Suppose $(X_1,Y_1),(X_2,Y_2),\ldots,(X_n,Y_n)$ are i.i.d random vectors with a continuous distribution. Let $R_i =\operatorname{Rank}(X_i)$ among $X_1,X_2,\ldots,X_n$ and $Q_i=\operatorname{Rank}(Y_i)$ among $Y_1,Y_2,\ldots,Y_n$, $\,i=1,2,\ldots,n$.
Spearman's rank correlation coefficient is then the sample quantity
$$r_S=\frac{\sum_{i=1}^n \left(R_i-\frac{n+1}2 \right)\left(Q_i-\frac{n+1}2 \right)}{\sqrt{\sum_{i=1}^n \left(R_i-\frac{n+1}2 \right)^2}\sqrt{\sum_{i=1}^n \left(Q_i-\frac{n+1}2\right)^2}}$$
It can be shown that
$$E(r_S)\to \rho_G \quad\text{ as }n\to \infty\,, \tag{$\star$}$$
where $\rho_G$ is the grade correlation coefficient defined as
$$\rho_G=\operatorname{Corr}(F(X_1),G(Y_1))$$
Here $F$ and $G$ are the distribution functions of $X$ and $Y$ respectively.
So $r_S$ is an asymptotically unbiased estimator of $\rho_G$, and at least in this sense $\rho_G$ is a parameter of interest and can be considered to be a population counterpart of $r_S$.
On the other hand, the statistic $$T_n=\frac1{\binom{n}{2}}\sum_{1\le i<j\le n}\operatorname{sgn}(X_i-X_j)\operatorname{sgn}(Y_i-Y_j)$$ is exactly unbiased for its population counterpart, Kendall's tau:
$$\tau=E\left[\operatorname{sgn}(X_1-X_2)\operatorname{sgn}(Y_1-Y_2)\right]$$
If you note that
$$\sum_{j:j\ne i}\operatorname{sgn}(X_i-X_j)=(R_i-1)-(n-R_i)=2\left(R_i-\frac{n+1}2\right)$$
and similarly
$$\sum_{j:j\ne i}\operatorname{sgn}(Y_i-Y_j)=2\left(Q_i-\frac{n+1}2\right)\,,$$
we have this relation between $r_S$ and $T_n$:
\begin{align}
r_S&=\frac{12}{n(n^2-1)}\sum_{i=1}^n \left(R_i-\frac{n+1}2\right)\left(Q_i-\frac{n+1}2\right)
\\&=\frac3{n(n^2-1)}\sum_{i=1}^n \left\{\sum_{j\ne i}\operatorname{sgn}(X_i-X_j)\right\}\left\{\sum_{k\ne i}\operatorname{sgn}(Y_i-Y_k)\right\}
\\&=\frac3{n+1}T_n+\frac{3(n-2)}{n+1}U_n\,, \tag{1}
\end{align}
where $$U_n=\frac1{n(n-1)(n-2)}\sum_{i\ne j\ne k}\operatorname{sgn}(X_i-X_j)\operatorname{sgn}(Y_i-Y_k)$$
Using the independence of $X_2$ and $Y_3$, we can write
\begin{align}
E(U_n)&=E\left[\operatorname{sgn}(X_1-X_2)\operatorname{sgn}(Y_1-Y_3)\right]
\\&=E \left[ E\left[\operatorname{sgn}(X_1-X_2)\operatorname{sgn}(Y_1-Y_3)\right]\mid X_1,Y_1 \right]
\\&=E \left[ E\left[\operatorname{sgn}(X_1-X_2)\mid X_1 \right] E\left[\operatorname{sgn}(Y_1-Y_3) \mid Y_1\right] \,\right]
\\&=E\left[F(X_1)-(1-F(X_1))\right] \left[G(Y_1)-(1-G(Y_1))\right]
\\&=4 E\left[F(X_1)-\frac12\right]\left[G(Y_1)-\frac12\right]
\\&=\frac13 \rho_G \tag{2}
\end{align}
Equations $(1)$ and $(2)$ then together imply $(\star)$.
Typically we are interested in testing the null hypothesis $$H_0: X \text{ and }Y \text{ are independently distributed}$$
Under $H_0$, we have $\rho_G=0$ as well as $\tau=0$, which implies $E_{H_0}(r_S)=0$. The variance under $H_0$ can be shown to be $\operatorname{Var}_{H_0}(r_S)=\frac1{n-1}$. A large sample test is then based on
$$\sqrt{n-1}\,r_S \stackrel{d}\longrightarrow N(0,1)\quad, \text{ under }H_0$$
Note however, that this is not a test for $\rho_G=0$ and it does not give confidence intervals for $\rho_G$ or $E(r_S)$ since the asymptotic distribution of $r_S$ is derived only under $H_0$.
Reference:
Nonparametric Statistical Inference (5th ed.) by Gibbons and Chakraborti, pages 416-421.
Nonparametric Statistical Methods (3rd ed.) by Hollander/Wolfe/Chicken, pages 427-440.
Statistical Inference Based on Ranks by T.P. Hettmansperger.
Related question: Spearman's correlation as a parameter.
