Is there such an identity about SO(3)?

[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?

 

[matt grime]

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

 

[George Jones]

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

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.

 

[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.