## The effect of adding an edge on Laplacian eigenvalues

Adding an edge to a connected graph (between two of its existing vertices) makes the graph “even more” connected. How to quantify this? The Poincaré inequality gives a way to do this: for any function ${f\colon V\to \mathbb R}$ with zero mean,

${\displaystyle \sum_V f(v)^2 \le C \sum_E (f(u)-f(v))^2 }$

where the sum on the left is over the vertex set ${V}$, the sum on the right is over the edge set ${E}$, and ${u, v}$ on the right are the endpoints of an edge. Suppose ${C}$ is the smallest possible constant in this inequality, the Poincaré constant of the graph. Adding an edge to the graph does not change the sum on the left but may increase the sum on the right. Thus, the new Poincaré constant ${C'}$ satisfies ${C'\le C}$: greater connectivity means smaller Poincaré constant.

It is more convenient to work with the reciprocal of ${C}$, especially because ${1/C = \lambda_2}$, the smallest positive eigenvalue of the graph Laplacian. Let ${\lambda_2'}$ be the corresponding eigenvalue after an edge is added. The above shows that ${\lambda_2 \le \lambda_2'}$. But there is also an upper bound on how much this eigenvalue (or any Laplacian eigenvalue) can grow. Indeed, adding an edge amounts to adding the matrix

${\displaystyle B = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}}$ (with a bunch of zero rows/columns)

to the Laplacian matrix ${L}$. The eigenvalues of ${B}$ are ${0}$ and ${2}$. As a consequence of the Courant-Fischer minimax formulas, the ordered eigenvalues ${\lambda_1 \le \dots \le \lambda_n}$ satisfy ${\lambda_k(L) \le \lambda_k(L+B) \le \lambda_k(L) + 2}$ for every ${k}$.

Both of the above bounds are sharp. If an diagonal edge is added to the 4-cycle,

the smallest positive eigenvalue remains the same (${2}$). The reason is that a typical eigenfunction for ${\lambda_2}$ takes on values ${-1, 0, 1, 0}$ (in cyclic order) and adding an edge between two vertices with the same value of such eigenfunction does not affect the Poincaré constant.

On the other hand, adding one more edge increases ${\lambda_2}$ from ${2}$ to ${4}$.

## Normalized Laplacian

Another form of the Poincaré inequality involves the vertex degree: when summing over vertices, we give each of them weight equal to the number of neighbors.

${\displaystyle \sum_V f(v)^2 d_v \le C \sum_E (f(u)-f(v))^2 }$ provided ${\displaystyle \sum_V f(v)d_v = 0}$

Here ${1/C = \mu_2}$, the smallest positive eigenvalue of the normalized Laplacian ${L_N}$, the matrix with ${1}$ on the diagonal, where off-diagonal entries are ${-1/\sqrt{d_ud_v}}$ if ${u\sim v}$, and ${0}$ otherwise. The sum of the normalized Laplacian eigenvalues is equal to ${n}$ (the number of vertices), regardless of how many edges are there. So we cannot expect all them to grow when an edge is added. But how do they change?

Let ${\mu_2'}$ be the corresponding eigenvalue after an edge is added.
Here are two examples for which ${\mu_2' - \mu_2 = 1/2}$:

Completing a 3-path to a 3-cycle increases the eigenvalue from 1 to 3/2.

Completing a 4-path to a 4-cycle increases the eigenvalue from 1/2 to 1.

This pattern does not continue: “5-path to 5-cycle” results in a smaller increase, which is not even largest among 5-vertex graphs. It appears that the above examples are the only two cases of ${\mu_2' - \mu_2 = 1/2}$, with all other graphs having ${\mu_2' - \mu_2 < 1/2}$. (I do not have a proof of this.)

## The opposite direction

Adding an edge can also make ${\mu_2}$ smaller (thus, the weighted Poincaré constant becomes larger, showing that it may not be as useful for quantifying the connectivity of a graph). The smallest example is adding an edge to the star graph on 4 vertices:

here ${\mu_2 = 1}$ but ${\mu_2' = 5/4 - \sqrt{33}/12 \approx 0.77}$.

