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I want to show that the Lie algebra of all trace-$0$ complex matrices acts transitively on $\mathbb{C}^n$, so the standard representation is irreducible. Let $v=\sum_i a_ie_i$, then the matrix $$ A_{i1}=a_i, A_{22}=-a_1 $$ has trace $0$ and maps $e_1$ to $v$. Is this a correct argument for the irreducibility of the standard representation of $\mathfrak{sl}(n,\mathbb{C})$ (defined via $X.v=Xv$)?

barbatos233
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    You can similarly map any basis vector $e_i$ to any vector $v$, but then you need to show that you can map any vector $u=\sum u_ie_i$ to any vector $v$, which doesn't follow simply from those $e_i \mapsto v$ mappings... – lisyarus Sep 21 '23 at 17:43
  • Another "argument" is that the standard representation is the highest weight representation $L(\omega_1)$, and these are all irreducible. – Dietrich Burde Sep 21 '23 at 18:19
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    This action is not transitive in the usual sense as, for example, the vector $0$ is only ever mapped to itself. Of course you do not need transitivity for irreducibility just that there are no invariant subspaces. – Callum Sep 21 '23 at 18:21
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    Here's a self-contained proof of irreducibility: take any proper non-trivial subspace $V\subset \mathbb{C}^n$ and choose a complementary subspace $W$. Pick a linear map $\phi$ on $\mathbb{C}^n$ such that $\mathrm{Im}(\phi) \subset W \subset \mathrm{Ker}(\phi)$. Then $\phi$ is nilpotent so it has trace $0$ and is thus in $\mathfrak{sl}(n,\mathbb{C})$. Then $\phi(V)\subset W$ and is non-trivial as long as $\phi \neq 0$ so $V$ is not invariant under $\mathfrak{sl}(n,\mathbb{C})$. – Callum Sep 22 '23 at 08:03
  • @Callum Thank you for your idea. If we choose two diagonalizable matrices $T,S$ who both have trace 0 (trace is invariant under similarity so one could change the basis without changing the trace(?)) and no common nontrivial proper invariant subspaces (if they have we need $v_1,...v_j$ (some eigenvectors of $T$) and $u_1,...,u_j$ (some eigenvectors of $S$) to have the same span, so we could let all the coefficient of $u_i$ under the eigenbasis of $T$ to be nonzero). Does this also show that there is no common invariant subspace? – barbatos233 Sep 22 '23 at 08:32
  • Firstly yes the trace is independent of the matrix representation (note I didn't actually mention matrices in my comment). If you could find two such elements then that would prove it but I'm not sure how you would construct them. Indeed, any two elements which share no invariant subspaces would prove it, they dont have to be diagonalisable. – Callum Sep 22 '23 at 13:15
  • @Callum Thank you for your patience. I'm choosing diagonalizable matrices because their invariant subspaces are easy to describe using their eigenvectors: they are the subspaces spanned by a subset of the eigenvectors. My argument in the previous comment was that if no eigenvectors of $S$ could be spanned by less than $n$ eigenvectors of $T$, then they would have no nontrivial proper common invariant subspaces. (Just for a clarification on my previous comment, although I am not sure if this would work. Your proof in the previous comment is very nice and I really appreciate your help.) – barbatos233 Sep 22 '23 at 13:52
  • You will want both of those elements to have all a full set of distinct eigenvalues (otherwise you will need to worry about larger eigenspaces possibly intersecting) to ensure that works but otherwise yes – Callum Sep 23 '23 at 09:17
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    What does it mean to be transitive in the context of a Lie algebra action? for a group action $\exists g:gx=y$ defines an equivalence relation. For a Lie algebra actions, this relation can fail to be reflexive / symmetric / transitive (just consider the Lie algebra action on a nonzero space given by $gx=0$ for all $x$: the first two fail). However for $n\ge 2$ the group $\mathrm{SL}_n(\mathbf{C})$ acts transitively on $\mathbf{C}^n-{0}$. – YCor Sep 23 '23 at 15:32

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