# Gromov’s torii are optimal

Anton Petrunin

###### Abstract

We give an optimal bound on normal curvatures of immersed $n$-torus in a
Euclidean ball of large dimension.

##  1 Introduction

Let us denote by $\mathbb{B}^{q}$ the unit ball in $\mathbb{R}^{q}$ centered
at the origin. Further, $\mathbb{T}^{n}$ will denote the $n$-dimensional torus
— the smooth manifold diffeomorphic to the product of $n$ circles.

This note is inspired by examples of embeddings
$\mathbb{T}^{n}\hookrightarrow\mathbb{B}^{q}$ for large $q$ with constant
normal curvatures $K_{n}=\sqrt{3{\hskip 0.5pt\cdot\hskip 0.5pt}n/(n+2)}$. In
other words, any geodesic in the torus has constant curvature $K_{n}$ as a
curve in $\mathbb{R}^{q}$. These examples were found by Michael Gromov among
geodesic subtorii in Clifford’s torii [5, 2.A], [4, 1.1.A.]. In particular,
Gromov’s torii have flat induced metric. (Recall that Clifford’s torus is a
product of $m$ circles of radius ${1}/{\sqrt{m}}$ in $\mathbb{R}^{2{\hskip
0.5pt\cdot\hskip 0.5pt}m}$; its normal curvatures lie in the range
$[1,\sqrt{m}]$.)

Gromov’s examples lead to the following surprising facts: any closed smooth
manifold $L$ admits a smooth embedding into $\mathbb{B}^{q}$ for large $q$
with normal curvatures less than $\sqrt{3}$; moreover, the induced Riemannian
metric on $L$ can be chosen to be proportional to any given metric $g$; see
[5, 1.D] and [4, 1.1.C].

The next theorem implies that Gromov’s torii have the best upper bound on
normal curvatures; in particular, the $\sqrt{3}$-bound is optimal.

1.1. Theorem. Suppose $\mathbb{T}^{n}$ is smoothly immersed in
$\mathbb{B}^{q}$. Then its maximal normal curvature is at least

$\sqrt{3{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{n}{n+2}}.$

To make the statement more exact, we need one more notation. Assume that $L$
is a smooth $n$-dimensional manifold immersed in $\mathbb{R}^{q}$; we will
always assume that $L$ is equipped with induced Riemannian metric. Let us
denote by $\mathrm{T}_{x}$ and $\mathrm{N}_{x}$ the tangent and normal spaces
of $L$ at $x$.

Recall that _second fundamental form_ $\mathrm{I}\\!\mathrm{I}$ at $x$ is a
symmetric quadratic form on $\mathrm{T}_{x}$ with values in $\mathrm{N}_{x}$.
It is uniquely defined by the identity
$\mathrm{I}\\!\mathrm{I}(\text{\sc{v}},\text{\sc{v}})\equiv\gamma^{\prime\prime}_{\text{\sc{v}}}(0)$,
where $\text{\sc{v}}\in\mathrm{T}_{x}$ and $\gamma_{\text{\sc{v}}}$ is an
$L$-geodesic that starts at $x$ with initial velocity vector v.

Given $x\in L$, denote by $\textsl{Ж}(x)$ the average value of
$|\mathrm{I}\\!\mathrm{I}(\text{\sc{u}},\text{\sc{u}})|^{2}$ for
$\text{\sc{u}}\in\mathrm{T}_{x}$ such that $|\text{\sc{u}}|=1$. Since
$K(\text{\sc{u}})=|\mathrm{I}\\!\mathrm{I}(\text{\sc{u}},\text{\sc{u}})|$ is
the normal curvature in the direction u, we have that $\textsl{Ж}(x)$ is the
average value $K^{2}(\text{\sc{u}})$. (The Cyrillic zhe Ж is used since it
resembles squared $K$.)

