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Consider the action functional


with $z(t)$ a complex path with end points $z_i=z(t_i),\; i=1,2$. $g(z,\bar{z})$ is a positive real function on $H=\{z\in C: Im(z)>0\}$.

After some easy questions the exercise asks me to determine (up to a multiplicative constant) the function $g(z,\bar{z})$ such that the action is symmetric under the transformation $z\rightarrow \gamma(z)= \frac{az+b}{cz+d}$, with $a,b,c,d \in R$.

Can somebody help me or give me some hints?

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1 Answer 1

If you took a course in GR or even remember the action for a point particle in SR, you should at least recognize the integral as the pathlength, the "function" $g$ you are looking for is a metric on the upper half plane and the transformations are called moebius transformations. This should give you enough search phrases and you should find an expression for the metric on wikipedia for example.

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Thanks for the answer. In fact I recognized that g is the metric, but I have no idea about the way to determine it. I simply determined the way it must transform under the moebius transformation. –  Gauge Jun 29 '12 at 11:14

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