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I'm pretty much stuck on the following question (taken from the book Lie groups and introduction to linear groups by Rossman W.):

Rossman W. Lie groups and introduction to linear groups 2.6.5

I've found some clues, but I think I lack proper understanding of what needs to be done here.

If I recall correctly, the relation between lie groups homomorphism $T$ and its lie algebra homomorphism $\tau$ is $\tau(X) = \frac{d}{dt} T(e^{tX})|_{t=0}$ when in this case $ X \in gl(n,\mathbb{R})$. But then, some factor involving the exponent function should appear. On the other hand, the derivative of $f(x)$ is $$\displaystyle \sum_{i} \xi_i ' \frac{\partial f}{\partial\xi_i}$$, so it should be some kind of chain rule?

In (b) I have similar problem - it seems like I need to "go down" to $so(3)$ and use its basis matrices, but I'm not sure how to justify it.

Any assistance will be highly appreciated.

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1 Answer 1

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It is exactly all about chain rule : $\tau(Y)f(x) = \dfrac{d}{dt}_{\vert t=0}T(e^{tY})f(x)=\dfrac{d}{dt}_{\vert t=0}f(e^{-tY}x)$.

If you write $\phi(t)=e^{-tY}x$ then $$\phi'(t)=-Yx=-\sum_{i,j=1}^nY_{ij}\xi_je_i$$ and you get $$\tau(Y)f(x) = df_{x}(-Yx)=-\sum_{i,j=1}^n\frac{\partial f}{\partial \xi_i}(x)Y_{ij}\xi_j.$$

You can see $\tau$ as the differential of $T:GL(n,\mathbb R)\rightarrow GL(F)$ at the identity element $I_n$.

Saying that $f\in \mathbb R[\xi_1,\xi_2,\xi_3]$ is invariant under the action of $SO(3,\mathbb R)$ means that the restriction of $\tau$ to the Lie algebra of $SO(3,\mathbb R)$ is the null linear map since $$\forall X\in so(3), \forall t\in \mathbb R, \quad T(e^{tX})f(x)=f(x).$$

And as you said, the Lie algebra $so(3)$ has a natural basis $$so(3)=<\begin{pmatrix} 0&0&1\\0&0&0\\-1&0&0\end{pmatrix},\begin{pmatrix} 0&1&0\\-1&0&0\\0&0&0\end{pmatrix},\begin{pmatrix} 0&0&0\\0&0&1\\0&-1&0\end{pmatrix}>=<A,B,[A,B]>$$ If you check fomula of question $(a)$ with one these three matrices, you will get exactly the expected three relations.

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  • $\begingroup$ Thanks for the answer and sorry it took me so much time to respond. I still can't understand, though, how $\xi_i'$ disappears in the last equation of part (a). $\endgroup$
    – Pavel
    Feb 9, 2013 at 11:31
  • $\begingroup$ The differential of $f$ at point $x$ is the linear form : $\displaystyle df_x= \sum_{i=1}^n \frac{\partial f}{\partial \xi_1}(x)d\xi_i$. It means that if $(\alpha_1,\cdots,\alpha_n)\in \mathbb R^n$ then $\displaystyle df_x(\alpha_1,\cdots,\alpha_n)=\sum_{i=1}^n \alpha_i\cdot \dfrac{\partial f}{\partial \xi_i}(x)$. Now, compute $df_x(-Y\cdot x)$ and check that you get the same thing that I wrote. $\endgroup$
    – Bebop
    Feb 9, 2013 at 12:10
  • $\begingroup$ Maybe I'm missing something, but how $d\xi_i$ turns to $\alpha_i$? My differentiation knowledge is a bit rusty... $\endgroup$
    – Pavel
    Feb 9, 2013 at 13:04
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    $\begingroup$ If you are working with a basis $(\xi_1,\cdots,\xi_n)$ on $\mathbb R^n$ then $d\xi_i$ is the linear form given by $d\xi_i(\xi_j)=\delta_{ij}$. So $\displaystyle d\xi_i(\sum_{i=1}^n\alpha_j \xi_j)=\sum_{i=1}^n\alpha_jd\xi_i(\xi_j)=\alpha_i$ and $(d\xi_1,\cdots,d\xi_n)$ is the dual basis associated to $(\xi_1,\cdots,\xi_n)$. In this basis, the matrix of $df_x$ is $J=\begin{pmatrix}\frac{\partial f}{\partial \xi_1}(x),\cdots,\frac{\partial f}{\partial \xi_n}(x)\end{pmatrix}\in M_{1,n}(\mathbb R)$ and for any vector $a=\begin{pmatrix}a_1\\\vdots\\a_n\end{pmatrix}$ of $\mathbb R^n$, $df_x(a)=J a$. $\endgroup$
    – Bebop
    Feb 9, 2013 at 16:55
  • $\begingroup$ So it stems to the Jacobian. Thanks for your help! $\endgroup$
    – Pavel
    Feb 10, 2013 at 19:20

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