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Math-1055 · mathematics · Graph theory

Are almost all graphs determined by their spectrum?

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Statement

Definitions and mathematical statement

For an integer $n\geq1$, let $\mathcal G_n$ be the set of all simple undirected graphs on the labelled vertex set $[n]=\{1,\ldots,n\}$. The adjacency matrix $A_G$ has entry $1$ at $(u,v)$ precisely when $uv\in E(G)$, and has zero diagonal. Its adjacency spectrum is the multiset of its $n$ real eigenvalues, including their multiplicities. Two graphs are cospectral when these multisets agree; equivalently, their characteristic polynomials $\det(tI-A_G)$ agree.

A graph $G$ is determined by its adjacency spectrum if every finite simple graph $H$ cospectral with $G$ is isomorphic to $G$. Here an isomorphism is a bijection between the vertex sets preserving adjacency in both directions. Relabelling a graph therefore does not count as a different spectral realization. Equality of spectra already forces $|V(H)|=|V(G)|$.

The question is whether \[ \lim_{n\to\infty} \frac{|\{G\in\mathcal G_n:G\text{ is determined by its spectrum}\}|} {2^{\binom n2}}=1. \] Equivalently, choose each possible edge independently with probability $1/2$; does the probability of a nonisomorphic cospectral mate tend to zero? This specifies the usual labelled random-graph meaning of "almost all." The data supplied are only the adjacency spectrum: neither eigenvectors nor the spectrum of the complement or Laplacian is included in the reconstruction requirement.

Short English statement

Does the adjacency spectrum uniquely identify a uniformly random large graph, up to renaming its vertices, with probability tending to one?

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Math-1055-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-1055

Are almost all graphs determined by their spectrum?

Definitions and mathematical statement

For an integer $n\geq1$, let $\mathcal G_n$ be the set of all simple undirected graphs on the labelled vertex set $[n]=\{1,\ldots,n\}$. The adjacency matrix $A_G$ has entry $1$ at $(u,v)$ precisely when $uv\in E(G)$, and has zero diagonal. Its adjacency spectrum is the multiset of its $n$ real eigenvalues, including their multiplicities. Two graphs are cospectral when these multisets agree; equivalently, their characteristic polynomials $\det(tI-A_G)$ agree.

A graph $G$ is determined by its adjacency spectrum if every finite simple graph $H$ cospectral with $G$ is isomorphic to $G$. Here an isomorphism is a bijection between the vertex sets preserving adjacency in both directions. Relabelling a graph therefore does not count as a different spectral realization. Equality of spectra already forces $|V(H)|=|V(G)|$.

The question is whether \[ \lim_{n\to\infty} \frac{|\{G\in\mathcal G_n:G\text{ is determined by its spectrum}\}|} {2^{\binom n2}}=1. \] Equivalently, choose each possible edge independently with probability $1/2$; does the probability of a nonisomorphic cospectral mate tend to zero? This specifies the usual labelled random-graph meaning of "almost all." The data supplied are only the adjacency spectrum: neither eigenvectors nor the spectrum of the complement or Laplacian is included in the reconstruction requirement.

Short English statement

Does the adjacency spectrum uniquely identify a uniformly random large graph, up to renaming its vertices, with probability tending to one?

