Character Table Spectracones & the $n=5$ SNIEP

1.1 Perron Similarities and Spectracones

The symmetric nonnegative inverse eigenvalue problem (SNIEP) asks which real lists are spectra of entrywise nonnegative symmetric matrices. For lists of length at most four, the Perron condition $\lambda_1=\max_i|\lambda_i|$ and nonnegative trace suffice. Dimension five introduces additional obstructions. Here $A\ge0$ always means entrywise nonnegative.

An invertible matrix $S$ is a Perron similarity if it diagonalizes an irreducible nonnegative matrix. Johnson and Paparella developed the real theory [JP16] and later its complex extension [JP25]. In the general theory spectral vectors may be complex; the plots below use real vectors. We place a positive Perron column first.

The entries of the realizing matrix depend linearly on $x$. Thus $\mathcal C(S)$ is polyhedral, and its normalized section is a polytope; the positive Perron column ensures $|x_i|\le1$ on this section. A Perron similarity is ideal when $\mathcal C(S)=\mathcal C_r(S)$, the conical hull of its rows. Character tables have the further useful property that one column of the realizing matrix tests nonnegativity of the entire matrix.

1.2 Character Tables Give Explicit Symmetric Realizations

For the trivial character, $L_1(x)=1+\sum_{k=2}^r c_kx_k$. Along $x_2=\cdots=x_r=t$, this gives $t\ge-1/(|G|-1)$; every nontrivial-character inequality becomes $d_i(1-t)\ge0$. These bounds explain the shared diagonal apex in each plotted family.

Assigning the repeated eigenvalues to different nonidentity classes changes the halfspaces. We call assignments equivalent when they define the same polytope in the ordered chamber. The groups studied here are a selected family, not an exhaustive list of groups with five classes: the dihedral group of order $14$, for example, also has a real $5\times5$ character table [D14].

2.1 The Loewy–McDonald Benchmark

Fix $1\ge a\ge b\ge-1$. The ordered triangle has area $2$. Imposing trace nonnegativity $1+2a+2b\ge0$ leaves a quadrilateral with vertices $(-1/4,-1/4),(1,1),(1,-1),(1/2,-1)$ and area $23/16$.

Write

The lower boundary is the trace-zero segment $b=-1/2-a$ for $-1/4\le a\le K_{1,a}$, the hyperbolic arc $b=-1/(4a)$ for $K_{1,a}\le a\le K_{2,a}$, and the line $b=-(1+a)/2$ for $K_{2,a}\le a\le1$. The remaining boundary follows $b=a$ and $a=1$.

Integrating the vertical width gives the exact area

2.2 Selected $5\times5$ Character-Table Pieces

There are $\binom42=6$ assignments of $(a,a,b,b)$ to the four nonidentity classes: $(a,a,b,b),(a,b,a,b),(a,b,b,a),(b,a,a,b),(b,a,b,a),(b,b,a,a)$.

The cyclic group $\mathbb Z_5$

The Fourier table $F_5=[\omega^{jk}]_{j,k=0}^4$, $\omega=e^{2\pi i/5}$, becomes real symmetric on the paired assignment $(1,a,b,b,a)$. With $\phi=(1+\sqrt5)/2$, the three nonnegativity tests are

The resulting triangle has vertices $(-1/4,-1/4),K_1,(1,1)$ and area $5\sqrt5/16\approx0.698771243$. It exactly follows the trace-zero segment down to $K_1$.

2.3 What the Adjoined Block Adds

Before adjoining a block, the selected union is $D_8\cup\mathbb Z_5$; the displayed $S_4,A_5$ pieces add no area. Its area is $1.292589478$, or $96.42\%$ of $\mathcal B$. It misses the small curved gap and a triangular wedge with vertices $P_*,Q_*,(1,-1)$.

The golden-ratio facet meets $1+a+2b=0$ at $P_*$ and $1+3a+4b=0$ at $Q_*$:

The wedge has area approximately $0.047611591$, or $3.55\%$ of the benchmark.

The condition supplies a reducible realization; the equation $1+a+2b=0$ alone does not force every realization to be reducible. For example, $(1,1/3,1/3,-2/3,-2/3)$ has a strictly positive symmetric circulant realization whose first row is

2.4 The Remaining Curved Gap

Once the adjoined block is present, the final selected union is already $\mathbb Z_5\cup(V_4\oplus[1])$. The $D_8/Q_8$, $S_4$, and $A_5$ pieces provide useful comparisons, but add no area once these two constructions are present: for $a\le0$ the $D_8$ pieces lie in the $\mathbb Z_5$ triangle, and for $a\ge0$ they lie in the adjoined quadrilateral.

The fraction covered is $99.97214\%$. The missing area is approximately $0.000373445$, or $0.027857\%$. It lies between the hyperbolic arc $K_1K_2$ and the two straight segments $K_1P_*$ and $P_*K_2$ shown in the figure below.

A finite collection of fixed similarities gives a finite polyhedral union. It therefore cannot equal this benchmark, whose boundary includes a nontrivial hyperbolic arc. If a separate necessity theorem identifies the benchmark with the entire SNIEP slice, the same observation applies to that full slice.

Decimals are approximations. Areas refer to the stated chamber; axis padding is only a display setting.

