Let be the vertex set of a graph; all graphs here are assumed connected. Write if are adjacent vertices. The Laplacian of a function is defined as . Its properties were discussed earlier in Laplacian spectrum of small graphs.
The normalized Laplacian involves the vertex degree (the number of vertices adjacent to ). It is defined by
The matrix of is obtained by multiplying the matrix of on both sides by the diagonal matrix with on the diagonal. Hence, is also positive semidefinite and has the same rank as . In particular, it has a simple eigenvalue 0 and the rest of eigenvalues are positive. Let us write for the eigenvalues of . In this notation, and for .
There are many graphs with integer Laplacian eigenvalues; they do not seem to follow any particular pattern. In contrast, the graphs with integer normalized Laplacian eigenvalues have a simple description.
The matrix of normalized Laplacian has in every diagonal entry. Its off-diagonal entries are if , and otherwise. Since the trace is , and there are positive eigenvalues, the way that a normalized Laplacian spectrum can consist of integers is .
It turns out that the largest normalized Laplacian eigenvalue equal to if and only if the graph is bipartite; otherwise it is less than (as a reminder, all graphs are assumed connected in this post). Indeed, is the norm of , which is the supremum of over all functions . A computation shows where the sum is taken over the edge set : each edge contributes one term to the sum, with being its endpoints. The sum becomes simpler if we write ; after this substitution,
The denominator can be rewritten as a sum over edges, namely . This shows that the introduction of vertex degree into the denominator achieves certain symmetry: each value of appears in the numerator as many times as in the denominator.
Since , the numerator is at most . Thus , and to achieve equality here we need whenever . This requires to be constant and the graph to be bipartite, its 2-coloring provided by . Conversely, any 2-coloring of a graph provides us with a function for which according to the above.
To achieve integer spectrum, we also need to be an eigenvalue of multiplicity for , which is equivalent to . Since the graph is bipartite, has the block form
where the (possibly rectangular) block has entries corresponding to . In order for to have rank 2, the matrix must have rank 1. Any zero entry of a rank-1 matrix lies in a zero row, but cannot have a zero row because the graph is connected. Thus, all entries of are positive, which means we have a complete bipartite graph.
The converse is immediate: for a complete bipartite graph the matrix has all entries equal to , hence its rank is 1.
As a final remark, the ordinary (un-normalized) Laplacian of a complete bipartite graph also has integer eigenvalues. Indeed, the complement of is the disjoint union of complete graphs and , with the spectra and . Passing to the complement means replacing each eigenvalue by , except that one stays (this is discussed in Laplacian spectrum of small graphs). Hence, the Laplacian spectrum of consists of: simple eigenvalues and , eigenvalue of multiplicity , and eigenvalue of multiplicity . Worth noting that the Laplacian spectrum determines a complete bipartite graph up to an isomorphism, while the normalized Laplacian spectrum does not.
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 with zero mean,
where the sum on the left is over the vertex set , the sum on the right is over the edge set , and on the right are the endpoints of an edge. Suppose 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 satisfies : greater connectivity means smaller Poincaré constant.
It is more convenient to work with the reciprocal of , especially because , the smallest positive eigenvalue of the graph Laplacian. Let be the corresponding eigenvalue after an edge is added. The above shows that . 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
Both of the above bounds are sharp. If an diagonal edge is added to the 4-cycle,
the smallest positive eigenvalue remains the same (). The reason is that a typical eigenfunction for takes on values (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 from to .
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.
Here , the smallest positive eigenvalue of the normalized Laplacian , the matrix with on the diagonal, where off-diagonal entries are if , and otherwise. The sum of the normalized Laplacian eigenvalues is equal to (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 be the corresponding eigenvalue after an edge is added.
Here are two examples for which :
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 , with all other graphs having . (I do not have a proof of this.)
The opposite direction
Adding an edge can also make 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 but .
Let us analyze the star graphs on vertices, for general . In an earlier post we saw that , so it remains to find . Let us order the vertices so that is the center of the star and is the added edge. Write for brevity. Then the normalized Laplacian matrix is
where all rows numbered consist of in the first column and in the th column. This shows that has rank , hence is an eigenvalue of multiplicity . Also, is a simple eigenvalue as for any connected graph. Less obviously, is an eigenvalue with the corresponding eigenvector – 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 . This means the characteristic polynomial of is of the form
where the constant remains to be found. I am going to skip to the answer here: . The quadratic formula delivers the last two eigenvalues:
The smaller of these is that we were looking for:
Thus, adding an edge to a star graph results in negative which decreases to as . Exhaustive search for shows that the star graph has the smallest value of among all graphs on vertices. It is reasonable to expect this pattern to continue, which would imply in all cases.
The algebraic connectivity of a graph is its smallest nontrivial Laplacian eigenvalue. Equivalently, it is the minimum of edge sums over all functions normalized by and . Here are vertex/edge sets, and means the difference of the values of at the vertices of the edge .