1.2. Theorem. Suppose $\mathbb{T}^{n}$ is smoothly immersed in
$\mathbb{B}^{q}$. Let us equip $\mathbb{T}^{n}$ with the induced Riemannian
metric; so we can take average values with respect to the induced volume.

  * (a)

If $n=2$, then the average value of Ж is at least $\tfrac{3}{2}$.

  * (b)

If the metric on $\mathbb{T}^{n}$ is flat, then the average value of Ж is at
least $3{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{n}{n+2}$.

  * (c)

If the image of $\mathbb{T}^{n}$ lies in $\partial\mathbb{B}^{q}$, then
$\textsl{Ж}\geqslant 3{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{n}{n+2}$ at some
point of $\mathbb{T}^{n}$.

  * (d)

If $n\leqslant 4$, then $\textsl{Ж}\geqslant 3{\hskip 0.5pt\cdot\hskip
0.5pt}\tfrac{n}{n+2}$ at some point of $\mathbb{T}^{n}$.

  * (e)

If the normal curvatures of $\mathbb{T}^{n}$ do not exceed $2$, then
$\textsl{Ж}\geqslant 3{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{n}{n+2}$ at some
point of $\mathbb{T}^{n}$.

Note that part 1 is a stronger version of 1. The remaining statements 1–1 are
stronger versions of 1 in some partial cases. All this follows since the
normal curvature in some direction at $x$ is at least $\sqrt{\textsl{Ж}(x)}$.

All proofs use our version of the Gauss formula; see below. The proofs of 1–1
use in addition that torus does not admit a metric with positive scalar
curvature [7, Corollary A].

1.3. Open question. Is it true that for any smooth immersion
$\mathbb{T}^{n}\looparrowright\mathbb{B}^{q}$ the inequality
$\textsl{Ж}\geqslant 3{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{n}{n+2}$ holds at
some point?

##  2 Gauss formula

Recall that $L$ is a smooth $n$-dimensional manifold immersed in
$\mathbb{R}^{q}$. Given $p\in L$, denote by $\mathrm{Sc}(p)$ and $H(p)$ the
scalar curvature, the mean curvature vector at $p$.

The following version of the Gauss formula plays a central role in all proofs;
it is used instead of the formula in [6, 5.B].

2.1. Gauss formula. The following identity

$\mathrm{Sc}=\tfrac{3}{2}{\hskip 0.5pt\cdot\hskip
0.5pt}|H|^{2}-\tfrac{n{\hskip 0.5pt\cdot\hskip 0.5pt}(n+2)}{2}{\hskip
0.5pt\cdot\hskip 0.5pt}\textsl{Ж}$

holds for any smooth $n$-dimensional immersed manifold in a Euclidean space.

Proof. Assume $\operatorname{\rm codim}L=1$. Choose $p\in L$; denote by
$k_{1},\dots,k_{n}$ the principal curvatures of $L$ at $p$. Note that

$|H|^{2}=\sum_{i}k_{i}^{2}+2{\hskip 0.5pt\cdot\hskip
0.5pt}\sum_{i<j}k_{i}{\hskip 0.5pt\cdot\hskip 0.5pt}k_{j}.$

Further,

$n{\hskip 0.5pt\cdot\hskip 0.5pt}(n+2){\hskip 0.5pt\cdot\hskip
0.5pt}\textsl{Ж}=3{\hskip 0.5pt\cdot\hskip 0.5pt}\sum_{i}k_{i}^{2}+2{\hskip
0.5pt\cdot\hskip 0.5pt}\sum_{i<j}k_{i}{\hskip 0.5pt\cdot\hskip 0.5pt}k_{j}.$

