# A Pair of Non-Isometric Potentials With the Same Semiclassical Invariants

Matthew West

###### Abstract

We show that there exist pairs of non-isometric potentials for the 1D
semiclassical Schrödinger operator whose spectra agree up to $O(h^{\infty})$,
yet their corresponding eigenvalues differ no less than exponentially. This
result was conjectured by Guillemen and Hezari in [GH12], where they prove a
very similar result, yet cannot remove the possibility of a subsequence
$h_{k}\to 0$ where the ground state eigenvalues may agree.

##  1 Introduction

Consider the semiclassical Schrödinger operator
$H=-h^{2}\frac{d^{2}}{dx^{2}}+V(x)$, with $h>0$, $V\in
C^{\infty}(\mathbb{R})$, and $\lim_{|x|\to\infty}V(x)=\infty$. Schrödinger
operators of this form admit discrete spectra, with corresponding rank one
eigenspaces. We denote the eigenvalues $E_{0}<E_{1}<E_{2}<\cdots$ and
corresponding $L^{2}$ normalized eigenfunctions
$\psi_{0},\psi_{1},\psi_{2},\cdots$. Note that both the eigenvalues and
eigenfunctions depend on the semiclassical parameter $h$, we will omit this
dependence in our notation. We say the spectra of semiclassical operators $H$
and $H^{\prime}$ agree up to order $h^{\infty}$,
$\text{spec}(H)=\text{spec}(H^{\prime})+O(h^{\infty})$, if for each
$E_{k}\in\text{spec}(H)$ and $E_{k}^{\prime}\in\text{spec}(H^{\prime})$, and
for each $N\in\mathbb{N}$ there exist constants $C_{k,N}$ such that
$|E_{k}-E_{k}^{\prime}|\leq C_{k,N}h^{N}$ for $h>0$. With this, we are ready
to state our main result (note that the potentials of Theorem 1 are plotted in
Figure 1).

######  Theorem 1.

There exist pairs of non-isometric potentials $V^{\pm}(x)\in
C^{\infty}(\mathbb{R})$ with $V(x)\geq 0$ defining Schrödinger operators
$H^{\pm}=-h^{2}\frac{d^{2}}{dx^{2}}+V^{\pm}(x)$ whose semiclassical spectra
agree modulo $O(h^{\infty})$, yet their eigenvalues $E_{j}^{\pm}$ differ for
all $h>0$, and $j\in\mathbb{Z}_{\geq 0}$. Moreover,
$D_{j}e^{-\frac{d_{j}}{h}}\leq E_{j}^{-}-E_{j}^{+}\leq
C_{j}e^{-\frac{c_{j}}{h}}$, for constants $c_{j},d_{j},C_{j},D_{j}>0$.

Figure 1: Two potential functions $V^{\pm}$ whose corresponding semiclassical
spectra agree up to $O(h^{\infty})$, yet whose eigenvalues are disctinct
($V^{-}$ left, $V^{+}$ right).

The above theorem was conjectured by Guillmen and Hezari in [GH12], where they
establish a similar result for the ground state eigenvalues, except they
cannot rule out the possibility of a subsequence $h_{k}\to 0$ where the ground
states agree. Moreover, they do not produce a lower bound for the difference
of ground state eigenvalues. Their methods rely on the Kato-Rellich theorem,
which guarantees the analyticity of the ground state eigenvalues in $h$, to
extend the nonequality of the ground state potentials at $h=1$, for all $h>0$
except for possibly a sequence $h_{k}\to 0$. Our methods are distinct from the
previously mentioned paper, we produce an explicit exponential lower bound for
the difference of ground state eigenvalues by adapting the Agmon bounding
methods employed by Simon in [Sim84].

To motivate our results, we consider the inverse spectral problem for the
semiclassical Schrödinger operator: if the spectra of two semiclassical
Schrödinger operators agree, are their potential functions isometric? Some
positive results exist, yet the methods employed can only distinguish spectra
up order $O(h^{N})$, for $N$ finite or $\infty$. To account for the coarseness
of these method one makes restrictions on the classes of potentials. Datchev-
Hezari-Ventura [DHV11] use order $o(h^{2})$ semiclassical trace invariants to
show that radial monotonic potentials are spectrally determined amongst all
other potentials. When working with analytic potentials with a unique global
minimum at the origin, order $O(h^{\infty})$ knowledge of the low lying
eigenvalues determine the potential, see [GU05] and [Hez09]. Colin de Verdiere
in [CdV11] studies the single well potential with a symmetry condition using
order $o(h^{2})$ techniques, and conjectures that the symmetry condition is
necessary. The main result of this paper serves as a counterexample which
establishes the necessity of the symmetry condition of [CdV11].

The corresponding limitation for the Laplacian over planar domains was
initially conjectured by Zelditch in [Zel04], and established by Fulling-
Kuchment in [FK05]. Here, two non-isometric planar domains with the same wave
trace invariants were show to exist. The so called Penrose-Lifshits mushroom
domains in Figure 2 have been shown to posses the property that
$\text{Tr}(\cos(t\sqrt{\Delta_{\Omega}}))-\text{Tr}(\cos(t\sqrt{\Delta_{\Omega^{\prime}}}))\in
C^{\infty}(\mathbb{R})$, that is the wave traces possess the same singular
structure, rendering wave trace methods unable to distinguish these domains.

Figure 2: Two non-isometric domains with the same wave trace invariants.

I would like to thank my advisor Hamid Hezari for the many insightful
conversations and the suggestion to study this problem.

##  2 Exponential Estimates for Eigenfunctions

In this section we will restrict our attention to perturbations of the
semiclassical harmonic oscillator:

$H=-h^{2}\frac{d^{2}}{dx^{2}}+x^{2}+\gamma(x),\quad\gamma\in
C_{0}^{\infty}(\mathbb{R}\setminus\\{0\\})$

and denote the semiclassical harmonic oscillator
$H_{0}=-h^{2}\frac{d^{2}}{dx^{2}}+x^{2}$. We will develop upper and lower
locally uniform estimates for the ground states of these operators, which we
will ultimately transfer to eigenvalue bounds using Hadamard’s variational
formula. Theorem 1 will follow from a judicious choice of $\gamma$, analogous
to the domains shown in Figure 2.

To begin we require the following characterization of the perturbed harmonic
oscillator spectra,

$\text{spec}(H)=\text{spec}(H_{0})+O(h^{\infty})=\\{h,3h,5h,7h,\dots\\}+O(h^{\infty}).$

Using the methods of quantom-birkhoff normal forms at the bottom of a
potential well established by Sjöstrand [Sjö92], or Borh-Sommerfeld
quantization to all orders given by Colin de Verdiere [CdV05], one can see
that the above holds. Moreover, any two perturbations of the harmonic
oscillator have spectra which agree up to $O(h^{\infty})$. This fact is
underpinned by the observation that the level sets of the Hamiltonians
corresponding to $H^{\pm}$ enclose the same area, see Figure 3.

Figure 3: Level sets of the Hamiltonians corresponding to $H^{\pm}$ ($H^{-}$
left, $H^{+}$ right).

