# How to map a complex plane Mobius transformation to 1+1D Minkowski real plane?

So consider the $$(x,t)$$ plane endowed with the minkowski metric, namely:

$$ds^2 = dx^2-dt^2.$$

It is well known that we can Wick rotate the time coordinate to get to the Euclidean metric. This can be seen heuristically as using the new coordinate $$t=i\tau$$, so that:

$$ds^2 = dx^2+d\tau^2.$$

If we consider the lightcone parametrisation $$w_+ = x+t$$ and $$w_- =x-t$$, these can be seen as complex coordinates $$z = x+i\tau$$ and $$\bar{z} = x-i\tau$$. The notation here of $$z$$ and $$\bar{z}$$ is a little bit misleading. Strictly speaking this is a bona fide change of coordinates only if we work on the complex two-plane, $$\mathbb{C}^2$$. This is only true if $$x$$ and $$\tau$$ are allowed to take complex values also.

However, in general we use this as a "trick" to lighten the notation, but in the end we want to work with $$x$$ and $$\tau$$ real. In that case, what we do is usually take that the aptly named $$\bar{z}$$ is actually equal to the complex conjugate of $$z$$, namely $$\bar{z}=z^*$$. This seem more or less reasonable, but in this analysis we forgot that initially $$\tau =- i t$$, and thus it should actually be purely imaginary. However usually we forget that fact and we go on, and usually, this goes well.

However I'm having trouble mapping the Mobius transformations of the complex plane to the corresponding global CFT transformation in the minkowski plane. Here is my semi-succesfull attempt. Consider a rotation in the complex plane : $$z\rightarrow e^{i\theta}z$$, $$\bar{z}\rightarrow e^{-i\theta}\bar{z}$$.

If we write then: $$x\rightarrow \frac{z+\bar{z}}{2}=cos(\theta)x-sin(\theta)\tau$$

$$\tau\rightarrow \frac{z-\bar{z}}{2i}=sin(\theta)x+cos(\theta)\tau.$$

Where I wrote $$z=x+i\tau$$. Now this is clearly problematic when we Wick-rotate back, as we should have that $$\tau = -i t$$. But this gives the transformation:

$$x\rightarrow \frac{z+\bar{z}}{2}=cos(\theta)x+isin(\theta)t$$

$$t\rightarrow \frac{z-\bar{z}}{2i}=i sin(\theta)x+cos(\theta)t$$

Now, if we redefine $$\theta = i\beta$$, we recover boosts in "real space":

$$x\rightarrow \frac{z+\bar{z}}{2}=cosh(\beta)x-sinh(\beta)t$$

$$\tau\rightarrow \frac{z-\bar{z}}{2i}=-sinh(\beta)x+cosh(\beta)t.$$

However, I have several question on this procedure and I don't really know how to extend it to the other transformation. The key point is the "redefinition" of $$\theta$$. This works here, but how could I have predicted that I needed to do this redefinition without knowing the end result?

Furthermore, I cheated a bit, because up until the end, I considered $$\theta$$ to be real, but then redefined it to be purely imaginary, $$\beta$$ being real.

EDIT : an easier way to see my problem is by going the other way. Consider a boost in real space, which can be written simply as:

$$(x+t)\rightarrow \gamma (x+t) ; (x-t)\rightarrow 1/\gamma (x-t).$$

This is mapped to $$z\rightarrow \gamma z$$, $$\bar{z}\rightarrow 1/\gamma \bar{z}$$, which in order to make sense, we need to impose $$\bar{\gamma}=1/\gamma$$, which is only true if $$\gamma=e^{i\theta}$$.

The conformal group of Minkowski space is $$SL(2,R) \otimes SL(2,R)$$, which rescales two light-cone coordinates independently. The conformal group of Euclidean space is $$SL(2,C)$$, which is NOT isomorphic $$SL(2,R) \otimes SL(2,R)$$.