The Structure Constants of $\mathfrak{su}(3)$

Contents

One trace for both arrays
Which triples can be nonzero at all
Family (1): three diagonal indices
Family (2): one diagonal index and one off-diagonal pair
Family (3): one index from each pair
The tables
A Jacobi check

This appendix completes Proposition 18.52 of Lie Groups, Lie Algebras, and Fibre Bundles: it computes every nonvanishing component of the totally antisymmetric array $f_{abc}$ and the totally symmetric array $d_{abc}$ of Equation (18.67) in the Gell-Mann basis of Definition 18.51, and exhibits the two complete tables. The chapter carries two entries as specimens, $f_{458}$ and $d_{247}$; the remaining twenty-three are obtained here, and — more to the point — so is the statement that there are no others. Nothing beyond Definition 18.51 and Equation (18.66) is used.

The naive route is $8^{3} = 512$ traces, or $\binom{8+2}{3} = 120$ after the symmetries are imposed. That is not what is done below. A single observation about which products of three matrix units have a nonzero trace cuts the work to three short families, and the classification of those families is what makes the tables exhaustive rather than merely checked.

One trace for both arrays

Lemma A34.1 (The generating trace).

With $T_{a} = \tfrac{1}{2}\lambda_{a}$ as in Definition 18.51,

\begin{equation}\tag{A34.1} \boxed{\tr\left(\lambda_{a}\lambda_{b}\lambda_{c}\right) = 2\left(d_{abc} + \ii\,f_{abc}\right)}\ec \end{equation}

so that $d_{abc} = \tfrac{1}{2}\Re\tr(\lambda_{a}\lambda_{b}\lambda_{c})$ and $f_{abc} = \tfrac{1}{2}\Im\tr(\lambda_{a}\lambda_{b}\lambda_{c})$; in particular both arrays are real. Rests on Equation (18.68), Definition 18.51 and Equation (18.66).

Proof.

Derives Lemma A34.1. Substituting $T_{a} = \tfrac{1}{2}\lambda_{a}$ into Equation (18.68), each of the three factors contributes a $\tfrac{1}{2}$, so

\begin{equation}\tag{A34.2} f_{abc} = -\frac{\ii}{4}\, \tr\left(\comm{\lambda_{a}}{\lambda_{b}}\lambda_{c}\right)\ec\qquad d_{abc} = \frac{1}{4}\, \tr\left(\acomm{\lambda_{a}}{\lambda_{b}}\lambda_{c}\right)\ep \end{equation}

Now $\lambda_{a}\lambda_{b} = \tfrac{1}{2}\acomm{\lambda_{a}}{\lambda_{b}} + \tfrac{1}{2}\comm{\lambda_{a}}{\lambda_{b}}$, so

\begin{equation*} \tr\left(\lambda_{a}\lambda_{b}\lambda_{c}\right) = \tfrac{1}{2}\tr\left(\acomm{\lambda_{a}}{\lambda_{b}}\lambda_{c}\right) + \tfrac{1}{2}\tr\left(\comm{\lambda_{a}}{\lambda_{b}}\lambda_{c}\right) = \tfrac{1}{2}\left(4d_{abc}\right) + \tfrac{1}{2}\left(4\ii f_{abc}\right)\ec \end{equation*}

which is Equation (A34.1). Reality: the $\lambda_{a}$ are Hermitian, so $\overline{\tr(\lambda_{a}\lambda_{b}\lambda_{c})} = \tr\left((\lambda_{a}\lambda_{b}\lambda_{c})^{\dagger}\right) = \tr(\lambda_{c}\lambda_{b}\lambda_{a})$, and reversing the order of three factors under the trace exchanges $a$ and $c$ in Equation (A34.1); by the total symmetry of $d$ and total antisymmetry of $f$ established in Proposition 18.52, that conjugates the right-hand side. Hence $d$ and $f$ are the real and imaginary parts of one complex number, and both are real numbers.

∎

Remark A34.2 (The symmetry conventions, and how many components there are).

