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$V_m=$Homogeneous polynomials in complex variable with total degree $m$,

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 $f(z)=a_0z_1^m+a_1z_1^{m-1}z_2+\dots +a_mz_2^m$, $z=(z_1,z_2)\in\mathbb{C}^2$

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 if $f(z) = z_1^{m_1} z_2^{m_2}$ and $U^{-1} = \pmatrix{a & b\cr c & d\cr}$, $f \circ U^{-1} = (a z_1 + b z_2)^{m_1} (c z_1 + d z_2)^{m_2}$.

Now I want to compute the corresponding Lie Algebra Representation $\pi_m$, According to the definition it can be computed as $$\pi_m(X)=\frac{d}{dt}\Pi_m(e^{tX})|_{t=0}$$ where $$X=\begin{pmatrix}X_{11}&X_{12}\\X_{21}&X_{22}\end{pmatrix}$$

is some matrix. So $(\pi_m(X)f)(z)=\frac{d}{dt}f(e^{-tXz})|_{t=0}$, I took a curve $z(t)\in\mathbb{C}^2$ as $z(t)=e^{-tX}z$, $z(t)=(z_1(t),z_2(t))$ by Chainrule $\pi_m(X)f=\partial f/\partial z_1 dz_1/dt+\partial f/\partial z_2 dz_2/dt|_{t=0}$ How ever $dz/dt|_{t=0}=-Xz$, so could any one tell me what will be next few steps so that I can get the representation of this Lie algebra? Thank you for the help!

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Your confusion lies in the fact that you have derivatives of $z_1$ and $z_2$ with respect to $t$ and then $dz/dt$. From your last step, you know that

$$\frac{dz_1}{dt}\bigg|_{t=0} = [-Xz]_{\text{$z_1$ component}},\hspace{3mm} \frac{dz_2}{dt}\bigg|_{t=0} = [-Xz]_{\text{$z_2$ component}}.$$

Now what does all the above meam? When you do $Xz$ you are left multiplying the vector $(z_1,z_2)$ by the matrix $X$. So when I write $[-Xz]_{\text{$z_1$ component}}$, I mean the first component of the vector $-Xz$. Unwinding all of this, $\pi_m(X)$ is thus the linear operator on $V_m$ given by $$\pi_m(X) = -(X_{11}z_1 + X_{12}z_2)\frac{\partial }{\partial z_1} - (X_{21}z_1 + X_{22}z_2) \frac{\partial}{\partial z_2}.$$

Let's see how all this works in practice. If you take $H= \left(\begin{array}{cc} 1 & 0 \\ 0 & - 1\end{array}\right)$, then

$$\pi_m(H) = -z_1 \frac{\partial}{\partial z_1} +z_2\frac{\partial}{\partial z_2}.$$

If for example you are looking at $V_1$ now then applying this to the polynomial $f(z_1,z_2) = z_1 + z_2$ gives you

$$\begin{eqnarray*} \pi_m(H)(f) &=& -z_1\frac{\partial }{\partial z_1} (z_1) +z_2\frac{\partial}{\partial z_2}(z_2)\\ &=& -z_1 + z_2. \end{eqnarray*}$$

In general if you are looking at $V_m$, you should get that

$$\pi_m(H)z_1^kz_2^{m-k} = (m-2k)z_1^kz_2^{m-k}.$$

Good Exercise to do: Prove that $V_m$ is irreducible for every $m \geq 1$. Even better, define a similar representation like $V_m$ for $\mathfrak{sl}_3$ and prove that that is irreducible too for every $m\geq 1$.

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Thank you @BenjaLim –  Bunuelian Trick Oct 10 '12 at 12:13
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@Flute Do you understand what I wrote above? Also, did you get it why the representation of $\textrm{SU}(3)$ are in $1-1$ correspondence with that of its Lie algebra? –  user38268 Oct 10 '12 at 12:18
    
I have understood what you have wrote above, but not why the representation of $SU(3)$ are in $1-1$ correspondense with its Lie Algebra –  Bunuelian Trick Oct 10 '12 at 12:51

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