You are completely correct in your analysis of the structure one obtains by considering the
Schrodinger equation for Z electrons in a central potential due to a nucleus of charge Ze
when the Coulomb interaction between electrons as well as relativistic effects such as spin-orbit coupling are ignored. In such a model the periods would indeed be 2, 8, 18 etc. for
exactly the reasons you have described. In physics jargon the energy in this model depends only on the principal quantum number $n \in {\mathbb Z}, n>0$ and the allowed $\ell$ values
are $\ell\le n-1$. Orbitals with $\ell=0,1,2,3,4, \cdots$ are labelled by $s,p,d,f \cdots$ for
historical reasons. Thus at $n=1$ one can have one or two states in the $(1s)$ orbital (accounting for spin), at $n=2$ one has the $(2s)$ and $(2p)$ orbitals with $2(1+3)=8$
states. And at $n=3$ one should have $18$ states by filling the $(3s),(3p),(3d)$ orbitals.
But in the real world this is not what happens. This simple model does not give a correct description of atoms in the real world once you get past Argon. I believe the main effect leading to the breakdown is the Coulomb interaction between the electrons.
So there is no simple mathematical model based say just on the representation theory of $SU(2)$ and simple solutions to the Schrodinger equation which will account for the structure of the periodic table past Argon. However one could ask whether including Coulomb interactions between electrons does gives a model which correctly reproduces the next few rows of the periodic table past Argon. I am not an expert on this, but since I doubt there are physical chemists on MO I'll just give my rough sense of things.
To approach this problem with some level of rigor probably requires difficult numerical work
and my impression is that this is beyond the current state of the art. However there are
rough models which try to approximate what is going on by assuming that the interactions between electrons can be replaced by a spherically symmetric potential which is no longer
of the $1/r$ form. This leaves the shell structure as is, but can change the ordering of
which shells are filled first. In such a model instead of filling the $(3d)$ shell after
Argon one starts to fill the $(4s)$ and $(3d)$ shells in a somewhat complicated order. Eventually one fills the $(4s),(3d),(4p)$ shells and this leads to the line of the periodic
table starting at K and ending at Krypton.
Added note: There is one nice piece of mathematics associated with this problem that I should have mentioned, even if it doesn't by itself explain the detailed structure of the periodic table. When Coulomb interactions between electrons and relativistic effects are ignored the energy levels of the Schrodinger equation with a central $1/r$ potential depend only the quantum number $n$, but not on the quantum number $\ell$ which determines the representation $V_\ell$ of $SO(3)$ referred to above. When screening is included this is no longer the case and the energies depend on both $n$ and $\ell$. Why is this?
With a $1/r$ central potential there is an additional vector $\vec D$ which commutes with the Hamiltonian. Classically this vector is the Runge-Lenz vector and its conservation explains why the perihelion of elliptical orbits in a $1/r$ potential do not precess. Quantum mechanically the
commutation relations of the operators $\vec D$ along with the angular momentum operators $\vec L$ are those of the Lie algebra of $SO(4)$ (for bound states with negative energy). There are two Casimir invariants, one vanishes and the other is proportional to the energy.
As a result the energy spectrum depends only on $n$ and can be computed using group theory without ever solving the Schrodinger equation explicitly. Perturbations due to screening, that is from some averaged effect of the Coulomb interactions between electrons,
change the $1/r$ potential to some more general function of $r$ and break the symmetry
generated by the Runge-Lenz vector $\vec D$. As a result the energy levels depend on both $n$ and $\ell$.
$\mathcal L(a-1,\{a,a\})$ 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.
Best Answer
Let me answer in terms of operator theory. The uncertainty principle can be interpreted as some particular inequality (as you say) such as $$ \frac{\hbar}{2}\Vert{u}\Vert^2\le \Vert{D_xu}\Vert\Vert{xu}\Vert, $$ inequality due to the identity $ 2\Re\langle\hbar\frac{1}{i}\partial_x u, ixu\rangle=\hbar\Vert{u}\Vert^2, $ and in fact to the non-commutation of the unbounded operators $D_x=\frac{\hbar}{i}\partial_x$ with $x$ (multiplication by $x$). Incidentally you can note that the non-commutation of bounded operators $A,B$ (say finite matrices) could never lead to $$ [A,B]=I\tag{NC} $$ as it is the case for $A=\frac{1}{i}\partial_x$, $B=ix$. In fact (NC) is not possible for matrices since the trace of a commutator is 0. This is one reason for which the quantization requires (necessarily infinite dimensional) unbounded operators.
Now this non-commutation property is at the source of quantum mechanics and produces the uncertainty principle inequalities. Each time you will have to deal with a non-commutative algebra of operators, you will have some type of uncertainty principle. In particular if your unknown quantities do not belong to a commutative algebra, you will have an uncertainty principle: the paradigmatic example remains the transfer from classical mechanics where the unknown quantities belong to a commutative algebra (functions on the phase space, such as position $x$ or momentum $\xi$), transfer to quantum mechanics where the unknown quantities are quantization of the previous quantities and are operators, such as $D_x,x$.
The uncertainty principle is simply the signature of non-commutativity and is not limited to quantum mechanics.