Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

3 Answers 3

up vote 2 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.....

share|improve this answer
    
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
1  
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.

share|improve this answer

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 $\frac{d}{dt}|_{t=0}\exp(tv)$ ... .

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.