Proposition 18.52 proves that $f_{abc}$ is totally antisymmetric and $d_{abc}$ totally symmetric in $(a,b,c)$. Those two rules fix every component from the tables below, and it is worth saying exactly how many components each independent entry generates.

A nonzero $f_{abc}$ necessarily has three distinct indices — $f_{aab} = -f_{aab} = 0$ — and its six permutations give six nonzero components, three equal to $+f_{abc}$ (the cyclic ones) and three equal to $-f_{abc}$. With nine independent entries that is $9\times6 = 54$ nonvanishing components of $f$.

For $d$ the count depends on the repetition pattern. An entry with three distinct indices generates six equal components; an entry of the shape $d_{aab}$ with $a\neq b$ generates three; and $d_{aaa}$ generates one. Of the sixteen independent entries below, four have distinct indices, eleven have exactly one repetition, and one is $d_{888}$; so $d$ has $4\times6 + 11\times3 + 1 = 58$ nonvanishing components.

Ordering within an entry is therefore free, and the tables list each independent entry once, in the index order used in Equations (18.69) and (18.70).

Which triples can be nonzero at all

Write $E_{jk}$ for the $3\times3$ matrix whose only nonzero entry is a $1$ in row $j$ and column $k$, so that $E_{jk}E_{lm} = \delta_{kl}E_{jm}$ and $\tr E_{jk} = \delta_{jk}$. Reading off Equations (18.63), (18.64) and (18.65), each Gell-Mann matrix is supported on one of four index sets:

\begin{equation}\tag{A34.3} \begin{aligned} \lambda_{1} &= E_{12} + E_{21}\ec & \lambda_{2} &= -\ii E_{12} + \ii E_{21}\ec\\ \lambda_{4} &= E_{13} + E_{31}\ec & \lambda_{5} &= -\ii E_{13} + \ii E_{31}\ec\\ \lambda_{6} &= E_{23} + E_{32}\ec & \lambda_{7} &= -\ii E_{23} + \ii E_{32}\ec\\ \lambda_{3} &= E_{11} - E_{22}\ec & \lambda_{8} &= \tfrac{1}{\sqrt{3}}\left(E_{11}+E_{22}-2E_{33}\right)\ep \end{aligned} \end{equation}

Read as a graph on the vertex set $\set{1,2,3}$, the first row lives on the edge $1$–$2$, the second on $1$–$3$, the third on $2$–$3$, and the fourth — $\lambda_{3}$ and $\lambda_{8}$, the diagonal ones — on loops. Call $\mathrm{A} = \set{1,2}$, $\mathrm{B} = \set{4,5}$, $\mathrm{C} = \set{6,7}$ the three off-diagonal pairs of generator labels and $\mathrm{D} = \set{3,8}$ the diagonal pair.

Lemma A34.3 (Selection rule).

$\tr\left(\lambda_{a}\lambda_{b}\lambda_{c}\right) = 0$ unless the unordered triple $\set{a,b,c}$ falls into one of the three families

all three indices in $\mathrm{D}$;
exactly one index in $\mathrm{D}$ and the other two in a single off-diagonal pair;
one index in each of $\mathrm{A}$, $\mathrm{B}$ and $\mathrm{C}$.

Rests on Equations (18.63), (18.64) and (18.65).

Proof.

Derives Lemma A34.3. Expanding each factor in the matrix units of Equation (A34.3),

\begin{equation}\tag{A34.4} \tr\left(\lambda_{a}\lambda_{b}\lambda_{c}\right) = \sum_{i,j,k}\left(\lambda_{a}\right)_{ij} \left(\lambda_{b}\right)_{jk}\left(\lambda_{c}\right)_{ki}\ec \end{equation}

so a nonzero contribution needs a closed walk $i \to j \to k \to i$ of length three in the index set $\set{1,2,3}$, whose three steps are supplied by $\lambda_{a}$, $\lambda_{b}$, $\lambda_{c}$ in that order. A $\lambda$ with index in $\mathrm{D}$ supplies only loops $i \to i$; a $\lambda$ in an off-diagonal pair supplies only the two steps along its own edge, and along no other.