The last identity follows since Ж is the average value of
$\left(\sum_{i}k_{i}{\hskip 0.5pt\cdot\hskip 0.5pt}x_{i}^{2}\right)^{2}$ on
the unit sphere $\mathbb{S}^{n-1}\subset\mathbb{R}^{n}=\mathrm{T}_{p}$; here
$(x_{1},\dots,x_{n})$ are the standard coordinates in $\mathbb{R}^{n}$. One
has to take into account that the following functions have unit average
values: $\tfrac{1}{3}{\hskip 0.5pt\cdot\hskip 0.5pt}n{\hskip 0.5pt\cdot\hskip
0.5pt}(n+2){\hskip 0.5pt\cdot\hskip 0.5pt}x_{i}^{4}$ and $n{\hskip
0.5pt\cdot\hskip 0.5pt}(n+2){\hskip 0.5pt\cdot\hskip 0.5pt}x_{i}^{2}{\hskip
0.5pt\cdot\hskip 0.5pt}x_{j}^{2}$ for $i\neq j$.

By the standard Gauss formula,

$\mathrm{Sc}=2{\hskip 0.5pt\cdot\hskip 0.5pt}\sum_{i<j}k_{i}{\hskip
0.5pt\cdot\hskip 0.5pt}k_{j}.$

It remains to rewrite the right-hand side using the expressions for $|H|^{2}$
and Ж.

If $\operatorname{\rm codim}L=k>1$, then the second fundamental form can be
presented as a direct sum of $k$ real-valued quadratic forms
$\mathrm{I}\\!\mathrm{I}_{1}\oplus\dots\oplus\mathrm{I}\\!\mathrm{I}_{k}$;
that is,

$\mathrm{I}\\!\mathrm{I}=e_{1}{\hskip 0.5pt\cdot\hskip
0.5pt}\mathrm{I}\\!\mathrm{I}_{1}+\dots+e_{k}{\hskip 0.5pt\cdot\hskip
0.5pt}\mathrm{I}\\!\mathrm{I}_{k},$

where $e_{1},\dots,e_{k}$ is an orthonormal basis of $\mathrm{N}_{p}$. Denote
by $\mathrm{Sc}_{i}$, $H_{i}$, and $\textsl{Ж}_{i}$ the values associated with
$\mathrm{I}\\!\mathrm{I}_{i}$. From above, we get

$\mathrm{Sc}_{i}=\tfrac{3}{2}{\hskip 0.5pt\cdot\hskip
0.5pt}|H_{i}|^{2}-\tfrac{n{\hskip 0.5pt\cdot\hskip 0.5pt}(n+2)}{2}{\hskip
0.5pt\cdot\hskip 0.5pt}\textsl{Ж}_{i}$

for each $i$.

Note that

$\mathrm{Sc}=\sum_{i}\mathrm{Sc}_{i},\qquad|H^{2}|=\sum_{i}|H_{i}|^{2},\quad\text{and}\quad\textsl{Ж}=\sum_{i}\textsl{Ж}_{i}.$

Hence the general case follows. ∎

Remark. A more direct proof of this formula can be obtained using the so-
called _extrinsic curvature tensor_ which is defined by
$\Phi(\text{\sc{x}},\text{\sc{y}},\text{\sc{v}},\text{\sc{w}})\mathrel{{:}{=}}\langle\mathrm{I}\\!\mathrm{I}(\text{\sc{x}},\text{\sc{y}}),\mathrm{I}\\!\mathrm{I}(\text{\sc{v}},\text{\sc{w}})\rangle$;
the necessary properties of this tensor are discussed in [8]. As a bonus, one
gets explicit expression for the second fundamental forms of all Gromov’s
torii.

##  3 Special cases

The following statement appears in the book of Yuri Burago and Viktor
Zalgaller [2, Theorem 28.2.5]; it generalizes the result of István Fáry about
average curvature of a curve in the unit ball [3, 11], but the proof is
essentially the same.

3.1. Lemma. Let $L$ be a closed $n$-dimension manifold that is smoothly
immersed in $\mathbb{B}^{q}$. Then the average value of $|H|$ on $L$ is at
least $n$.

