Consider the Gell-Mann matrices, with

$$ \lambda_3 = \operatorname{diag}(1, -1,0), \quad \lambda_8 = \frac{1}{\sqrt{3}}\operatorname{diag}(1, 1, -2), \quad, ... \ , $$

they span the Lie algebra $\mathfrak{su}(3)$. The Lie algebra transforms under the Lie group (not Lie algebra) $SU(3)$ in the adjoint representation, which is irreducible. By irreducibility, I would like to make the following statement:

Take a non-zero vector $T$ in $\mathfrak{su}(3)$, and consider the adjoint action $Ad_g$ (where $g$ is a group element in $SU(3)$) on $T$ and define the linear space $V_T:=\operatorname{span} \{Ad_g T | \forall g \in SU(3)\}$. $V_T$ is a non-trivial subspace of $\mathfrak{su}(3)$ and also closed under the adjoint action, thereby forming a representation of $SU(3)$, hence $V_T$ must be equal to $\mathfrak{su}(3)$ by irreducibility.

Now, I would like to confirm the above statement, by explicitly playing with the Gell-mann matrices. I can take $T = \lambda_8$, and hope to find a $g$ such that $Ad_g \lambda_8 \propto \lambda_3$ with a non-zero proportionality. The adjoint action can be written as matrix multiplication $g \lambda_8 g^\dagger$, hence $$ g \lambda_8 \propto \lambda_3 g \ . $$ But this proportionality is impossible, since it forces $$ \frac{1}{\sqrt{3}}\begin{bmatrix} g_{11} & g_{12} & - 2 g_{13}\\ g_{21} & g_{22} & -2 g_{23}\\ g_{31} & g_{32} & -2 g_{33} \end{bmatrix} \propto \begin{bmatrix} g_{11} & g_{12} & g_{13}\\ -g_{21} & - g_{22} & -g_{23} \\ 0 & 0 & 0 \end{bmatrix} \ . $$ This forces $g_{3i} = 0$, which is not good condition for $g \in SU(3)$.

I wonder which part of the above reasoning is wrong?

  • $\begingroup$ You are certainly wrong in working in the Lie Algebra and not appreciating the adjoint action as the Lie-Algebraic commutator with g instead of the product or group commutator with it. You cannot connect two elements in the Cartan Algebra this way, since $f_{i38}=0$. Why don't you post this in the MSE where it belongs? $\endgroup$ Commented Feb 28 at 15:36
  • $\begingroup$ WP. $\endgroup$ Commented Feb 28 at 15:49
  • $\begingroup$ @CosmasZachos Maybe I misunderstand. I am specifically interested in $\mathfrak{su}(3)$ as a representation of the Lie group $SU(3)$ (not as a rep of the Lie algebra $\mathfrak{su}(3)$). The Lie group $SU(3)$ acts on the vector space $\mathfrak{su}(3)$ by "matrix multiplication/group commutator". $\endgroup$ Commented Feb 28 at 16:55
  • $\begingroup$ Yes. Perhaps you might use the WP standard convention, then? $\endgroup$ Commented Feb 28 at 17:12

2 Answers 2


The question fails to interpret its own definition correctly: It is correct that every vector $v$ in an irreducible representation is cyclic, which is the technical term for the span of the orbit being the full representation.

However, this does not mean that for any two vectors $v,w$ in the representation you can find some $g$ such that $w = \rho(g)v$ - it only means that there exist finitely many elements $g_i$ in the group and constants $c_i$ such that $$w = \sum_{i=1}^N c_i g_i v,$$ whereas the question makes the mistake of assuming $N=1$.

  • $\begingroup$ Thank you for your sharp assessment. I still lack an intuitive understanding. Suppose I consider the $S^7 \subset \mathfrak{su}(3)$ centered at the origin. I start with $\lambda_8$ and color the radial projection of the adjoint orbit of $\lambda_8$ onto $S^7$. What subspace/portion of $S^7$ would be colored? Maybe this is more suited for MSE but I'd like to hear your comment, if any. $\endgroup$ Commented Feb 29 at 1:47
  • $\begingroup$ @user31415926 Consider that the action of the group on the algebra, considered as traceless Hermitian 3-by-3 matrices, is by conjugation by a unitary matrix. So the orbit of an algebra element is all matrices with the same eigenvalues. $\endgroup$
    – ACuriousMind
    Commented Feb 29 at 6:54

So, the perhaps underwhelming answer is that the action of a group on an irreducible representation need not be transitive. As a trivial example, consider the action of $SU(2)$ on $v=|\uparrow\rangle|\uparrow\rangle$ in $V_3\subset V_2\otimes V_2$ (the "vector" irreducible representation of $SU(2)$.) By the definition of the group action on tensor products, $g|\uparrow\rangle|\uparrow\rangle = (g|\uparrow\rangle)(g|\uparrow\rangle)$, and so $g(v)$ will always factor into a tensor product. Hence, $\not\exists g\in SU(2)$ such that $g(v) \propto |\uparrow\rangle |\downarrow\rangle + |\downarrow\rangle|\uparrow\rangle \in V_3$.

  • $\begingroup$ Thank you, your example is interesting, but also confuses me a bit more. So can I say that the orbit of $g$ action on $|\uparrow\rangle |\downarrow\rangle$ spans a sub-representation within $V_3$? Would that contradict with the irreducibility of $V_3$? $\endgroup$ Commented Feb 28 at 17:01
  • $\begingroup$ Is it surprising that a vector space can be spanned by a proper subset? For example, if $V = \mathbb C^2$, $V\otimes V$ is spanned by $\{w\otimes w|w\in V\}$, but there are plenty of vectors in $V\otimes V$ (e.g. entangled states) that cannot be factored into a tensor product. $\endgroup$
    – TLDR
    Commented Feb 29 at 19:46
  • $\begingroup$ You are right. The step going from the $G$-orbit to the spanned vector space does add a lot of new states, a lot more than I previously imagined. $\endgroup$ Commented Mar 1 at 3:11

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