# Fundamental Representation of $SU(3)$ is a complex representation

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?

-
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
This link might be useful: motls.blogspot.com/2013/04/complex-real-and-pseudoreal.html – Anne O'Nyme Jul 8 '14 at 9:48

First of all, we are dealing with unitary representations, so that the $T^a$s are always self-adjoint and the representations have the form $$U(v) = e^{i \sum_{a=1}^Nv^a T_a}$$ with $v \in \mathbb R^N$. When you say that $U$ is real you just mean that the representation is made of the very real, unitary, $n\times n$ matrices $U$. This way, the condition $(T_a)^* = -T_a$ is equivalent to the reality (in the proper sense) requirement.

Let us come to your pair of questions.

(1). You are right on your point:

PROPOSITION. A unitary finite dimensional representation is complex (i.e. it is neither real nor pseudoreal) if and only if at least one self-adjoint generator $T_a$ has an eigenvalue $\lambda$ such that $-\lambda$ is not an eigenvalue.

PROOF

Suppose that $$(T_a)^* = -V T_a V^{-1}\tag{1}$$ for some unitary matrix $V$ and every $a=1,2,3,\ldots, N$. Since we also know that $T_a$ is self-adjoint, there is an orthogonal basis of eigenvectors $u_j^{(a)}\neq 0$, $j=1,\ldots, n$ and the eigenvalues $\lambda_j^{(a)}$ are real. Therefore: $$T_au_j^{(a)}= \lambda_j^{(a)} u_j^{(a)}\:.$$ Taking the complex conjugation and using (1) $$VT_aV^{-1}u_j^{(a)*}= -\lambda_j^{(a)} u_j^{(a)*}$$ so that $V^{-1}u_j^{(a)*}\neq 0$ is an eigenvector of $T_a$ with eigenvalue $-\lambda_j$. We conclude that $\lambda$ is an eigenvalue if and only if $-\lambda$ is (consequently, if the dimension of the space is odd, $0$ must necessarily be an eigenvalue as well).

Soppose,vice versa, that for the self-adjoint matrix $T^a$ its (real) eigenvalues satisfy the constraint that $\lambda$ is an eigenvalue if and only if $-\lambda$ is. As $T^a$ is self adjoint, there is a unitary matrix such that: $$T_a = U diag(\lambda_1, -\lambda_1, \lambda_2, -\lambda_2,..., \lambda_{n/2},-\lambda_{n/2}) U^{-1}$$ when $n$ is even, otherwise there is a further vanishing last element on the diagonal. Thus $$T^*_a = U^* diag(\lambda_1, -\lambda_1, \lambda_2, -\lambda_2,..., \lambda_{n/2},-\lambda_{n/2}) U^{-1*}$$ Notice that $U^*$ is unitary if $U$ is such. Let us indicate by $e_1,e_2, \ldots, e_n$ the orthonormal basis of eigenvectors of $T^a$ where the matrix takes the above diagonal form. If $W$ is the (real) unitary matrix which swaps $e_1$ with $e_2$, $e_3$ with $e_4$ and so on (leaving $e_n$ fixed if $n$ is odd), we have that $$W diag(\lambda_1, -\lambda_1, \lambda_2, -\lambda_2,..., \lambda_{n/2},-\lambda_{n/2}) W^{-1} = - diag(\lambda_1, -\lambda_1, \lambda_2, -\lambda_2,..., \lambda_{n/2},-\lambda_{n/2})$$ and thus $$T^*_a = U^*W^{-1}(- UT_aU^{-1}) WU^{-1*} = -S T_a S^{-1}$$ with $S= U^*W^{-1}U$, which is unitary because composition of unitary matrices.

We can conclude that, as you claim, a way to show that a representation is complex (i.e. it is not real) is to show that at least one generator matrix $T_a$ has (non-vanishing) eigenvalues that do not come in plus-minus pairs.

QED

(2). In view of (1), if the list of eigenvalues you presented is correct, the considered representation is obviously complex.

-

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)

-
– Angie38750 Feb 3 '14 at 4:16