# Is there such an identity about SO(3)?

1. May 28, 2006

### kakarukeys

$$T^i K_{ij} = K T^j K^{-1}$$
repeated indices imply summation.
$$T^i$$ are the generators (Lie algebra elements) of SO(3).
i.e.$$T^i_{jk} = - \epsilon_{ijk}$$
$$T^i \in so(3)$$
$$K \in SO(3)$$

How to show it's true?
Is there a universal formula for all Lie group?

Last edited: May 28, 2006
2. May 28, 2006

### matt grime

Those elements do not generate SO(3). SO(3) is uncountable, it cannot have a finte set of generators.

3. May 28, 2006

### George Jones

Staff Emeritus
For physicists, generators of a Lie group are elements of the corresponding Lie algebra, or maybe such elements multiplied by i. I don't know what the K's are.

4. May 28, 2006

### kakarukeys

Yes. I have just clarified!
I also checked a special case, and it's true.
I don't know how to prove this identity besides brute-force calculation which gives no insight at all.