Let us analyze the star graphs on ${n}$ vertices, for general ${n}$. In an earlier post we saw that ${\mu_2 = 1}$, so it remains to find ${\mu_2'}$. Let us order the vertices so that ${1}$ is the center of the star and ${(2,3)}$ is the added edge. Write ${\beta = -1/\sqrt{n-1}}$ for brevity. Then the normalized Laplacian matrix ${L_N}$ is

${\displaystyle \begin{pmatrix} 1 & \beta/\sqrt{2} & \beta/\sqrt{2} & \beta & \cdots & \beta \\ \beta/\sqrt{2} & 1 & -1/2 & 0 & \cdots & 0 \\ \beta/\sqrt{2} & -1/2 & 1 & 0 & \cdots & 0 \\ \beta & 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ \beta & 0 & 0 & 0 & \cdots & 1 \\ \end{pmatrix}}$

where all rows numbered ${i > 3}$ consist of ${\beta}$ in the first column and ${1}$ in the ${i}$th column. This shows that ${L_N-I}$ has rank ${4}$, hence ${1}$ is an eigenvalue of multiplicity ${n-4}$. Also, ${0}$ is a simple eigenvalue as for any connected graph. Less obviously, ${3/2}$ is an eigenvalue with the corresponding eigenvector ${(0, 1, -1, 0, \dots, 0)}$ – this eigenvalue comes from the submatrix in row/columns 2 and 3, where the surrounding values in these row/columns are nice enough to cancel out. We are left with two eigenvalues to find, and we know their sum is ${n - (n-4) - 3/2 = 5/2}$. This means the characteristic polynomial of ${L_N}$ is of the form

${\displaystyle p(t) = t (t-1)^{n-4} (t-3/2) (t^2 - 5t/2 + \gamma)}$

where the constant ${\gamma}$ remains to be found. I am going to skip to the answer here: ${\gamma = n/(n-1)}$. The quadratic formula delivers the last two eigenvalues:

${\displaystyle \frac14 \left( 5 \pm \sqrt{\frac{9n-25}{n-1}}\right) }$

The smaller of these is ${\mu_2'}$ that we were looking for:

${\displaystyle \mu_2' = \frac14 \left( 5 - \sqrt{\frac{9n-25}{n-1}}\right) \to \frac12 }$ as ${n\to\infty}$

Thus, adding an edge to a star graph results in negative ${\mu_2' - \mu_2}$ which decreases to ${-1/2}$ as ${n\to\infty}$. Exhaustive search for ${n\le 9}$ shows that the star graph has the smallest value of ${\mu_2' - \mu_2}$ among all graphs on ${n}$ vertices. It is reasonable to expect this pattern to continue, which would imply ${\mu_2' - \mu_2 > -1/2}$ in all cases.

## Maximal algebraic connectivity among graphs of bounded degree

The algebraic connectivity ${ a(G) }$ of a graph ${G }$ is its smallest nontrivial Laplacian eigenvalue. Equivalently, it is the minimum of edge sums ${\sum_{e\in E} df(e)^2}$ over all functions ${f\colon V\to\mathbb R}$ normalized by ${ \sum_{v\in V} f(v) = 0 }$ and ${\sum_{v\in V} f(v)^2 = 1 }$. Here  ${ V, E }$ are vertex/edge sets, and ${ df(e)}$ means the difference of the values of ${f }$ at the vertices of the edge ${ e}$.

It is clear from the definition that adding edges to ${ G}$ cannot make  ${a(G) }$ smaller; thus, among all graphs on ${n }$ vertices the maximal value of ${a(G) }$ is attained by the complete graph, for which ${a(G) = n}$. For any other graph ${ a(G)\le n-2}$ which can be shown as follows. Pick two non-adjacent vertices ${u }$ and ${v }$ and let ${f(u)=1/\sqrt{2} }$, ${ f(v) = -1/\sqrt{2}}$, and ${f=0 }$ elsewhere. This function is normalized as required above, and its edge sum ${\sum_{e\in E} df(e)^2}$ is at most ${ n-2}$ since there are at most ${ 2(n-2)}$ edges with a nonzero contribution.

What if we require the graph to have degree vertex at most ${d}$, and look for maximal connectivity then? First of all, only connected graphs are under consideration, since ${a(G)=0}$ for non-connected graphs. Also, only the cases ${n\ge d+2 }$ are of interest, otherwise the complete graph wins. The argument in the previous paragraph shows that ${ a(G) \le d}$ but is this bound attained?

