I am trying to learn about Lie Theory and I am trying to solve some problems from various lectures notes I have found online. But I have completely stuck on this question:
We work with $A_2=\mathfrak{sl}(3)$. Let's denote an irred by the Dynkin labels of its highest weight. Then $\textbf{3}=(10)$ and $\bar{\textbf{3}}=(01)$. Show by using tensors that $\textbf{3}\otimes{\textbf{3}}$ contains two irred of which $(01)$ is one of them. Draw a picture of the weights in the other one, using the additivity of weights in a tensor product. What is its highest weight? Is it always true, that the tensor product of a representation with itself, with the highest weight $\lambda$, contains the rep with the highest weight $2\lambda$?
My attempt is the follow:
First I thought that I could use Young tableaux so $$ {\textbf{3}}\otimes{\textbf{3}}= (10)\otimes(10) = \square \otimes \square = \square\square\oplus\stackrel{\square}{\square}=(20)\otimes(01) $$ But then I thought that it says "with tensors", so it maybe not the wanted solution. So I thought maybe it wants me to write: $$\psi_a\otimes\psi_b = \frac{1}{2}(\psi_{(a}\psi_{b)}+\psi_{[a}\psi_{b]})$$
Then it asks for the picture of the weights(I don't really understand which picture it wants me to draw). So I draw the weight space diagram(which I found in some other more advanced note)
$(11)$" />
$(01)$" />And as it asks for the $(11)$, we see using the weight space diagram(and its rules) that the highest weight is in the center and it is $2$.
So for the final question, I can then again draw the weight space diagram and conclude that yes, the tensor product of a rep with highest weight $\lambda$ with itself contains the representation with $2\lambda$ highest weight (and it will lie in the center).
That is what I have done until now, I hope that something makes sense or is not entirely wrong! Any help and hints would be greatly appreciated, thanks!