\section{The lithium dianion}\label{sec:07-lithium-dianion}

We prove that a lithium nucleus does not bind five electrons. As for the
helium dianion, the proof applies
Proposition~\ref{prop:quadratic-removal}. The upper bound on the threshold
$\AtomE{4}{3}$ comes from Proposition~\ref{prop:lithium-upper}. The lemma
below supplies the exterior gaps. Positivity of the weighted form is the
computer-assisted Lemma~\ref{lem:li5-weighted-form} of
Appendix~\ref{sec:09-geometric-certificates}.

\begin{lemma}[Exterior gaps]\label{lem:lithium-dianion-exterior-gaps}
Set
\[
 U_0=-\frac{37431243}{5000000},\qquad
 (b_1,\delta_1)=\left(\frac{26}{25},\frac{129}{40}\right),\qquad
 (b_2,\delta_2)=\left(\frac{103}{50},\frac{3469}{5000}\right).
\]
Let $a\in\{1,2\}$ and let $u$ belong to the fermionic form domain of
$\AtomH{5}{3}$. If $u$ vanishes almost everywhere on $\{q_a\le b_a\}$,
then
\[
 \mathfrak q_{5,3}[u]-U_0\|u\|^2\ge\delta_a\|u\|^2.
\]
\end{lemma}

\begin{proof}
Let $\mathfrak h$ denote the Coulomb integral of
Set-up~\ref{setup:atoms} with $N=5$ and $Z=3$ on the full space
$H^1((\mathbb R^3)^5;\mathbb C^{32})$ without a symmetry restriction.
It agrees with $\mathfrak q_{5,3}$ on fermionic vectors. We apply
Lemma~\ref{lem:regional-bound} with $N=5$, $Z=3$, so that a core electron
contributes $-9/2$ and $A_j=3-\kappa-jR/(R+b)$. The three parameter
choices used are
\begin{equation}\label{eq:lithium-dianion-regional-values}
 \begin{aligned}
 M_0(26/25,26/25,25/39,5/3)&=\frac{25925}{6084},\\
 M_0(3/5,103/50,125/128,8/5)&=\frac{604295}{98304},\\
 M_1(103/50,103/50,50/103,2)&=\frac{131831}{21218}.
 \end{aligned}
\end{equation}
Indeed the corresponding $A_j$ are $428/351$, $1/2$ and $115/206$, so
the maximum in $M_j$ is attained at $h=2$ in each case, and the values
follow by rational arithmetic.

\emph{The first exterior gap.}
If $u=0$ on $\{q_1\le26/25\}$, then all five radii exceed $26/25$ on the
support of $u$. The first value in
\eqref{eq:lithium-dianion-regional-values} gives
\[
 \mathfrak q_{5,3}[u]-U_0\|u\|^2
 \ge\left(\frac{129}{40}
       +\frac{545603}{7605000000}\right)\|u\|^2.
\]

\emph{The second exterior gap.}
Suppose $u=0$ on $\{q_2\le b_2\}$.
Let $\theta(r)$ be zero for $r\le3/5$, affine from zero to $\pi/2$
on $[3/5,b_2]$, and equal to $\pi/2$ for $r\ge b_2$.
For $S\subset\{1,\ldots,5\}$ put
\[
 \chi_S=\prod_{i\in S}\cos\theta(r_i)
              \prod_{i\notin S}\sin\theta(r_i).
\]
Then $\sum_S\chi_S^2=1$, and the quadratic-form localization identity is
\[
 \mathfrak h[u]=\sum_S\mathfrak h[\chi_Su]
       -\frac12\int\sum_i|\theta'(r_i)|^2|u|^2.
\]
At most one radius lies below $b_2$ on the support of $u$. Hence the
localization error is at most
$\pi^2\|u\|^2/[8(b_2-3/5)^2]$, and $\chi_Su=0$ when $|S|\ge2$.
The term with $S=\varnothing$ has all five radii above $3/5$.
For $S=\{i\}$, the radius $r_i$ is below $b_2$ and the other four
are above $b_2$. The localized spinors are global $H^1$ functions. Their
vanishing conditions, sliced in the other variables, give the Dirichlet
traces required by Lemma~\ref{lem:regional-bound}. No symmetry of the
localized spinors is required.
The last two bounds in \eqref{eq:lithium-dianion-regional-values}, together
with
\[
 \frac{604295}{98304}<\frac{123}{20}<\frac{31}{5}
                      <\frac{131831}{21218},
\]
therefore give
\[
 \mathfrak q_{5,3}[u]
 \ge-\left(\frac{131831}{21218}
           +\frac{\pi^2}{8(b_2-3/5)^2}\right)\|u\|^2.
\]
Using $\pi<22/7$, the remaining comparisons are
\[
 \frac{131831}{21218}<\frac{310659}{50000},\qquad
 \frac{\pi^2}{8(b_2-3/5)^2}
 <\frac{151250}{261121}<\frac{14481}{25000},
\]
and
\[
 -U_0-\frac{310659}{50000}-\frac{14481}{25000}
 =\frac{3469143}{5000000}>\frac{3469}{5000}.
\]
This proves the second gap and completes the two cases.
\end{proof}

\begin{theorem}\label{thm:lithium-dianion}
Under Set-ups~\ref{setup:atoms} and~\ref{setup:execution},
$\AtomE{5}{3}=\AtomE{4}{3}$, and $\AtomH{5}{3}$ has no nonzero $L^2$
eigenfunction at any energy at or below $\AtomE{4}{3}$.
\end{theorem}

\begin{proof}
Use $U_0,b_a,\delta_a$ from
Lemma~\ref{lem:lithium-dianion-exterior-gaps}, and set
\[
 \beta=2,\qquad J=\{1,2\},\qquad
 \lambda_1=\frac{83}{250},\qquad \lambda_2=\frac{527}{1000}.
\]
Proposition~\ref{prop:lithium-upper} supplies the required upper bound on
the threshold:
\[
 \AtomE{4}{3}<-7.497020240<U_0.
\]
With $N=5$, $Z=3$, $U=U_0$ and $d_a=\delta_a$, let $B_a$ and
$M_{2,U_0}$ be the moment functions and weighted form defined in
Proposition~\ref{prop:quadratic-removal}. Its exterior hypotheses hold by
Lemma~\ref{lem:lithium-dianion-exterior-gaps}.
Lemma~\ref{lem:li5-weighted-form} in
Appendix~\ref{sec:09-geometric-certificates} uses precisely these constants
and moment functions, and gives
\[
 M_{2,U_0}[u]+\sum_{a=1}^2\lambda_a\int B_a(q_a)|u|^2
 \ge\frac1{100000}\|u\|^2
\]
for every compactly supported fermionic $u\in H^1$.
Thus all hypotheses of Proposition~\ref{prop:quadratic-removal} hold with
$\varepsilon=1/100000$. Its conclusion gives both the energy equality and
the exclusion of every $L^2$ eigenfunction at or below $\AtomE{4}{3}$.
\end{proof}