Fix an integer $n\geq 0$, let $V_n$ be the complex vector space of polynomials in two complex variables $z_1$ and $z_2$ homogeneous of degree n. Define a representation
$$\phi_n:SL(2,\mathbb{C})\to GL(V_n)$$by
\begin{equation}\label{e5}
\phi_n(A)P\left(\begin{bmatrix}
z_1 \\
z_2
\end{bmatrix}\right)=P\left(A^{-1}\begin{bmatrix}
z_1 \\
z_2
\end{bmatrix}\right)
\end{equation}
Now consider the differential $d\phi_n$ of the representation $\phi_n$. Then $d\phi_n:\mathfrak{sl}(2,\mathbb{C})\to\mathfrak{gl}(V_n)$ is defined by $$d\phi_n(\gamma'(0))=\frac{d}{dt}(\phi_n\circ\gamma)(t)\bigg|_{t=0}$$ where $\gamma(t)$ is a smooth curve on $SL(2,\mathbb{C})$ with $\gamma(0)=Id:=\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$.
Question: Show that $d\phi_n$ is a Lie algebra homomorphism
Attempt: Let $\alpha,\beta$ be any two differentiable curves on $SL(2,\mathbb{C})$ with $\alpha(0)=\beta(0)=Id$
- Well-definedness:
If $\alpha'(0)=\beta'(0)$ then \begin{align*} \frac{d}{dt}(\phi_n\circ\alpha)(t)\bigg|_{t=0} =\phi_n'(\alpha(t)).\alpha'(t)\big|_{t=0}=\phi_n'(\alpha(0)).\alpha'(0)=\phi_n'(\beta(0)).\beta'(0)=\frac{d}{dt}(\phi_n\circ\beta)(t)\bigg|_{t=0} \end{align*}.
- We show that $d\phi_n$ preserves the addition: \begin{equation}d\phi_n(\alpha'(0)+\beta'(0))=d\phi_n((\alpha+\beta)'(0))=\frac{d}{dt}(\phi_n\circ (\alpha+\beta))(t)\bigg|_{t=0}=\phi_n'((\alpha+\beta)(0)).(\alpha+\beta)'(0)\end{equation} On the other hand $$d\phi_n(\alpha'(0))+d\phi_n(\beta'(0))=\frac{d}{dt}\Big((\phi_n\circ \alpha)(t)+(\phi_n\circ \beta(t)\Big)\bigg|_{t=0}=\phi_n'(Id)\big(\alpha'(0)+\beta'(0)\big)$$
But I couldn't show that $$\phi_n'(2.Id)=\phi_n'(Id)$$
I think there must be something wrong here. But where?
- We show $d\phi_n([\alpha'(0),\beta'(0)])=[d\phi_n(\alpha'(0)),d\phi_n(\beta'(0)]$
\begin{align*} d\phi_n([\alpha'(0),\beta'(0)]) &=d\phi_n\big(\alpha'(0)\beta'(0)-\beta'(0)\alpha'(0)\big)=d\phi_n\big(\alpha'(0)\beta'(0)\big)-d\phi_n\big(\beta'(0)\alpha'(0)\big) \end{align*} On the other hand, \begin{align*} [d\phi_n(\alpha'(0)),d\phi_n(\beta'(0)]&=d\phi_n(\alpha'(0))d\phi_n(\beta'(0))-d\phi_n(\beta'(0))d\phi_n(\alpha'(0)) \\ &=\frac{d}{dt}\phi_n(\alpha(t)\beta(t))\bigg|_{t=0}-\frac{d}{dt}\phi_n(\beta(t)\alpha)(t)\bigg|_{t=0} \end{align*}
But I couldn't show the equality $$d\phi_n\big(\alpha'(0)\beta'(0)\big)=\frac{d}{dt}\phi_n(\alpha(t)\beta(t))\bigg|_{t=0}$$