######  Proposition 1.

Let $H=-h^{2}\frac{d^{2}}{dx^{2}}+V(x)$ be a semiclassical Schrödinger
operator with real valued potential of the form $V(x)=x^{2}+\gamma(x)$, for
some $\gamma\in C_{0}^{\infty}(\mathbb{R})$ supported away from the origin.
Let $\psi_{j}(x)$ denote the $L^{2}$ normalized eigenfunctions of $H$. Then,
for any $\delta,r,R\in(0,\infty)$ with $r<R$, there exist constants
$C_{j},D_{j}>0$ such that for all $h>0$ sufficiently small, the following
estimates hold uniformly on $r<|x|<R$

$\displaystyle\qquad(i)$ $\displaystyle\qquad|\psi_{j}(x)|\leq
C_{j}e^{-\frac{1}{h}(1-\delta)^{2}\left|\int_{0}^{x}\sqrt{V(x)}\,dx\right|}$
$\displaystyle\qquad(ii)$ $\displaystyle\qquad|\psi_{j}(x)|\geq
D_{j}e^{-\frac{1}{h}(1+\delta)^{2}\left|\int_{0}^{x}\sqrt{V(x)}\,dx\right|}$

######  Remark 1.

A similar proposition appears in the work of Simon [Sim84], with some
important differences. Simon studies the ground state eigenfunctions of the
double well potential, in contrast to this paper where we study the single
well potential. Additionally, the above result holds for each eigenfunction,
whereas in the paper of Simon, the analogous result was only established for
the ground state.

###### Proof of Proposition 1.(i).

Define $\tilde{\varphi}(x)=\left|\int_{0}^{x}\sqrt{V(t)}\,dt\right|$. Note
that for each choice of positive constants $\sigma,r,R,\delta$ there exists a
smooth mollifier $f$ which satisfies the following:

$\displaystyle\qquad(1)$ $\displaystyle\text{ on the annulus
}r<|x|<R,\;f*\tilde{\varphi}\leq\tilde{\varphi}+\sigma\text{ and
}f*(\tilde{\varphi}^{\prime})\leq\tilde{\varphi}^{\prime}+\sigma$
$\displaystyle\qquad(2)$ $\displaystyle\text{ there exists positive constants
}a,k\text{ with }a<r,\text{ such that for all }|y|\leq a\text{ and
}r<|x|<R,\text{ we have }$ $\displaystyle\text{
}\delta(f*\tilde{\varphi})(x)-(f*\tilde{\varphi})(y)\geq k>0.$

Let $\epsilon>0$ and define
$A=\inf\left\\{(1-\epsilon)\frac{\sqrt{V(x)}}{\sqrt{V(x)}+\sigma}:r<|x|<R\right\\}$.
Finally, define $\varphi=(Af)*\tilde{\varphi}$. We may choose
$\epsilon,\sigma$ sufficiently small so that

$(1-\delta)\tilde{\varphi}\leq(1-\delta)(f*\tilde{\varphi})\leq
A(f*\tilde{\varphi}),$

and

$A(f*\tilde{\varphi})\leq
A(\tilde{\varphi}+\sigma)\leq\tilde{\varphi}+\sigma\leq\tilde{\varphi}+\delta\tilde{\varphi}=(1+\delta)\tilde{\varphi}.$

Thus,

$(1-\delta)\tilde{\varphi}\leq\varphi\leq(1+\delta)\tilde{\varphi}.$ (1)

Also,

$|\varphi^{\prime}(x)|\leq A(\sqrt{V(x)}+\sigma)\leq(1-\epsilon)\sqrt{V(x)}.$
(2)

Without loss of generality we will smoothly re-define $\varphi$ such that for
$|x|\geq 2R$, $\varphi^{\prime}(x)=0$. Recall that $E_{j}$ is the
$j^{\text{th}}$ eigenvalue of $H$, and let $\psi$ be any smooth function. We
will now introduce the key object which gives us the upper bound of
Proposition 1.(i):

$\displaystyle\left\langle
e^{\frac{\varphi}{h}}\psi,(H-E_{j})e^{-\frac{\varphi}{h}}\psi\right\rangle$
$\displaystyle=\left\langle\psi,(V-E_{j})\psi\right\rangle-\left\langle\psi,\left(h\frac{d}{dx}-\varphi^{\prime}\right)^{2}\psi\right\rangle.$

By the fundamental theorem of calculus, and that $\varphi^{\prime}$ is
eventually 0, we have that

$\left\langle\psi,\left[\left(h\frac{d}{dx}\right)\varphi^{\prime}+\varphi^{\prime}\left(h\frac{d}{dx}\right)\right]\psi\right\rangle=0.$

So

$\displaystyle\left\langle
e^{\frac{\varphi}{h}}\psi,(H-E_{j})e^{-\frac{\varphi}{h}}\psi\right\rangle$
$\displaystyle=\left\langle\psi,\left[\left(-h^{2}\frac{d^{2}}{dx^{2}}-(\varphi^{\prime})^{2}\right)+V-E_{j}\right]\psi\right\rangle.$

By bound (2), and since $-\frac{d^{2}}{dx^{2}}$ is positive definite we arrive
at the following bound:

$\displaystyle\left\langle
e^{\frac{\varphi}{h}}\psi,(H-E_{j})e^{-\frac{\varphi}{h}}\psi\right\rangle\geq\left\langle\psi,\left[\epsilon
V-E_{j}\right]\psi\right\rangle.$

Recall that the eigenvalue $E_{j}=h(2j+1)+O(h^{\infty})$, so for all $|x|>a/2$
and $h$ sufficiently small, there exist a $C>0$ satisfying $\epsilon
V(x)-E_{j}\geq\frac{1}{C}>0$. Taking $\psi$ supported in $|x|>a/2$, we get

$||\psi||_{2}^{2}\leq C\left\langle
e^{\frac{\varphi}{h}}\psi,(H-E_{j})e^{-\frac{\varphi}{h}}\psi\right\rangle.$

Let $\psi=e^{\frac{\varphi}{h}}\eta\psi_{j}$, where $\eta(x)=0$ if
$|x|<\frac{a}{2}$ and $\eta(x)=1$ if $|x|>a$, and is smooth. Using the above
bound,

$\displaystyle\int_{|x|>a}e^{\frac{2\varphi}{h}}\psi_{j}^{2}\,dx$
$\displaystyle\leq C\left\langle
e^{\frac{2\varphi}{h}}\psi_{j},(H-E_{j})\eta\psi_{j}\right\rangle.$

Using the eigenvalue equation, bounding the derivatives of $\eta$ on
$a/2<|x|<a$, and using monotonicity of $\varphi$ on $|x|>a/2$, we get

$\displaystyle\int_{|x|>a}e^{\frac{2\varphi}{h}}\psi_{j}^{2}\,dx$
$\displaystyle\leq
h^{2}Ce^{\frac{2\varphi(a)}{h}}\int_{\frac{a}{2}<|x|<a}(K\psi_{j}^{\prime}\psi_{j}+D\psi_{j}^{2}).$

Applying Cauchy-Schwarz, using that $E_{j}=h(2j+1)+O(h^{\infty})$ and the
eigenvalue equation to see that for some constant $K_{j}$,
$||\psi_{j}^{\prime}||=\frac{K_{j}}{h}+O(h^{\infty})$, along with the fact
that the eigenfunctions are $L^{2}$ normalized we have

$\displaystyle\int_{|x|>a}e^{\frac{2\varphi}{h}}\psi_{j}^{2}\,dx$
$\displaystyle\leq Ce^{\frac{2\varphi(a)}{h}}(K_{j}h+h^{2}D).$

Finally, for $h$ sufficiently small, we get the following estimate:

$\int_{|x|>a}e^{\frac{2\varphi}{h}}\psi_{j}^{2}\,dx\leq
e^{\frac{2\varphi(a)}{h}}C.$ (3)

We will now use bound (3) to achieve the desired pointwise bound on $r<|x|<R$
by using convexity of $|\psi_{j}|$ in the forbidden region
$|x|>\sqrt{h(2j+1)+O(h^{\infty})}$. Without loss of generality assume that
$r<x<R$, then for all $b$ satisfying $0<b<r-a$ and $h$ sufficiently small we
have

