Veronese minimizes normal curvatures

Let $M\subset\mathbb{R}^{d}$ be a closed smooth $n$-dimensional submanifold. Assume $d$ is large and $M$ lies in an $r$-ball. What can we say about the normal curvatures of $M$?

First note that the curvatures cannot be smaller than $\tfrac{1}{r}$ at all points. Moreover, the average value of $|H|$ must be at least $n{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{1}{r}$; here $H$ denotes the mean curvature vector [2, 28.2.5], [8, 3.1]. This statement is a straightforward generalization of the result of István Fáry about closed curves in a ball [3, 12].

On the other hand, the $n$-dimensional torus can be embedded into an $r$-ball with all normal curvatures $\sqrt{3{\hskip 0.5pt\cdot\hskip 0.5pt}n/(n+2)}{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{1}{r}$. This embedding was found by Michael Gromov [6, 2.A], [5, 1.1.A]. The bound is optimal; that is, any smooth $n$-dimensional torus in an $r$-ball has normal curvature at least $\sqrt{3{\hskip 0.5pt\cdot\hskip 0.5pt}n/(n+2)}{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{1}{r}$ at some point [8]. Gromov’s examples easily imply the following: any closed smooth manifold $M$ admits a smooth embedding into an $r$-ball of sufficiently large dimension with normal curvatures less than $\sqrt{3}{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{1}{r}$ [6, 1.D], [5, 1.1.C]. But what happens between $\tfrac{1}{r}$ and $\sqrt{3}{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{1}{r}$?

In this note, we consider embeddings in an $r$-ball with normal curvatures at most $\tfrac{2}{\sqrt{3}}{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{1}{r}$. We show that if the inequality is strict, then the manifold must be homeomorphic to a sphere (see § 2). For the nonstrict inequality, in addition to spheres, we get real, complex, quaternionic, and octonionic planes mapped by rescaled Veronese embeddings (see § 4).

Let $M$ be a closed smooth $n$-dimensional submanifold in a closed $r$-ball in $\mathbb{R}^{d}$. Suppose that the normal curvatures of $M$ are strictly less than $\tfrac{2}{\sqrt{3}}{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{1}{r}$. Then $M$ is homeomorphic to the $n$-sphere.

Denote the $r$-ball by $\mathbb{B}^{d}$. We can assume that $r=\tfrac{1}{\sqrt{3}}$; that is, $r$ is the circumradius of an equilateral triangle with side 1. Therefore, the normal curvatures of $M$ are smaller than $2$.

Choose a unit-speed geodesic $\gamma\colon[0,\tfrac{\pi}{2}]\to M$; let $x=\gamma(0)$ and $y=\gamma(\tfrac{\pi}{2})$. By the assumption, the curvature of $\gamma$ in $\mathbb{R}^{d}$ is less than $2$. Applying Schur’s bow lemma, we get $|x-y|>1$.

Let $\Pi$ be the perpendicular bisector to $[x,y]$. Since the curvature of $\gamma$ is smaller than 2, if $t\neq t_{0}$, then

$$\measuredangle(\gamma^{\prime}(t_{0}),\gamma^{\prime}(t))<2{\hskip 0.5pt\cdot\hskip 0.5pt}|t-t_{0}|\quad\text{and}\quad\langle\gamma^{\prime}(t_{0}),\gamma^{\prime}(t)\rangle>\cos(2{\hskip 0.5pt\cdot\hskip 0.5pt}|t-t_{0}|).$$

Therefore,

$$\langle y-x,\gamma^{\prime}(t_{0})\rangle>\int\limits_{0}^{\frac{\pi}{2}}\cos(2{\hskip 0.5pt\cdot\hskip 0.5pt}|t-t_{0}|){\hskip 0.5pt\cdot\hskip 0.5pt}dt\geqslant 0.$$

In particular, the derivative of function $f\colon t\mapsto\langle y-x,\gamma(t)\rangle$ is positive. Therefore, $\gamma$ intersects $\Pi$ transversely at a single point; denote it by $s$.

Choose a unit vector ${\textsc{u}}\in\mathrm{T}_{x}$; let $\gamma_{{\textsc{u}}}\colon[0,\tfrac{\pi}{2}]\to M$ be the unit-speed geodesic that starts from $x$ in the direction u, and let $z=\gamma_{{\textsc{u}}}(\tfrac{\pi}{2})$. The argument above shows that $|x-z|>1$.

Denote by $H_{x}$ and $H_{y}$ the closed half-spaces bounded by $\Pi$ that contain $x$ and $y$ respectively. Assume $z\in H_{x}$, then we have $|y-z|\geqslant|x-z|>1$. Since $|x-y|>1$, the triangle $[xyz]$ has all sides larger than $1$, which is impossible since $x,y,z\in\mathbb{B}^{d}$. Therefore, $\gamma_{{\textsc{u}}}$ meets $\Pi$ before in $\tfrac{\pi}{2}$; denote by $r({{\textsc{u}}})$ be the first such time moment.

