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Here's the setup: I have $SL(2;\mathbb{C})$ acting on $V = \mathbb{C}[z,w] = \oplus_d V_d$, where $V_d$ is the homogeneous complex polynomials of degree $d$. The action is precomposition: $\pi(g)f(z,w) = f(g^{-1}(z,w))$. This is a representation with $(\pi,V_d)$ an irreducible subrepresentation for each $d$.

As a representation, it induces an $\mathfrak{sl}(2;\mathbb{C})$ action on $V_d$ via $d\pi_\mathbb{1}:\mathfrak{sl}(2;\mathbb{C})\to \operatorname{End}(V)$. I want to compute $d\pi$ in coordinates so I can explicitly write down the action of the generators $H,E$, and $F$ on each $V_d$.

Naively, I would do this: choose $z^kw^{d-k}$ as a basis of $V_d$, compute $$\pi\begin{pmatrix}a & b \\ c & d\end{pmatrix}\in \operatorname{Aut}(V_d)\cong GL(d+1;\mathbb{C}),$$ differentiate each coordinate function of $\pi$ in each direction $\mathbb{C}^4$, and evaluate at the identity.

But I'm worried about this strategy. Since $SL(2;\mathbb{C})$ is a submanifold of $GL(2;\mathbb{C})$ defined by $\ker(\det)$, $\partial_a$, $\partial_b$, $\partial_c$, and $\partial_d$ are not tangent to $SL(2;\mathbb{C})$. I don't see any reason this will actually give me the $\mathfrak{sl}(2;\mathbb{C})$ action.

So, question:

Will this naive strategy work for reasons I don't see, or should I choose some other coordinate system about the identity --- say, $e^H,e^E,e^F$?

More generally,

If $N\subset M$ is given by $N = F^{-1}(0)$, where $F:M\to X$ is a smooth map and $0\in X$ is a regular value, and $f:\nu N\to Y$ is a smooth map defined on a neighborhood of $N$ in $M$, can one compute $d(f|_N)$ by choosing a coordinate system on $M$ which does not necessarily restrict to coordinates on $N$ and computing $df$?

I'm not sure why I'm confused today about this, but I am. (Perhaps I need more coffee.) I'd be grateful for some clarity. (Also, as an aside, if I'm wrong about any of the representation theory I outlined or, god forbid, the smooth manifold theory, I'd appreciate correction.)

By the way, this is from a homework assignment, so please DO NOT answer with the explicit $\mathfrak{sl}(2;\mathbb{C})$ action!

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up vote 3 down vote accepted

I will try my best not to give away everything. I will use your notation above: $\pi$ is the representation of $SL_2(\Bbb{C})$ that you defined above and we have the induced representation $d\pi$ on the Lie algebra $\mathfrak{sl}_2(\Bbb{C})$. Recall that

$$d\pi(X) = \frac{d}{dt}\pi(e^{tX})\bigg|_{t=0}.$$

From which we get that if $f \in V_d$, $(d\pi(X)f)(z,w) = \frac{d}{dt}f(e^{-tX}z,e^{-tX}w)\bigg|_{t=0}.$ You now have a formula for computing the action of any $X \in \mathfrak{sl}_2(\Bbb{C})$ on any $f \in V_d$. Now because the usual matrices $H,E,F$ form a basis for $\mathfrak{sl}_2(\Bbb{C})$ you only need to compute the matrices for $d\pi(E),d\pi(F),d\pi(H)$.

Now you mentioned in you naive approach above that you want to compute $\pi$ of any matrix in $\mathfrak{sl}_2(\Bbb{C})$ on the canonical basis for $V_d$. Instead, what I suggest you try is instead of working on basis elements for $V_d$ just work with a general homogeneous polynomial $f \in V_d$.

As a start, from the formula that I gave you above you can use the chain rule to simplify things. Try working out the action of $H$ in the Cartan subalgebra on a general $f \in V_d$. You should get that

$$(d\pi(H)f)(z,w) = - \frac{\partial f}{\partial z} z + \frac{\partial f}{\partial w}w$$

from which it follows that $d\pi(H)$ is the linear operator on $V_d$ given by.....

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I don't want to accept until I've worked through the answer, and I won't be able to work through the answer until later in the week. Just commenting here so you know I haven't forgotten it. – Neal Oct 3 '12 at 1:23
Okay, this looks good! Thank you! – Neal Oct 20 '12 at 22:18
@Neal Thanks for accepting my answer. – user38268 Oct 20 '12 at 23:02
It's a good answer! A pleasure to accept it. – Neal Oct 20 '12 at 23:35

Even though this question has been answered, let me add something more to your general question. The general result you inquire about is true even under less restrictive circumstances. Suppose $N$ is a subset of $M$ which has a smooth structure (not necessarily one induced by the smooth structure on $M$) and such that the inclusion map $i_N:N\hookrightarrow M$ is smooth. Then for $f:M\rightarrow Y$ smooth, $d(f|_N) = d(f\circ i_N) = df\circ di_N$, implying that $d(f|_N):TN\rightarrow TY$ is determined by $df:TM\rightarrow TY$. In your case, since $0$ is a regular point of $F$, the smooth structure on $N=F^{-1}(0)$ induced by that on $M$ makes $N$ an embedded submanifold of $M$, with tangent space $TN \simeq di_N(TN) = \ker dF$, as you say.

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BenjaLim's answer is correct. I just want to add a few general statements.

  1. As illustrated in BenjaLim's answer, the utilitarian approach is to use a fact that, for some reason, I completely forgot: the go-to definition of the image of a vector under a differential. If $\gamma'(0) = v$, then $$df(v) = \frac{d}{dt}|_{t=0} f(\gamma(t)).$$

  2. The answer to the general question is as follows. Let $p\in N$; then $\ker dF_p \cong T_pN\subset T_pM$. To compute $d(f|_N)$, then, one need only compute $\ker dF_p$ and then its image under $df$. That is, $d(f|_N) = df|_{\ker dF}$.

  3. In this case, I needed only to compute $\ker d(\det)$ and then restrict the differential of the action to the kernel. Of course, the kernel is spanned by $E,F,H$, and the best way to compute the image of $v$ under the differential is by taking $\frac{d}{dt}|_{t=0}\exp(tv)$ ... .

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