Count the loops among the three steps. If all three steps are loops, all three indices lie in $\mathrm{D}$: family (1). If exactly two are loops, the third is an edge step, which cannot close a walk that has returned to its starting vertex twice — a walk $i\to i\to i\to j$ ends at $j \neq i$ — so this case contributes nothing. If exactly one step is a loop, the other two are edge steps $u \to v$ and $v \to u$ on one and the same edge, since the walk must return; hence one index in $\mathrm{D}$ and two in a single pair, which is family (2). If no step is a loop, the walk visits three vertices along three distinct edges of the triangle on $\set{1,2,3}$ — two of the three steps on one edge would force the third to be a loop — so one index comes from each of $\mathrm{A}$, $\mathrm{B}$, $\mathrm{C}$: family (3).

∎

Everything is now a finite enumeration inside three small families, and the products needed are the ten diagonal matrices

\begin{equation}\tag{A34.5} \begin{aligned} \lambda_{1}\lambda_{1} &= \lambda_{2}\lambda_{2} = \diag(1,1,0)\ec & \lambda_{1}\lambda_{2} &= \diag(\ii,-\ii,0)\ec\\ \lambda_{4}\lambda_{4} &= \lambda_{5}\lambda_{5} = \diag(1,0,1)\ec & \lambda_{4}\lambda_{5} &= \diag(\ii,0,-\ii)\ec\\ \lambda_{6}\lambda_{6} &= \lambda_{7}\lambda_{7} = \diag(0,1,1)\ec & \lambda_{6}\lambda_{7} &= \diag(0,\ii,-\ii)\ec \end{aligned} \end{equation}

each read straight off Equation (A34.3): for instance $\lambda_{6}\lambda_{7} = \left(E_{23}+E_{32}\right)\left(-\ii E_{23}+\ii E_{32}\right) = \ii E_{23}E_{32} - \ii E_{32}E_{23} = \ii E_{22} - \ii E_{33}$.

Family (1): three diagonal indices

Here $\lambda_{3}$ and $\lambda_{8}$ commute, so every commutator vanishes and $f_{abc} = 0$ throughout the family. For $d$ the four independent triples are $333$, $338$, $388$, $888$, and Equation (A34.1) needs only a trace of a product of diagonal matrices:

\begin{align*} \tr\left(\lambda_{3}^{3}\right) &= \tr\diag(1,-1,0) = 0\ec\\ \tr\left(\lambda_{3}^{2}\lambda_{8}\right) &= \tr\left[\diag(1,1,0)\,\tfrac{1}{\sqrt3}\diag(1,1,-2)\right] = \tfrac{1}{\sqrt3}\left(1+1+0\right) = \tfrac{2}{\sqrt3}\ec\\ \tr\left(\lambda_{3}\lambda_{8}^{2}\right) &= \tr\left[\tfrac{1}{3}\diag(1,-1,0)\diag(1,1,4)\right] = \tfrac{1}{3}\left(1-1+0\right) = 0\ec\\ \tr\left(\lambda_{8}^{3}\right) &= \tfrac{1}{3\sqrt3}\tr\diag(1,1,-8) = \tfrac{-6}{3\sqrt3} = -\tfrac{2}{\sqrt3}\ep \end{align*}

Halving each, the family contributes

\begin{equation}\tag{A34.6} d_{333} = d_{388} = 0\ec\qquad d_{338} = \frac{1}{\sqrt3}\ec\qquad d_{888} = -\frac{1}{\sqrt3}\ep \end{equation}

Family (2): one diagonal index and one off-diagonal pair

Put the diagonal generator last, so that the trace is that of $\left(\lambda_{a}\lambda_{b}\right)\lambda_{\delta}$ with $a,b$ in one pair and $\delta \in \set{3,8}$; the first factor is one of the six diagonal matrices Equation (A34.5). There are $3\times3\times2 = 18$ independent triples — three pairs, three choices of $(a,b)$ up to the symmetries, two diagonal generators — and each is one line. Writing $\lambda_{3} = \diag(1,-1,0)$ and $\lambda_{8} = \tfrac{1}{\sqrt3}\diag(1,1,-2)$:

