Maggiore's QFT textbook error?

[MathematicalPhysicist]

In page 15 of the first edition of this textbook, in equations 2.6 and 2.7, he writes:

(2.6)$$e^{i\alpha_a T^a_R} e^{i\beta_a T^a_R}=e^{i\delta_a T^a_R}$$
where $$T^a_R$$ is the generator of the group represented by R.
Now in equation (2.7) he take the logarithm:
(2.7)$$i\delta_a T^a_R=log{[1+i\alpha_aT^a_R+0.5(i\alpha_aT^a_R)^2][1+i\beta_a T^a_R+0.5(i\beta_a T^a_R)^2]}=log[1+i(\alpha_a+\beta_a)T^a_R-0.5(\alpha_a T^a_R)^2-0.5(\beta_a T^a_R)^2-\alpha_a \beta_b T^a_R T^b_R]$$

and I don't understand from where did he get the term with the b's, I guess it should a's instead of b's, but then again he writes that he uses the taylor expansion of log(1+x) upto second order to get to equation (2.8)$$\alpha_a \beta_b [T^a_R,T^b_R]=i\gamma_c(\alpha,\beta)T^c_R$$, I don't understnad why did he change indexes in equation 2.7, can anyone enlighten me with this?

Thanks.

 

[strangerep]

In page 15 of the first edition of this textbook, in equations 2.6 and 2.7, he writes:

(2.6)$$e^{i\alpha_a T^a_R} e^{i\beta_a T^a_R}=e^{i\delta_a T^a_R}$$
where $$T^a_R$$ is the generator of the group represented by R.
Now in equation (2.7) he take the logarithm:
(2.7)$$i\delta_a T^a_R=log{[1+i\alpha_aT^a_R+0.5(i\alpha_aT^a_R)^2][1+i\beta_a T^a_R+0.5(i\beta_a T^a_R)^2]}=log[1+i(\alpha_a+\beta_a)T^a_R-0.5(\alpha_a T^a_R)^2-0.5(\beta_a T^a_R)^2-\alpha_a \beta_b T^a_R T^b_R]$$

I don't understand why did he change indexes in equation 2.7,
can anyone enlighten me with this?

Your latex's not quite right...

Magiorre's eq(2.7) is

$$i\delta_a T^a_R ~=~ \log\big\{[1+i\alpha_aT^a_R+0.5(i\alpha_aT^a_R)^2][1+i\beta_a T^a_R+0.5(i\beta_a T^a_R)^2]\big\} ~=~ \log[1+i(\alpha_a+\beta_a)T^a_R-0.5(\alpha_a T^a_R)^2-0.5(\beta_a T^a_R)^2-\alpha_a \beta_b T^a_R T^b_R]$$

which involves abuses of the summation convention. (Actually, even (2.6) should use another
dummy index like b in the second exponential.)

Basically, he uses the b dummy index so that you can correctly keep track of what's
being summed with what...

 

[MathematicalPhysicist]

OK, thanks.
That clears this matter.