Suppose $C$ is a positive semidefinite (PSD) matrix with unit diagonals $C_{ii}=1$. Can we find upper and lower bounds for the determinant $\det(C)$ based on $\min_{i\neq j} |C_{ij}|$ and $\max_{i\neq j} |C_{ij}|$?
Let's define $X$ such that $C = X^\top X$. The matrix $C$ encodes the angles between columns of $X$: $C_{ij} = X_i^\top X_j = \cos(\theta_{ij})$, where $\theta_{ij}$ is the angle between columns $i$ and $j$.
Determinant lower bound: By the Gershgorin circle theorem, the eigenvalues of $C$ are not smaller than $1-n\max_{i\neq j}|C_{ij}|$, implying a lower bound of $\det(C)\ge (1-n \max_{i\neq j}|C_{ij}|)^n$.
Determinant upper bound: Let $S_i$ denote the span of columns up to the $i$-th column: $S_i:=\text{span}(X_1,\dots,X_i)$. We have $\det(C) = \prod_i^n \text{dist}(X_i,S_{i-1})^2$. If $Y_i$ is the projection of $X_i$ onto $S_{i-1}$, then it must be closer to it than all $X_1,\dots, X_{i-1}$: $\text{dist}(X_i,S_{i-1})\le \text{dist}(X_i,X_1),\dots,\text{dist}(X_i,X_{i-1})$. Note that $\text{dist}(X_i,X_j) = \sqrt{1-C_{ij}}$. Therefore, $\text{dist}(X_i,S_{i-1})\le \min_{j < i} \sqrt{1-C_{ij}^2}$. Putting this together with the previous equality: $\det(C) \le \prod_{i=1}^n\min_{j < i} (1-C_{ij}^2)^{1/2}$, which implies $\det(C) \le (1-\min_i \max_{j< i}C_{ij}^2)^{n/2} \le (1-\min_{i\neq j}C_{ij}^2)^{n/2}$.
Combining the two bounds, we have: $$ (1-n\max_{i\neq j}|C_{ij}|)^n\le \det(C)\le (1-\min_{i\neq j}C_{ij}^2)^{n/2} $$
For the special case when $n=2$, if $C = \begin{bmatrix} 1 , \epsilon\\ \epsilon, 1\end{bmatrix}$ then $\det(C) = 1 - \epsilon^2$, and there is only one off-diagonal value $\epsilon$. The inequality thus becomes: $$ (1-2\epsilon)^2 \le 1 - \epsilon^2 \le (1-\epsilon^2) $$
Questions: Are the conclusions drawn correct? Are they related to some known property of determinants? How tight are the conclusions? Can they be made tighter?