Kronecker's delta from the solution to the final project in Peskin's and Schroeder's book

[MathematicalPhysicist]

TL;DR Summary

The reference is the final project on pages 775-777 of Peskin's and Schroeder's textbook on QFT.
And the solution from here:
https://zzxianyu.files.wordpress.com/2017/01/peskin_problems.pdf

on page 194.

My problem is on page 194 of the solution, where he writes: $\frac{1}{2}\delta^{ab}\frac{1}{2}\delta^{ab}=2$.
I assume there are three colours and thus $a,b \in \{ 0,1,2 \}$.
So I get: $\delta^{ab}\delta^{ab} = \delta^{00}\delta^{11}+\delta^{11}\delta^{00}+\delta^{11}\delta^{22}+\delta^{22}\delta^{11}+\delta^{22}\delta^{00}+\delta^{00}\delta^{22}+\delta^{22}\delta^{22}+\delta^{11}\delta^{11}+\delta^{00}\delta^{00}=9$, so in the above equation in the solution shouldn't it be: $9/4=2.25$ and not $2$ as it's written in this solution?

 

[PeroK]

$\delta^{ab}\delta^{ab} = \delta^{00}\delta^{11}+\dots$

I can stop you there and ask what the specific values of $a$ and $b$ are in that first term?

 

[MathematicalPhysicist]

I can stop you there and ask what the specific values of $a$ and $b$ are in that first term?

Yes, you are quite right.
If there's a summation convention here then it should be $2(\delta^{00}\delta^{00}+\delta^{11}\delta^{11}+\delta^{22}\delta^{22})$, the factor two is because we are counting twice.
But I still get 6 and not 8.
Where did I get it wrong?

 

[PeroK]

Yes, you are quite right.
If there's a summation convention here then it should be $2(\delta^{00}\delta^{00}+\delta^{11}\delta^{11}+\delta^{22}\delta^{22})$, the factor two is because we are counting twice.
But I still get 6 and not 8.
Where did I get it wrong?

How are $a$ and $b$ defined? If they run from $0$ to$3$ then there are only four cases where $\delta_{ab} \ne 0$. The answer should be $1$.

 

[MathematicalPhysicist]

How are $a$ and $b$ defined? If they run from $0$ to$3$ then there are only four cases where $\delta_{ab} \ne 0$. The answer should be $1$.

I said from what I understand we have three colours, so $a,b\in \{ 0,1,2\}$ so the sum should be from 0 to 2.

 

[PeroK]

I said from what I understand we have three colours, so $a,b\in \{ 0,1,2\}$ so the sum should be from 0 to 2.

I'm not sure how he gets $8$ from that.

 

[nrqed]

I said from what I understand we have three colours, so $a,b\in \{ 0,1,2\}$ so the sum should be from 0 to 2.

The sum is over all the generators of SU(3). There are eight of them.