It is clear from the definition that adding edges to cannot make smaller; thus, among all graphs on vertices the maximal value of is attained by the complete graph, for which . For any other graph which can be shown as follows. Pick two non-adjacent vertices and and let , , and elsewhere. This function is normalized as required above, and its edge sum is at most since there are at most edges with a nonzero contribution.
What if we require the graph to have degree vertex at most , and look for maximal connectivity then? First of all, only connected graphs are under consideration, since for non-connected graphs. Also, only the cases are of interest, otherwise the complete graph wins. The argument in the previous paragraph shows that but is this bound attained?
The case is boring: the only two connected graphs are the path and the cycle . The cycle wins with versus .
When , one might suspect a pattern based on the following winners:
The structure of these two is the same: place points on a circle, connect each of them to nearest neighbors.
But this pattern does not continue: the 8-vertex winner is completely different.
This is simply the complete bipartite graph . 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 , the complete bipartite graph achieves and therefore maximizes the algebraic connectivity. The fact that follows by considering graph complement, as discussed in Laplacian spectrum of small graphs. The complement of is the disjoint union of two copies of the complete graph , for which the maximal eigenvalue is . Hence .
When is odd, we have a natural candidate in for which the argument from the previous paragraph shows . This is indeed a winner when :
The winner is not unique since one can add another edge between two of the vertices of degree 2. This does not change the , however: there is a fundamental eigenfunction that has equal values at the vertices of that added edge.
Same for : the complete bipartite graph shares the maximum value of algebraic connectivity with two other graphs formed by adding edges to it:
However, the family does not win the case in general: we already saw a 4-regular graph on 7 vertices with , beating . Perhaps wins when is odd?
I do not have any other patterns to conjecture, but here are two winners for : 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.
Let be a graph with vertices . The degree of vertex is denoted . Let be the Laplacian matrix of , so that , is when the vertices are adjacent, and is otherwise. The eigenvalues of are written as .
The graph is regular if all vertices have the same degree: . How can this property be seen from its Laplacian eigenvalues ?
Since the sum of eigenvalues is equal to the trace, we have . Moreover, is the trace of , which is equal to the sum of the squares of all entries of . This sum is because the th row of contains one entry equal to and entries equal to . In conclusion, .
The Cauchy-Schwarz inequality says that with equality if and only if all numbers are equal, i.e., the graph is regular. In terms of eigenvalues, this means that the difference
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, is an even integer. Indeed, the sum is even because it double-counts the edges. Hence the number of vertices of odd degree is even, which implies that is even for every positive integer .
Up to a constant factor, is simply the degree variance: the variance of the sequence . What graph maximizes it for a given ? We want to have some very large degrees and some very small ones.
Let be the union of the complete graph on vertices and isolated vertices. The sum of degrees is and the sum of squares of degrees is . Hence,
For the maximum is attained by , that is there is one isolated vertex. For the maximum is . In general it is attained by .
The graph is disconnected. But any graph has the same degree variance as its complement. And the complement is always connected: it consists of a “center”, a complete graph on vertices, and “periphery”, a set of vertices that are connected to each central vertex. Put another way, is obtained from the complete bipartite graph by connecting all vertices of the group together.
Tom A. B. Snijders (1981) proved that and are the only graphs maximizing the degree variance; in particular, is the unique maximizer among the connected graphs. It is pictured below for .
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 be the vertex set of a graph. Write if are adjacent vertices. Given a function , define .
This is a linear operator (the graph Laplacian) on the Euclidean space of all functions with the norm . It is symmetric: and positive semidefinite: . Since equality is attained for constant , 0 is always an eigenvalue of .
This is the standard setup, but I prefer to change things a little and replace by the smaller space of functions with zero mean: . Indeed, maps to anyway, and since it kills the constants, it makes sense to focus on . It is a vector space of dimension where .
One advantage is that the smallest eigenvalue is 0 if and only if the graph is disconnected: indeed, is equivalent to being constant on each connected component. We also gain better symmetry between and the Laplacian of the graph complement, denoted . Indeed, since , it follows that for every . So, the identity holds on (it does not hold on ). Hence the eigenvalues of are obtained by subtracting the eigenvalues of from . As a corollary, the largest eigenvalue of is at most , with equality if and only if the graph complement is disconnected. More precisely, the multiplicity of eigenvalue is one less than the number of connected components of the graph complement.
Let denote the diameter of the graph. Then the number of distinct Laplacian eigenvalues is at least . Indeed, let be two vertices at distance from each other. Define and elsewhere. Also let for . Note that for all . One can prove by induction that when the distance from to is greater than , and when the distance from to is equal to . In particular, when and . This shows that is not a linear combination of . Since , it follows that is not a linear combination of . Hence the minimal polynomial of has degree at least , which implies the claim.
Let’s consider a few examples of connected graphs.
