$K_{n}$ graph - crossing lemma given a complete graph $K_{n}$, i not able to understand why for $n >> 5$, we know that there are $\Omega(n^4) = \Omega(m^2)$ crossings in $K_{n}$.
i not able to see why its $\Omega(n^4)$ , for example in $K_{4}$ graph we dont have any crossings at all:

in my opinion, for any subgraph $K_{4}$ in $K_{n}$ , we will not get any crossings.
about $\Omega(m^2)$ , i not sure, but is it related that for every node we have edge to all others nodes?
 A: You didn't say but I guess you're asking about planar drawings of $K_n$. For $n\ge5$ any planar drawing of $K_n$ contains at least
$$\frac{\binom n5}{n-4}=\frac{n(n-1)(n-2)(n-3)}{120}$$
crossings. This is because $K_n$ contains $\binom n5$ copies of $K_5$, each $K_5$ contains at least one crossing, and each crossing occurs in at most $n-4$ of the  $K_5$'s.
A: You may enjoy learning about the famous crossing lemma, which was conjectured by Erdös and Guy in 1973 where $1/64$ was replaced by some absolute constant $c$.
Theorem (Leighton 1982, Ajtai-Chvátal-Newborn-Szemerédi 1982) Let $G$ be a simple graph with $n$ vertices and $m$ edges, where $m\geq 4n$. Then
$$
\operatorname{cr}(G) \geq \frac{1}{64} \frac{m^3}{n^2}
$$
where $\operatorname{cr}(G)$ is the crossing number of $G$, i.e. the minimum number of crossings needed in any plane drawing of $G$.
Chapter 45 of Proofs From The Book (6th edition) by Aigner and Ziegler contains a magnificent proof of this theorem using the probabilistic method. In fact, this is the very last proof presented in the book.
Using this theorem, and the fact that $K_n$ has $m=\binom{n}{2}=\frac{n(n-1)}{2}$ many edges, we can apply the above lemma to get for $n\geq 9$:
$$
\operatorname{cr}(K_n) \geq \frac{1}{64} \frac{n^3(n-1)^3}{8n^2} = \frac{1}{512} n(n-1)^3
$$
On the other hand, it is clear that one can have at most $C n^4$ many possible crossings, because there can be at most $\binom{m}{2}\approx \frac{n^4}{4}$ many crossings between different edges. Thus,  $\operatorname{cr}(K_n)=\Omega(n^4)$.
