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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!

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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 $U=V_n$.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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