Pair $\mathrm{A} = \set{1,2}$, on $\diag(\cdot,\cdot,0)$.

\begin{align*} \tr\left(\lambda_{1}\lambda_{1}\lambda_{3}\right) &= \tr\diag(1,-1,0) = 0 &&\Rightarrow\quad d_{113} = d_{223} = 0\ec\\ \tr\left(\lambda_{1}\lambda_{1}\lambda_{8}\right) &= \tfrac{1}{\sqrt3}\tr\diag(1,1,0) = \tfrac{2}{\sqrt3} &&\Rightarrow\quad d_{118} = d_{228} = \tfrac{1}{\sqrt3}\ec\\ \tr\left(\lambda_{1}\lambda_{2}\lambda_{3}\right) &= \tr\diag(\ii,\ii,0) = 2\ii &&\Rightarrow\quad d_{123} = 0\ec\ f_{123} = 1\ec\\ \tr\left(\lambda_{1}\lambda_{2}\lambda_{8}\right) &= \tfrac{1}{\sqrt3}\tr\diag(\ii,-\ii,0) = 0 &&\Rightarrow\quad d_{128} = f_{128} = 0\ep \end{align*}

The second line covers $\lambda_{2}\lambda_{2}$ as well, since $\lambda_{2}^{2} = \lambda_{1}^{2}$ by Equation (A34.5).

Pair $\mathrm{B} = \set{4,5}$, on $\diag(\cdot,0,\cdot)$.

\begin{align*} \tr\left(\lambda_{4}\lambda_{4}\lambda_{3}\right) &= \tr\diag(1,0,0) = 1 &&\Rightarrow\quad d_{344} = d_{355} = \tfrac{1}{2}\ec\\ \tr\left(\lambda_{4}\lambda_{4}\lambda_{8}\right) &= \tfrac{1}{\sqrt3}\tr\diag(1,0,-2) = -\tfrac{1}{\sqrt3} &&\Rightarrow\quad d_{448} = d_{558} = -\tfrac{1}{2\sqrt3}\ec\\ \tr\left(\lambda_{4}\lambda_{5}\lambda_{3}\right) &= \tr\diag(\ii,0,0) = \ii &&\Rightarrow\quad d_{345} = 0\ec\ f_{345} = \tfrac{1}{2}\ec\\ \tr\left(\lambda_{4}\lambda_{5}\lambda_{8}\right) &= \tfrac{1}{\sqrt3}\tr\diag(\ii,0,2\ii) = \ii\sqrt3 &&\Rightarrow\quad d_{458} = 0\ec\ f_{458} = \tfrac{\sqrt3}{2}\ep \end{align*}

The last line reproduces the specimen computed in Proposition 18.52.

Pair $\mathrm{C} = \set{6,7}$, on $\diag(0,\cdot,\cdot)$.

\begin{align*} \tr\left(\lambda_{6}\lambda_{6}\lambda_{3}\right) &= \tr\diag(0,-1,0) = -1 &&\Rightarrow\quad d_{366} = d_{377} = -\tfrac{1}{2}\ec\\ \tr\left(\lambda_{6}\lambda_{6}\lambda_{8}\right) &= \tfrac{1}{\sqrt3}\tr\diag(0,1,-2) = -\tfrac{1}{\sqrt3} &&\Rightarrow\quad d_{668} = d_{778} = -\tfrac{1}{2\sqrt3}\ec\\ \tr\left(\lambda_{6}\lambda_{7}\lambda_{3}\right) &= \tr\diag(0,-\ii,0) = -\ii &&\Rightarrow\quad d_{367} = 0\ec\ f_{367} = -\tfrac{1}{2}\ec\\ \tr\left(\lambda_{6}\lambda_{7}\lambda_{8}\right) &= \tfrac{1}{\sqrt3}\tr\diag(0,\ii,2\ii) = \ii\sqrt3 &&\Rightarrow\quad d_{678} = 0\ec\ f_{678} = \tfrac{\sqrt3}{2}\ep \end{align*}

Every entry follows from Equation (A34.1) by halving the real part for $d$ and the imaginary part for $f$; the index order has been rearranged into the one used in Equations (18.69) and (18.70), which is legitimate by Remark A34.2 — note that $f_{367} = f_{673}$ and $f_{345} = f_{453}$, both cyclic and therefore sign-preserving.