$\displaystyle\psi_{j}^{2}(x)$
$\displaystyle\leq\frac{1}{2b}\int_{x-b}^{x+b}e^{-2\frac{\varphi}{h}}e^{2\frac{\varphi}{h}}\psi_{j}^{2}\,dt.$

By the monotonicity of $\varphi$ in $|x|>a$ and applying bound (3) we get

$\displaystyle\psi_{j}^{2}(x)$
$\displaystyle\leq\frac{1}{2b}e^{-2\frac{\varphi(x-b)}{h}}e^{\frac{2\varphi(a)}{h}}C.$

Let $M=\max_{a<|x|<R}\varphi^{\prime}(x)$. Then using
$-\varphi(x-b)\leq-\varphi(x)+bM$ we get

$\displaystyle\psi_{j}^{2}(x)$
$\displaystyle\leq\frac{C}{2b}e^{-2\frac{\varphi(x)}{h}}e^{2\frac{bM}{h}}e^{\frac{2\varphi(a)}{h}}.$

By our choice of $a$ which guarantees $\delta\varphi(x)-\varphi(a)\geq k>0$,
we get

$\displaystyle\psi_{j}^{2}(x)$
$\displaystyle\leq\frac{C}{2b}e^{-2(1-\delta)\frac{\varphi(x)}{h}}e^{2\frac{bM-k}{h}}.$

Taking $b<k/M$ and using bound (1) establishes the upper bound of Proposition
1:

$|\psi_{j}(x)|\leq Ce^{-(1-\delta)^{2}\frac{\tilde{\varphi}(x)}{h}}.$

∎

The proof of Proposition 1.(ii) is more difficult. We will first need to
establish a lower bound for $\psi_{j}$ on the boundary of the allowed region
$\\{-\sqrt{E_{j}},\sqrt{E_{j}}\\}$. To achieve this we need a fact about the
$L^{2}(\mathbb{R})$ convergence of rescaled eigenfunctions of $H$.

######  Lemma 2.

Let $\tilde{H}=-\frac{d^{2}}{dx^{2}}+x^{2}+h^{-1}\gamma(\sqrt{h}x)$ be the
rescaled perturbed semiclassical harmonic oscillator, and
$\tilde{\psi}_{j}(x)$ be the eigenfunctions of $\tilde{H}$. Let
$\tilde{H}_{0}=-\frac{d^{2}}{dx^{2}}+x^{2}$ be the harmonic oscillator, and
let $\tilde{\kappa}_{j}$ denote its eigenfunctions. Then
$\tilde{\psi}_{j}(x)\to\tilde{\kappa}_{j}(x)$ in $L^{2}(\mathbb{R})$ as $h\to
0$.

###### Proof.

Define $(U\phi)(x):=h^{-\frac{1}{2}}\phi(h^{-\frac{1}{2}}x)$, and notice that
$\tilde{H}=h^{-1}U^{-1}HU$ (and $\tilde{H}_{0}=h^{-1}U^{-1}H_{0}U$). Let
$\tilde{E}_{j}$ denote the eigenvalues of $\tilde{H}$. The rescaling operator
$U$ gives us a relationship between the eigenfunctions and eigenvalues of
$\tilde{H}$ and $H$:

$E_{j}=h\tilde{E}_{j},\quad\psi_{j}=U\tilde{\psi}_{j}.$

From this, and the fact that
$\text{spec}(H)=\\{h,3h,5h,\dots\\}+O(h^{\infty})$, we get that

$\text{spec}(\tilde{H})=\\{1,3,5,\dots\\}+O(h^{\infty}).$

Note by a similar rescaling, we get that
$\text{spec}(\tilde{H}_{0})=\\{1,3,5,7,\dots\\}$.

We define the projections $P_{\tilde{H}}$ and $P_{H_{0}}$ as follows:

$P_{\tilde{H}}=\frac{1}{2\pi i}\oint_{\partial
D(\tilde{E}_{j},\epsilon)}(z-\tilde{H})^{-1}\,dz,\qquad
P_{H_{0}}=\frac{1}{2\pi i}\oint_{\partial
D(2j+1,\epsilon)}(z-H_{0})^{-1}\,dz,$

where $D(z,\epsilon)$ is the disc centered at $z$ of radius $\epsilon$. Note
that these projections are rank one, onto the subspaces spanned by
$\tilde{\psi}_{j}$ and $\tilde{\kappa}_{j}$ respectively, see [BS12]. Now, for
all $\epsilon>0$ and $h>0$ sufficiently small, we have that
$|\tilde{E}_{j-1}-(2j-1)|<\epsilon$, $|\tilde{E}_{j}-(2j+1)|<\epsilon$, and
$|\tilde{E}_{j+1}-(2j+3)|<\epsilon$ so we may write the difference of the
projections under the same integrand:

$P_{\tilde{H}_{0}}-P_{\tilde{H}}=\frac{1}{2\pi i}\oint_{\partial
D(2j+1,\epsilon)}(z-\tilde{H}_{0})^{-1}(\tilde{H}_{0}-\tilde{H})(z-\tilde{H})^{-1}\,dz.$

Now, applying this difference to the eigenfunction $\tilde{\psi}_{j}$ yields:

$\displaystyle\left|\left|(P_{\tilde{H}_{0}}-P_{\tilde{H}})\tilde{\psi}_{j}\right|\right|_{2}$
$\displaystyle\leq\frac{1}{2\pi}\left|\left|\oint_{\partial
D(2j+1,\epsilon)}(z-\tilde{E}_{1}(h))^{-1}(z-H_{0})^{-1}\,dz\right|\right|_{op}\left|\left|(h^{-1}\gamma(\sqrt{h}x))\tilde{\psi}_{j}\right|\right|_{2}.$

Bounding by the distance to the spectrum along the contour, and that
$\gamma\in C_{0}^{\infty}(\mathbb{R}\setminus\\{0\\})$, we get

$\displaystyle\left|\left|(P_{\tilde{H}_{0}}-P_{\tilde{H}})\tilde{\psi}_{j}\right|\right|_{2}$
$\displaystyle\leq
C\left|\left|h^{-1}\tilde{\psi}_{j}\right|\right|_{L^{2}(\text{supp}(\gamma(\sqrt{h}x)))}.$

One can see that $\lim_{h\to
0}\left|\left|h^{-1}\tilde{\psi}_{j}\right|\right|_{L^{2}(\text{supp}(\gamma(\sqrt{h}x)))}=0$
by Proposition 1.(i), and that $\lim_{h\to
0}\inf\text{supp}(\gamma(\sqrt{h}x))=\infty$. Some algebra yields the desired
results:

$1-\langle\tilde{\psi}_{j},\tilde{\kappa}_{j}\rangle^{2}=\left|\left|(P_{\tilde{H}_{0}}-P_{\tilde{H}})\tilde{\psi}_{j}\right|\right|_{2}^{2},$

so $\lim_{h\to 0}\langle\tilde{\psi}_{j},\tilde{\kappa}_{j}\rangle=1$. Finally

$\left|\left|\tilde{\psi}_{j}-\tilde{\kappa}_{j}\right|\right|_{2}^{2}=2(1-\langle\tilde{\psi}_{j},\tilde{\kappa}_{j}\rangle),$

which tends to 0 in the limit as $h\to 0$. ∎

The above lemma tells us that in the semiclassical limit, the $L^{2}$ mass of
the perturbed eigenfunctions are distributed in the same way as the non-
perturbed eigenfunctions. In fact, a stronger result holds, the perturbed
eigenfunctions converge locally uniformly to the non-perturbed eigenfunctions.
This is proven along the way to establishing lemma 4. However, before we can
prove the lemma, we require a bound on the roots of $\tilde{\kappa}_{j}$.