Proof. Consider the function $u\colon x\mapsto\tfrac{1}{2}{\hskip
0.5pt\cdot\hskip 0.5pt}|x|^{2}$ on $L$. Note that

$(\Delta u)(x)=n+\langle H(x),x\rangle.$

It follows that the average value of $\langle H(x),x\rangle$ is $-n$. Since
$|x|\leqslant 1$, we get the result. ∎

Proof of 1. By 3, the average value of $|H|^{2}$ is at least 4. Further, by
the Gauss–Bonnet formula, scalar curvature (which is twice the Gauss curvature
in this case) has zero average. Therefore 2 implies the statement. ∎

Proof of 1. By 3, the average value of $|H|^{2}$ is at least $n^{2}$. Since
$\mathrm{Sc}\equiv 0$, it remains to apply 2. ∎

Proof of 1. Since the image lies in the unit sphere, we have that $|H|^{2}$ is
at least $n^{2}$ at each point. Since $\mathbb{T}^{n}$ does not admit a metric
with positive scalar curvature [7, Corollary A], we have
$\mathrm{Sc}(x)\leqslant 0$ at some point $x$. It remains to apply 2 at $x$. ∎

##  4 Main case

The following lemma is an easy corollary of the bow lemma of Axel Schur [10,
9]. It tells how we use the bound on normal curvatures in 1. If $n\leqslant
4$, then the proof works without this assumption.

4.1. Lemma. Let $L$ be a manifold smoothly immersed in $\mathbb{B}^{q}$.
Suppose its normal curvatures are at most $2$. Given $x\in L$, denote by
$\beta=\beta(x)$ the angle between vector $x$ and the normal space
$\mathrm{N}_{x}$. Then $|x|\leqslant\cos\beta$.

Proof. Let $\xi$ be a tangent direction at $x$ such that
$\measuredangle(x,\xi)=\nobreak\tfrac{\pi}{2}-\beta$. In the plane spanned by
$x$ and $\xi$, choose a unit-speed circle arc $\sigma$ from $0$ to $x$ that
comes to $x$ in the direction opposite to $\xi$; extend $\sigma$ after $x$ by
a unit-speed semicircle $\tilde{\gamma}$ with curvature $2$ in such a way that
the concatenation $\sigma*\tilde{\gamma}$ is an arc of a smooth convex plane
curve; see the figure.

Observe that if $|x|>\cos\beta$, then $\tilde{\gamma}$ leaves
$\mathbb{B}^{q}$; that is, $|\tilde{\gamma}(t_{0})|>1$ for some $t_{0}$.

Let $\gamma$ be the unit-speed geodesic in $L$ that runs from $x$ in the
direction $\xi$. Note that curvatures of $\sigma*\gamma$ do not exceed the
curvatures of $\sigma*\tilde{\gamma}$ at the corresponding points. Applying
the bow lemma for $\sigma*\gamma$ and $\sigma*\tilde{\gamma}$, we get
$|\gamma(t_{0})|\geqslant|\tilde{\gamma}(t_{0})|$. It follows that $L$ does
not lie in $\mathbb{B}^{q}$ — a contradiction. ∎

Let $g$ be a Riemannian metric on $\mathbb{T}^{n}$. Suppose $n\geqslant 3$,
and $u\colon\mathbb{T}^{n}\to\mathbb{R}$ is a smooth positive function. Recall
that

$\left(\mathrm{Sc}{\hskip 0.5pt\cdot\hskip 0.5pt}u-4{{\hskip 0.5pt\cdot\hskip
0.5pt}}\tfrac{n-1}{n-2}{{\hskip 0.5pt\cdot\hskip 0.5pt}}\Delta u\right){\hskip
0.5pt\cdot\hskip 0.5pt}u^{\frac{n-2}{n+2}}$

is the scalar curvature of the metric $u^{\frac{4}{n-2}}{\hskip
0.5pt\cdot\hskip 0.5pt}g$; see for example [1, 6.3]. Since _any Riemannian
metric on $\mathbb{T}^{n}$ has nonpositive scalar curvature at some point_ [7,
Corollary A], we get the following.