The case ${d=2}$ is boring: the only two connected graphs are the path ${P_n}$ and the cycle ${ C_n}$. The cycle wins with  ${a(C_n) = 2(1-\cos(2\pi/n)) }$ versus ${ a(P_n) = 2(1-\cos(\pi/n))}$.

When ${ d=4}$, one might suspect a pattern based on the following winners:

The structure of these two is the same: place ${n }$ points on a circle, connect each of them to ${4 }$ nearest neighbors.

But this pattern does not continue: the 8-vertex winner is completely different.

This is simply the complete bipartite graph ${ K_{4, 4} }$. And it makes sense that the “4 neighbors” graph loses when the number of vertices is large: there is too much “redundancy” among its edges, many of which connect the vertices that were already connected by short paths.

In general, when ${n = 2d }$, the complete bipartite graph ${ K_{d, d}}$ achieves ${a = d }$ and therefore maximizes the algebraic connectivity. The fact that ${a(K_{d, d}) = d }$ follows by considering graph complement, as discussed in Laplacian spectrum of small graphs. The complement of ${K_{d, d} }$ is the disjoint union of two copies of the complete graph ${K_d }$, for which the maximal eigenvalue is ${d }$. Hence ${a(G) = n - \lambda_{\max}(G^c) = n-d = d }$.

When ${ n = 2d-1}$ is odd, we have a natural candidate in ${ K_{d,d-1}}$ for which the argument from the previous paragraph shows ${a(G) = d-1 }$. This is indeed a winner when ${n=5, d=3 }$:

The winner is not unique since one can add another edge between two of the vertices of degree 2. This does not change the ${a(G)}$, however: there is a fundamental eigenfunction that has equal values at the vertices of that added edge.

Same for ${n=9, d=5 }$: the complete bipartite graph shares the maximum value of algebraic connectivity with two other graphs formed by adding edges to it:

However, the family ${ K_{d,d-1}}$ does not win the case ${ n = 2d-1}$ in general: we already saw a 4-regular graph on 7 vertices with ${a(G)\approx 3.2 }$, beating ${ a(K_{4, 3}) = 3}$. Perhaps ${ K_{d,d-1}}$ wins when ${ d}$ is odd?

I do not have any other patterns to conjecture, but here are two winners for ${ n = 8, d= 3}$: the cube and the “twisted cube”.

The cube is twisted by replacing a pair of edges on the top face with the diagonals. This is still a 3-regular graph and the algebraic connectivity stays the same, but it is no longer bipartite: a 5-cycle appears.

## Measuring the regularity of a graph by its Laplacian eigenvalues

Let ${G}$ be a graph with vertices ${1, 2, \dots, n}$. The degree of vertex ${i}$ is denoted ${d_i}$. Let ${L}$ be the Laplacian matrix of ${G}$, so that ${L_{ii}=d_i}$, ${L_{ij}}$ is ${-1}$ when the vertices ${i, j}$ are adjacent, and is ${0}$ otherwise. The eigenvalues of ${L}$ are written as ${\lambda_1\le \dots \le \lambda_n}$.

The graph is regular if all vertices have the same degree: ${d_1=\cdots = d_n}$. How can this property be seen from its Laplacian eigenvalues ${\lambda_1, \dots, \lambda_n}$?

Since the sum of eigenvalues is equal to the trace, we have ${\sum \lambda_i = \sum d_i}$. Moreover, ${\sum \lambda_i^2}$ is the trace of ${L^2}$, which is equal to the sum of the squares of all entries of ${L}$. This sum is ${\sum d_i^2 + \sum d_i}$ because the ${i}$th row of ${L}$ contains one entry equal to ${d_i}$ and ${d_i}$ entries equal to ${-1}$. In conclusion, ${\sum d_i^2 = \sum \lambda_i^2 - \sum\lambda_i}$.

The Cauchy-Schwarz inequality says that ${n\sum d_i^2 \ge \left(\sum d_i \right)^2}$ with equality if and only if all numbers ${d_i}$ are equal, i.e., the graph is regular. In terms of eigenvalues, this means that the difference
${\displaystyle D =n\sum d_i^2 - \left(\sum d_i \right)^2 = n\sum (\lambda_i^2 - \lambda_i) - \left( \sum\lambda_i \right)^2 }$
is always nonnegative, and is equal to zero precisely when the graph is regular. This is how one can see the regularity of a graph from its Laplacian spectrum.