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.
One graph of diameter 3, the path. Its spectrum is .
One graph of diameter 1, the complete graph. Its spectrum is . This pattern continues for other complete graphs: since the complement is the empty graph ( components), all eigenvalues are equal to .
Four graphs of diameter 2, which are shown below, with each caption being the spectrum.
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.
One graph of diameter 4, the path. Its spectrum is related to the golden ratio: it consists of .
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):
1.382, 1.382, 3.618, 3.618
1.382, 2.382, 3.618, 4.618
One has both integers and non-integers in its spectrum, the smallest such graph. Its eigenvalues are .
Two have eigenvalues of multiplicity 3, indicating a high degree of symmetry (spectrum shown on hover).
1, 1, 1, 5
3, 5, 5, 5
Two have all eigenvalues integer and distinct:
1, 2, 4, 5
2, 3, 4, 5
The 5-cycle and the complete graph are the only regular graphs on 5 vertices.
This is where we first encounter isospectral graphs: the Laplacian spectrum cannot tell them apart.
Both of these have spectrum but they are obviously non-isomorphic (consider the vertex degrees):
Both have these have spectrum 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.
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.
Regular graphs, excluding the cycle (spectrum 0.585786, 0.585786, 2, 2, 3.414214, 3.414214, 4) and the complete one.
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 , where is its (angular) frequency. The function solves the wave equation : the wave vibrates without moving, hence the name. In mathematics, these are the (Dirichlet) eigenfunctions of the Laplacian.
Subject to boundary conditions (fixed ends), all standing waves on the interval are of the form for . 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 . Eigenfunctions with zero value on the boundary are of the form for positive integers . 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 is a union of equally spaced horizontal lines, and 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 and . Here are two examples from Partial Differential Equations by Walter Strauss:
When , the square is divided by nodal lines into 12 nodal domains:
After slight perturbation 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 , with eigenfunctions such as
This is too good not to replicate: the eigenfunctions naturally extend as doubly periodic functions with anti-period .
The dreaded calculus torture device that works for exactly two integrals, and .
Actually, no. A version of it (with one integration by parts) works for :
hence (assuming )
Yes, this is more of a calculus joke. A more serious example comes from Fourier series.
The functions , , are orthogonal on , in the sense that
This is usually proved using a trigonometric identity that converts the product to a sum. But the double integration by parts give a nicer proof, because no obscure identities are needed. No boundary terms will appear because the sines vanish at both endpoints:
All integrals here must vanish because . As a bonus, we get the orthogonality of cosines, , with no additional effort.
The double integration by parts is also a more conceptual proof, because it gets to the heart of the matter: eigenvectors of a symmetric matrix (operator) that correspond to different eigenvalues are orthogonal. The trigonometric form is incidental, the eigenfunction property is essential. Let’s try this one more time, for the mixed boundary value problem , . Suppose that and satisfy the boundary conditions, , and . Since and vanish at both endpoints, we can pass the primes easily:
The signature of a Hermitian matrix can be defined either as the pair (number of positive eigenvalues, number of negative eigenvalues), or simply as the difference
The function hides some information when the matrix is degenerate, but this will not be of concern here. For a matrix of size , the number is between and and has the same parity as .
Given any square matrix with real entries and a complex number , we can form , which is a Hermitian matrix. Then is an integer-valued function of . Restricting attention to the unit circle , we obtain a piecewise constant function with jumps at the points where is degenerate.
When is a Seifert matrix of a knot, the function is the Tristram-Levine signature of the knot. To each knot there are infinitely many Seifert surfaces , and to each Seifert surface there are infinitely many Seifert matrices , depending on how we choose the generators of . Yet, depends on alone.
Below I plot for a few knots using Scilab for computation of signature and the knot data from
Technical point: since Scilab enumerates colors using positive integers, the colors below correspond to the number of positive eigenvalues rather than the signature. Namely, black for 0, blue (1), green (2), and cyan (3).
As always, first comes the trefoil:
One of its Seifert matrices is
and the Tristram-Levine signature is
Next, the knot
with Seifert matrix
and the signature
And finally, the knot :
with Seifert matrix
and the signature
I experimented with a few more, trying to create more colors. However, guessing the complexity of the signature by looking at the Seifert matrix is one of many skills I do not have. So I conclude with the simple code used to plot the signatures.
[n,n] = size(A);
Npoints = 200;
r = 2*%pi/Npoints;
arcs = zeros(6,Npoints);
colors = zeros(Npoints);
for m = 1:Npoints
x = cos(2*%pi*(m-1/2)/Npoints);
y = sin(2*%pi*(m-1/2)/Npoints);
omega = complex(x,y);
B = (1-omega)*A+(1-conj(omega))*A';
signature = sum(sign(spec(B)));
colors(m) = (signature+n)/2+1;
arcs(:,m) = [x-r,y-r,2*r,2*r,0,360*64]';