######  Lemma 3.

For each $j\in\mathbb{Z}_{\geq 0}$,
$\max\\{|r|:\tilde{\kappa}_{j}(r)=0\\}\leq\sqrt{2j-2}$.

###### Proof.

Let $A^{*}:=(x-\frac{d}{dx})$ and $A:=(x+\frac{d}{dx})$ be the creation and
annihilation operators associated to $\tilde{H}_{0}$, satisfying
$AA^{*}-\text{Id}=A^{*}A+\text{Id}=\tilde{H}_{0}$. One can see that
$\tilde{\kappa}_{j}=\frac{1}{\sqrt{2^{j}j!}}{A^{*}}^{j}\tilde{\kappa}_{0}$,
where $\tilde{\kappa}_{0}=\pi^{-\frac{1}{4}}e^{-\frac{x^{2}}{2}}$. Moreover,

$\tilde{\kappa}_{j}(x)=\frac{1}{\sqrt{2^{j}j!\sqrt{\pi}}}P_{j}(x)e^{-\frac{x^{2}}{2}},$

where $P_{j}(x)$ are the physicist’s Hermite polynomials. From the creation
operator definition of $\tilde{\kappa}_{j}$, one can see that the $P_{j}$
satisfy the recurrence relation $P_{j+1}(x)=2xP_{j}(x)-\frac{d}{dx}P_{j}(x)$.
An inductive argument shows that $\frac{d}{dx}P_{j}(x)=2jP_{j-1}(x)$, and so
we have the modified recurrence relation $P_{j+1}(x)=2xP_{j}(x)-2jP_{j-1}(x)$.
Define the tridiagonal symmetric matricies

$C_{j}=\begin{pmatrix}0&\sqrt{\frac{1}{2}}&0&0&0&\cdots&0&0\\\
\sqrt{\frac{1}{2}}&0&\sqrt{\frac{2}{2}}&0&0&\cdots&0&0\\\
0&\sqrt{\frac{2}{2}}&0&\sqrt{\frac{3}{2}}&0&\cdots&0&0\\\
\vdots&&&\ddots&&&\vdots&\vdots\\\ 0&0&0&0&0&\cdots&0&\sqrt{\frac{j-1}{2}}\\\
0&0&0&0&0&\cdots&\sqrt{\frac{j-1}{2}}&0\end{pmatrix}.$

Through induction and using the above recurrence relation, one can see that
$P_{j}(x)=2^{j}\text{det}(xI-C_{j})$, and so the roots of $P_{j}(x)$ are
exactly the eigenvalues of $C_{j}$. Applying the Gershgorin circle theorem, we
may bound the eigenvalues of $C_{j}$, hence the roots of $P_{j}$ which are
exactly the roots of $\tilde{\kappa}_{j}$:

$\max\\{|r|:\tilde{\kappa}_{j}(r)=0\\}\leq\sqrt{\frac{j-2}{2}}+\sqrt{\frac{j-1}{2}}\leq\sqrt{2j-2}.$

∎

######  Lemma 4.

For all $h$ sufficiently small,
$|\psi_{j}(\pm\sqrt{E_{j}})|\geq\frac{D_{j}}{\sqrt{h}}\geq 1$.

###### Proof.

We will first get the bound on the rescaled eigenfunction $\tilde{\psi}_{j}$,
then transfer this bound to $\psi_{j}$. Let
$\theta=\tilde{\psi}_{j}-\tilde{\kappa}_{j}$. Lemma 2 establishes that
$\theta\to 0$ in $L^{2}(\mathbb{R})$. Now, let $I\subseteq\mathbb{R}$ be any
bounded interval. Then $\theta^{\prime\prime}\to 0$ in $L^{2}(I)$, since

$\displaystyle\left|\left|\theta^{\prime\prime}\right|\right|_{L^{2}(I)}$
$\displaystyle\leq\left|\left|h^{-1}\gamma(\sqrt{h}x)\tilde{\psi}_{j}(x)\right|\right|_{L^{2}(\mathbb{R})}+\left|\left|x^{2}(\tilde{\psi}_{j}(x)-\tilde{\kappa}_{j}(x))\right|\right|_{L^{2}(I)}+\left|\left|\tilde{E}_{j}\tilde{\psi}_{j}(x)-\tilde{\kappa}_{j}(x)\right|\right|_{L^{2}(\mathbb{R})},$

where the first term vanishes by Proposition 1.(i) and that $\lim_{h\to
0}\inf\text{supp}(\gamma(\sqrt{h}x))=\infty$, the second term vanishes due to
Lemma 2, and the third term vanishes due to lemma 2 and since $\lim_{h\to
0}\tilde{E}_{j}=2j-1$.

Let $\eta\in C_{0}^{\infty}(\mathbb{R})$ and consider the following identity:
$\eta(\theta^{\prime})^{2}+\eta\theta\theta^{\prime\prime}-\frac{1}{2}\theta^{2}\eta^{\prime\prime}=\frac{d}{dx}(\eta\theta\theta^{\prime}-\frac{1}{2}\theta^{2}\eta^{\prime})$.
Integrating yields

$\int\eta(\theta^{\prime})^{2}\,dx=\frac{1}{2}\int\eta^{\prime\prime}\theta^{2}\,dx-\int\eta\theta\theta^{\prime\prime}\,dx,$

which tends to 0 as $h\to 0$ since $\theta,\theta^{\prime\prime}\to 0$ in
$L^{2}(I)$, thus $\theta^{\prime}\to 0$ in $L^{2}(I)$. Now let $x\in[b,c]=I$ .
Then, for each $a<b$, by the fundamental theorem of calculus

$|\theta(x)-\theta(a)|\leq||\theta^{\prime}||_{L^{2}([a,c])},$

which tends to zero as $h\to 0$. Since $\lim_{a\to-\infty}\theta(a)=0$, we
must have that $\lim_{h\to 0}||\theta(x)||_{L^{\infty}(I)}=0$, or equivalently
$\tilde{\psi}_{j}\to\tilde{\kappa}_{j}$ in $L^{\infty}(I)$.

Now, we pass to the unscaled semiclassical operator. Working in the allowed
region $C=\left[-\sqrt{E_{j}},\sqrt{E_{j}}\right]$:

$\displaystyle\left|\left|\psi_{j}(x)-h^{-\frac{1}{2}}\tilde{\kappa}_{j}(h^{-\frac{1}{2}}x)\right|\right|_{L^{\infty}(C)}$
$\displaystyle=h^{-\frac{1}{2}}\left|\left|\tilde{\psi}_{j}(x)-\tilde{\kappa}_{j}(x)\right|\right|_{L^{\infty}\left(\left[-\sqrt{\tilde{E}_{j}},\sqrt{\tilde{E}_{j}}\right]\right)}.$

Multiplying the above equation by $\sqrt{h}$, and recalling that
$\sqrt{\tilde{E}_{j}}=\sqrt{2j+1}+O(h^{\infty})$ yields

$\displaystyle\lim_{h\to
0}\left|\left|\sqrt{h}\psi_{j}(x)-\tilde{\kappa}_{j}(h^{-\frac{1}{2}}x)\right|\right|_{L^{\infty}(C)}$
$\displaystyle=0.$

Lemma 3 establishes that $|\tilde{\kappa}_{j}(\pm\sqrt{2j+1})|>0$, and since
$\sqrt{E_{j}}=\sqrt{h(2j+1)}+O(h^{\infty})$ for all $h$ sufficiently small
there is a constant $D_{j}>0$ satisfying,

$|\tilde{\kappa}_{j}(\pm h^{-\frac{1}{2}}\sqrt{E_{j}})|\geq D_{j}.$

Thus by the above limit, for sufficiently small $h$,
$|\psi_{j}(\pm\sqrt{E_{j}})|\geq\frac{D_{j}}{\sqrt{h}}$.