4.2. Claim. For any Riemannian metric on $\mathbb{T}^{n}$ and any positive
smooth function $u\colon\mathbb{T}^{n}\to\mathbb{R}$, the function

$\mathrm{Sc}{\hskip 0.5pt\cdot\hskip 0.5pt}u-4{{\hskip 0.5pt\cdot\hskip
0.5pt}}\tfrac{n-1}{n-2}{{\hskip 0.5pt\cdot\hskip 0.5pt}}\Delta u$

returns a nonpositive value at some point.

Proof of 1 and 1. The case $n=2$ follows from 1; so we can assume that
$n\geqslant 3$. Consider the function $u\colon
x\mapsto\exp(-\tfrac{k}{2}{\hskip 0.5pt\cdot\hskip 0.5pt}|x|^{2})$ on the
torus.

We will apply the following formula

$\Delta(\varphi\circ f)=(\varphi^{\prime}\circ f){\hskip 0.5pt\cdot\hskip
0.5pt}\Delta f+(\varphi^{\prime\prime}\circ f){\hskip 0.5pt\cdot\hskip
0.5pt}|\nabla f|^{2}$

to $f\colon x\mapsto\tfrac{1}{2}{\hskip 0.5pt\cdot\hskip 0.5pt}|x|^{2}$ and
$\varphi\colon y\mapsto\exp(-k{\hskip 0.5pt\cdot\hskip 0.5pt}y)$; so
$u=\varphi\circ f$.

Set $r(x)=|x|$, $\alpha(x)=\measuredangle(H(x),x)$, and $\beta(x)$ as in 4.
Note that

$\beta\leqslant\alpha\leqslant\pi-\beta.$ ➊

Observe that

$\Delta f=|H|{\hskip 0.5pt\cdot\hskip 0.5pt}r{\hskip 0.5pt\cdot\hskip
0.5pt}\cos\alpha+n,\quad|\nabla f|=r{\hskip 0.5pt\cdot\hskip
0.5pt}\sin\beta,\quad\varphi^{\prime}=-k{\hskip 0.5pt\cdot\hskip
0.5pt}\varphi,\quad\varphi^{\prime\prime}=k^{2}{\hskip 0.5pt\cdot\hskip
0.5pt}\varphi.$

Therefore

$\displaystyle\Delta u$ $\displaystyle=u{\hskip 0.5pt\cdot\hskip
0.5pt}[-k{\hskip 0.5pt\cdot\hskip 0.5pt}|H|{\hskip 0.5pt\cdot\hskip
0.5pt}r{\hskip 0.5pt\cdot\hskip 0.5pt}\cos\alpha-k{\hskip 0.5pt\cdot\hskip
0.5pt}n+k^{2}{\hskip 0.5pt\cdot\hskip 0.5pt}r^{2}{\hskip 0.5pt\cdot\hskip
0.5pt}\sin^{2}\beta].$

Recall that $\mathrm{Sc}=-\tfrac{n{\hskip 0.5pt\cdot\hskip
0.5pt}(n+2)}{2}{\hskip 0.5pt\cdot\hskip 0.5pt}\textsl{Ж}+\tfrac{3}{2}{\hskip
0.5pt\cdot\hskip 0.5pt}|H|^{2}$; see 2. By 4, the function

