Undergrad The Lie algebra of ##\frak{so}(3)## without complexification

Click For Summary
The discussion centers on deriving the Lie algebra of so(3) using the Cartan subalgebra and root vectors without resorting to complexification. It highlights that existing literature typically involves complexification to relate so(3) to su(2) and sl(2,C). The necessity of breaking symmetry to achieve a representation through the root system is emphasized, indicating that complex numbers are essential for this isomorphism. The conversation also references external resources for further understanding of basis transformations and the root system. Ultimately, the need for a non-complexified approach to so(3) remains a key inquiry.
redtree
Messages
335
Reaction score
15
TL;DR
Can the Lie algebra of ##\frak{so}(3)## (or ##\frak{su}(2)##) be formulated without complexification utilizing the Cartan subalgebra?
All of the formulations of the Lie algebra of ##\frak{so}(3)## (or ##\frak{su}(2)##) utilizing raising/lowering operators that I have seen in the literature involve complexification to ##\frak{su}(2) + i \frak{su}(2) \cong \frak{sl}(2,\mathbb{C})##. I have found explicit derivations in a particular representation, but none from the Cartan subalgebra.

Can one derive a formulation of the Lie algebra of ##\frak{so}(3)## utilizing the Cartan subalgebra and root vectors without complexification? If so, where can I find it?
 
Physics news on Phys.org
Moderator's note: Thread moved to the linear algebra math forum.
 
$$\mathfrak{so}(3)\cong \left(\mathbb{R}^3,\times\right)=\bigl\langle U,V,W\,|\,[U,V]=W,[V,W]=U,[W,U]=V \bigr\rangle $$
We need to break that symmetry in order to get the representation via the root system that specifies the generator of the Cartan subalgebra which is not symmetric. We therefore need complex numbers for the isomorphism. You can find the basis transformations at
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/

For a description of how the root system works, see
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-structures/
and the general basis of real orthogonal Lie algebras here:
https://www.physicsforums.com/insig...hogonal-Lie-Algebra-On-Odd-Dimensional-Spaces
 
Last edited:
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 1 ·
Replies
1
Views
4K