∎

The final lemma that we require gives us a way to transfer the lower bound of
Lemma 4 outward from the boundary of the allowed region
$[-\sqrt{E_{j}},\sqrt{E_{j}}]$.

######  Lemma 5.

Consider the interval $I=[x_{1},x_{2}(1+\epsilon)]$ with $\sqrt{E_{j}}\leq
x_{1}$. Let $v=\sup_{t\in I}\sqrt{V(t)}$. Then

$|\psi_{j}(x_{2})|\geq
e^{-\frac{v}{h}(x_{2}-x_{1})}\left(1-e^{-2\epsilon\frac{v}{h}x_{2}}\right)|\psi_{j}(x_{1})|.$

###### Proof.

Define
$\phi(x)=\left(e^{-\frac{v}{h}(x-x_{1})}-e^{-2\frac{v}{h}(x_{2}(1+\epsilon)-x_{1})}e^{\frac{v}{h}(x-x_{1})}\right)|\psi_{j}(x_{1})|$.
Notice that $\phi(x_{2}(1+\epsilon))=0$, and

$\phi(x_{1})=(1-e^{-2\frac{v}{h}(x_{2}(1+\epsilon)-x_{1})})|\psi_{j}(x_{1})|\leq|\psi_{j}(x_{1})|.$

Thus, on $\partial I$, $\phi\leq|\psi_{j}|$. We claim that on the interior of
$I$, $\phi\leq|\psi_{j}|$. Let $\eta=|\psi_{j}|-\phi$ and define
$I_{0}=\\{x:\eta(x)<0\\}\subseteq I$. Now, since $\eta\geq 0$ on $\partial I$,
$\eta=0$ on $\partial I_{0}$. Let $W(x)=V(x)-E_{j}$ and note by choice of
lower bound on $I$ we have that $W\geq 0$ on $I$. Then on $I_{0}$:

$\frac{d^{2}}{dx^{2}}\eta=W\eta-\left(\left(\frac{v}{h}\right)^{2}-W\right)\phi,$

which is less than zero for all $h$ sufficiently small. Thus, $\eta$ is
concave on $I_{0}$, and so it attains is minimum on $\partial I_{0}$. So
$\eta\geq 0$ on $I_{0}$, and $I_{0}$ is empty, i.e. $\phi\leq|\psi_{j}|$ on
$I$. With this, evaluating $\phi$ at $x_{2}$ yields $|\psi_{j}(x_{2})|\geq
e^{-\frac{v}{h}(x_{2}-x_{1})}(1-e^{-2\epsilon\frac{v}{h}x_{2}})|\psi_{j}(x_{1})|$.
∎

Finally, we are able to prove Proposition 1.(ii), which will involve
propagating the bound from Lemma 4, using Lemma 5.

###### Proof of Proposition 1.(ii).

Let $\sqrt{E_{j}}<|x|<R$, and without loss of generality, assume $x$ is
positive. Let $0=x_{0}<x_{1}=\sqrt{E_{j}}<x_{2}<\cdots<x_{n}=x$ be a partition
of $[0,x]$ such that the upper Darboux sum of $\int_{0}^{x}\sqrt{V(x)}\,dx$
satisfies

$\sum_{i=1}^{n}\sup_{t\in[x_{i-1},x_{i}]}\sqrt{V(t)}|x_{i}-x_{i-1}|\leq\left(1+\frac{\delta}{2}\right)\int_{0}^{x}\sqrt{V(t)}\,dt.$

for some $\delta>0$. Let $\epsilon>0$ be small enough such that if we define
$D_{i}=[x_{i-1},x_{i}(1+\epsilon))]$, and $v_{i}=\sup_{t\in
D_{i}}\sqrt{V(t)}$, we get

$\sum_{i=1}^{n}v_{i}|x_{i}-x_{i-1}|\leq\left(1+\delta\right)\int_{0}^{x}\sqrt{V(t)}\,dt.$
(4)

Next, applying Lemma 5 iteratively on this partition, we get that

$\displaystyle|\psi_{j}(x)|$ $\displaystyle\geq
e^{-\sum_{i=2}^{n}\frac{v_{i}}{h}(x_{i}-x_{i-1})}\prod_{i=2}^{n}\left(1-e^{-2\epsilon\frac{v_{i}}{h}x_{i}}\right)|\psi_{j}(x_{1})|.$

Lemma 4 guarantees (for $h>0$ sufficiently small),
$|\psi_{j}(x_{1})|=|\psi_{j}(\sqrt{E_{j}})|\geq 1$, so we recover the full sum

$\displaystyle|\psi_{j}(x)|$ $\displaystyle\geq
e^{-\sum_{i=1}^{n}\frac{v_{i}}{h}(x_{i}-x_{i-1})}\prod_{i=1}^{n}\left(1-e^{-2\epsilon\frac{v_{i}}{h}x_{i}}\right).$

Since $v_{i}=O(1)$ for $i\neq 1$, and $v_{1}=O(\sqrt{h})$, for all $h$
sufficiently small, we estimate
$\prod_{i=1}^{n}\left(1-e^{-2\epsilon\frac{v_{i}}{h}x_{i}}\right)\geq 2^{-n}$.
Using this along with bound (4), we get the final desired bound:

$\displaystyle|\psi_{j}(x)|$ $\displaystyle\geq
De^{-(1+\delta)^{2}\frac{1}{h}\left|\int_{0}^{x}\sqrt{V}\,dt\right|}.$

To get uniformity in $r<|x|<R$, extend the partition. ∎

##  3 Proof of the Main Result

To prove Theorem 1 we will construct the desired potential functions
$V^{\pm}$. Let $\alpha,\beta\in C_{0}^{\infty}(\mathbb{R})$ with
$\text{supp}(\alpha)\subseteq(1,2)$ and $\text{supp}(\beta)\subseteq(3,4)$ and
set $V^{\pm}(x)=x^{2}+\alpha(x)+\beta(\pm x)$. To recover the sub
$O(h^{\infty})$ differences between the eigenvalues of the associated
operators $H^{\pm}=-h^{2}\frac{d^{2}}{dx^{2}}+x^{2}+\alpha(x)+\beta(\pm x)$,
we will apply a variation in $\beta(\pm x)$. Define the following family of
operators

$\displaystyle H_{t}^{\pm}$
$\displaystyle=-h^{2}\frac{d^{2}}{dx^{2}}+x^{2}+\alpha(x)+t\beta(\pm x),$

and notice that $H_{1}^{\pm}=H^{\pm}$. Denote the corresponding eigenfunctions
and eigenvectors $\psi_{j}^{\pm}(t,x),\,E_{j}^{\pm}(t)$.