As an aside, ${D }$ is an even integer. Indeed, the sum ${\sum d_i}$ is even because it double-counts the edges. Hence the number of vertices of odd degree is even, which implies that ${\sum d_i^k }$ is even for every positive integer  ${k }$.

Up to a constant factor, ${D}$ is simply the degree variance: the variance of the sequence ${d_1, \dots, d_n}$. What graph maximizes it for a given ${n}$? We want to have some very large degrees and some very small ones.

Let ${G_{m, n}}$ be the union of the complete graph ${K_m}$ on ${m}$ vertices and ${(n-m)}$ isolated vertices. The sum of degrees is ${m(m-1)}$ and the sum of squares of degrees is ${m(m-1)^2}$. Hence,

${D = nm(m-1)^2 - (m(m-1))^2 = m(m-1)^2(n-m)}$

For ${n=3, 4, 5, 6}$ the maximum is attained by ${m=n-1}$, that is there is one isolated vertex. For ${n=7, 8, 9, 10}$ the maximum is ${m=n-2}$. In general it is attained by ${m^*=\lfloor (3n+2)/4 \rfloor}$.

The graph ${G_{m, n}}$ is disconnected. But any graph has the same degree variance as its complement. And the complement ${G^c(m, n)}$ is always connected: it consists of a “center”, a complete graph on ${n-m}$ vertices, and “periphery”, a set of ${m}$ vertices that are connected to each central vertex. Put another way, ${G^c(m, n)}$ is obtained from the complete bipartite graph ${K_{m, n-m}}$ by connecting all vertices of the ${n-m}$ group together.

Tom A. B. Snijders (1981) proved that ${G(m^*, n)}$ and ${G^c(m^*, n)}$ are the only graphs maximizing the degree variance; in particular, ${G^c(m^*, n)}$ is the unique maximizer among the connected graphs. It is pictured below for ${n=4, \dots, 9}$.

## Laplacian spectrum of small graphs

This is a collection of entirely unoriginal remarks about Laplacian spectrum of graphs. For an accessible overview of the subject I recommend the M.S. thesis The Laplacian Spectrum of Graphs by Michael William Newman. It also includes a large table of graphs with their spectra. Here I will avoid introducing matrices and enumerating vertices.

Let ${V}$ be the vertex set of a graph. Write ${u\sim v}$ if ${u, v}$ are adjacent vertices. Given a function ${f\colon V\to \mathbb R}$, define ${L f(v) = \sum_{u\colon u\sim v}(f(v)-f(u))}$.
This is a linear operator (the graph Laplacian) on the Euclidean space ${\ell^2(V)}$ of all functions ${f\colon V\to \mathbb R}$ with the norm ${\|f\|^2 = \sum_{v\in V} f(v)^2}$. It is symmetric: ${\langle L f, g\rangle = \langle f, L g\rangle }$ and positive semidefinite: ${\langle L f, f\rangle = \frac12 \sum_{u\sim v}(f(u)-f(v))^2\ge 0}$. Since equality is attained for constant ${f}$, 0 is always an eigenvalue of ${L}$.

This is the standard setup, but I prefer to change things a little and replace ${\ell^2(V)}$ by the smaller space ${\ell^2_0(V)}$ of functions with zero mean: ${\sum_{v\in V}f(v)=0}$. Indeed, ${L}$ maps ${\ell^2(V)}$ to ${\ell^2_0(V)}$ anyway, and since it kills the constants, it makes sense to focus on ${\ell^2_0(V)}$. It is a vector space of dimension ${n-1}$ where ${n=|V|}$.

One advantage is that the smallest eigenvalue is 0 if and only if the graph is disconnected: indeed, ${\langle L f, f\rangle=0}$ is equivalent to ${f}$ being constant on each connected component. We also gain better symmetry between ${L}$ and the Laplacian of the graph complement, denoted ${L'}$. Indeed, since ${L' f(v) = \sum_{u\colon u\not \sim v}(f(v)-f(u))}$, it follows that ${(L+L')f(v) = \sum_{u\colon u\ne v} (f(v)-f(u)) = n f(v)}$ for every ${f\in \ell^2_0(V)}$. So, the identity ${L+L' = nI}$ holds on ${\ell^2_0(V)}$ (it does not hold on ${\ell^2(V)}$). Hence the eigenvalues of ${L'}$ are obtained by subtracting the eigenvalues of ${L}$ from ${n}$. As a corollary, the largest eigenvalue of ${L}$ is at most ${n}$, with equality if and only if the graph complement is disconnected. More precisely, the multiplicity of eigenvalue ${n}$ is one less than the number of connected components of the graph complement.

