1
$\begingroup$

I want to check if the fundamental representation of $SU(3)$ is irreducible. The algebra is

$$\mathbb{su}(3) = \{ m \in Mat(3,\mathbb{C} )\ |\ m = -m^+,\ Tr[m] = 0 \}$$ and I've found the generators. So, I've checked if the algebra is Abelian. For example, taken the generators $t_1,t_2$ defined as $$ t_1= \left( \begin{matrix} 0& 1 & 0\\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right)\qquad t_2= \left( \begin{matrix} 0& 0 & 1\\ 0 & 0 & 0 \\ -1 & 0 & 0 \end{matrix} \right) $$ the commutator isn't equal to 0, so the algebra isn't Abelian. I can restore the group elements through exponential map and taken $G,H \in SU(3)$ ($G,H$ are MATRICES) $$ G = e^{\ a t_1}\\ H = e^{\ b t_2} $$ I see that $[G,H]\neq 0$ because $[t_1,t_2]\neq 0$ and the $G,H$ doesn't admit a common basis of eigenvectors and I can't diagonalize them contemporary. So in $\mathbb{C}^3$ the fundamental representation of $SU(3)$ is irreducible.

Is it a right proof?

$\endgroup$
  • 1
    $\begingroup$ The fundamental representation (as used in physics) is by definition the smallest irreducible representation. I don't understand the question. Also, note that check-my-work questions are off-topic. $\endgroup$ – ACuriousMind Feb 7 '15 at 19:44
  • $\begingroup$ I know that by definition it is the smallest irreducible representation… but i want to prove it. Replace $SU(3)$ with a generic subgroup of $GL(n)$ and use the same demonstration. If the algebra isn't abelian, can i say that the fundamental representation of this subgroup is irreducible on $\mathbb{C}^n$? $\endgroup$ – apt45 Feb 7 '15 at 20:28
  • $\begingroup$ A representation of a Lie algebra is by definition not abelian, since the representation must have the same commutators. The way to check irreducibility on explicit matrices is to check whether the elements of the algebra/group can be written in block-matrix form. $\endgroup$ – ACuriousMind Feb 7 '15 at 20:33
  • $\begingroup$ In block-matrix form in a particular basis. But if i demonstrate that the general matrices of the subgroup doesn't commute can't I conclude they aren't diagonalizable contemporary ? $\endgroup$ – apt45 Feb 7 '15 at 20:42
  • $\begingroup$ You can, but simultaneous block-matrix form is a weaker condition than simultaneous diagonal form. $\endgroup$ – ACuriousMind Feb 7 '15 at 20:49
1
$\begingroup$

The Lie algebra of a non-Abelian Lie group can't be commutative, or the group itself would have to be Abelian. What you need to check is, e.g., that the only matrices that commute with any other matrix of $SU(3)$ is a multiple of the identity matrix, as required by Schur's lemma.

$\endgroup$
  • $\begingroup$ Phoenix87, thank you. The reason of my question is that i want to prove that the fundamental representation of O(3) is irreducible on $\mathbb{R}^3$ and i'm looking for a method to demonstrate this fact. Do you know? $\endgroup$ – apt45 Feb 7 '15 at 22:41
  • $\begingroup$ what I wrote in the answer applies to the representation of almost any group. More generally you have to verify that there are no proper subrepresentations of the fundamental one (which are given by central projections) $\endgroup$ – Phoenix87 Feb 7 '15 at 22:45
  • $\begingroup$ But what is the easiest way to prove that there are no proper subrepresentations? Maybe with 2x2 matrices is simple… but with 3x3 matrices, the calculations begin to be very tedious . $\endgroup$ – apt45 Feb 7 '15 at 22:49
  • $\begingroup$ Maybe with a suggestion I can try... but I think that will be a easiest way to do it. $\endgroup$ – apt45 Feb 7 '15 at 23:33
1
$\begingroup$

I interpret the question as follows. Let $SU(3)$ be the group of complex $3\times 3$ matrices with $U^\dagger U=I$ and $\det U=1$ acting as linear operators in $\mathbb C^3$. Is there a subspace $M \subset \mathbb C^3$ with $M \neq \{0\}$, $M \neq \mathbb C^3$ such that $U(M) \subset M$ for every $U \in SU(3)$?

