$\mathcal L(a-1,\{a,a\})$ is the hypersurface of the determinant $0$ matrices in the rank $a \times a$. I'm afraid that this hypersurface is not a topological manifold, and hence also not a smooth one.

To see this, it is sufficient to look in a small neighborhood of a rank $a-2$ matrix. Were the hypersurface a manifold, that space would have to be topologically a ball, as we see in the neighborhood of a cuspidal curve singularity like the singularity $(x,y,z) = (0,0,1)$ of $y^2z-x^3=0$. However, our neighborhood will not be a ball.

The topology of an algebraic variety in the neighborhood of a point is determined only by the leading terms of the polynomial equation that cuts it out. In the neighborhood of the point:

$\left(\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right)$

(here depicted with $a=6$) the leading terms come are the terms involving all $a-2$ nonzero entries. Since there is one term for each permutation, there are only two leading terms : the identity permutation and the transposition of the last two entries.

This means that locally, the hypersurface looks like the equation $x_1x_2-x_3x_4$ in the variables $x_1,x_2,\dots, x_{a^2-1}$, where $x_1$, $x_2$, $x_3$, $x_4$ are the entries of the bottom-right $2 \times 2$ submatrix. The vanishing set of the equation $x_1 x_2 -x_3x_4$ in just the variables $x_1,x_2,x_3,x_4$ is topologically isomorphic to the cone on an $S^1$ bundle on $S^2 \times S^2$, which is the cone on a non-sphere, so is not a topological manifold. Since we can detect this failure homologically, adding more variables, which just corresponds to taking the product with $\mathbb R^{2(a^2-5)}$, does not make it a manifold.