###### Proof of Theorem 1.

The following equation, known as Hadamard’s variational formula, will be
useful

$\frac{d}{dt}E_{j}^{\pm}(t)=\int\beta(\pm x)(\psi_{j}^{\pm}(t,x))^{2}\,dx.$

The proof of this formula for the one dimensional case is elementary, see
[GH12]. Now, with Hadamard’s variational formula and the fundamental theorem
of calculus, we have

$E_{j}^{\pm}(t)-E_{j}^{\pm}(0)=\int_{0}^{t}\int\beta(\pm
x)(\psi_{j}^{\pm}(s,x))^{2}\,dx\,ds.$

Using Proposition 1 and the above equation, and that
$E_{j}^{\pm}(1)=E_{j}^{\pm}$ gives us the following bounds:

$E_{j}^{\pm}-E_{j}^{\pm}(0)\geq\int_{0}^{1}\int\beta(\pm
x)De^{-\frac{(1+\delta)^{2}}{h}\left|\int_{0}^{x}\sqrt{a^{2}+\alpha(a)+s\beta(\pm
a)}\,da\right|}\,dx\,ds$ (5)
$E_{j}^{\pm}-E_{j}^{\pm}(0)\leq\int_{0}^{1}\int\beta(\pm
x)Ce^{-\frac{(1-\delta)^{2}}{h}\left|\int_{0}^{x}\sqrt{a^{2}+\alpha(a)+s\beta(\pm
a)}\,da\right|}\,dx\,ds$ (6)

Notice that the minus version of lower bound (5) avoids integrating in the
exponent through the support of $\alpha$. This allows us to express the
difference of the minus version of (5) with the plus version of (6) as

$\displaystyle E_{j}^{-}-E_{j}^{+}$
$\displaystyle\geq\int_{0}^{1}\int\beta(x)\left(De^{-\frac{(1+\delta)^{2}}{h}\int_{0}^{x}\sqrt{a^{2}+s\beta(a)}\,da}-Ce^{-\frac{(1-\delta)^{2}}{h}\int_{0}^{x}\sqrt{a^{2}+\alpha(a)+s\beta(a)}\,da}\right)\,dxds$

Factoring, recalling that $\beta$ is supported in $(3,4)$, and recognizing
that the integrals in the exponent are monotonic in $x$ gives

$\displaystyle E_{j}^{-}-E_{j}^{+}$
$\displaystyle\geq\int_{0}^{1}e^{-\frac{(1+\delta)^{2}}{h}\int_{0}^{4}\sqrt{a^{2}+s\beta(a)}\,da}\int\beta(x)\left(D-Ce^{-\frac{1}{h}\int_{0}^{x}(1-\delta)^{2}\sqrt{a^{2}+\alpha(a)+s\beta(a)}-(1+\delta)^{2}\sqrt{a^{2}+s\beta(a)}\,da}\right)\,dxds.$

The bounds of Proposition 1 hold for any choice of $\delta$, so we may take
$\delta$ small enough to satisfy for each $x\in(3,4)$

$\int_{0}^{x}(1-\delta)^{2}\sqrt{a^{2}+\alpha(a)+s\beta(a)}-(1+\delta)^{2}\sqrt{a^{2}+s\beta(a)}\,da>0.$

With this choice of delta, and again using monotonicity of the integral in the
exponent, we have

$\displaystyle E_{j}^{-}-E_{j}^{+}$ $\displaystyle\geq
e^{-\frac{(1+\delta)^{2}}{h}\int_{0}^{4}\sqrt{a^{2}+\beta(a)}\,da}\left(D-Ce^{-\frac{1}{h}\int_{0}^{3}(1-\delta)^{2}\sqrt{a^{2}+\alpha(a)}-(1+\delta)^{2}\sqrt{a^{2}}\,da}\right)\int\beta(x)\,dx.$

For all $h$ sufficiently small, this then reduces to the inequality
$E_{j}^{-}-E_{j}^{+}\geq De^{\frac{-d}{h}}$, for some positive constants
$c,C>0$. To prove the upper bound $E_{j}^{-}-E_{j}^{+}\leq Ce^{\frac{-c}{h}}$,
a similar argument is used where we instead take the difference of the minus
version of (6) with the positive version of (5). ∎

## References

  * [BS12] Feliks Aleksandrovich Berezin and Mikhail Shubin.  The Schrödinger Equation, volume 66.  Springer Science & Business Media, 2012.
  * [CdV05] Yves Colin de Verdiere.  Bohr-sommerfeld rules to all orders.  Ann. Henri Poincaré, 6(5):925–936, 2005.
  * [CdV11] Yves Colin de Verdière.  A semi-classical inverse problem ii: reconstruction of the potential.  In Geometric aspects of analysis and mechanics, pages 97–119. Springer, 2011.
  * [DHV11] Kiril Datchev, Hamid Hezari, and Ivan Ventura.  Spectral uniqueness of radial semiclassical Schrödinger operators.  Math. Res. Lett., 18(3):521–529, 2011.
  * [FK05] Stephen A Fulling and Peter Kuchment.  Coincidence of length spectra does not imply isospectrality.  Inverse problems, 21(4):1391, 2005.
  * [GH12] Victor Guillemin and Hamid Hezari.  A fulling–kuchment theorem for the 1d harmonic oscillator.  Inverse problems, 28(4):045009, 2012.
  * [GU05] Victor Guillemin and Alejandro Uribe.  Some inverse spectral results for semi-classical schrödinger operators.  Mathematical Research Letters, 14, 10 2005.
  * [Hez09] Hamid Hezari.  Inverse spectral problems for schrödinger operators.  Communications in Mathematical Physics, 288(3):1061–1088, 2009\.
  * [Sim84] Barry Simon.  Semiclassical analysis of low lying eigenvalues, ii. tunneling.  Annals of Mathematics, pages 89–118, 1984.
  * [Sjö92] Johannes Sjöstrand.  Semi-excited states in nondegenerate potential wells.  Asymptotic analysis, 6(1):29–43, 1992.
  * [Zel04] Steve Zelditch.  The inverse spectral problem.  Surveys in differential geometry, 9(1):401–467, 2004.


LSBC120pt

# The bishop and knight checkmate on a large chessboard

Johan Wästlund

###### Abstract

In 1983 the chess periodical EG published a summary of a letter from Julius
Telesin outlining how a king, a bishop and a knight can checkmate a lonely
king on an arbitrarily large chessboard. The Telesin checkmating procedure
doesn’t seem to be widely known, and the published account left out a number
of details. We describe the method precisely and show that it works against
every defense. We also discuss the open question of the asymptotics of the
largest distance to mate on a large square board.

##  1 Introduction

The purpose of this article is to demonstrate rigorously a fact that doesn’t
seem to be widely known and that hasn’t been explained in much detail in the
literature: A king, a bishop and a knight can force checkmate against a lonely
king on an arbitrarily large chessboard!

For this to be possible, the board must have a corner of the color that the
bishop can reach. But provided this is the case, and excluding situations
where there is an immediate problem of stalemate or a piece being lost,
checkmate can always be achieved. Naturally the 50-move rule must be waived
throughout this discussion.

The checkmating procedure was described by Julius Telesin in a letter
summarized in the journal EG (a journal devoted to studies and theoretical
aspects of the endgame) in 1983 [6]. There is no doubt about the correctness,
but in some aspects the exposition is not very clear.

More recently the question has been discussed at Chess Stack Exchange [7]
without a conclusive answer, and apparently without anyone being aware of
Telesin’s solution. In the thread there is an answer from 2015 by Dan
Stronger, who gives data from a tablebase program showing that mate can be
achieved in at most 93 moves on a 16 by 16 board. He says that (presumably by
playing around with the program) he has become convinced that checkmate is
possible on arbitrarily large boards, but only briefly describes the strategy.
Stronger’s longest winning line is cited in the Wikipedia article [9] without
mention of Telesin’s earlier result.

Václav Kotěšovec pushed the tablebase calculations to board size $24\times 24$
in 2017, and gives the longest distance to mate in [4, p. 349], see also [5],
linking to the Telesin article. Another interesting paper is [3], which
discusses the extraction of useful strategies from tablebases for the standard
board. The KBN vs K checkmate on large boards has also been briefly discussed
on MatPlus.Net [8].

