Trace identities for $\text{SO}(n)$

Question

The Green-Schwarz mechanism in Type I string theory involves certain identities relating traces in the vector and adjoint representations of $\text{SO}(n)$ of dimension $n$ and $n(n - 1)/2$ respectively. If $\text{Tr}_\text{a} T^n$ indicates the trace of the adjoint for the $n$th power of a generator $T$ of the Lie algebra of $\text{SO}(n)$ and $\text{Tr}_\text{v} T^n$ is the trace in the vector representation, how do I see that$$\text{Tr}_\text{a} T^2 = (n - 2)\text{Tr}_\text{v}T^2,$$$$\text{Tr}_\text{a} T^4 = (n - 8)\text{Tr}_\text{v}T^4 + 3\text{Tr}_\text{v}T^2\text{Tr}_\text{v}T^2,$$$$\text{Tr}_\text{a}T^6 = (n - 32)\text{Tr}_\text{v} T^6 + 15\text{Tr}_\text{v} T^2\text{Tr}_\text{v}T^4?$$Progress. I've tried figuring out why we can take $T$ to be in the Cartan subalgebra, but I have not been very successful at figuring out why. Could anybody help?

Answer 1

It is well-known that the Killing form of $\mathfrak{so}(n)$ satisfies $$ Tr (ad(x)ad(y))=(n-2)Tr(xy), $$ see here. The first formula comes form setting $T=ad(x)=ad(y)$ in the Killing form above. For elements $x$ in the Cartan subalgebra we know that $ad(x)$ is a diagonal matrix. This makes a computation easy. We also obtain $$ Tr(ad(x)^2ad(y)^2)=(n-6)Tr(x^2y^2)-2Tr(xyxy)+Tr(x^2)Tr(y^2)+2(Tr(xy)^2). $$ Letting $x=y$ and $T=ad(x)$ we obtain the second formula. Similarly we get the third one. One can prove this by making an "Ansatz", i.e., of the form $Tr(ad(x)^2ad(y)^2)=aTr(x^2y^2)+bTr(xyxy)+cTr(x^2)Tr(y^2)+d(Tr(xy)^2)$ and solving a linear system of equations in $a,b,c,d$.

Answer 2

Solution 1. Take any generator of $\text{SO}(N)$. It corresponds, in the fundamental representation, to rotating a plane defined by two of the coordinates. By rotating or relabeling these coordinates appropriately, we can redefine the generator to rotate $x^1$ and $x^2$. The generator can then be taken as $\sigma_2$, in the vector representation, with zeros in all other positions. Since $\sigma_2^2 = \mathbb{1}$, we have $\text{Tr}_\text{v}T^{2N} = 2$ for all integer $N$. Note that the normalization of the generators is irrelevant for the identities we want to show, as they have the same power in the generators in each term.

To find this generator in the adjoint representation, recall that in the adjoint representation $(T_i^{\text{adj}})_{jk} = -if_{ijk}$, so we need to find the commutator of the generator in question with all other generators. There are $N - 2$ other generators $T_{1i}$ rotating $x^1$ and $x^i$ with $i \neq 1$, $2$, with $[T_{12}, T_{1i}] = iT_{2i}$, and similarly $N - 2$ generators $T_{2i}$ with $[T_{12}, T_{2i}] = -iT_{1i}$. From this we can construct the matrices and explicitly find the trace,$$\text{Tr}_\text{a}T^2 = (T^{\text{adj}})_{jk}(T^{\text{adj}})_{kj} = 2\sum_i 1^2 = 2(N - 2),\quad \text{Tr}_\text{a}T^{2N} = 2(N - 2),$$where the factor of $2$ behind the sum comes from $j = (1i)$, $k = (2i)$ and $j = (2i)$, $k = (1i)$.

We can then check\begin{align*} 2(N - 2) & = \text{Tr}_\text{a}T^2 = (N - 2)\,\text{Tr}_\text{v}T^2 = 2(N - 2), \\ 2(N - 2) & = \text{Tr}_\text{a}T^4 = (N - 8)\,\text{Tr}_\text{a}T^4 + 3(\text{Tr}_\text{v}T^2)^2 = 2(N - 8) + 12 = 2(N - 2), \\ 2(N - 2) & = \text{Tr}_\text{a}T^6= (N - 32)\,\text{Tr}_\text{v}T^6 + 15 \text{Tr}_\text{v}T^2 \text{Tr}_\text{v}T^4 = 2(N - 32) + 60 = 2(N - 2). \end{align*}Solution 2. We first prove a lemma.

