The empty component is the highest weight crystal $B({\mathrm{\Lambda }}_0)$

The ${\mathrm{\Lambda }}_0$ isotypic part. The weight ${\mathrm{\Lambda }}_0$ weight space of $F$ is one-dimensional, spanned by the vacuum; since ${\mathrm{\Lambda }}_0$ is a maximal weight, the top isotypic submodule is generated by the vacuum and equals the simple module $M=V({\mathrm{\Lambda }}_0)$ by [F2], whose crystal is the empty component $B({\mathrm{\Lambda }}_0)$ by [F1].

All highest crystal vertices. In decreasing-content order a block of $a$ equal rows of length $L$, starting at row $r$, contributes its addable corner of content $L-r+1$ followed by its removable corner of content $L-r-a+1$. These are consecutive in the full boundary list: the next block's addable content is strictly smaller than this removable content. For the top block, its only preceding plus is this addable corner, so its minus can cancel exactly when $a\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}e)$. If not, that minus survives: $+-$ reduction cannot pair it with a later plus. If $e\mid a$, the two same-residue signs cancel and can be deleted from the signature; deleting the top block from the diagram shifts all remaining contents by $a$, preserving their residues. Induction over the blocks proves that all minuses cancel exactly when every positive part multiplicity is divisible by $e$. Such a shape repeats every row of a unique ordinary partition $\rho$ exactly $e$ times. In each stack of $e$ equal rows, every column has one cell of each residue, so $m_i=\mathrm{\mid }\rho \mathrm{\mid }$ for all $i$. Its weight is ${\mathrm{\Lambda }}_0-\mathrm{\mid }\rho \mathrm{\mid }\delta$, and the number of these highest labels at weight ${\mathrm{\Lambda }}_0-k\delta$ is $p(k)$.

The decomposition. The full Fock module has finite label spaces and support in ${\mathrm{\Lambda }}_0-Q^{+}$, so [F3] applies and identifies the field constituents of the highest crystal labels of step 1.2; combined with step 1.1 this gives $F={⨁}_{k\ge 0}V({\mathrm{\Lambda }}_0-k\delta {)}^{\oplus p(k)}$ as a locally weight-finite indexed sum, and taking characters gives the displayed product formula. The raw standard vectors attached to the non-regular highest labels are not asserted to be field-highest vectors, and the ambient basis is not asserted to be component-compatible.

Conclusion. The empty component is the canonical $B({\mathrm{\Lambda }}_0)$ with the $e$-regular labels, its module and lower lattice are $M$ and $M\cap F_0$, it occurs once, and the full decomposition has $p(k)$ copies of $V({\mathrm{\Lambda }}_0-k\delta )$, with the Axiom of Choice used only through the declared decomposition [F3].