underlying graph, given by $g: \mathbb{R}_{\ge 0}^N \rightarrow \{0,1\}^N$, a semi-ring morphism.

A qubit lives in the Bloch sphere, $\mathbb{S}^3$. Define $e_{ij}:=e_{ij}^0 \lvert0\rangle + e_{ij}^1 \rvert1\rangle$, with $\lvert e_{ij}^0\rvert^2 + \lvert e_{ij}^1\rvert^2 = 1$, and $e_{ij}^k \in \mathbb{C}$. $f$ is given by applying $e_{ij} \rightarrow \lvert e_{ij}^1\rvert^2$ for $1 \le i < j \le n$.

To get unlabeled networks, quotient by the permutation group $\Sigma_n$ where $\sigma \cdot e_{ij} = e_{\sigma(i)\sigma(j)}$.

EDIT 1: As pointed out in the comments, this map has fibers which are tori over a generic point, and some of those tori are replaced by circles when the coordinates are 0 (or perhaps 1).

Thus, $f$ is a singular toric fibration. Is it possible to apply a modified Serre spectral sequence to compute the cohomology?

It has been shown by myself and Stephen Rosenberg that the cohomology of the base space is trivial here with code on GitHub.

EDIT 2: Let $Y_n=(\mathbb{S}^3)^N / \Sigma_n$ and $X_N=(\mathbb{S}^3)^N/\Sigma_N$. Claim: $H^*(Y_n)$ is non-trivial. $H^*(X_N)$ sits in $H^*(Y_n)$ as an invariant subspace, since $Y_n \rightarrow X_N$ as $\Sigma_n \subset \Sigma_N$ and $H^*$ is a contravariant functor. Note that $X_N$ is a symmetric space. $H^*(\mathbb{S}^3)$ is non-trivial, so we expect $H^*(X_N)$ to be nontrivial.

ChatGPT gives a helpful outline of how to compute $H^*(X_N)$ when $N=4$, noting that $H^*(S^3)=\mathbb{R}[1,x]$, with 1 in degree 0 and $x$ in degree 3. $H^*((S^3)^4)=\mathbb{R}[1,x]^{\otimes 4}$. Then it states "the action of $\Sigma_4$ will induce certain relations in the cohomology", and suggest using tools from equivariant cohomology and representation theory.

EDIT 3: From Siu-Cheong: Note that $(P^1)^N/\Sigma_N \cong P^N$ by mapping inhomogeneous coordinates $(Z_1,...,Z_N)$ to coefficients of the symmetric polynomial $(x-Z_1)...(x-Z_N)$. $(S^3)^N/\Sigma_N$ has a $T^N$ fibration over $(P^1)^N/\Sigma_N$ by Hopf $S^3 \rightarrow P^1$. Then there is a spectral sequence on this fiber bundle. The $E_2$ page is the tensor product between $H^*(\text{base}=P^N)$ and $H^*(\text{fiber}=T^N)$. So $H^*((S^3)^N/\Sigma_N)$ has dim. not greater than $H^*(\text{base}=P^N) \otimes H^*(\text{fiber}=T^N)$.

Acknowledgements: Thank you to Siu-Cheong Lau for helpful comments.

I am investigating a particular map from a product of three-spheres to the moduli space of (non-negative, real edge weight) networks. The map in question is

$$f: \smash{\left( \mathbb{S}^3 \right)}^N \rightarrow \mathbb{R}_{\ge 0}^N$$

where $N=\binom{n}{2}$ is the number of edges of a complete graph. We are placing a qubit on each edge of the underlying graph, given by $g: \mathbb{R}_{\ge 0}^N \rightarrow \{0,1\}^N$, a semi-ring morphism.

A qubit lives in the Bloch sphere, $\mathbb{S}^3$. Define $e_{ij}:=e_{ij}^0 \lvert0\rangle + e_{ij}^1 \rvert1\rangle$, with $\lvert e_{ij}^0\rvert^2 + \lvert e_{ij}^1\rvert^2 = 1$, and $e_{ij}^k \in \mathbb{C}$. $f$ is given by applying $e_{ij} \rightarrow \lvert e_{ij}^1\rvert^2$ for $1 \le i < j \le n$.

To get unlabeled networks, quotient by the permutation group $\Sigma_n$ where $\sigma \cdot e_{ij} = e_{\sigma(i)\sigma(j)}$.

EDIT 1: As pointed out in the comments, this map has fibers which are tori over a generic point, and some of those tori are replaced by circles when the coordinates are 0 (or perhaps 1).

Thus, $f$ is a singular toric fibration. Is it possible to apply a modified Serre spectral sequence to compute the cohomology?

It has been shown by myself and Stephen Rosenberg that the cohomology of the base space is trivial here with code on GitHub.

EDIT 2: Let $Y_n=(\mathbb{S}^3)^N / \Sigma_n$ and $X_N=(\mathbb{S}^3)^N/\Sigma_N$. Claim: $H^*(Y_n)$ is non-trivial. $H^*(Y_n)$ sits in $H^*(X_N)$ as an invariant subspace, since $X_N \rightarrow Y_n$ as $\Sigma_n \subset \Sigma_N$ and $H^*$ is a contravariant functor. Note that $X_N$ is a symmetric space. $H^*(\mathbb{S}^3)$ is non-trivial, so we expect $H^*(X_N)$ to be nontrivial.

EDIT 3: Note that $(P^1)^N/\Sigma_N \cong P^N$ by mapping inhomogeneous coordinates $(Z_1,...,Z_N)$ to coefficients of the symmetric polynomial $(x-Z_1)...(x-Z_N)$. $(S^3)^N/\Sigma_N$ has a $T^N$ fibration over $(P^1)^N/\Sigma_N$ by Hopf $S^3 \rightarrow P^1$. Then there is a spectral sequence on this fiber bundle. The $E_2$ page is the tensor product between $H^*(\text{base}=P^N)$ and $H^*(\text{fiber}=T^N)$. So $H^*((S^3)^N/\Sigma_N)$ has dim. not greater than dim. $$H^*(\text{base}=P^N) \otimes H^*(\text{fiber}=T^N)$$

Acknowledgements: Thank you to Siu-Cheong Lau for helpful comments.