Introduction

There are two major conventions for the metric signature: (+,,,)\p{+,-,-,-} (west coast) and (,+,+,+)\p{-,+,+,+} (east coast). However, the first convention that I have met in my journey of learning physics is neither of them: the imaginary time. Shortly after, I started using the west coast convention, so I never really used the imaginary time convention seriously. I personally dislike the imaginary time convention, and so do most people in the physics community and history, which is why most modern textbooks use either the west coast or the east coast convention. One of my past physics teachers deemed the imaginary time convention to be a heresy (异端邪说).

The teacher’s writing

However, in some cases, the imaginary time convention can be convenient due to the use of multi-index notation (which is more concise and feature-rich than the Einstein notation). Here is one of such cases: the derivation of the metric in Poincaré coordinates for the anti-de Sitter space.

The dd-dimensional anti-de Sitter space AdSd\mrm{AdS}_d of scale ll is defined as the hyperboloid l2=T12T22+i=1d1(Xi)2-l^2=-T_1^2-T_2^2+\sum_{i=1}^{d-1}\p{X^i}^2 in Md1,2M^{d-1,2} (the analogue of the Minkowski space, but with signature d1,2d-1,2). The Poincaré coordinates are defined as zl2T1+Xd1,tlT2T1+Xd1,xilXiT1+Xd1,i=1,,d2.\begin{align*} z&\ceq\fr{l^2}{T_1+X^{d-1}},\\ t&\ceq\fr{lT_2}{T_1+X^{d-1}},\\ x^i&\ceq\fr{lX^i}{T_1+X^{d-1}},&i=1,\ldots,d-2. \end{align*}

The derivation

Define TT1T\ceq T_1 and XXd1X\ceq X^{d-1} just for fun. Then, define two (d1)\p{d-1}-dimensional multi-indices Y(iT2,X1,,Xd2),y(it,x1,,xd2).Y\ceq\p{\i T_2,X^1,\ldots,X^{d-2}},\quad y\ceq\p{\i t,x^1,\ldots,x^{d-2}}.

The hyperboloid constraint and the metric (east coast convention) are then X2T2+Y2=l2,ds2=dX2dT2+dY2,X^2-T^2+Y^2=-l^2,\quad \d s^2=\d X^2-\d T^2+\d Y^2, which are equivalently (X+T)(XT)=l2Y2,ds2=(dX+dT)(dXdT)+dY2.\p{X+T}\p{X-T}=-l^2-Y^2,\quad\d s^2=\p{\d X+\d T}\p{\d X-\d T}+\d Y^2.(1)\p{1} The definition of the Poincaré coordinates can be written as z=l2X+T,y=zlY,z=\fr{l^2}{X+T},\quad y=\fr zlY, or equivalently X+T=l2z,Y=lyz.X+T=\fr{l^2}z,\quad Y=\fr{ly}z.(2)\p{2}

Substitute Equation 2 into the first equation in Equation 1. Then, we have XT=zy2z.X-T=-z-\fr{y^2}z.(3)\p{3} Differentiate Equation 2 and 3, and we have dX+dT=l2z2dz,dXdT=dz+y2z2dz2yzdy,dY=l(dyzyz2dz).\d X+\d T=-\fr{l^2}{z^2}\,\d z,\quad \d X-\d T=-\d z+\fr{y^2}{z^2}\,\d z-\fr{2y}z\,\d y,\quad \d Y=l\p{\fr{\d y}z-\fr y{z^2}\,\d z}. Substitute this into the second equation in Equation 1, and we have ds2=l2z2dz(dz+y2z2dz2yzdy)+l2(dyzyz2dz)2=l2z2(dy2+dz2).\d s^2=-\fr{l^2}{z^2}\d z\p{-\d z+\fr{y^2}{z^2}\,\d z-\fr{2y}z\,\d y}+l^2\p{\fr{\d y}z-\fr y{z^2}\,\d z}^2 =\fr{l^2}{z^2}\p{\d y^2+\d z^2}.

Finally, substitute back the definition of yy, and we have the result ds2=l2z2(dt2+i=1d2(dxi)2+dz2).\d s^2=\fr{l^2}{z^2}\p{-\d t^2+\sum_{i=1}^{d-2}\p{\d x^i}^2+\d z^2}.