Warp products and (2+1) dimensional spacetimes

Abstract

We investigate the geometrical aspects of the extended (2+1) dimensional Banados–Teitelboim–Zanelli spacetimes in the multiply warped product scheme. To do this, we analyze the interior physical properties by constructing the explicit warp functions in these regions.

Appendices

Appendix 1: Mathematical aspects of a quartic equation

Now, we consider four roots of the lapse function in (2.48) which produces a quartic equation of the form

$$\begin{aligned} x^{4}+a_{1}x^{2}+a_{2}x+a_{3}=0, \end{aligned}$$

(5.1)

where

$$\begin{aligned} a_{1}=-Ml^{2},\quad a_{2}=MBl^{2},\quad a_{3}=\frac{1}{4}J^{2}l^{2}. \end{aligned}$$

(5.2)

Following the algorithm for treating the quartic equation [28], we obtain the four roots $(x_{1},x_{2},x_{3},x_{4})$ of the Eq. (5.1) given by

$$\begin{aligned} x_{1}&= \frac{1}{2}p_{1}+\frac{1}{2}p_{2},\nonumber \\ x_{2}&= \frac{1}{2}p_{1}-\frac{1}{2}p_{2},\nonumber \\ x_{3}&= -\frac{1}{2}p_{1}+\frac{1}{2}p_{3},\nonumber \\ x_{4}&= -\frac{1}{2}p_{1}-\frac{1}{2}p_{3}, \end{aligned}$$

(5.3)

where

$$\begin{aligned} p_{1}=\left( y_{0}-a_{1}\right) ^{1/2}. \end{aligned}$$

(5.4)

For $p_{1}\ne 0$ we obtain ¹

$$\begin{aligned} p_{2}&= \left( -p_{1}^{2}-2a_{1}-\frac{2a_{2}}{p_{1}}\right) ^{1/2},\nonumber \\ p_{3}&= \left( -p_{1}^{2}-2a_{1}+\frac{2a_{2}}{p_{1}}\right) ^{1/2}, \end{aligned}$$

(5.5)

and for $p_{1}=0$ we arrive at

$$\begin{aligned} p_{2}&= \left( -2a_{1}+2\left( y_{0}^{2}-4a_{3}\right) ^{1/2}\right) ^{1/2},\nonumber \\ p_{3}&= \left( -2a_{1}-2\left( y_{0}^{2}-4a_{3}\right) ^{1/2}\right) ^{1/2}. \end{aligned}$$

(5.6)

Here $y_{0}$ is defined as a real root of the following cubic equation

$$\begin{aligned} y^{3}+b_{1}y^{2}+b_{2}y+b_{3}=0, \end{aligned}$$

(5.7)

where

$$\begin{aligned} b_{1}=-a_{1},\quad b_{2}=-4a_{3},\quad b_{3}=4a_{1}a_{3}- a_{2}^{2}. \end{aligned}$$

(5.8)

The three roots of the cubic equation (5.7) are then readily given by (see “Appendix 2” for details)

$$\begin{aligned} y_{1}&= q_{1}+q_{2}-\frac{1}{3}b_{1},\nonumber \\ y_{2}&= -\frac{1}{2}(q_{1}+q_{2})+\frac{i\sqrt{3}}{2}(q_{1}-q_{2})-\frac{1}{3}b_{1},\nonumber \\ y_{3}&= -\frac{1}{2}(q_{1}+q_{2})-\frac{i\sqrt{3}}{2}(q_{1}-q_{2})-\frac{1}{3}b_{1}, \end{aligned}$$

(5.9)

where

$$\begin{aligned} q_{1}&= \left( q_{3}+(q_{3}^{2}+q_{4}^{3})^{1/2}\right) ^{1/3},\nonumber \\ q_{2}&= \left( q_{3}-(q_{3}^{2}+q_{4}^{3})^{1/2}\right) ^{1/3},\nonumber \\ q_{3}&= \frac{9b_{1}b_{2}-27b_{3}-2b_{1}^{2}}{54},\nonumber \\ q_{4}&= \frac{3b_{2}-b_{1}^{2}}{9}. \end{aligned}$$

(5.10)

These three roots in (5.9) can be also rewritten in terms of trigonometric functions as follows

$$\begin{aligned} y_{1}&= 2(-q_{4})^{1/2}\cos \left( \frac{\psi }{3}\right) -\frac{1}{3}b_{1},\nonumber \\ y_{2}&= 2(-q_{4})^{1/2}\cos \left( \frac{\psi }{3}+\frac{2\pi }{3}\right) -\frac{1}{3}b_{1},\nonumber \\ y_{3}&= 2(-q_{4})^{1/2}\cos \left( \frac{\psi }{3}+\frac{4\pi }{3}\right) -\frac{1}{3}b_{1}, \end{aligned}$$

(5.11)

where $\psi$ satisfies the identities

$$\begin{aligned} \cos \psi&= \frac{q_{3}}{(-q_{4}^{3})^{1/2}},\nonumber \\ \sin \psi&= \left( 1+\frac{q_{3}^{2}}{q_{4}^{3}}\right) ^{1/2}. \end{aligned}$$

(5.12)

Now, we consider the specific case of the static BTZ limit where the coefficients of the quartic equation (5.1) are given as

$$\begin{aligned} a_{1}=-Ml^{2},\quad a_{2}=0,\quad a_{3}=0. \end{aligned}$$

(5.13)