Lemma. The Chern character factorization property$$\text{tr}_{\rho_1 \times \rho_2} e^{iF} = (\text{tr}_{\rho_1}e^{iF})(\text{tr}_{\rho_2}e^{iF})$$allows us to deduce that, for $\text{SO}(N)$,$$\text{Tr}\,e^{iF} = {1\over2}(\text{tr}\,e^{iF})^2 - {1\over2}\text{tr}\,e^{2iF}.$$The symbol $\text{Tr}$ is used to refer to the adjoint representation, whereas the symbol $\text{tr}$ is used to refer to the $n$-dimensional fundamental representation.

Proof. Before studying the relevant Chern characters, recall that the adjoint representation of $\text{SO}(N)$ is equivalent to the second exterior power of the fundamental representation. We can therefore choose the following basis vectors for the adjoint representation,$$|e_{ij}\rangle = {1\over{\sqrt{2}}}(|e_i\rangle \otimes |e_j\rangle - |e_j\rangle \otimes |e_i\rangle),$$where $\{|e_i\rangle\}$ is a basis for the fundamental representation. With this choice of basis, the $\text{SO}(N)$ generators in the adjoint representation $\{\lambda_\text{A}^a\}$ are related to the $\text{SO}(N)$ generators in the fundamental representation $\{\lambda^a\}$ by$$\lambda_\text{A}^a = \lambda^a \otimes 1 + 1 \otimes \lambda^a.$$Now, let us expand the Chern character for the adjoint representation,\begin{align*} \text{Tr}\,e^{iF_\text{A}} & = {1\over2} \sum_{i, j = 1}^N \sum_{m = 0}^\infty {1\over{m!}} i^m \langle e_{ij}| (F \otimes 1 + 1 \otimes F)^m |e_{ij}\rangle \\ & = {1\over2} \sum_{i, j = 1}^N \sum_{m = 0}^\infty {1\over{m!}}i^m \sum_{n = 0}^m \binom{m}{n} \langle e_{ij} |F^n \otimes F^{m - n} |e_{ij}\rangle.\end{align*}Notice that the double summation $\sum_{m = 0}^\infty \sum_{n = 0}^m \ldots$ is equivalent to $\sum_{n = 0}^\infty \sum_{m = n}^\infty \ldots$. Therefore,\begin{align*} \text{Tr}\,e^{iF_\text{A}} & = {1\over2} \sum_{i, j = 1}^N \sum_{n = 0}^\infty \sum_{m = n}^\infty {1\over{n!(m - n)!}}i^m\langle e_i|F^n|e_i\rangle \langle e_j|F^{m - n}|e_j\rangle \\ & \phantom{=} - {1\over2} \sum_{i, j = 1}^N \sum_{n = 0}^\infty \sum_{m = n}^\infty {1\over{n!(m - n)!}} i^m\langle e_i|F^n|e_j\rangle \langle e_j|F^{m - n}|e_i\rangle \\ & = {1\over2}(\text{tr}\,e^{iF})^2 - {1\over2}\text{tr}\,e^{2iF}.\end{align*}$$\tag*{$\square$}$$We begin with the identity for Chern characters in the adjoint and fundamental representation of $\text{SO}(N)$,$$\text{Tr}\,e^{iF_\text{A}} = {1\over2}(\text{tr}\,e^{iF})^2 - {1\over2}\text{tr}\,e^{2iF}.$$Remember that the trace of an odd power of $F_\text{A}$ and an odd power of $F$ vanish, since these matrices are antisymmetric. The Chern characters therefore have the following expansion,\begin{align*} \text{Tr}\,e^{iF_\text{A}} & = {1\over2} N(N - 1) - {1\over2}\text{Tr}\,F_\text{A}^2 + {1\over{24}}\text{Tr}\,F_\text{A}^4 - {1\over{720}}\text{Tr}\,F_\text{A}^6 + \ldots, \\\text{tr}\,e^{iF} & = N - {1\over2}\text{tr}\,F^2 + {1\over{24}}\text{tr}\,F^4 - {1\over{720}}\text{tr}\,F^6 + \ldots.\end{align*}Substituting these expansions into our identity, we find that\begin{align*} \text{Tr}\,F_\text{A}^2 & = (N - 2)\, \text{tr}\,F^2, \\ \text{Tr}\,F_\text{A}^4 & = (N - 8)\,\text{tr}\,F^4 + 3(\text{tr}\,F^2)^2, \\ \text{Tr}\,F_\text{A}^6 & = (N - 32)\,\text{tr}\,F^6 + 15\text{tr}\,F^2 \text{tr}\,F^4,\end{align*}where we have extracted the forms of degree $4$, $8$, and $12$ respectively.