Let ${D}$ denote the diameter of the graph. Then the number of distinct Laplacian eigenvalues is at least ${D}$. Indeed, let ${u, v}$ be two vertices at distance ${D}$ from each other. Define ${f_0(u) = 1}$ and ${f_0=0}$ elsewhere. Also let ${f_k=L^k f_0}$ for ${k=1, 2, \dots}$. Note that ${f_k\in \ell_0^2(V)}$ for all ${k\ge 1}$. One can prove by induction that ${f_k(w)=0}$ when the distance from ${w}$ to ${u}$ is greater than ${k}$, and ${(-1)^k f_k(w) > 0}$ when the distance from ${w}$ to ${u}$ is equal to ${k}$. In particular, ${f_k(v) = 0}$ when ${k and ${f_D(v)\ne 0}$. This shows that ${f_D}$ is not a linear combination of ${f_1, \dots, f_{D-1}}$. Since ${f_k = L^{k-1}f_1}$, it follows that ${L^{D-1}}$ is not a linear combination of ${L^0, L^1, \dots, L^{D-2}}$. Hence the minimal polynomial of ${L}$ has degree at least ${D}$, which implies the claim.

Let’s consider a few examples of connected graphs.

## 3 vertices

There are two connected graphs: the 3-path (D=2) and the 3-cycle (D=1). In both cases we get D distinct eigenvalues. The spectra are [1, 3] and [3, 3], respectively.

## 4 vertices

• One graph of diameter 3, the path. Its spectrum is ${[2-\sqrt{2}, 2, 2+\sqrt{2}]}$.
• One graph of diameter 1, the complete graph. Its spectrum is ${[4, 4, 4]}$. This pattern continues for other complete graphs: since the complement is the empty graph (${n}$ components), all ${n-1}$ eigenvalues are equal to ${n}$.
• Four graphs of diameter 2, which are shown below, with each caption being the spectrum.

Remarks:

• The graph [1, 3, 4] has more distinct eigenvalues than its diameter.
• The graph [2, 2, 4] is regular (all vertices have the same degree).
• The smallest eigenvalue of graphs [1, 1, 4] and [2, 2, 4] is multiple, due to the graphs having a large group of automorphisms (here rotations); applying some of these automorphisms to an eigenfunctions for the smallest eigenvalue yields another eigenfunction.
• [1, 3, 4] and [2, 4, 4] also have automorphisms, but their automorphisms preserve the eigenfunction for the lowest eigenvalue, up to a constant factor.

## 5 vertices

• One graph of diameter 4, the path. Its spectrum is related to the golden ratio: it consists of ${(3\pm \sqrt{5})/2, (5\pm \sqrt{5})/2}$.
• One graph of diameter 1, the complete one: [5, 5, 5, 5]
• Five graphs of diameter 3. All have connected complement, with the highest eigenvalue strictly between 4 and 5. None are regular. Each has 4 distinct eigenvalues.
• 14 graphs of diameter 2. Some of these are noted below.

Two have connected complement, so their eigenvalues are less than 5 (spectrum shown on hover):

One has both integers and non-integers in its spectrum, the smallest such graph. Its eigenvalues are ${3\pm \sqrt{2}, 3, 5}$.

Two have eigenvalues of multiplicity 3, indicating a high degree of symmetry (spectrum shown on hover).

Two have all eigenvalues integer and distinct:

The 5-cycle and the complete graph are the only regular graphs on 5 vertices.

## 6 vertices

This is where we first encounter isospectral graphs: the Laplacian spectrum cannot tell them apart.

Both of these have spectrum ${3\pm \sqrt{5}, 2, 3, 3}$ but they are obviously non-isomorphic (consider the vertex degrees):

Both have these have spectrum ${3\pm \sqrt{5}, 3, 3, 4}$ and are non-isomorphic.

Indeed, the second pair is obtained from the first by taking graph complement.

Also notable are regular graphs on 6 vertices, all of which have integer spectrum.