Family (3): one index from each pair

This family is eight triples $(a,b,c)$ with $a \in \set{1,2}$, $b \in \set{4,5}$, $c \in \set{6,7}$, and all eight fall out of a single observation: in Equation (A34.4) only one closed walk survives. The first factor lies on the edge $1$–$2$, the second on $1$–$3$, the third on $2$–$3$; a walk $i \to j \to k \to i$ must therefore start with a step of the first edge, and the only choice compatible with the second step lying on $1$–$3$ is $j = 1$, hence $i = 2$, $k = 3$. So

\begin{equation}\tag{A34.7} \tr\left(\lambda_{a}\lambda_{b}\lambda_{c}\right) = \left(\lambda_{a}\right)_{21}\left(\lambda_{b}\right)_{13} \left(\lambda_{c}\right)_{32}\ec \end{equation}

and Equation (A34.3) supplies the three factors:

\begin{equation}\tag{A34.8} \left(\lambda_{1}\right)_{21} = 1\ec\quad \left(\lambda_{2}\right)_{21} = \ii\ec\qquad \left(\lambda_{4}\right)_{13} = 1\ec\quad \left(\lambda_{5}\right)_{13} = -\ii\ec\qquad \left(\lambda_{6}\right)_{32} = 1\ec\quad \left(\lambda_{7}\right)_{32} = \ii\ep \end{equation}

Multiplying out the eight products of Equation (A34.8) and halving real and imaginary parts as in Equation (A34.1):

\begin{equation}\tag{A34.9} \begin{aligned} \tr\left(\lambda_{1}\lambda_{4}\lambda_{6}\right) &= 1 &&\Rightarrow\ d_{146} = \tfrac{1}{2}\ec\ f_{146} = 0\ec & \tr\left(\lambda_{2}\lambda_{4}\lambda_{6}\right) &= \ii &&\Rightarrow\ f_{246} = \tfrac{1}{2}\ec\\ \tr\left(\lambda_{1}\lambda_{4}\lambda_{7}\right) &= \ii &&\Rightarrow\ f_{147} = \tfrac{1}{2}\ec & \tr\left(\lambda_{2}\lambda_{4}\lambda_{7}\right) &= -1 &&\Rightarrow\ d_{247} = -\tfrac{1}{2}\ec\\ \tr\left(\lambda_{1}\lambda_{5}\lambda_{6}\right) &= -\ii &&\Rightarrow\ f_{156} = -\tfrac{1}{2}\ec & \tr\left(\lambda_{2}\lambda_{5}\lambda_{6}\right) &= 1 &&\Rightarrow\ d_{256} = \tfrac{1}{2}\ec\\ \tr\left(\lambda_{1}\lambda_{5}\lambda_{7}\right) &= 1 &&\Rightarrow\ d_{157} = \tfrac{1}{2}\ec & \tr\left(\lambda_{2}\lambda_{5}\lambda_{7}\right) &= \ii &&\Rightarrow\ f_{257} = \tfrac{1}{2}\ec \end{aligned} \end{equation}

the unlisted partner of each entry vanishing. The value $d_{247} = -\tfrac{1}{2}$ reproduces the second specimen of Proposition 18.52.

The tables

Collecting Equation (A34.6), the three blocks of Family (2): one diagonal index and one off-diagonal pair and Equation (A34.9) gives Tables A34.1 and A34.2. By Lemma A34.3 no triple outside the three families can contribute, and inside them the enumeration was exhaustive; the tables are therefore complete, and reproduce Equations (18.69) and (18.70).

