Continuous symmetry transformations Taylor expansion Continuous symmetry transformations form a Lie group. The product of two such transformations is also a symmetry transformation: $T(\theta_1^a)T(\theta_2^a) = T(\theta_3^a)$ where $\theta_3^a=f^a(\theta_1^a,\theta_2^a)$. I now would like to to perform a Taylor expansion of $f$ around $\theta^a=0$:
$$\theta_1^a=f(\theta_1^a,\theta_2^a=0)=f^a(0,0)+\frac{\partial f^a}{\partial \theta_2^b}\theta_1^b+\frac{\partial f^a}{\partial \theta_2^b}\theta_2^b+\frac{1\partial^2 f^a}{2\partial \theta_1^b \partial \theta_1^c}\theta_1^b \theta_1^c + \frac{1\partial^2 f^a}{2\partial \theta_2^b \partial \theta_2^c}\theta_2^b \theta_2^c + \frac{\partial^2 f^a}{\partial \theta_1^b \partial \theta_1^c}\theta_1^b \theta_2^c + ...$$ They then go on and write: $$f(\theta_1^a,\theta_2^a)=\theta_1^a+\theta_2^a+\sum_{b,c}f_{bc}^a \theta_1^b\theta_2^c+...$$ 
My question is now what exactly is meant with the different indices and how do you get from the first expression the the second? I think that $\theta_1$ etc. are matrices in the Lie group and $f$ denotes the operation on that Lie group. But that's as far as I get. 
 A: A Lie group can be parametrized by a set of continuous parameters. $\theta^a$s are these group parameters. $a=1,..,n$ where $n$ is the number of parameters need to specify the group elements uniquely. 
The group transformations obey a composition law $T(\theta_1)T(\theta_2)=T(f(\theta_1,\theta_2))\equiv T(\theta_3)$, where I am denoting the set of $\theta_1^a$ etc.. as $\theta$ when they appear inside the brackets. 
If $\theta_1^a=0$, then the first transformation is just identity. $$T(0)T(\theta_2)=T(\theta_2)=T(f(\theta_1=0,\theta_2))$$
Therefore $f^a(\theta_1=0,\theta_2)=\theta_2^a$. Similarly  $f^a(\theta_1,\theta_2=0)=\theta_1^a$.
Therefore we can expand $\theta_3^a=f^a(\theta_1,\theta_2)$ aroung $\theta_1=0,\theta_2=0$ as 
$$\theta_3^a=f^a(\theta_1,\theta_2)=\theta_1^a+\theta_2^a+f^a_{bc}\theta_1^b\theta_2^c+...$$ 
upto second order. Clearly there can not be any $\theta_2^b\theta_2^c$ term because then we can not satisfy $f^a(\theta_1=0,\theta_2)=\theta_2^a$
Edit: Let us see why we can not have a $\theta_2^b\theta_2^c$ term. First of all, we are working upto second order in infinitesimals. In fact $T(\theta_1=0)T(\theta_2)=T(\theta_2)$ is true even for finite transformations. Therefore  $f^a(\theta_1=0,\theta_2)=\theta_2^a$ should be satisfied to all orders. But suppose we have 
$$\theta_3^a=f^a(\theta_1,\theta_2)=\theta_1^a+\theta_2^a+f^a_{bc}\theta_1^b\theta_2^c+g^a_{bc}\theta_2^b\theta_2^c+h^a_{bc}\theta_1^b\theta_1^c+...$$ 
then $f^a(\theta_1=0,\theta_2)=\theta_2^a+h^a_{bc}\theta_2^b\theta_2^c$ which implies that $h^a_{bc}=0$. Therefore there can not be any $\theta_2^b\theta_2^c$ term. Same argument applies for $\theta_1^b\theta_1^c$ term
