Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Let in a $D(R)$ dimensional representation of $SU(N)$ the generators, $T^a$s follow the following commutation rule:
$\qquad \qquad \qquad [T^a_R, T^b_R]=if^{abc}T^c_R$.

Now if $-(T^a_R)^* = T^a_R $, the representation $R$ is real. Again if we can find a unitary matrix, $V(\neq I)$ such that

$ \qquad \qquad \qquad -(T^a_R)^*=V^{-1} T^a_R V \quad \forall a $

then the representation $R$ is pseudoreal.

If a representation is neither real nor pseudoreal, the representation $R$ is complex.

Claim: One way to show that a representation is complex is to show that at least one generator matrix $T^a_R$ has eigenvalues that do not come in plus-minus pairs.

Now let us consider $SU(3)$ group. The generators in the fundamental representation are given by

$T^a =\lambda^a/2; \quad a=1,...8$,
where $\lambda^a$s are the Gell-Mann matrices. We see that $T^8$ has eigenvalues $(1/\sqrt{12}, 1/\sqrt{12}, -1/\sqrt{3} )$.

My doubt is:

According to the claim, is the fundamental representation of $SU(3)$ a complex representation?

share|improve this question
An element of the fundamental representation of $SU(N)$ is a $n*n$ complex matrix $M$ such that $y^\dagger M^\dagger Mx = y^\dagger x$, for every complex vectors $x, y$. –  Trimok Jun 28 '13 at 16:39
add comment

1 Answer

The N-dimensional fundamental representation of SU(N) for N greater than two is a complex representation whose complex conjugate is often called the antifundamental representation.

Thus SU(3) fundamental representation is a complex representation.

(see for example: Wiki)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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