In other words, is the natural action of $SU(3)$ on $\mathbb C^3$ irreducible?

Your proof is wrong, because the fact that $SU(3)$ is not Abelian does not imply that it is irreducible.

Actually the irreducibility property is true and it holds for the natural action of $SU(N)$ in $\mathbb C^N$. I sketch a proof in the rest of my answer. Consider $x,y\in \mathbb C^N\setminus \{0\}$. It therefore holds $y = cx/||x|| + dy'$, where $y' \perp x$ and $||y'||=1$ and $c,d \in \mathbb C$.

Working in the plane generated by the orthonormal vectors $x/||x||$ and $y'$, it is easy to construct $U \in SU(N)$ such that $U x/||x|| = y/||y||$ using elementary computations (*).

Therefore $y= s Ux$ for some $s\in \mathbb C$ and $U \in SU(N)$.

We have established the following lemma.

If $x,y\in \mathbb C^N\setminus \{0\}$, there is $U \in SU(N)$ and $s \in \mathbb C$ such that $sUx= y$.

This fact implies that the action of $SU(N)$ on $\mathbb C^N$ is irreducible. Let us prove it. Suppose that $M \subset \mathbb C^N$ is a subspace invariant under $SU(N)$. If $M\ni x\neq 0$ and $y \in \mathbb C^N$, there is $U\in SU(N)$ such that $sUx =U(sx) =y$ for some constant $s$. As $sx \in M$ and $M$ is invariant, we have $U(sx) = y \in M$. Since $y$ is arbitrary, we finally obtain $M= \mathbb C^N$. This means that $SU(N)$ is irreducible on $\mathbb C^N$.


(*) $U$ can be costructed as follows. If $x/||x||$ and $y'$ are parallel everything is trivial. In the other case, $x/||x||, y', e_3,\ldots, e_N$ is an orthonormal basis of $\mathbb C^N$ for suitable orthogonal unit vectors $e_3,\ldots, e_N$. The unique linear map such that $$U : x/||x|| \mapsto \frac{cx/||x|| + dy'}{\sqrt{|c|^2+|d|^2}}$$ $$U: y' \mapsto ax/||x|| + by'$$ $$U: e_j \mapsto e_j $$ is unitary if and only if $$|a|^2+|b|^2 =1\tag{1}$$ and $$a\overline{c}+ b \overline{d}=0\:.\tag{2}$$ Here $d$ and $c$ are fixed and the system of equations (1)-(2) in $a, b \in \mathbb C$ always admit solutions.

Notice that if $(a,b)$ is a solution, $(\lambda a, \lambda b)$ is such, for $|\lambda|=1$.

Representing $U$ with respect to the said basis, since $U$ is represented by a unitary matrix, its determinant must be a unit complex. In other words $$|\det U| =|cb -da|/\sqrt{|c|^2+|d|^2}=1$$ so we can always fix the multiplier $\lambda$ in order to have $\det U=1$. However it does not automatically imply that $U \in SU(N)$, because the determinant is computed with respect to the wrong basis. Indeed, the relevant determinant used in the definition of $SU(N)$ is the one computed with respect to the canonical basis of $\mathbb C^N$. It is $${\det}_C U = \det M U M^{-1}$$ where $M$ is the unitary matrix relating the canonical basis of $\mathbb C^N$ to $x/||x||, y', e_3,\ldots, e_N$. Therefore $${\det}_C U = \det M \det U (\det M)^{-1} = \det U=1\:.$$ Summarizing, we have obtained that the found operator $U$, represented with respect to the canonical basis of $\mathbb C$ is a unitary matrix with determinant $1$. In other words $U$ belongs to $SU(N)$ as wanted.

$\endgroup$

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.