$\displaystyle\mathrm{Sc}{\hskip 0.5pt\cdot\hskip 0.5pt}u-4{{\hskip
0.5pt\cdot\hskip 0.5pt}}\tfrac{n-1}{n-2}{\hskip 0.5pt\cdot\hskip 0.5pt}\Delta
u$ $\displaystyle=u{\hskip 0.5pt\cdot\hskip 0.5pt}\biggl{[}-\tfrac{n{\hskip
0.5pt\cdot\hskip 0.5pt}(n+2)}{2}{\hskip 0.5pt\cdot\hskip
0.5pt}\textsl{Ж}+\tfrac{3}{2}{\hskip 0.5pt\cdot\hskip 0.5pt}|H|^{2}+4{{\hskip
0.5pt\cdot\hskip 0.5pt}}\tfrac{n-1}{n-2}{\hskip 0.5pt\cdot\hskip
0.5pt}k{\hskip 0.5pt\cdot\hskip 0.5pt}|H|{\hskip 0.5pt\cdot\hskip
0.5pt}r{\hskip 0.5pt\cdot\hskip 0.5pt}\cos\alpha+$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad+4{{\hskip 0.5pt\cdot\hskip
0.5pt}}\tfrac{n-1}{n-2}{\hskip 0.5pt\cdot\hskip 0.5pt}(k{\hskip
0.5pt\cdot\hskip 0.5pt}n-k^{2}{\hskip 0.5pt\cdot\hskip 0.5pt}r^{2}{\hskip
0.5pt\cdot\hskip 0.5pt}\sin^{2}\beta)\biggr{]}$

returns a nonpositive value at some point $x\in\mathbb{T}^{n}$.

Choose

$k=\tfrac{3}{4}{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{n-2}{n-1}{\hskip
0.5pt\cdot\hskip 0.5pt}n,\quad\text{so}\quad n=\tfrac{4}{3}{\hskip
0.5pt\cdot\hskip 0.5pt}\tfrac{n-1}{n-2}{\hskip 0.5pt\cdot\hskip 0.5pt}k.$

At the point $x$, we have

$\displaystyle\tfrac{n{\hskip 0.5pt\cdot\hskip 0.5pt}(n+2)}{2}{\hskip
0.5pt\cdot\hskip 0.5pt}\textsl{Ж}$ $\displaystyle\geqslant\tfrac{3}{2}{\hskip
0.5pt\cdot\hskip 0.5pt}(|H|+n{\hskip 0.5pt\cdot\hskip 0.5pt}r{\hskip
0.5pt\cdot\hskip 0.5pt}\cos\alpha)^{2}-\tfrac{3}{2}{\hskip 0.5pt\cdot\hskip
0.5pt}n^{2}{\hskip 0.5pt\cdot\hskip 0.5pt}r^{2}{\hskip 0.5pt\cdot\hskip
0.5pt}\cos^{2}\alpha+$ ➋ $\displaystyle\qquad\qquad\qquad\quad+3{\hskip
0.5pt\cdot\hskip 0.5pt}n^{2}-\tfrac{9}{4}{\hskip 0.5pt\cdot\hskip
0.5pt}\tfrac{n-2}{n-1}{\hskip 0.5pt\cdot\hskip 0.5pt}n^{2}{\hskip
0.5pt\cdot\hskip 0.5pt}r^{2}{\hskip 0.5pt\cdot\hskip
0.5pt}\sin^{2}\beta\geqslant\tfrac{3}{2}{\hskip 0.5pt\cdot\hskip 0.5pt}n^{2}.$

Indeed, by ➊ ‣ 4, $\cos^{2}\alpha+\sin^{2}\beta\leqslant\nobreak 1$. If
$n\leqslant 4$, then $\tfrac{3}{2}\geqslant\tfrac{9}{4}{\hskip
0.5pt\cdot\hskip 0.5pt}\tfrac{n-2}{n-1}$; therefore the last inequality
follows, and it proves 1.

Further, if $n\geqslant 5$, then for the last inequality in ➋ ‣ 4 we need to
use in addition that $r^{2}+\sin^{2}\beta\leqslant 1$ which follows 4. Hence 1
follows. ∎

Acknowledgments. I want to thank Michael Gromov and Nina Lebedeva for help and
encouragement. This work was done at Euler International Mathematical
Institute of PDMI RAS; it was partially supported by the National Science
Foundation, grant DMS-2005279, and the Ministry of Education and Science of
the Russian Federation, grant 075-15-2022-289.

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