Turán's theorem

(Induction on n) An example of sets ${\displaystyle A}$ and ${\displaystyle B}$ for ${\displaystyle r=3}$.

(Maximal Degree Vertex) Deleting edges within ${\displaystyle A}$ and drawing edges between ${\displaystyle A}$ and ${\displaystyle B}$.

This was Turán's original proof. Take a ${\displaystyle K_{r+1}}$-free graph on ${\displaystyle n}$ vertices with the maximal number of edges. Find a ${\displaystyle K_{r}}$ (which exists by maximality), and partition the vertices into the set ${\displaystyle A}$ of the ${\displaystyle r}$ vertices in the ${\displaystyle K_{r}}$ and the set ${\displaystyle B}$ of the ${\displaystyle n-r}$ other vertices.

Now, one can bound edges above as follows:

• There are exactly ${\displaystyle {\binom {r}{2}}}$ edges within ${\displaystyle A}$.
• There are at most ${\displaystyle (r-1)|B|=(r-1)(n-r)}$ edges between ${\displaystyle A}$ and ${\displaystyle B}$, since no vertex in ${\displaystyle B}$ can connect to all of ${\displaystyle A}$.
• The number of edges within ${\displaystyle B}$ is at most the number of edges of ${\displaystyle T(n-r,r)}$ by the inductive hypothesis.

Adding these bounds gives the result.[1][3]

This proof is due to Paul Erdős. Take the vertex ${\displaystyle v}$ of largest degree. Consider the set ${\displaystyle A}$ of vertices not adjacent to ${\displaystyle v}$ and the set ${\displaystyle B}$ of vertices adjacent to ${\displaystyle v}$.

Now, delete all edges within ${\displaystyle A}$ and draw all edges between ${\displaystyle A}$ and ${\displaystyle B}$. This increases the number of edges by our maximality assumption and keeps the graph ${\displaystyle K_{r+1}}$-free. Now, ${\displaystyle B}$ is ${\displaystyle K_{r}}$-free, so the same argument can be repeated on ${\displaystyle B}$.

Repeating this argument eventually produces a graph in the same form as a Turán graph, which is a collection of independent sets, with edges between each two vertices from different independent sets. A simple calculation shows that the number of edges of this graph is maximized when all independent set sizes are as close to equal as possible.[3][4]

This proof, as well as the Zykov Symmetrization proof, involve reducing to the case where the graph is a complete multipartite graph, and showing that the number of edges is maximized when there are ${\displaystyle r}$ independent sets of size as close as possible to equal. This step can be done as follows:

Let ${\displaystyle S_{1},S_{2},\ldots ,S_{r}}$ be the independent sets of the multipartite graph. Since two vertices have an edge between them if and only if they are not in the same independent set, the number of edges is

$${\displaystyle \sum _{i\neq j}\left|S_{i}\right|\left|S_{j}\right|={\frac {1}{2}}\left(n^{2}-\sum _{i}\left|S_{i}\right|^{2}\right),}$$

where the left hand side follows from direct counting, and the right hand side follows from complementary counting. To show the ${\displaystyle \left(1-{\frac {1}{r}}\right){\frac {n^{2}}{2}}}$ bound, applying the Cauchy–Schwarz inequality to the ${\textstyle \sum \limits _{i}\left|S_{i}\right|^{2}}$ term on the right hand side suffices, since ${\textstyle \sum \limits _{i}\left|S_{i}\right|=n}$.

To prove the Turán Graph is optimal, one can argue that no two ${\displaystyle S_{i}}$ differ by more than one in size. In particular, supposing that we have ${\displaystyle \left|S_{i}\right|\geq \left|S_{j}\right|+2}$ for some ${\displaystyle i\neq j}$, moving one vertex from ${\displaystyle S_{j}}$ to ${\displaystyle S_{i}}$ (and adjusting edges accordingly) would increase the value of the sum. This can be seen by examining the changes to either side of the above expression for the number of edges, or by noting that the degree of the moved vertex increases.

