The best known lower bound on the determinantal complexity of the $4\times4$ permanent is 10. In this short note, I prove that this value is exactly 15: the $4\times4$ permanent is the determinant of a $15\times15$ matrix of affine linear forms, and cannot be that of any smaller matrix. To do so, I must establish both a lower and upper bound. To establish the upper bound, I use Grenet's construction of size $2^n-1$. Hence that construction is optimal at $n=4$ (and $n=3$). This now constitutes the first known exact value of the determinantal complexity of the permanent at $n=4$, improving over prior knowledge in which the exact value was known only for $n=3$. Recall that the general quadratic bound only gives 8 for $n=4$.

To establish the lower bound, I work from a normal form at a zero of the permanent. In the first row (and column) of the normal form, the linear forms vanish on a linear space inside the singular locus of the permanent. A hessian inequality bounds the size of the matrix in terms of that space. Then, a matrix of size 14 forces the space into any of nine coordinate types. The forcing is either up to symmetry, or else into a family that degenerates to one of them. I then rule out every case. All involved computations are exact.

In respect of the claim, I have made all certificates public.

\textbf{This note was hand-written by the human author.}