Telesin’s solution (as summarized in [6], presumably by an editor) is somewhat
difficult to verify. In some cases it is explained that, for instance, the
knight “approaches”, but before understanding the overall plan in enough
detail, it’s not entirely clear where exactly the knight wants to go. Neither
is it obvious that all defenses have been considered. I therefore want to
share my thoughts on the KBN vs K checkmate, making some points more precise.

It may be argued that to the ordinary chess player, mating procedures on
boards larger than 8 by 8 are of limited interest. But the reason that many
players find the KBN vs K checkmate rather tricky might be precisely that one
doesn’t see how it would work in general. It appears ad-hoc, and therefore
becomes hard to understand and remember. Some interesting similar remarks are
made in [1, p.168].

##  2 Background

We first summarize the standard chess knowledge about the KBN vs K endgame. We
assume throughout that White has the two pieces and Black the lonely king. To
force checkmate, the black king has to be driven into a corner that the white
bishop can reach (otherwise there will be a stalemate problem).

defgh$1$$2$$3$$4$$5$defgh$1$$2$$3$$4$$5$ Figure 1: Left: Once the black king
is constrained to the immediate vicinity of a corner that the bishop can
reach, checkmating is relatively straightforward. Right: The white king
approaches, and Black is finally checkmated either in the corner or on a
square next to the corner.

Once we obtain a situation as in Figure 1, the finish is relatively easy. Here
the white pieces constrain the black king to a set of six squares near the
corner. White can now maneuver their king to e2, then (in some order) the
bishop via h3 to f1 and the king to f2. Finally the knight reroutes via g5 for
instance, and checkmate is given by Bg2 and Nf3 in the correct order, giving
check in the penultimate move.

On the standard 8 by 8 board, once White has centralized their own king, the
black king is already fairly close to the edge of the board. Knowing that only
two of the corners allow checkmate, the defending player often heads for one
of the “safe” corners. The tricky part is then to force the black king over to
another corner. This is what seems ad-hoc and leaves the impression that the
method just barely works.

For instance, the textbook instructions generated by [3] explicitly state that
step 1 of the procedure is to drive the black king to the edge of the board,
and step 2 to force the king to the appropriate corner. As we shall see, step
2 is achievable, but on a large board it seems we cannot drive the black king
from a safe corner to another corner along the edge, and therefore it’s not
clear whether step 1 brings us any closer to achieving step 2.

abcdefg$1$$2$$3$$4$$5$$6$ Figure 2: White has to force the black king out of a
“safe” corner.

Suppose as in Figure 2 that we have a light-squared bishop and want to drive
the king from a1 to h1, or in case of a larger board, to a more distant light
square corner on the first rank. To dig the black king out of the corner, the
knight has to control a1. And if we get the black king to c1, some white piece
must control b1 to stop Black from going back. Whether we play Na3, Ba2, or
Ka2, some white piece will be misplaced.

On an 8 by 8 board, we can still shove the king more or less directly from one
corner to another, since the opposite edge stops the black king from running
further: After 1. Nc2+ Kb1 2. Bb3 (waiting) Kc1 3. Ba2 Kd1 4. Nd4 Ke1 5. Kd3
Kf2, we have 6. Ne2! (this move has famously been missed even by strong
grandmasters). In combination with Be6 (stopping by on b3 if Black tries to go
back), this prevents the black king from escaping.

But on a much larger board, this would not work: 6. - Kg2 7. Be6 Kh2 followed
by 8. - Ki3, and Black runs towards the north-east corner.

As was pointed out by a reviewer, there is another method of pushing the black
king to the right corner known as the Delétang triangles method [2]. This
method consists in restricting the black king diagonal by diagonal to ever
smaller triangles, which is also how the Telesin method works. But it requires
the knight to hold part of the barrier and the white king to move into the
triangle, and this doesn’t seem to generalize to a large board.

##  3 Telesin’s method

It seems that on a large board, White cannot force the black king from one
corner to another along the edge of the board. But Telesin shows, modifying
the triangles method, that we can constrain the black king not along the edge,
but diagonal by diagonal.

We assume for simplicity that the board is square with an even number of
squares along each side, although everything should work in the same way on
any rectangular board with at least one even side. Suppose we start from the
position of Figure 2 but on a large board. Still numbering from a dark a1
corner, Telesin [6] instead suggests (after 1. Nc2+ Kb1 2. Bb3 Kc1 3. Ba2 Kd1
4. Nd4 Ke1) the line 5. Bd5 Kf2 6. Nf5 Kg1 7. Kd4 Kh2 8. Bf3, leading to the
position of Figure 3.

14pt14pt14pt14ptabcdef$\dots$$1$$2$$3$$4$$5$$6$$\vdots$ Figure 3: Telesin’s
method on a large chess board.

It seems we have let the black king out, but as Telesin shows, White can
restrict Black to the south-east side of a diagonal barrier as indicated.
Black can run almost all the way up to the north-east corner, but must at some
point turn back and allow White to decrease the triangular zone.

### The simple enclosure

The basic idea of the checkmating procedure is that the white king and bishop
can keep the black king constantly on one side of a certain diagonal. We begin
by demonstrating how this works, formalizing Telesin’s concept of “zone
restriction” [6], and later turn to how to make progress and actually drive
the black king towards a corner. Suppose that the kings and the white bishop
are in the position of Figure 4.

Figure 4: The simple enclosure

In this situation, the white king and bishop can keep the black king on the
south-east side of the indicated barrier. This can be done without moving the
bishop, and without the help of the knight. White just has to follow Black
along the barrier, making sure to always put the king on a file at least as
far to the west as Black’s king, and on a rank at least as far north, while
also being in time to defend the bishop if Black attacks it. In order for this
to work, the knight should be on the north-west side of the barrier so as not
to allow Black to gain time by attacking it, and it should also be a couple of
squares away from the barrier in order not to stand in the way of the king or
bishop.

### The dented enclosure

In Figure 4, as soon as Black makes a move, White can put their king on a dark
square right next to the bishop and on the opposite side to the black king
with respect to the bishop’s north-west to south-east diagonal. White can then
maintain what we will call a _dented enclosure_ (Figure 5), even further
restricting Black’s movement. Here we number the ranks and files without any
assumptions on how far away the corners are.

abcdefgh$2$$3$$4$$5$$6$$7$$2$$3$$4$$5$$6$$7$ Figure 5: The dented enclosure.

By applying a certain strategy, White can make sure that Black never crosses
this “dented” barrier. If Black tries to escape to the north-east, White
follows along with the king just as when defending the simple enclosure. If on
the other hand Black goes via f4 to e3, White plays Kc4 creating what Telesin
calls a “valve” (more about that in a moment), and unless Black goes back,
White can install the bishop on e4, reducing the problem of containing the
black king to a smaller simple enclosure.

An important point is that we can’t allow Black to obtain a repetition of
moves. If we are constantly kept busy, we can’t make progress towards
checkmate. If in Figure 5 Black plays 1. - Kf4, we can’t answer 2. Kd4,
because on 2. - Kf5 we would have to go back with 3. Kc5 to be in time if
Black tries Kf5-f6-e7. Since Black could then repeat with 3. - Kf4, it’s clear
that we can’t make progress by meeting Kf4 with Kd4. Similarly we can’t meet
Kf4 with a bishop move, since the bishop too would have to return on the
following move. Therefore on 1. Kf4, the king and bishop must wait, and the
knight make a move towards its destination (which we discuss in a moment).

