# $8 \otimes 8$ in $SU(3)$, dimension of the Young-tableau corresponding to the $\bar{10}$

In Georgi's Lie Algebras in Particle Physics, he calculates the decomposition of $$8\otimes 8$$ in $$SU(3)$$, and obtains $$8\otimes 8 = 27 \oplus 10 \oplus \bar{10} \oplus 8 \oplus 8 \oplus 1,$$

corresponding to a Young-tableaux decomposition that looks like this:

I can follow the procedure for obtaining this decomposition in diagram form, but I'm struggling to see why the tableau above the "$$\bar{10}$$" does indeed correspond to $$\bar{10}$$.

Calculating the dimension of the representation using hook lengths we obtain $$\frac{3\times 4\times 5\times2\times 3\times 4}{4\times 3\times 2\times 2\times 2\times 1}=15$$ where the numerator is the product of the numbers obtained by placing a $$3$$ (for $$SU(3)$$) in the top-left box, and adding $$1$$ as we move to the right along the row, then subtracting $$1$$ from each of the numbers in these boxes and placing those in the next row down. The denominator is the product of the Hooks for each box. Following that exact procedure for all the other tableaux, I find the correct dimensions. So why does that give the wrong dimension for this particular diagram? Also, What about this diagram means that it corresponds to $$\bar{10}$$, and not just $$10$$? Isn't the antifundamental representation of $$SU(N)$$ simply a diagram with $$N-1$$ rows and $$1$$ column?

• Comment to the post (v2): One of the hooklengths in the denominator should be 3 rather than 2. – Qmechanic Oct 7 '20 at 19:29
• Aha! Thanks, that was a silly mistake. And what is it about the diagram that means it corresponds to $\bar{10}$ rather than just $10$? – Ali Oct 7 '20 at 19:53

Convince yourself that the correct denominator, instead, gives you $$\frac{3\times 4\times 5\times2\times 3\times 4}{4\times 3\times 3\times 2\times 2\times 1}=10.$$