# Lie group multiplication and Lie algebra commutation

#### spaghetti3451

I've heard it said that the commutation relations of the generators of a Lie algebra determine the multiplication laws of the Lie group elements.

I would like to prove this statement for $SO(3)$.

I know that the commutation relations are $[J_{i},J_{j}]=i\epsilon_{ijk}J_{k}$.

Can you suggest a possible next step for showing how this can be used to determine the multiplication law for $SO(3)$?

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#### jk22

I'm not used to Lie but I think the steps are : You could find a set of matrices that satisfy these commutation relation, $$J_k$$

Then it builds a basis for the Lie algebra.

By exponentiating we get elements of the group SO (3)

Thus we need to compute $$exp (aJ_x+bJ_z)$$ for exemple.

#### spaghetti3451

Does the multiplication law refer to the fact that if we multiply two elements $e^{i\theta_{1}J_{1}}$ and $e^{i\theta_{2}J_{2}}$ from the Lie group, we get an element which is also in the Lie group?

Or does the group multiplication law refer to the fact that $e^{i\theta_{1}J_{1}}e^{i\theta_{2}J_{2}}=e^{i(\theta_{1}J_{1}+\theta_{2}J_{2})}$?

#### jk22

Does the multiplication law refer to the fact that if we multiply two elements $e^{i\theta_{1}J_{1}}$ and $e^{i\theta_{2}J_{2}}$ from the Lie group, we get an element which is also in the Lie group?

Or does the group multiplication law refer to the fact that $e^{i\theta_{1}J_{1}}e^{i\theta_{2}J_{2}}=e^{i(\theta_{1}J_{1}+\theta_{2}J_{2})}$?
The last is in general not true see Baker Campbell Hausdorff formula

#### spaghetti3451

Alright then, how would you define the multiplication law for $SO(3)$?

#### jk22

The operation could be matrix multiplication but the exponential comes from writing a rotation out of an infinitesimal one.

Infinitesimal rotations are commutative but rotations are not.

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