Here you can find my notes to my presentation on the paper
Note on the Turán number of the 3-linear hypergraph

I presented this paper in a talk for the Spring School of Combinatorics 2023.

Here is the handout for the talk.

The authors of the paper use

The whole paper deals with *hypergraphs* but I am going to refer to them simply
as *graphs* for brevity.

Lastly, the proof of the lemma is slightly confusing in the original paper so I changed the notation little bit to make it easier to digest. Nonetheless, the proof otherwise corresponds to the one in the paper.

*The notes follow.*

I'd like to welcome you to my talk. If you should have any questions during the talk, do not hesitate to ask.

I am going to talk about the paper by: Chaoliang Tang, Hehui Wu, Shengtong Zhang, Zeyu Zheng.

The paper, I'd like to discuss is called Note on the Turán number of the

Let's start with definitions.

I will also use

So what do we already know about the

This was recently improved by Fletcher
who showed that

Here we will show a stronger result which is

So how can we prove all of this?

Let's start with the first theorem and prove

Then we need

- Proof by contradiction. Minimal counter-example.
- Observation using
$\frac{\chi(v)}{d(v)}$ - From the observation, by the pigeonhole principle we will see the existence of some special edge
- Analysis of
$d(v)$ s' of the edge - Only possible case satisfies assumption in the lemma. We can use the lemma. Lemma will create a contradiction with the minimality of the counter example.

Now, we are ready to do the proof.
Suppose the contrary. That there is a crown-free graph with more than

First, let's specify the

The core of this proof is the observation that we can express *the number of vertices of
degree six or more*.

Now, from our initial assumption, we get that

Because the sum of sums is equal to

Now, without the loss of generality assume that degrees are non-decreasing

Now we will need to use the lemma I mentioned earlier.

This lemma sounds complicated. But in fact, the one important thing it tells us
is in the last sentence and that is that all *elements of $E_S$ are
subsets of $S$.*
This tells us that the graph induced by vertices in

Now when we proved the first theorem, we will quickly prove the second.

The proof is similar to the first one.
The only real difference is in the

So again, suppose the contrary. Let

We again define

From this, we get a similar formula involving sums as in the previous case.

We assumed that

We again do a few observations on the degrees of vertices of the edge.
Without loss of generality let degrees be non-decreasing.
Surely

Any questions? :)

Now I would like to attempt the proof of the lemma.

But first, let me give you some definitions.
I'm going to use

- WLOG
$d(y), d(z)\geq 5$ and$d(x)\geq 4$ . - From this we will show that
$d(y) = d(z) = 5$ . - Show that
$S \setminus \{x,y,z\} = G(y) = G(z) \supset G(x)$ - Define
$F$ to be the set of all edges containing vertex from$S$ but disjoint from$e$ - Start the proof that
$F = \emptyset$ - Create an auxiliary bipartite graph
$H$ , where vertices are some edges of$G$ . - Show that
$H$ looks like$2$ squares - Assume that some
$f\in F$ . Show that no matter what vertices are in$f$ we get a contradiction. Use the symmetry of$H$ to only consider a few cases.

Without loss of generality,

I claim that

Let's start with the first claim.
Let's say it is false and

Now to the second claim. Again suppose the contrary and let there be an edge

Now we have

Remember, the lemma tells us that

We will define an auxiliary bipartite graph as follows

What do we know about this bipartite graph?
Surely there are

I claim that it cannot be

Now, let's take some edge intersecting

Now, we would like to show that

Before we do that we need to do one last thing and that is to
show that all edges containing

Now, from the linearity, it is not possible to have edges

From now on we need to use the symmetry of the problem and the fact that we
know all possible edges containing

With this I'd like to end and thank you for your attention. If you have any questions you can ask them now or talk to me later during the week. Thank you.