# Commutator of Lie Group Generators

This is from Maggiore's "A Modern Introduction to Field Theory", Page 15.

I have a Lie group with matrix generators

$$T^{a}_{R}$$

Where $$a$$ takes values from 1 to the dimension of the Lie group.

I want to show that $$\alpha_{a} \beta_{b}[T^{a}_{R}, T^{b}_{R} ] = i \gamma_{c}(\alpha, \beta) T^{c}_{R}$$

(Einstein summation is implied.)

Where $$\gamma_{c} = - 2 (\delta_{c}(\alpha, \beta) - \alpha_c - \beta_{c}).$$

In the matrix representation,

$$\exp(i \alpha_{a} T^{a}_{R}) \exp(i \beta_{b} T^{b}_{R})= \exp(i \delta_{c}(\alpha, \beta) T^{c}_{R}) .$$

If I expand the exponentials (up to quadratic order) and take the natural logarithm of both sides, I have

$$i \delta_{a} T^{a}_{R} = \log \Big(1 + i(\alpha_a + \beta_a)T^{a}_{R} - \frac{1}{2}(\alpha_a T^{a}_{R})^2 - \frac{1}{2}(\beta_a T^{a}_{R})^2 - \alpha_{a} \beta_{b} T^{a}_{R} T^{b}_{R}\Big).$$

I used the fact that $$\log(1 + x) \approxeq x - \frac{x^2}{2}$$ to get (up to quadratic T)

$$i \delta_{a} T^{a}_{R} = i(\alpha_a + \beta_a)T^{a}_{R} - \frac{1}{2}(\alpha_a T^{a}_{R})^2 - \frac{1}{2}(\beta_a T^{a}_{R})^2 - \alpha_{a} \beta_{b} T^{a}_{R} T^{b}_{R} + \frac{(\alpha_{a} + \beta_{a})(\alpha_{b} + \beta_{b}) T^{a}_{R} T^{b}_{R}}{2}$$

Which implies that

$$- 2i (\delta_{a} - \alpha_{a} - \beta_{a})T^{a}_{R} = (\alpha_a T^{a}_{R})^{2} + (\beta_{a} T^{a}_{R})^2 - (\alpha_{a} + \beta_{a})(\alpha_{b} + \beta_{b})T^{a}_{R} T^{b}_{R}.$$

The left hand side is just as required. However, I cannot get the commutator. The only surviving term after cancellation on the right hand side is

$$-(\alpha_a \beta_b + \beta_a \alpha_b) T^{a}_{R} T^{b}_{R}.$$

Is it possible to get the commutator from here? Where have I gone wrong?

EDIT:

In the second last step I missed a term: $$2 \alpha_{a} \beta_{b} T^{a}_{R} T^{b}_{R}$$. The RHS reads

$$(\alpha_{a} \beta_{b} - \beta_{a} \alpha_{b}) T^{a}_{R} T^{b}_{R}$$

In the second term, we relabel $$a\leftrightarrow b$$ to get the commutator $$\alpha_{a}\beta_{b} [T^{a}_{R}, T^{b}_{R} ]$$.

Thank you Mane.andrea for pointing out my error.

• In your last equation there should be a minus between the two terms, you lost a $2\alpha_a \beta_b T^a_RT^b_R$ from the third to last equation to the second to last equation (I think). Then relabel $a\leftrightarrow b$ in the second term of the last equation and there you go. – MannyC Feb 13 at 13:59