Let $V_n = \mathbb{C}^n$ the representations of $SU(n)$ given by matrix multiplication $SU(n) \times V_n \rightarrow V_n, (A, v) \mapsto A \cdot v .\,$ Show that $V_n$ is irreducible.

I tried to prove this by induction using the fact that the representation is completely reducible, because $SU(n)$ is a compact Lie group, but I'm not sure is the right way to do it.

Any suggestions? Thanks in advance!


Let $v\in V_n\setminus\{0\}$. For each $w\in V_n$ such that $\|w\|=\|v\|$, there is a $M\in SU(n)$ such that $M.v=w$. Therefore, if $U\subset V_n$ is a vector subspace such that

  • $v\in U$;
  • $SU(n).U\subset U$,

then $U\supset\{w\in V_n\,|\,\|w\|=\|v\|\}$. But then $U=V_n$.

  • $\begingroup$ Sorry, I'm not sure anymore to have all understood... How can you affirm the existence of a matrix $M \in SU(n)$ such that $ M \cdot v = w$? It follows maybe from the fact that for all vectors $w \in V_n$ such that $ ||w|| = ||v|| $ we have that $||A \cdot v|| = ||w||$ for all $A \in SU(n) $? $\endgroup$ – userr777 Nov 3 '18 at 15:39
  • 1
    $\begingroup$ No. Let $v_1=\frac v{\|v\|}$, and consider vectors $v_2,\ldots,v_n$ such that $(v_1,v_2,\ldots,v_n)$ is an orthonormal basis. Then the matrix $P$ whose columns are the $v_i$'s belongs to $SU(n)$ and $P.e_1=v_1$. Now, you construct a matrix $Q$ from $w$ by the same process. Then $Q.P^{-1}\in SU(n)$ and $Q(P^{-1}.v_1)=Q.e_1=w_1$. Therefore, $Q(P^{-1}.v)=w$. $\endgroup$ – José Carlos Santos Nov 3 '18 at 16:08
  • $\begingroup$ Ok now it's very clear, thanks! $\endgroup$ – userr777 Nov 3 '18 at 17:04

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.