$abc$	$f_{abc}$	family	generating trace
$123$	$1$	(2), pair $\set{1,2}$	$\tr(\lambda_{1}\lambda_{2}\lambda_{3}) = 2\ii$
$147$	$\tfrac{1}{2}$	(3)	$\tr(\lambda_{1}\lambda_{4}\lambda_{7}) = \ii$
$156$	$-\tfrac{1}{2}$	(3)	$\tr(\lambda_{1}\lambda_{5}\lambda_{6}) = -\ii$
$246$	$\tfrac{1}{2}$	(3)	$\tr(\lambda_{2}\lambda_{4}\lambda_{6}) = \ii$
$257$	$\tfrac{1}{2}$	(3)	$\tr(\lambda_{2}\lambda_{5}\lambda_{7}) = \ii$
$345$	$\tfrac{1}{2}$	(2), pair $\set{4,5}$	$\tr(\lambda_{4}\lambda_{5}\lambda_{3}) = \ii$
$367$	$-\tfrac{1}{2}$	(2), pair $\set{6,7}$	$\tr(\lambda_{6}\lambda_{7}\lambda_{3}) = -\ii$
$458$	$\tfrac{\sqrt3}{2}$	(2), pair $\set{4,5}$	$\tr(\lambda_{4}\lambda_{5}\lambda_{8}) = \ii\sqrt3$
$678$	$\tfrac{\sqrt3}{2}$	(2), pair $\set{6,7}$	$\tr(\lambda_{6}\lambda_{7}\lambda_{8}) = \ii\sqrt3$

The nine independent nonvanishing components of the antisymmetric structure constants $f_{abc}$ of $\mathfrak{su}(3)$ in the Gell-Mann basis of Definition 18.51, with the family of Lemma A34.3 each belongs to and the trace from which it was read. Total antisymmetry generates six nonzero components from each entry, fifty-four in all; every other component vanishes.

$abc$	$d_{abc}$	$abc$	$d_{abc}$
Family (1): indices in $\set{3,8}$
$338$	$\tfrac{1}{\sqrt3}$	$888$	$-\tfrac{1}{\sqrt3}$
Family (2): one index in $\set{3,8}$, two in one pair
$118$	$\tfrac{1}{\sqrt3}$	$448$	$-\tfrac{1}{2\sqrt3}$
$228$	$\tfrac{1}{\sqrt3}$	$558$	$-\tfrac{1}{2\sqrt3}$
$344$	$\tfrac{1}{2}$	$668$	$-\tfrac{1}{2\sqrt3}$
$355$	$\tfrac{1}{2}$	$778$	$-\tfrac{1}{2\sqrt3}$
$366$	$-\tfrac{1}{2}$	$377$	$-\tfrac{1}{2}$
Family (3): one index from each pair
$146$	$\tfrac{1}{2}$	$247$	$-\tfrac{1}{2}$
$157$	$\tfrac{1}{2}$	$256$	$\tfrac{1}{2}$

The sixteen independent nonvanishing components of the symmetric tensor $d_{abc}$ of $\mathfrak{su}(3)$, grouped by the family of Lemma A34.3. Total symmetry generates six equal components from each of the four entries with distinct indices, three from each of the eleven with one repetition, and one from $d_{888}$: fifty-eight nonvanishing components in all.

Two entries of Equation (18.70) deserve a word, because the chapter lists them in an order that hides where they come from. $d_{338} = \tfrac{1}{\sqrt3}$ is a family-(1) entry, computed from three diagonal matrices, while $d_{118}$ and $d_{228}$ are family-(2) entries computed from $\lambda_{1}^{2} = \lambda_{2}^{2} = \diag(1,1,0)$; the three come out equal because $\lambda_{3}^{2}$ equals that same matrix. That coincidence is the reason the chapter can write $d_{118} = d_{228} = d_{338}$ in one line.