This proof is due to Motzkin & Straus (1965). They begin by considering a ${\displaystyle K_{r+1}}$ free graph with vertices labelled ${\displaystyle 1,2,\ldots ,n}$, and considering maximizing the function

$${\displaystyle f(x_{1},x_{2},\ldots ,x_{n})=\sum _{i,j\ {\text{adjacent}}}x_{i}x_{j}}$$

over all nonnegative ${\displaystyle x_{1},x_{2},\ldots ,x_{n}}$ with sum ${\displaystyle 1}$. This function is known as the Lagrangian of the graph and its edges. The idea behind their proof is that if ${\displaystyle x_{i},x_{j}}$ are both nonzero while ${\displaystyle i,j}$ are not adjacent in the graph, the function

$${\displaystyle f(x_{1},\ldots ,x_{i}-t,\ldots ,x_{j}+t,\ldots ,x_{n})}$$

is linear in ${\displaystyle t}$. Hence, one can either replace ${\displaystyle (x_{i},x_{j})}$ with either ${\displaystyle (x_{i}+x_{j},0)}$ or ${\displaystyle (0,x_{i}+x_{j})}$ without decreasing the value of the function. Hence, there is a point with at most ${\displaystyle r}$ nonzero variables where the function is maximized.

Now, the Cauchy–Schwarz inequality gives that the maximal value is at most ${\displaystyle {\frac {1}{2}}\left(1-{\frac {1}{r}}\right)}$. Plugging in ${\displaystyle x_{i}={\frac {1}{n}}}$ for all ${\displaystyle i}$ gives that the maximal value is at least ${\displaystyle {\frac {|E|}{n^{2}}}}$, giving the desired bound.[3][5]

The key claim in this proof was independently found by Caro and Wei. This proof is due to Noga Alon and Joel Spencer, from their book The Probabilistic Method. The proof shows that every graph with degrees ${\displaystyle d_{1},d_{2},\ldots ,d_{n}}$ has an independent set of size at least

$${\displaystyle S={\frac {1}{d_{1}+1}}+{\frac {1}{d_{2}+1}}+\cdots +{\frac {1}{d_{n}+1}}.}$$

The proof attempts to find such an independent set as follows:

• Consider a random permutation of the vertices of a ${\displaystyle K_{r+1}}$-free graph
• Select every vertex that is adjacent to none of the vertices before it.

A vertex of degree ${\displaystyle d}$ is included in this with probability ${\displaystyle {\frac {1}{d+1}}}$, so this process gives an average of ${\displaystyle S}$ vertices in the chosen set.

(Zykov Symmetrization) Example of first step.

Applying this fact to the complement graph and bounding the size of the chosen set using the Cauchy–Schwarz inequality proves Turán's theorem.[3] See Method of conditional probabilities § Turán's theorem for more.

(Zykov Symmetrization) Example of second step.

Aigner and Ziegler call the final one of their five proofs "the most beautiful of them all". Its origins are unclear, but the approach is often referred to as Zykov Symmetrization as it was used in Zykov's proof of a generalization of Turán's Theorem [6]. This proof goes by taking a ${\displaystyle K_{r+1}}$-free graph, and applying steps to make it more similar to the Turán Graph while increasing edge count.

In particular, given a ${\displaystyle K_{r+1}}$-free graph, the following steps are applied:

• If ${\displaystyle u,v}$ are non-adjacent vertices and ${\displaystyle u}$ has a higher degree than ${\displaystyle v}$, replace ${\displaystyle v}$ with a copy of ${\displaystyle u}$. Repeat this until all non-adjacent vertices have the same degree.
• If ${\displaystyle u,v,w}$ are vertices with ${\displaystyle u,v}$ and ${\displaystyle v,w}$ non-adjacent but ${\displaystyle u,w}$ adjacent, then replace both ${\displaystyle u}$ and ${\displaystyle w}$ with copies of ${\displaystyle v}$.

All of these steps keep the graph ${\displaystyle K_{r+1}}$ free while increasing the number of edges.

Now, non-adjacency forms an equivalence relation. The equivalence classes give that any maximal graph the same form as a Turán graph. As in the maximal degree vertex proof, a simple calculation shows that the number of edges is maximized when all independent set sizes are as close to equal as possible.[3]

Turán's theorem shows that the largest number of edges in a ${\displaystyle K_{r+1}}$-free graph is ${\displaystyle \left(1-{\frac {1}{r}}+o(1)\right){\frac {n^{2}}{2}}}$. The Erdős–Stone theorem finds the number of edges up to a ${\displaystyle o(n^{2})}$ error in all other graphs:

(Erdős–Stone) Suppose ${\displaystyle H}$ is a graph with chromatic number ${\displaystyle \chi (H)}$. The largest possible number of edges in a graph where ${\displaystyle H}$ does not appear as a subgraph is

$${\displaystyle \left(1-{\frac {1}{\chi (H)-1}}+o(1)\right){\frac {n^{2}}{2}}}$$

where the ${\displaystyle o(1)}$ constant only depends on ${\displaystyle H}$.

