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?


1 Answer 1


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!


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).

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