A Jacobi check

The tables are a computation, and a computation deserves an independent test. The Jacobi identity for the brackets Equation (18.67) is such a test, and it is nontrivial on a triple that involves several different entries at once. Take $(T_{1},T_{4},T_{5})$:

\begin{equation}\tag{A34.10} \comm{T_{1}}{\comm{T_{4}}{T_{5}}} + \comm{T_{4}}{\comm{T_{5}}{T_{1}}} + \comm{T_{5}}{\comm{T_{1}}{T_{4}}} = 0\ep \end{equation}

The first inner bracket has two nonzero components, which is the trap: $\comm{T_{4}}{T_{5}} = \ii f_{45c}T_{c} = \ii\left(\tfrac{1}{2}T_{3} + \tfrac{\sqrt3}{2}T_{8}\right)$, by $f_{453} = f_{345} = \tfrac{1}{2}$ and $f_{458} = \tfrac{\sqrt3}{2}$ from Table A34.1. Since $f_{18c} = 0$ for every $c$ — no entry of Table A34.1 contains both $1$ and $8$ — the $T_{8}$ piece drops, and with $f_{132} = -f_{123} = -1$,

\begin{equation*} \comm{T_{1}}{\comm{T_{4}}{T_{5}}} = \frac{\ii}{2}\comm{T_{1}}{T_{3}} = \frac{\ii}{2}\left(\ii f_{132}T_{2}\right) = \frac{\ii}{2}\left(-\ii T_{2}\right) = \frac{1}{2}\,T_{2}\ep \end{equation*}

For the second term, $f_{516} = -f_{156} = \tfrac{1}{2}$ gives $\comm{T_{5}}{T_{1}} = \tfrac{\ii}{2}T_{6}$, and $f_{462} = f_{246} = \tfrac{1}{2}$ gives

\begin{equation*} \comm{T_{4}}{\comm{T_{5}}{T_{1}}} = \frac{\ii}{2}\comm{T_{4}}{T_{6}} = \frac{\ii}{2}\left(\frac{\ii}{2}T_{2}\right) = -\frac{1}{4}\,T_{2}\ep \end{equation*}

For the third, $f_{147} = \tfrac{1}{2}$ gives $\comm{T_{1}}{T_{4}} = \tfrac{\ii}{2}T_{7}$, and $f_{572} = f_{257} = \tfrac{1}{2}$ gives

\begin{equation*} \comm{T_{5}}{\comm{T_{1}}{T_{4}}} = \frac{\ii}{2}\comm{T_{5}}{T_{7}} = \frac{\ii}{2}\left(\frac{\ii}{2}T_{2}\right) = -\frac{1}{4}\,T_{2}\ep \end{equation*}

The three add to $\left(\tfrac{1}{2} - \tfrac{1}{4} - \tfrac{1}{4}\right)T_{2} = 0$, so Equation (A34.10) holds. The check consumes $f_{123}$, $f_{147}$, $f_{156}$, $f_{246}$, $f_{257}$, $f_{345}$ and $f_{458}$ — seven of the nine independent entries, drawn from both families that carry a nonzero $f$ — and it fails if any one of them is altered. Had the $f_{453}$ contribution to $\comm{T_{4}}{T_{5}}$ been overlooked, leaving only the $f_{458}$ piece that the chapter's specimen computes, the identity would have returned $-\tfrac{1}{2}T_{2}$ instead of zero.

Remark A34.4.

Tables A34.1 and A34.2 discharge the tabulation asserted in Proposition 18.52 and used throughout Lie Groups, Lie Algebras, and Fibre Bundles: in Proposition 18.53, where the entry $\kappa_{33} = 3$ is checked against the $f_{3cd}$ read off Table A34.1; in Proposition 18.54, whose cubic Casimir $C_{3} = d_{abc}T_{a}T_{b}T_{c}$ is built from Table A34.2 and whose invariance identity Equation (18.76) couples the two tables; and in Proposition 18.55, where the adjoint generators are $\left(T_{a}^{\mathrm{ad}}\right)_{bc} = -\ii f_{abc}$. The completeness established by Lemma A34.3 is what licenses the sums over $c$ and $d$ in those places to be evaluated by listing the entries of the tables and stopping.