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The element of $SU(N)$ can be expressed as $g=\exp(W)$, where $W=i∑_{a} _a T_a$ lives in the Lie algebra of the group and $[T_a,T_b]= f_{a,b,c} T_c$ (see chapter 8.5 in A Altland and B. Simons' book). As I am not familiar with Lie group, is there any simple way to get Jacobian of $dg = J()d$, without using too much knowledge of Lie group?

Chuizhen Chen
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  • The Wikipedia page has two elementary-enough demonstrations. –  Aug 30 '17 at 08:31
  • Thanks for your suggestion. The page explain the derivative of the exponential map. In the page, the derivative of matrix $d\exp(X)$ gives a matrix $\exp(X)\frac{ 1 - e^{ - ad_{x} } }{ad_{X}} dX$. But I still don't known how to get Jacobian using the above relation? – Chuizhen Chen Sep 01 '17 at 09:36
  • If you apply the formula on Wikipedia page to $\partial g/\partial \pi_a$, for $a=1,2,3$, you get your Jacobian, or am I missing something? –  Sep 01 '17 at 09:40
  • I see. For $SU(2)$, Using $\frac{\partial g}{\partial \pi_a} = \exp(X)(1-\frac{ad_X}{ad_X})\frac{\partial X}{\partial \pi_a}$, we can get every elements of matrix $\partial g_{nm}/\partial\pi_a$, and then we have the Jacobian matrix $J_F=\frac{\partial (g_{11},g_{12},g_{21},g_{22})}{\partial(\pi_1,\pi_2,\pi_3)}$. But it is not a square matrix, how can I get the determinant of Jacobian matrix $J_F$ which are usually related to the final result $dg = J(\pi)d\pi $? – Chuizhen Chen Sep 02 '17 at 11:26
  • I am afraid I misunderstood your question then: when you wrote "Jacobian" I thought you mean the linear operator $J$ that transforms an infinitesimal variation of $\pi$ into an infinitesimal variation of $g$ (rigorously, $g(\pi+\delta)g(\pi)^{-1}=J(\pi)\delta + O(\delta^2)$). My suggestions were geared toward computing that operator. But $SU(N)$ is not a vector space, so you can't meaningfully speak of a matrix for $J$, least of its determinant. May I ask about the context of your question? What is the higher goal your pursue? –  Sep 02 '17 at 21:36
  • Well, of course, $J\delta$ is a $N\times N$ matrix, so by taking a basis of the set of $N\times N$ matrices, you can find a matrix for $J$ along the line of your last comment. But then it will be have $N^2$ rows for 3 columns: not square as you noticed. –  Sep 02 '17 at 21:41
  • Thanks. My starting point is to do renormalization goup analysis of the nonlinear $\sigma$-model, which is defined by the partition function $Z=\int dg e^{-S[g]}$. $S[g] = \frac{1}{\lambda} \int d^{d}r tr[\nabla g\nabla g^{-1}]$, with $g\in SU(N)$. To do RG analysis, I need to do the functional integral explicitly. $g$ can be written as $g = e^{W}$ where $W=i\sum_{a}\pi^a T^a\in \mathcal{G}$ lives in the Lie algebra of the group. Then we can replace the integral over the group $dg$ with parameters $d\pi$ explicitly, by using $dg=J({\pi})d\pi$. My question is how to get $J({\pi})$ here? – Chuizhen Chen Sep 03 '17 at 02:16
  • I see. What you want is a parametrisation of the Haar measure of $SU(N)$. Most commonly one does not use the coordinates $\pi_a$ but a generalisation of Euler angles. This has already been expounded on this site: see this answer please. –  Sep 03 '17 at 12:35
  • If you absolutely want to use the $\pi_a$'s, there is a classic formula by Weyl involving the eigenvalues of $W$. Here is a reference covering it all and reasonably readable by a physicist. Don't get mislead by the title and the historical introduction: it is just that the parametrisation with Euler angles were already known to Hurwitz at the end of the 19th century. –  Sep 03 '17 at 12:42
  • That's wonderful. Thanks for your help. – Chuizhen Chen Sep 05 '17 at 02:06
  • Sorry, it took me so long to get the point! –  Sep 05 '17 at 11:38

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