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?
You merely miscalculated the hook rule.
Convince yourself that the correct denominator, instead, gives you $$\frac{3\times 4\times 5\times2\times 3\times 4}{4\times 3\times 3\times 2\times 2\times 1}=10.$$
Note that the 10 in the previous line is really a singlet (the first column, detachable!) and a 3-box 10.
So, its complex conjugate, your diagram, is what needs to be appended below it, to make all columns of height 3, so, precisely your tableau in question!