TITLE: Extremal properties of the determinant for matrices with entries in a fixed subset of $[-1,1]^{n^2}$? 
QUESTION [9 upvotes]: Given a multiset $S\subset [-1,1]^{n^2}$, we set
$$m(S)=\min\vert \det(M)\vert$$
where the minimum is over all matrices with entries forming the multiset $S$
and 
$$a(n)=\max m(S)$$
where the maximum is over all multisets with $n^2$ elements in $[-1,1]$.
Obviously $a(2)=2$ by considering $S=\lbrace 1,1,1,-1\rbrace$.
I know nothing else (except for the trivial bounds $0 < a(n)\leq n^{n/2}$).
Even the computation of $a(3)$ (or of a good lower bound on $a(3)$) 
seems quite a feat to me.

REPLY [2 votes]: I don't know the answer to your question, but the following is a bit long for a comment.
We can always assume one of the matrix elements is 1 since rescaling all elements so that the largest element equals 1 increases the minimal determinant by the scaling factor raised to the power $n$.  It seems likely that any solution will have many 1s—of course less than $2n$, but possibly order $n$ of them—although I can't see how to prove this.  For $n=3$ the matrix with elements in $[-1,1]$ of largest determinant is 
$$
\begin{bmatrix}
-1 & 1 & 1\\
1 & -1 & 1\\
1 & 1 & -1
\end{bmatrix}.
$$
This provides an upper bound of 4, which is slightly better than Hadamard's bound of about 5.2, but still apparently far too high.
It's possible to get a very slight improvement on Kevin Costello's lower bound for $n=3$.  Consider the multiset $\{1,1,1,1,1,a,b,c,d\}$ with
$$
a=0.19552006830186067389,
$$
$$
b=-0.47998412524185001,
$$
$$
c=-0.898326460649234689,
$$
$$
d=-0.248885944004550461.
$$
It has minimal determinant 0.185913849057346968.  A hill-climbing procedure repeatedly arrives at solutions with five 1s.  If the procedure is modified to assume five 1s, it readily finds solutions very close to the one above.  The hill-climbing solutions (approximately) satisfy the property that the four matrices
$$
\begin{bmatrix}1 & 1 & a\\ 1 & d & 1\\b & 1 & c\end{bmatrix},\quad
\begin{bmatrix}1 & 1 & 1\\ 1 & 1 & a\\d & b & c\end{bmatrix},\quad
\begin{bmatrix}1 & 1 & a\\ 1 & c & 1\\d & 1 & b\end{bmatrix},\quad
\begin{bmatrix}1 & 1 & a\\ 1 & 1 & d\\b & c & 1\end{bmatrix},
$$
all have (minus the) minimal determinant.  Imposing equality of these determinants, and maximizing their magnitude, yields the precise values above.  (They are some messy algebraic numbers.)