Im getting crazy trying to derive this simple expression.

Say $f^{abc}$ are structure constants of a Lie algebra of $SU(N)$ with $[T^a, T^b]=i f^{abc}T^c$. Then chosing normalization such that


Show that for fundamental rep of $SU(N)$: $$tr(T^aT^b)=\frac{1}{2}\delta^{ab}$$

And for adjoint rep: $$tr(T^aT^b)=N\delta^{ab}$$

Let me know if you have any suggestions, all much appreciated.

EDIT: Adjoint representation case is simply a matter of definition $(T_{adj}^a)^{bc}=-if^{abc}$


  • 2
    $\begingroup$ This appears to be a pure math question (and also potentially off-topic by our homework policy), might it be better suited for Mathematics? $\endgroup$ – ACuriousMind Jul 12 '17 at 15:12
  • 1
    $\begingroup$ This comes from Schwartz QFT book (p.487)(Gives me the impression that its 1 line proof). These should be basic things in QFT so I thought that most people should know about, apologies if posting in the wrong section. $\endgroup$ – Wint Jul 12 '17 at 15:14
  • $\begingroup$ Dear Cosmas Zachos, than you very much for the reply! Indeed adjoint is clear to me now just by using the definition. Could you elaborate more on the fundamental case? What do you mean about not writing the algebra? $\endgroup$ – Wint Jul 12 '17 at 15:45

It is a one-line proof, but you are meant to be familiar with the language, or some equivalent thereof. Let's call F the fundamental representation N×N matrices, A the adjoint (N²—1)×(N²—1) ones, and 0 the 1-dim singlet rep ones.

The Kronecker product fundamental-antifundamental (N×N)⊗(N×N) representation matrices are given by the coproduct, $$ F^a\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^a = A^a \oplus 0 = A^a, $$ which you must check satisfies the same Lie algebra!

Now multiply this with its a ⟶ b analog, $$ (F^a\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^a) (F^b\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^b) = A^a A^b= \\ F^a F^b\otimes 1\!\!1 + 1\!\!1 \otimes \bar F^a \bar F^b + F^a\otimes \bar F^b + F^b\otimes \bar F^a, $$ and trace. The trace of the direct product amounts to a plain product of the separate traces of each tensor factor, of course, so the last two terms vanish, as the F s are traceless.

You thus establish in a blink that $$ 2N\operatorname{tr} F^a F^b = \operatorname{tr} A^a A^b . $$ If the trace on the right is N, the trace on the left must be 1/2.

You know the trace of all irreps must be fixed w.r.t. the adjoint normalization, as scaling up the generators of the algebra results in an identical rescaling of the structure constants, which we already fixed.

The reason for these choices, despised by mathematicians (who normalize the linchpin f s with 1), is because we want all SU(N) s to include an isospin SU(2) with its conventional spin 1/2 normalizations... The mayhem started with flavor SU(3) and lives on...

To make sure you are in control of the algorithm, try composing ${\bf 3}\otimes {\bf 3}= {\bf 6}\oplus \bar{\bf 3}$ in SU(3) and compute this trace for the sextet. You should find $3 - 1/2$=5/2. It is half the representation index you find in math texts.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.