From (5.9), one readily obtains the three roots of the cubic equation (5.7)

$$\begin{aligned} y_{1}=a_{1},\quad y_{2}=0,\quad y_{3}=0. \end{aligned}$$

(5.14)

If we choose $y_{0}$ as in $y_{0}=y_{1}=a_{1}=-Ml^{2}$, exploiting (5.3) we obtain the four roots of the quartic equation (5.1) as follows

$$\begin{aligned} x_{1}=0,\quad x_{2}=0,\quad x_{3}=M^{1/2}l,\quad x_{4}=-M^{1/2}l, \end{aligned}$$

(5.15)

which imply that $r_{+}=x_{3}$, $r_{-}=x_{4}$, $r_{1}=x_{1}$ and $r_{2}=x_{2}$. Similarly for the choice of $y_{0}=y_{2}=y_{3}=0$, we again obtain the four roots of the Eq. (5.1)

$$\begin{aligned} x_{1}=M^{1/2}l,\quad x_{2}=0,\quad x_{3}=0,\quad x_{4}=-M^{1/2}l, \end{aligned}$$

(5.16)

to conclude that $r_{+}=x_{1}$, $r_{-}=x_{4}$, $r_{1}=x_{2}$ and $r_{2}=x_{3}$. The mapping of $(r_{+},r_{-},r_{1},r_{2})$ onto $(x_{1},x_{2},x_{3},x_{4})$ depends on the choice of $y_{0}$ in $(y_{1},y_{2},y_{3})$.

Appendix 2: Algorithm for roots of a cubic equation

Next, we recapitulate the algorithm for finding solutions of a cubic equation appeared in [28]. To do this, we start with a cubic equation of the form

$$\begin{aligned} y^{3}+b_{1}y^{2}+b_{2}y+b_{3}=0. \end{aligned}$$

(6.1)

Introducing a new variable

$$\begin{aligned} z=y+\frac{1}{3}b_{1}, \end{aligned}$$

(6.2)

we find the following form

$$\begin{aligned} z^{3}+3q_{4}z-2q_{3}=0, \end{aligned}$$

(6.3)

where $q_{3}$ and $q_{4}$ are given by (5.10). Next, we readily check that

$$\begin{aligned} z_{1}=q_{1}+q_{2}, \end{aligned}$$

(6.4)

with $q_{1}$ and $q_{2}$ being defined as in (5.10), is a root of the cubic equation (6.3). Moreover, we find that (6.3) is factorized as

$$\begin{aligned} (z-z_{1})(z^{2}+z_{1}z+3q_{4})=0, \end{aligned}$$

(6.5)

from which we obtain the other two roots of the from

$$\begin{aligned} z_{2}&= -\frac{1}{2}(q_{1}+q_{2})+\frac{i\sqrt{3}}{2}(q_{1}-q_{2}),\nonumber \\ z_{3}&= -\frac{1}{2}(q_{1}+q_{2})-\frac{i\sqrt{3}}{2}(q_{1}-q_{2}). \end{aligned}$$

(6.6)

Even though $z_{2}$ and $z_{3}$ possess an $i$ as shown in (6.6), this does not indicate anything about the numbers of real and complex roots, since $q_{1}$ and $q_{2}$ are themselves complex in general. In order to determine which roots are real or complex, we need to introduce a criterion parameter, namely the discriminant $D$ defined as

$$\begin{aligned} D=q_{3}^{2}+q_{4}^{3}. \end{aligned}$$

(6.7)

If $D>0$, we have one real root and two complex conjugates; if $D=0$, we have three real roots and at least two roots are equal; if $D<0$, we have three real roots and all roots are different. For our case of interest associated with (2.49), we need to have three different roots in the cubic equation. From now on, we will thus consider $D<0$ case only. Keeping these roots $(z_{1},z_{2},z_{3})$ in mind and using the definition (6.2), we readily find the roots (5.9) for the cubic equation in (5.7) or in (6.1). Moreover, we obtain the identities:

$$\begin{aligned} z_{1}+z_{2}+z_{3}&= 0,\nonumber \\ z_{1}z_{2}+z_{1}z_{3}+z_{2}z_{3}&= 3q_{4},\nonumber \\ z_{1}z_{2}z_{3}&= 2q_{3},\nonumber \\ z_{1}^{2}+z_{2}^{2}+z_{3}^{2}&= -6q_{4},\nonumber \\ z_{1}^{3}+z_{2}^{3}+z_{3}^{3}&= 6q_{3},\nonumber \\ z_{1}^{4}+z_{2}^{4}+z_{3}^{4}&= 18q_{4}^{2},\nonumber \\ z_{1}^{5}+z_{2}^{5}+z_{3}^{5}&= -30q_{3}q_{4}. \end{aligned}$$

(6.8)

Finally, we formulate the above roots in terms of trigonometric function. To do this, we exploit the angle $\psi$ defined in (5.12). Here we again consider the negative discriminant case: $D<0$ which is relevant to our cubic equation at hand. Exploiting the identities

$$\begin{aligned} q_{3}\pm i(-q_{3}^{2}-q_{4}^{3})^{1/2}=(-q_{4}^{3})^{1/2}e^{\pm i\psi }, \end{aligned}$$

(6.9)

we arrive at the desired forms in (5.11).