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$V_m=$Homogeneous polynomials in complex variable with total degree $m$, could any one tell me how is that Linear Transformation look like explicitly?And would you please tell me how this is an representation of $SU(2)$? I need to know in detail.

Let $U\in SU(2)$ is just a linear map on $\mathbb{C}^2$, Define a Linear Transformation $\Pi_m:V_m\rightarrow V_m$ given by $[\Pi_m(U)f](z)=f(U^{-1}z)$ where $f(z)=a_0z_1^m+a_1z_1^{m-1}z_2+\dots +a_mz_2^m$, $z=(z_1,z_2)\in\mathbb{C}^2$

What I know about a representation of $SU(2)$ by definition is, it must be a vector space $V$ together with a homomorphism $\phi:SU(2)\rightarrow GL(V)$

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Are you sure you copied the problem exactly as it's given? First, it seems that $\Pi$ should also be $\Pi_m$, but worse, the definition of $\Pi_m$ defines a transformation of polynomials whereas you write that $\Pi$ maps $\mathbb C^2$ to itself. It seems it should be $\Pi_m\colon V_m\to V_m$. – joriki Oct 10 '12 at 6:33
@joriki, Thank you for noticing – Un Chien Andalou Oct 10 '12 at 7:05
up vote 1 down vote accepted

The representation is the map $\Pi_m: SU(2) \to GL(V_m)$ where for $U \in SU(2)$, $\Pi_m(U)$ is the transformation that takes $f \in V_m$ to $f \circ U^{-1}$, i.e. $ (\Pi_m(U)(f))(z) = f(U^{-1}(z))$. Note that if $f$ is a homogeneous polynomial of degree $m$ then $f \circ U^{-1}$ is also a homogeneous polynomial of degree $m$, e.g. if $f(z) = z_1^{m_1} z_2^{m_2}$ and $U^{-1} = \pmatrix{a & b\cr c & d\cr}$, $f \circ U^{-1} = (a z_1 + b z_2)^{m_1} (c z_1 + d z_2)^{m_2}$.

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Well, Can I say $\Pi_m:SU(2)\times V_m\rightarrow V_m$ as a map? – Un Chien Andalou Oct 10 '12 at 7:23
Not if you want $\Pi_m$ to be the representation. – Robert Israel Oct 10 '12 at 18:26

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