Question on N-dimensional Lie Groups

[antibrane]

I'm currently learning Lie groups/algebras and I am trying to find the infinitesimal generators of the special orthogonal group SO(n). It is the n-dimensions that throws me off. I know that the answer is $n(n-1)/2$ generators of the form,

$$X_{\rho,\sigma}=-i\left(x_{\rho}\frac{\partial}{\partial x_{\sigma}}-x_{\sigma}\frac{\partial}{\partial x_{\rho}}\right)$$

where $1\leq x_{\rho}\leq n$ such that $\sigma > \rho$, but how do I get this? The only way I could think of is somehow trying to find the n-dimensional rotation matrix in general and then going from there (I have actually tried this and it gets me something similar)--there must be a simpler way though.

I also run into this problem in other areas of the book I am using. For example attempting to establish that the Lie algebra sl(n,C) is an ideal of gl(n,C). I guess I am looking for some advice on how to compute these algebras in n-dimensions. Thanks in advance for any help/suggestions, and let me know if I can make anything more clear.

 

[adriank]

Can't you prove that sl(n) is an ideal of gl(n) directly? You know that sl(n) is the subset of gl(n) consisting of matrices of trace zero. So prove that for A in sl(n) and B in gl(n), [A,B] is in sl(n). Use the fact that tr(AB) = tr(BA). (You could also note that the trace is a Lie algebra homomorphism gl(n) -> C with kernel sl(n).)

 

[Simon_Tyler]

Adriank is correct about the approach to sl(n) being an ideal of gl(n).

As for the dimensions of so(n), the representation you give for the generators is clearly antisymmetric in $$\rho$$ and $$\sigma$$. This means you can group your generators $$X_{\rho \sigma}$$ into a matrix where $$X_{\sigma\rho}=-X_{\rho \sigma}$$, ie a $$n\times n$$ antisymmetric matrix. The only independent elements in such a matrix are (say) the upper right triangle. This has $$1 + 2 + \ldots + n-1 = \tfrac12 n(n-1)$$ elements.

 

[antibrane]

thanks very much to both of you. the answers to both questions actually seem quite obvious after the fact. hah

 

[blue_raver22]

Hi there,

First of all, congratulations on delving into the world of Lie groups and algebras! It can definitely be a challenging subject, but it is also incredibly powerful and useful in many areas of mathematics and physics.

To answer your question about finding the infinitesimal generators of SO(n), it is helpful to first understand what a Lie group is. A Lie group is a group that is also a smooth manifold, meaning it has both group operations (such as multiplication and inversion) and a smooth structure (meaning it can be described by smooth functions). In the case of SO(n), it is a group of n-dimensional rotations that also has a smooth structure.

Now, the infinitesimal generators of a Lie group are the elements that, when exponentiated, give you the group elements. In other words, they are the "infinitely small" versions of the group elements. In the case of SO(n), these infinitesimal generators are the elements of the Lie algebra of SO(n), denoted as so(n). The Lie algebra is essentially the tangent space of the Lie group at the identity element, and it can be thought of as the "linearized version" of the Lie group.

So, how do we find the infinitesimal generators of SO(n)? As you mentioned, one way is to try to find the n-dimensional rotation matrix and then linearize it. However, there is a simpler and more elegant way to do this using the concept of Lie derivatives.

Without going into too much detail, the infinitesimal generators of SO(n) can be found using the Lie derivatives of the coordinate functions on SO(n). These coordinate functions are essentially the components of the rotation matrix, and the Lie derivatives give us a way to "twist" these components to get the infinitesimal generators.

Now, in the case of SO(n), the coordinate functions are just the entries of the rotation matrix, which we can denote as x_{ij} (where i and j range from 1 to n). Using the Lie derivatives, we can then construct the infinitesimal generators of the form that you mentioned: X_{\rho,\sigma}=-i\left(x_{\rho}\frac{\partial}{\partial x_{\sigma}}-x_{\sigma}\frac{\partial}{\partial x_{\rho}}\right). This formula may look intimidating, but it essentially just "twists" the coordinate functions in a certain way to give us the infinitesimal generators.

As