Let us show that the function ${{\textsc{u}}}\mapsto r({{\textsc{u}}})$ is smooth. In other words, $\gamma_{{\textsc{u}}}$ intersects $\Pi$ transversely at time $r({{\textsc{u}}})$. Assume this is not the case, so $\gamma_{{\textsc{u}}}$ is tangent to $\Pi$ at $r({{\textsc{u}}})$. Let $\hat{\gamma}_{{\textsc{u}}}$ be the concatenation of the reflection of $\gamma_{{\textsc{u}}}|_{[0,r({{\textsc{u}}})]}$ across $\Pi$ and $\gamma_{{\textsc{u}}}|_{[r({{\textsc{u}}}),\frac{\pi}{2}]}$. Note that $\hat{\gamma}_{{\textsc{u}}}$ is $C^{1}$-smooth, and it is $C^{\infty}$-smooth everywhere except $r({{\textsc{u}}})$. Therefore, Schur’s bow lemma is applicable to $\hat{\gamma}_{{\textsc{u}}}$, and hence, $|y-z|>1$. Again, all sides of triangle $[xyz]$ are larger than $1$; hence, it cannot lie in $\mathbb{B}^{d}$ — a contradiction.

It follows that the set

$$V_{x}=\left\{\,t{\hskip 0.5pt\cdot\hskip 0.5pt}{{\textsc{u}}}\in\mathrm{T}_{x}\,:\,|{{\textsc{u}}}|=1,\quad 0\leqslant t\leqslant r({{\textsc{u}}}),\,\right\}$$

is diffeomorphic to the closed $n$-disc. Denote by $W_{x}$ the connected component of $x$ in $M\cap H_{x}$.

From the Gauss formula [9, Lemma 5], the sectional curvatures of $M$ are less than $4$. In particular, the exponential map $\exp_{x}\colon\mathrm{T}_{x}\to M$ is a local diffeomorphism in the $\tfrac{\pi}{2}$-ball centered at the origin of $\mathrm{T}_{x}$.

It follows that $\exp_{x}\colon V_{x}\to W_{x}$ is a local diffeomorphism; in particular, $W_{x}$ is a smooth manifold with boundary. Since $V_{x}$ is simply connected, $\exp_{x}$ defines a diffeomorphism $V_{x}\to W_{x}$. In particular, $W_{x}$ is a closed topological $n$-disc and $\partial W_{x}$ is a smooth hypersurface in $M$.

Let us swap the roles of $x$ and $y$, and repeat the construction. We get another closed topological $n$-disc $W_{y}\subset M$ bounded by a smooth hypersurface $\partial W_{y}$.

Observe that $\partial W_{x}$ intersects $\partial W_{y}$ at $s$. Furthermore, both $\partial W_{x}$ and $\partial W_{y}$ are connected components of $s$ in $M\cap\Pi$. Therefore, $\partial W_{x}=\partial W_{y}$. That is, $M$ can be obtained by gluing two $n$-discs by a diffeomorphism between their boundaries. Hence $M$ is homeomorphic to the $n$-sphere. ∎

The real, complex, quaternionic projective spaces of dimension $n$, and the octonionic projective plane will be denoted by $\mathbb{R}\mathrm{P}^{n}$, $\mathbb{C}\mathrm{P}^{n}$, $\mathbb{H}\mathrm{P}^{n}$, and $\mathbb{O}\mathrm{P}^{2}$ respectively. We assume that each of these spaces is equipped with the canonical metric; in particular, all the spaces have closed geodesics of length $\pi$.

Moreover,

• (i)

  all normal curvatures of the images of these embeddings are equal to $2$

• (ii)

  the images of these embeddings lie in a sphere of radius $r_{n}=\sqrt{n/(2{\hskip 0.5pt\cdot\hskip 0.5pt}n+2)}$ (for $\mathbb{O}\mathrm{P}^{2}$, we assume that $n=2$).

The proposition can be extracted from two theorems in [11, § 2]. The embeddings provided by the proposition will be called Veronese embeddings. Note that $r_{n}$ is the circumradius of a regular $n$-simplex with edge length $1$. Note that $r_{2}=1/\sqrt{3}$; therefore, the second part of proposition shows that our sphere theorem has the optimal bound.

The Veronese embeddings have a very explicit algebraic description and many nice geometric properties. In particular, these embeddings are equivariant, and their images are minimal submanifolds in the $r_{n}$-spheres. All of this is discussed in the cited paper by Kunio Sakamoto.

The following lemma is also closely related to the result of Kunio Sakamoto; it implies that the properties in the proposition uniquely describe Veronese embeddings up to motion of the ambient space.

Recall that the extrinsic curvature tensor $\Phi$ of a submanifold is defined as

$$\Phi({{\textsc{x}}},{{\textsc{y}}},{{\textsc{v}}},{{\textsc{w}}})=\langle\mathrm{I}\!\mathrm{I}({{\textsc{x}}},{{\textsc{y}}}),\mathrm{I}\!\mathrm{I}({{\textsc{v}}},{{\textsc{w}}})\rangle,$$

here x, y, v, w are tangent vectors to the submanifold at some point; see [7]. Note that the $\Phi$-tensor describes the second fundamental form $\mathrm{I}\!\mathrm{I}$ up to motion of the ambient space. Therefore, once we show that the $\Phi$-tensors of $M$ and $M^{\prime}$ coincide at one point, we get that $M$ and $M^{\prime}$ are congruent.