Here [3, 3, 3, 3, 6] (complete bipartite) and [2, 3, 3, 5, 5] (prism) are both regular of degree 3, but the spectrum allows us to tell them apart.

The prism is the smallest regular graph for which the first eigenvalue is a simple one. It has plenty of automorphisms, but the relevant eigenfunction (1 on one face of the prism, -1 on the other face) is compatible with all of them.

## 7 vertices

There are four regular graphs on 7 vertices. Two of them are by now familiar: 7-cycle and complete graph. Here are the other two, both regular of degree 4 but with different spectra.

There are lots of isospectral pairs of graphs on 7 vertices, so I will list only the isospectral triples, of which there are five.

Spectrum 0.676596, 2, 3, 3.642074, 5, 5.681331:

Spectrum 0.726927, 2, 3.140435, 4, 4, 6.132637:

Spectrum 0.867363, 3, 3, 3.859565, 5, 6.273073:

Spectrum 1.318669, 2, 3.357926, 4, 5, 6.323404:

All of the triples mentioned so far have connected complement: for example, taking the complement of the triple with the spectrum [0.676596, 2, 3, 3.642074, 5, 5.681331] turns it into the triple with the spectrum [1.318669, 2, 3.357926, 4, 5, 6.323404].

Last but not least, an isospectral triple with an integer spectrum: 3, 4, 4, 6, 6, 7. This one has no counterpart since the complement of each of these graphs is disconnected.

## 8 vertices

Regular graphs, excluding the cycle (spectrum 0.585786, 0.585786, 2, 2, 3.414214, 3.414214, 4) and the complete one.

#### Degree 6 regular

Credits: the list of graphs by Brendan McKay and NetworkX library specifically the methods read_graph6 (to read the files provided by Prof. McKay), laplacian_spectrum, diameter, degree, and draw.

## Nodal lines

Wikipedia article on nodes offers this 1D illustration: a node is an interior point at which a standing wave does not move.

(At the endpoints the wave is forced to stay put, so I would not count them as nodes despite being marked on the plot.)

A standing wave in one dimension is described by the equation ${f''+\omega^2 f=0}$, where ${\omega}$ is its (angular) frequency. The function ${u(x,t) = f(x)\cos \omega t}$ solves the wave equation ${u_{tt}=u_{xx}}$: the wave vibrates without moving, hence the name. In mathematics, these are the (Dirichlet) eigenfunctions of the Laplacian.

Subject to boundary conditions ${f(0)=0 = f(\pi)}$ (fixed ends), all standing waves on the interval ${(0,\pi)}$ are of the form ${\sin nx}$ for ${n=1,2,3,\dots}$. Their eigenvalues are exactly the perfect squares, and the nodes are equally spaced on the interval.

Things get more interesting in two dimensions. For simplicity consider the square ${Q=(0,\pi)\times (0,\pi)}$. Eigenfunctions with zero value on the boundary are of the form ${f(x,y) = \sin mx \sin ny}$ for positive integers ${m,n}$. The set of eigenvalues has richer structure, it consists of the integers that can be expressed as the sum of two positive squares: 2, 5, 8, 10, 13, 17,…

The zero sets of eigenfunctions in two dimensions are called nodal lines. At a first glance it may appear that we have nothing interesting: the zero set of ${\sin mx \sin ny}$ is a union of ${n-1}$ equally spaced horizontal lines, and ${m-1}$ equally spaced vertical lines:

But there is much more, because a sum of two eigenfunctions with the same eigenvalue is also an eigenfunction. To begin with, we can form linear combinations of ${\sin mx \sin ny}$ and ${\sin nx \sin my}$. Here are two examples from Partial Differential Equations by Walter Strauss:

When ${f(x,y) = \sin 12x \sin y+\sin x \sin 12y }$, the square is divided by nodal lines into 12 nodal domains:

After slight perturbation ${f(x,y) = \sin 12x \sin y+0.9\sin x \sin 12y }$ there is a single nodal line dividing the square into two regions of intricate geometry:

And then there are numbers that can be written as sums of squares in two different ways. The smallest is ${50=1^2+7^2 = 5^2+5^2}$, with eigenfunctions such as

$\displaystyle f(x,y) = \sin x\sin 7y +2\sin 5x \sin 5y+\sin 7x\sin y$

pictured below.

This is too good not to replicate: the eigenfunctions naturally extend as doubly periodic functions with anti-period ${\pi}$.