One can see that the Turán graph ${\displaystyle T(n,\chi (H)-1)}$ cannot contain any copies of ${\displaystyle H}$, so the Turán graph establishes the lower bound. As a ${\displaystyle K_{r+1}}$ has chromatic number ${\displaystyle r+1}$, Turán's theorem is the special case in which ${\displaystyle H}$ is a ${\displaystyle K_{r+1}}$.

The general question of how many edges can be included in a graph without a copy of some ${\displaystyle H}$ is the forbidden subgraph problem.

Another natural extension of Turán's theorem is the following question: if a graph has no ${\displaystyle K_{r+1}}$s, how many copies of ${\displaystyle K_{a}}$ can it have? Turán's theorem is the case where ${\displaystyle a=2}$. Zykov's Theorem answers this question:

(Zykov's Theorem) The graph on ${\displaystyle n}$ vertices with no ${\displaystyle K_{r+1}}$s and the largest possible number of ${\displaystyle K_{a}}$s is the Turán graph ${\displaystyle T(n,r)}$

This was first shown by Zykov (1949) using Zykov Symmetrization[1][3]. Since the Turán Graph contains ${\displaystyle r}$ parts with size around ${\displaystyle {\frac {n}{r}}}$, the number of ${\displaystyle K_{a}}$s in ${\displaystyle T(n,r)}$ is around ${\displaystyle {\binom {r}{a}}\left({\frac {n}{r}}\right)^{a}}$. A paper by Alon and Shikhelman in 2016 gives the following generalization, which is similar to the Erdos-Stone generalization of Turán's theorem:

(Alon-Shikhelman, 2016) Let ${\displaystyle H}$ be a graph with chromatic number ${\displaystyle \chi (H)>a}$. The largest possible number of ${\displaystyle K_{a}}$s in a graph with no copy of ${\displaystyle H}$ is ${\displaystyle (1+o(1)){\binom {\chi (H)-1}{a}}\left({\frac {n}{\chi (H)-1}}\right)^{a}.}$[9]

As in Erdős–Stone, the Turán graph ${\displaystyle T(n,\chi (H)-1)}$ attains the desired number of copies of ${\displaystyle K_{a}}$.

Turan's theorem states that if a graph has edge homomorphism density strictly above ${\displaystyle 1-{\frac {1}{r-1}}}$, it has a nonzero number of ${\displaystyle K_{r}}$s. One could ask the far more general question: if you are given the edge density of a graph, what can you say about the density of ${\displaystyle K_{r}}$s?

An issue with answering this question is that for a given density, there may be some bound not attained by any graph, but approached by some infinite sequence of graphs. To deal with this, weighted graphs or graphons are often considered. In particular, graphons contain the limit of any infinite sequence of graphs.

For a given edge density ${\displaystyle d}$, the construction for the largest ${\displaystyle K_{r}}$ density is as follows:

Take a number of vertices ${\displaystyle N}$ approaching infinity. Pick a set of ${\displaystyle {\sqrt {d}}N}$ of the vertices, and connect two vertices if and only if they are in the chosen set.

This gives a ${\displaystyle K_{r}}$ density of ${\displaystyle d^{k/2}.}$ The construction for the smallest ${\displaystyle K_{r}}$ density is as follows:

Take a number of vertices approaching infinity. Let ${\displaystyle t}$ be the integer such that ${\displaystyle 1-{\frac {1}{t-1}}<d\leq 1-{\frac {1}{t}}}$. Take a ${\displaystyle t}$-partite graph where all parts but the unique smallest part have the same size, and sizes of the parts are chosen such that the total edge density is ${\displaystyle d}$.

For ${\displaystyle d\leq 1-{\frac {1}{r-1}}}$, this gives a graph that is ${\displaystyle (r-1)}$-partite and hence gives no ${\displaystyle K_{r}}$s.

The lower bound was proven by Razborov (2008)[10] for the case of triangles, and was later generalized to all cliques by Reiher (2016)[11]. The upper bound is a consequence of the Kruskal–Katona theorem [12].