After 1. - Kf4 2. Nxx Ke3, the only response that defends the barrier is 3.
Kc4. And if Black then continues 3. - Kd2, we have to play 4. Be4 or the black
king will get out. We see how several of the moves, including the “valve”, are
forced by the outline of the barrier. These moves are given in [6], but become
harder to understand without an exact description of the zone to which the
black king is restricted.

If we get to play 4. Be4, we have established a simple enclosure which is
smaller than the dented enclosure we just had. And as soon as we have
decreased the enclosure, we have made progress. Notice also that if Black runs
to the north-east and White follows, then as soon as Black makes a move that
doesn’t go north, White can establish a smaller dented enclosure by moving the
bishop along the barrier, to e6 or higher.

### The knight’s action if Black waits

We have seen that a simple enclosure always leads to a smaller dented
enclosure. Moreover, from a dented enclosure, as soon as the black king tries
to run anywhere, the white king and bishop can decrease the enclosure one way
or another. It’s only if Black lingers near the white king and bishop that the
knight will have to take action. Therefore it doesn’t matter how far away the
knight is.

abcdefgh$2$$3$$4$$5$$6$$7$$2$$3$$4$$5$$6$$7$ Figure 6: The knight takes action
if the black king waits around the white king and bishop.

In the situation of Figure 6, assuming that Black waits on the 4:th and 5:th
ranks, for instance in the three squares indicated, White will maneuver the
knight to c4. Notice that the knight on c4 doesn’t get in the way of the king
or bishop, and that the enclosure can be maintained whatever Black does. Since
the black king now can’t go to e3, the white king doesn’t need access to c4.

Suppose first that White gets to play 1. Nc4 while Black is on the 4:th rank
or higher. Black can now no longer move to the 3:rd rank without allowing
White to shift the bishop to e4 and establish a smaller simple enclosure.

If Black stays on ranks 4–6, White plays 2. Kd6 and tries to decrease the
enclosure by getting the bishop either to e4 or to e6. The only way Black can
try to prevent this is by playing 2. - Kf4/f5. But then White can force Black
away and get the bishop to e4 anyway, for instance 2. - Kf5 3. Ke7 Kg4 4. Kf6
Kf4 5. Ke6, forcing Black to the g-file, and 6. Be4 establishing the simple
enclosure.

There is one slight problem that needs to be treated separately: If just
before the knight is about to go to c4, Black plays 1. - Ke3, then White is
forced to play 2. Kc4. If Black now returns with 2. - Kf4, White can’t play 3.
Nc4 because the king is in the way (if Black instead continues south, White
shifts the bishop to e4, making progress without the knight). But White then
triangulates with 3. Kd4 Kf5/g5 (otherwise the bishop gets to e4) 4. Kc5, and
unless Black goes to the 6:th rank, 5. Nc4.

In some situations, other knight maneuvers can be faster and just as effective
(the knight just needs to control some key dark squares), but going to c4 (or
the corresponding square relative to the bishop) always works.

### Conclusion: White can force checkmate

Playing White from any position, and provided we can avoid immediate stalemate
or losing a piece, we can centralize our pieces and establish an enclosure in
some direction. Basically we just put our bishop next to our king and notice
where the black king is relative to the bishop’s diagonals. Possibly we also
have to let the knight run away from the black king. Then, using the maneuvers
already described, we can make sure to eventually decrease the enclosure.
There is only one way that this process can end: We finally constrain the king
to the immediate vicinity of a mating corner, and the standard technique
applies.

##  4 Number of moves required for checkmate

In endgames like rook or two bishops against a bare king, checkmate can be
forced in time (number of moves) proportional to the side of the board, let’s
call it $n$. The Telesin checkmating procedure for KBN vs K on the other hand
requires time proportional to the area $n^{2}$ of the board. When White
reduces to a simple enclosure, Black can choose in which direction to run, and
can therefore run back and forth along the barrier, eventually visiting a
positive proportion of the squares within the original enclosure.

Telesin states that a 1000 by 1000 board allows checkmate in at most 1,028,000
moves, indicating that his calculations yield a main term of $n^{2}$ plus some
overhead. In [5] it is conjectured that from the worst possible starting
configuration that allows a forced win at all, the checkmate requires about
$n^{2}/6+3n$ moves for even $n$, and $n^{2}/6+2n$ for odd $n$. The conjecture
seems to be due to Kotěšovec, but it’s not entirely clear from [4, 5] how
strong the evidence is.

### Forcing along an edge anyway?

We believe that Telesin’s method is generally necessary even if we have forced
Black to a safe corner, since it seems we cannot get the black king out of
such a corner without also letting it escape almost to the opposite safe
corner. If this is the case, then a position with the black king trapped in a
safe corner is, for large $n$, essentially the worst case, since the king is
at maximum distance from a mating corner and requires the maximum number of
rounds of zone restriction.

But this is still not entirely clear. As if just to make things even more
confusing, it _is_ possible to force the black king along an edge of the
board, if only we can get it away from the safe corner without misplacing our
pieces! This is another (minor) point that makes the Telesin article a bit
hard to understand: In the main line the black king runs immediately to an
edge of the board, from where it would have been possible (in [6, §7]) to
shove it directly to a mating corner, checkmating faster than in the line
given.

Suppose we arrive at the position below, with Black to move and assuming that
White tries to drive the black king to a distant light square corner to the
right.

abcdefghi$1$$2$$3$$4$$5$$6$$1$$2$$3$$4$$5$$6$

White can then force the same relative position of the pieces to occur again,
but with the entire position shifted two steps to the right, so that all
pieces are on the g-file. Moreover, this will happen in at most 12 moves.

If Black starts 1. - Kf1, White responds 2. Nf4. If on the other hand Black
chooses 1. - Kd1, we can play 2. Bb3+ (this is what is not possible two files
further left, when digging Black out of the corner!) Ke1 3. Nf4 Kf1 4. Be6, so
that in any case we reach the following position:

abcdefghi$1$$2$$3$$4$$5$$6$$1$$2$$3$$4$$5$$6$

Now on 4. - Kg1 White plays 5. Kf3 leading to the position below, while if
Black tries to go back, we have 4. Ke1 5. Nd3+! Here on 5. - Kd1 Black gets
checkmated immediately with 6. Bb3# which would not be possible two files
further left. Therefore 5. - Kf1 is forced, and after 6. Kf3 Kg1 7. Nf4 we
reach the same position:

abcdefghi$1$$2$$3$$4$$5$$6$$1$$2$$3$$4$$5$$6$

Next White plays 8. Ng2 no matter where Black goes. Since the bishop already
controls h3 and i2, Black cannot get further than to h2. White can then
maneuver the king to g3 and the bishop to g6 while making sure Black never
escapes to e2 or i2. No later than after White’s move 13 we arrive at:

abcdefghi$1$$2$$3$$4$$5$$6$$1$$2$$3$$4$$5$$6$

The pattern then repeats, and when we approach the south-east corner, the
standard checkmating method applies. Therefore as soon as we reach a position
with the black king trapped in a controlled way along an edge and not too
close to a safe corner, we can obtain checkmate in linear time.

We have to leave the question of the “time complexity” of the KBN vs K
checkmate open. An interesting challenge would be to describe a strategy for
Black that survives from some “stable” position for $c\cdot n^{2}$ moves, for
some fixed $c>0$. But perhaps White can restrict the black king in some
slightly more efficient way, nudging it towards a mating corner at a small but
constant speed, so that checkmate is forced in linear time after all.

## References

  * [1] Alexandre V. Borovik. Mathematics under the microscope - Notes on Cognitive Aspects of Mathematical Practice, AMS 2010.
  * [2] Daniel Delétang. _Mat avec le fou et le cavalier_ , La Stratégie 56(2):25–32 (1923).
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