Quartiles of local time of brownian bridge at origin Let $(L_t)_{0\leq t \leq 1}$ be the local time at $0$ of a brownian bridge. Let $(T_l)_{0\leq l \leq L_1}$ be its generalized inverse (as in $T_l := \inf\{t\geq 0 : L_t \geq l\}$). What is the joint distribution of
$$(T_{L_1/4}, T_{L_1/2}, T_{3L_1/4}) ?$$
In particular, I am interested in the joint distribution of $(T_{L_1/2}, T_{L_1/4} - T_{L_1/2}+T_{3L_1/4})$, which I conjecture to have both marginals uniform in $[0,1]$.
 A: Joint Law of Brownian Motion and Its Local Time.
In what follows, we will use the joint law of Brownian motion $B_t$ with $B_0=x \in \mathbb{R}$ and its local time accumulated at zero $\ell_t$, which can be found, e.g., in (1.3.8) on pg. 140 of:
A. Borodin and P. Salminen (1996). Handbook of Brownian Motion, Springer. 
This formula states that: 
$$
\mathbb{P}(\ell_t \in dy , B_t \in dz) = p(x,y,z,t) dy dz \tag{1}
$$ where we have introduced the joint density function:
$$
p(x,y,z,t) = \frac{1}{t \sqrt{2 \pi t}} (|x| + y + |z|) \exp\left(-\frac{(|x| + y + |z|)^2}{2t}  \right) 
$$  
Law of $T_{L_1/2}$.
Let $X_t$ denote a Brownian bridge between $X_0=a$ and $X_1=b$, and let $L_t$ be its local time accumulated at zero over $[0,t]$. Given $ t_1 \in (0,1)$, note that the event $\{ T_{L_1/2}>t_1 \}$ is the same as the event $\{ L_{t_1} < L_1/2 \}$. To determine this latter probability, we use the law of the local time of a Brownian bridge.  This is obtained from (1) by conditioning on the endpoints of the Brownian motion. Doing this we get:
\begin{align*}
\mathbb{P}^{a,b}\left( T_{L_1/2} > t_1 \right)&=
\mathbb{P}^{a,b}\left( L_{t_1} < \frac{L_1}{2} \right) \\
&=Z^{-1} \int_{R} p(a,y_1,z_1,t_1) p(z_1,y_2,b,1-t_1) dy_1 dy_2  dz_1
\end{align*} where $R$ is the region:
$$
R=\left\{ (y_1,y_2,z_1) \in \mathbb{R}^3: \begin{array}{c} 0 \le y_1 \le (y_1+y_2)/2 \\ 0 \le y_2 \le \infty \end{array} \right\}
$$ and $Z$ is a normalization constant:
$$
Z=\int_{-\infty}^{\infty} \int_0^{\infty} \int_0^{\infty}  p(a,y_1,z_1,t_1) p(z_1,y_2,b,t_2-t_1) dy_1 dy_2  dz_1
$$  One can check from this formula that when $a=b=0$, the law of $T_{L_1/2}$ is indeed uniform over $(0,1)$, as conjectured by the OP.
Joint Law of $(T_{L_1/4}, T_{L_1/2}, T_{3L_1/4})$.
Similar to the previous calculation, but this time given $0 < t_1 < t_2 < t_3 < 1$, 
\begin{align*}
 \mathbb{P}^{a,b}&\left(T_{L_1/4} > t_1 , T_{L_1/2}>t_2,  T_{3L_1/4} > t_3 \right) \\ &= \mathbb{P}^{a,b}\left(L_{t_1} < \frac{L_1}{4} , L_{t_2} < \frac{L_1}{2} ,  L_{t_3} < \frac{3 L_1}{4} \right) \\
 &= 
Z^{-1} \int_{R} p(a,y_1,z_1,t_1) p(z_1,y_2,z_2,t_2-t_1) \\
& \times p(z_2,y_3,z_3,t_3-t_2) p(z_3,y_4,b,1-t_3) dz_1 dz_2 dz_3 dy_1 dy_2 dy_3 dy_4 
\end{align*} where $R$ is the region:
$$
R=\left\{ (y_1,y_2,y_3,y_4,z_1,z_2,z_3) \in \mathbb{R}^7: 
\begin{array}{c}
0 \le y_1 \le S/4 \\  0 \le y_2 \le S/2 \\ 0 \le y_3 \le 3 S/4 \\ 0 \le y_4 \le \infty \\ S= y_1+y_2+y_3+y_4 \end{array} \right\}
$$ and $Z$ is a normalization constant.
Law of $(T_{L_1},T_{L_1/4}-T_{L_1/2}+ T_{3L_1/4})$.
One can use the joint CDF given above to compute this law.  For $a=b=0$, the OP's conjecture that the marginals are uniform seems to be correct.
