# Questions regarding the 'adjoint of $SO(N)$' section in Zee's group theory

I am reading A. Zee's book about group theory and I am confused about several parts of his discussion in chapter 4.1 about the adjoint of $$SO(n)$$.

First question:

The first thing mentioned is that an antisymmetric tensor $$T^{ij}$$ in $$SO(N)$$ furnished (still don't fully understand how that word is used in this book) a $$\frac{1}{2}N(N-1)$$-dimensional representation. He also says there are $$\frac{1}{2}N(N-1)$$ generators of $$SO(N)$$ which are the $$N$$-dimensional matrices $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$

He then renames the $$(mn)$$ index $$1 \leq a \leq \frac{1}{2}N(N-1)$$. But from this part he immediately states:

Then we have $$\frac{1}{2}N(N-1)$$ matrices $$\mathcal{J}_{a}^{i j}$$, each of which is a $$\frac{1}{2}N(N-1)$$-by-$$\frac{1}{2}N(N-1)$$ dimensional matrix.

How does this happen? $$\mathcal{J}_{(m n)}^{i j}$$ was $$N$$-dimensional, but $$\mathcal{J}_{a}^{i j}$$ is $$\frac{1}{2}N(N-1)$$-dimensional just from reindexing?

Second question

This follows immediately from the first question.

Zee then states

We can also regard the antisymmetric tensor $$T^{ij}$$ as an $$N$$-by-$$N$$ matrix, and hence write it as a linear combination of the $$\mathcal{J}_a$$'s, with coefficients denoted by $$A_a$$: $$T^{ij} = A_a \mathcal{J}^{ij}_a$$

(where summation over $$a$$'s is implied and $$i=1,2,...,N$$ and $$a=1,2,...,\frac{1}{2}N(N-1)$$

This is just another question of dimensionality. How can we switch between $$T^{ij}$$ being $$\frac{1}{2}N(N-1)$$-dimensional to $$N$$-dimensional?

I don't have that text, but your question is standard stuff. I suspect what you are missing is the fast interplay of D-dimensional vectors and the D×D matrices acting on them, sometimes regarded as vector elements themselves, of a larger vector!

The dimension D of the representation is the dimensionality of the "vectors" on which the D×D representation matrices act. But all kinds of objects "furnish" D-dimensional vectors on which the suitable representation matrices act to transform them. "Furnish" means provide components of a large D-dimensional vector specifying a possibly new representation.

Antisymmetric tensors $$T^{ij}$$ in $$SO(N)$$ with indices $$j=1,...,N$$ comprise a set of $$\frac{1}{2}N(N-1)$$ independent antisymmetric matrices, so they furnish vectors of an irreducible D-dimensional representation, with $$D=\frac{1}{2}N(N-1)$$: they define the relevant vector space. Each such matrix is a component of the D-vector.

It is also true that there are $$\frac{1}{2}N(N-1)$$ generators of $$SO(N)$$, indexed by the (mn) pairs, $$\mathcal{J}_{(m n)}^{i j}=\left(\delta^{m i} \delta^{n j}-\delta^{m j} \delta^{n i}\right)$$ which act on the space of N-dimensional js to rotate them to such is.

• However, these N×N antisymmetric matrices may also be regarded as the $$D=\frac{1}{2}N(N-1)$$ components of a new vector, acted upon by some larger $$\frac{1}{2}N(N-1) \times \frac{1}{2}N(N-1)$$ matrices, (actually $$\frac{1}{2}N(N-1)$$ independent such, indexed by (mn). This is the adjoint representation.)

So the D=N generators became the $$D=\frac{1}{2}N(N-1)$$-vector elements of the adjoint (which they furnished). Fortuitously, these two dimensions are the same for N=3, so the fundamental of SO(3) happens to also be its adjoint!

Now

We can also regard the antisymmetric tensor $$T^{ij}$$ as an $$N$$-by-$$N$$ matrix, and hence write it as a linear combination of the $$\mathcal{J}_a$$'s, with coefficients denoted by $$A_a$$: $$T^{ij} = A_a \mathcal{J}^{ij}_a.$$

This object is an N×N antisymmetric matrix, but also an $$\frac{1}{2}N(N-1)$$-dimensional vector in the adjoint representation vector space, introduced above, itching to be rotated by the suitable $$\frac{1}{2}N(N-1)\times\frac{1}{2}N(N-1)$$ adjoint representation matrices. (These matrices are gotten/furnished by the structure constants of the Lie algebra you'd get from your $$\cal J$$s in the D=N irrep., but don't worry about them.)

• The dimension of the adjoint representation is also the dimension of the corresponding Lie algebra (regarded as a vector space as you saw).