Research attempt

Scope and process.
This is a GPT-6 Astra Ultra research attempt dated 1 October 2026, using primary-source browsing and elementary mathematical checks. The canonical definition above is retained. Results below are restricted results or reductions, not a claim of a new solution to the entire catalogue question. No formal proof assistant or external mathematical referee checked this text.
Literature and chosen obstruction.
Koval and Kwan [R1] construct exponentially many graphs determined by their adjacency spectrum, without claiming the almost-all conclusion. Abiad, Brouwer and Haemers [R2] analyze Godsil--McKay switching and the distinction between cospectrality and nonisomorphism. We check a basic switching construction explicitly, then prove that one fixed-size switching mechanism occurs with vanishing probability in $G(n,1/2)$. Eliminating one mechanism is not a proof that all spectral mates vanish.
A matrix proof of switching.
Let $X$ be a set of $k$ vertices, with $k$ even. Assume its induced graph is regular and each vertex outside $X$ has either $0$, $k/2$ or $k$ neighbours in $X$. Write the adjacency matrix in blocks as \[ A=\begin{pmatrix}B&N\\N^{\mathsf T}&D\end{pmatrix},\qquad Q=\frac2k J-I_k. \] Since $J^2=kJ$, we have $Q^2=I_k$ and $Q^{\mathsf T}=Q$. Regularity and symmetry of $B$ imply $BJ=JB=rJ$, whence $QBQ=B$. A column $v$ of $N$ of weight zero or $k$ is fixed by $Q$, and a column of weight $k/2$ is sent to ${\bf1}-v$. Thus conjugation by $\operatorname{diag}(Q,I)$ replaces each half-neighbourhood by its complement, leaves the other edges alone, and yields a simple graph adjacency matrix $A'$. It is an orthogonal similarity, so $A,A'$ have identical characteristic polynomials.
A verified nonisomorphic pair.
Take independent $X=\{1,2,3,4\}$ and independent outside vertices $a,b,c$, whose neighbourhoods in $X$ are respectively $\{1,2\},\{1,3\},\{1,4\}$. The graph is a three-armed star with each edge subdivided once. After switching, vertex $1$ is isolated and the other six vertices form a cycle. The graphs are nonisomorphic because their degree multisets are respectively \[ \{3,2,2,2,1,1,1\},\qquad\{2,2,2,2,2,2,0\}. \] The $3\times3$ matrix $N^{\mathsf T}N$ has diagonal entries two and off-diagonal entries one, so equals $I_3+J_3$ and has eigenvalues $4,1,1$. The bipartite adjacency spectrum therefore consists of $\pm2$, two copies of each of $\pm1$, and one zero. Both characteristic polynomials are $t(t^2-4)(t^2-1)^2$. The exact rational-matrix check in scripts/check_attempt_batch_algebra_2026_10_01.py also verifies the conjugation and differing degree lists.
Why this local mechanism is rare in a random graph.
Fix a particular four-vertex set $X$ in $G(n,1/2)$. For any outside vertex, its four independent adjacency bits to $X$ have allowed weights zero, two or four with probability \[ (1+6+1)/16=1/2. \] The bits for different outside vertices are independent. Therefore the probability all outside vertices satisfy the switching condition is $2^{-(n-4)}$. Requiring regularity inside $X$ can only decrease this probability. A union bound gives \[ \mathbb P(\text{some four-vertex switching set exists}) \le\binom n4\,2^{-(n-4)}\longrightarrow0. \] For a fixed even $k\ge4$, the same argument replaces $1/2$ by $\theta_k=(2+\binom k{k/2})/2^k<1$, giving the bound $\binom nk\theta_k^{n-k}\to0$. The case $k=2$ is different: $Q$ just swaps the two vertices, so this operation produces an isomorphic graph and is not a spectral ambiguity.
Verification and exact logical gap.
The nonisomorphic example proves that spectra do not determine every graph, but the conjecture concerns a probability tending to one. The probability argument concerns each fixed size $k$; it cannot be summed over unbounded $k$ without additional uniform estimates. More fundamentally, a cospectral mate need not arise from a single switching set of this form. Even ruling out all such switchings would leave other orthogonal similarities and global constructions. Adjacency cospectrality alone is used throughout; no eigenvectors, Laplacian data or spectrum of the complement have been supplied.
Final status.
UNRESOLVED. A genuine cospectral ambiguity and a rigorous vanishing-probability estimate for each fixed-size switching mechanism are established. Controlling every possible nonisomorphic cospectral mate is the first missing universal step.

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