Self-dual $126$ and $126^*$ in SO(10) or Spin(10) Irreps

Question

We are looking at 126 or $126^*$ in SO(10) or Spin(10) Irreps.

126 is known as a complex total "anti-symmetric" and "self-dual" 5-index tensor irreps in Spin(10). This means the component of 126 is from the counting of anti-symmetric 5-index tensor $$\frac{1}{2}{10 \choose 5}=126. $$ I am puzzled by some statements of the Georgi book Lie Algebras in Particle Physics 2nd ed -- From Isospin to Unified Theories.

My questions are:

1. If 126 in Spin(10) irrep is self-dual, would 126 and $126^*$ be the same or different?

2. But it looks that in Georgi book (23.41)and (24.12), it said that $$ 16 x 16 = 10 + 120 + 126, $$ which is different from https://arxiv.org/abs/1912.10969 Table A.79, said $$ 16 x 16 = 10 + 120 + 126^*. $$ So are 126 and $126^*$ the same or different? Why Gerogi seemed to suggest 126 and $126^*$ are the same?

3. The 126 is a complex total "anti-symmetric self-dual" 5-index tensor. But I am not sure why Gerogi said that it is symmetric in some way, saying the $(5)$ as 5-index tensor of the symmetry of $\Gamma_j Rs$ of SU(11)? See below. What does this mean?

Answer 1

1. The modules $126$ and $126^*$ are different. The one has highest weight $2\lambda_4$, the other $2\lambda_5$. The graph automorphism interchanges them.

2. $16$ is different from $16^*$, which are $\lambda_4$ and $\lambda_5$. Indeed, we see $L(2\lambda_4)$ in the symmetric square of $L(\lambda_4)$.

3. The tensor square of a module breaks up as the exterior square and the symmetric square. The symmetric square of $16$ is $10\oplus 126$, i.e., $L(\lambda_1)\oplus L(2\lambda_4)$, and the exterior square is $120=L(\lambda_3)$. It appears symmetrically presumably means it is a submodule of the symmetric square.