when $\gamma=1$.

At this point, we have all of the requirements to evaluate on-shell action for the gray part of figure 1. There are six different GHY terms and four different joint terms:

\[fig:fig1\]

![ Initial time for $d>3$. Gravitational action is evaluated on the gray region. ](a1.png)

-   Both of the surfaces at $r^{\ast}=\epsilon_{0}$ are spacelike. Using we find $$\begin{aligned}
    I_{GHY}^{XW} &=   \frac{-r^{d-1} ~\Omega_{d-1}}{16 \pi G_{n}} ~\left( f'(r) +\frac{2(d-1)}{r} f(r) \right) ~\left(t(W)-t(X)\right),
    \\
    I_{GHY}^{YZ} &=   \frac{-r^{d-1} ~\Omega_{d-1}}{16 \pi G_{n}} ~\left( f'(r) +\frac{2(d-1)}{r} f(r) \right) ~\left(t(Z)-t(Y)\right).\end{aligned}$$ Summing these two terms results in the contribution of $r^{\ast} = r^{\ast}(\epsilon_{0})$ surfaces, $$I_{GHY}^{sing}= \frac{r^{d-1} ~\Omega_{d-1}}{16 \pi G_{n}} ~\left( f'(r) +\frac{2(d-1)}{r} f(r)\right) ~ (4 r^{\ast}(\epsilon_{0})+ 2 \mu^+-2\mu^-).$$ Using and we have $$\frac{d I_\text{GHY}^\text{sing}}{d \mu}=0.$$ Thus, in the symmetric case, the GHY terms of the near singularities cancel each other and are independent of the boundary time.

-   Null joint terms take place at $P$ and $Q$, $$\begin{aligned}
    \label{pointP}
    I_{J}^{P} &= \frac{-\Omega_{d-1}~ r_P^{d-1}}{8 \pi G_{N}} ~\log \frac{\mathfrak{a}^2}{|f(r_P)|},
    \\
    \label{pointQ}
    I_{J}^{Q} &= \frac{-\Omega_{d-1} ~r_Q^{d-1}}{8 \pi G_{N}} ~\log \frac{\mathfrak{a}^2}{|f(r_Q)|}.\end{aligned}$$ Using $r_P^{\ast}=r_Q^{\ast}=0$ and we have $\frac{d~ r_P}{d \mu}=\frac{d~ r_Q}{d \mu}=0$. Therefore the time derivative of null-joint terms at these points vanishes $$\frac{d I_{J}^\text{null}}{d \mu}=0.$$

-   There are four 