3.1 The Ambient Trace Polytope

Split the second repeated pair and impose $1\ge a\ge b\ge c\ge-1$. The natural necessary container is

It has five facets, six vertices, and volume $55/48\approx1.145833333$. Its three positive-trace sign vertices are $(1,1,1),(1,1,-1),(1,-1,-1)$. The other three vertices lie on the trace-zero face:

The character constructions below certify realizability of their own regions; the ambient trace polytope is not itself a realizability criterion.

3.2 The Three-Dimensional Assignment Pieces

There are $4!/2!=12$ assignments of $(a,a,b,c)$ to the nonidentity classes. They give three distinct $D_8/Q_8$ pieces, seven $S_4$ pieces, and seven $A_5$ pieces. These are intersections of spectratopes with the ordered slice, so all are convex polytopes.

The three $D_8/Q_8$ tetrahedra

The following inequalities are additional to those defining $\Delta_3$:

Set $E=(-1/7,-1/7,-1/7)$ and $O=(1,1,1)$. Each piece has four vertices:

Their volumes are $4/21$, $2/63$, and $4/63$, respectively. Since $T_2\subseteq T_3$ and $\operatorname{Vol}(T_1\cap T_3)=292/6825$,

Containment of the selected $S_4$ and $A_5$ pieces

All seven $S_4$ pieces lie inside $T_1$. The selected $A_5$ pieces are contained in $T_1$ as well. Their union volumes are approximately $0.163113364$ and $0.022785841$. These groups illustrate different eigenvector choices, but contribute no additional volume to the selected nonabelian union.

The adjoined $V_4\oplus[1]$ polytope

Apply $H_4$ to $(1,a,b,c)$ and adjoin $[a]$. Inside $\Delta_3$ the active conditions are

The other two $H_4$ inequalities are redundant here. The five vertices are $(0,-1/2,-1/2),(0,0,-1),(0,0,0),(1,-1,-1),(1,1,1)$. Its volume is $5/12$, or $4/11\approx36.36\%$ of $\Delta_3$. Dropping the final inequality would give a different, larger region.

Why $\mathbb Z_5$ has zero three-dimensional volume

A real symmetric circulant has equal eigenvalues on each conjugate frequency pair. For the spectrum $(1,a,a,b,c)$ this forces $b=c$. Thus the Fourier construction occupies a two-dimensional triangle on that wall, with vertices $(-1/4,-1/4,-1/4)$, $(K_{1,a},K_{1,b},K_{1,b})$, and $(1,1,1)$. Adding it changes the set but does not change its three-dimensional volume.

3.3 Interactive Three-Dimensional Explorer

Select a group to inspect its assignment pieces against the ambient polytope. The surfaces below use the same character-table inequalities as the cross-sections.

3.4 Cross-Sections and the $b=c$ Wall

At fixed $a$, slice the triangle $a\ge b\ge c\ge-1$ by $b+c\ge-1-2a$. The slider spans $-1/4\le a\le1$. The companion plot shows where the slice meets the repeated-pair wall $b=c$.

• Negative $a$: the selected nonabelian pieces begin at $a=-1/7$ for $D_8$, $a=-1/23$ for $S_4$, and $a=-1/59$ for $A_5$. Their absence below those thresholds reflects the limitations of these fixed similarities. In fact, the whole $a\le0$ part of $\Delta_3$ is already symmetrically realizable by the Suleĭmanova theorem: there is at most one positive eigenvalue and the trace is nonnegative [GLP26, §1].
• The adjoined block at $a=0$: it covers the triangle $0\ge b\ge c\ge-1$, $b+c\ge-1$, with vertices $(0,0),(0,-1),(-1/2,-1/2)$ in $(b,c)$ coordinates.
• Containment persists in each slice: every $S_4$ cross-section lies inside the first $D_8$ cross-section. Overlapping translucent fills should not be interpreted as new spectral coverage.

The meshes and clipped polygons are numerical visualizations of the displayed halfspaces. The reported areas and volumes are computed separately from the corresponding polytopes or analytic boundary formulas; sampled hyperbola points are used only for drawing.

1. [GLP26] D. Z. Gershnik, A. J. Lewis, and P. Paparella, Character tables are ideal Perron similarities, Journal of Algebra 694 (2026), 782–800. Published paper; arXiv version 2. The character inequalities are Theorem 4.3.
2. [JP16] C. R. Johnson and P. Paparella, Perron spectratopes and the real nonnegative inverse eigenvalue problem, Linear Algebra and its Applications 493 (2016), 281–300. Preprint.
3. [JP25] C. R. Johnson and P. Paparella, Perron similarities and the nonnegative inverse eigenvalue problem, Transactions of the American Mathematical Society 378 (2025), 8361–8389. Published paper; Preprint.
4. [MN00] J. J. McDonald and M. Neumann, The Soules approach to the inverse eigenvalue problem for nonnegative symmetric matrices of order $n\le5$, Contemporary Mathematics 259 (2000), 387–390. Author’s publication list.
5. [LM04] R. Loewy and J. J. McDonald, The symmetric nonnegative inverse eigenvalue problem for $5\times5$ matrices, Linear Algebra and its Applications 393 (2004), 275–298. Published paper.
6. [D14] GroupNames, character table of the dihedral group of order $14$ (labelled $D_7$ in the convention used by that database). Character table and class sizes.