The tensor $\Phi$ can be written as

$$\Phi({{\textsc{x}}},{{\textsc{y}}},{{\textsc{v}}},{{\textsc{w}}})=\mathrm{E}({{\textsc{x}}},{{\textsc{y}}},{{\textsc{v}}},{{\textsc{w}}})+\tfrac{1}{3}{\hskip 0.5pt\cdot\hskip 0.5pt}(\operatorname{\rm Rm}({{\textsc{x}}},{{\textsc{v}}},{{\textsc{y}}},{{\textsc{w}}})+\operatorname{\rm Rm}({{\textsc{x}}},{{\textsc{w}}},{{\textsc{y}}},{{\textsc{v}}}))$$

where $\mathrm{E}$ is the total symmetrization of $\Phi$; that is,

$$\mathrm{E}({{\textsc{x}}},{{\textsc{y}}},{{\textsc{v}}},{{\textsc{w}}})=\tfrac{1}{3}{\hskip 0.5pt\cdot\hskip 0.5pt}(\Phi({{\textsc{x}}},{{\textsc{y}}},{{\textsc{v}}},{{\textsc{w}}})+\Phi({{\textsc{y}}},{{\textsc{v}}},{{\textsc{x}}},{{\textsc{w}}})+\Phi({{\textsc{v}}},{{\textsc{x}}},{{\textsc{y}}},{{\textsc{w}}})),$$

and

$$\operatorname{\rm Rm}({{\textsc{x}}},{{\textsc{y}}},{{\textsc{v}}},{{\textsc{w}}})=\Phi({{\textsc{x}}},{{\textsc{v}}},{{\textsc{y}}},{{\textsc{w}}})-\Phi({{\textsc{x}}},{{\textsc{w}}},{{\textsc{y}}},{{\textsc{v}}})$$

is the Riemannian curvature tensor of $M$.

Since $M$ is isometric to $M^{\prime}$, they have the same Riemannian curvature tensors. It remains to show that the $\mathrm{E}$-tensors are the same. But

$$f({{\textsc{x}}})=\mathrm{E}({{\textsc{x}}},{{\textsc{x}}},{{\textsc{x}}},{{\textsc{x}}})=\nobreak|\mathrm{I}\!\mathrm{I}({{\textsc{x}}},{{\textsc{x}}})|^{2}$$

is a homogeneous polynomial of degree $4$ on the tangent space and it describes $\mathrm{E}$ completely.

The geodesics in $M$ and $M^{\prime}$ are closed and have the same length. Since each of these geodesic forms a circle in $\mathbb{R}^{d}$, all these circles have the same curvature, say $\kappa$. Therefore, $\mathrm{I}\!\mathrm{I}({{\textsc{x}}},{{\textsc{x}}})=\nobreak\kappa{\hskip 0.5pt\cdot\hskip 0.5pt}|{{\textsc{x}}}|^{2}$ and $\mathrm{E}({{\textsc{x}}},{{\textsc{x}}},{{\textsc{x}}},{{\textsc{x}}})=\kappa^{2}{\hskip 0.5pt\cdot\hskip 0.5pt}|{{\textsc{x}}}|^{4}$ for both submanifolds and for any tangent vector x. This finishes the proof. ∎

Let $M$ be a closed smooth $n$-dimensional submanifold in a closed $r$-ball in $\mathbb{R}^{d}$. Suppose that the normal curvatures of $M$ are at most $\tfrac{2}{\sqrt{3}}{\hskip 0.5pt\cdot\hskip 0.5pt}\tfrac{1}{r}$. If $M$ is not homeomorphic to a sphere, then up to rescaling, it is congruent to an image of the Veronese embedding of a projective plane $\mathbb{R}\mathrm{P}^{2}$, $\mathbb{C}\mathrm{P}^{2}$, $\mathbb{H}\mathrm{P}^{2}$, or $\mathbb{O}\mathrm{P}^{2}$.

This result is an application of the following theorem; its weaker form was proved by Detlef Gromoll and Karsten Grove [4], and the final step was made by Burkhard Wilking [13].

Recall that a compact rank-one symmetric space is isometric to a rescaled copy of one of the following spaces: $\mathbb{R}\mathrm{P}^{n}$, $\mathbb{C}\mathrm{P}^{n}$, $\mathbb{H}\mathrm{P}^{n}$, $\mathbb{O}\mathrm{P}^{2}$, and unit spheres $\mathbb{S}^{n}$; see for example, [14, 8.12.2]. As before we assume that these spaces are equipped with the canonical metrics; in particular, all the projective spaces have closed geodesics of length $\pi$.

By the proposition in § 3, the images of Veronese embedding satisfy the assumption of the theorem. It remains to show that there are no other embeddings of that type.

Choose a unit-speed geodesic $\gamma\colon[0,\tfrac{\pi}{2}]\to M$; let $x=\gamma(0)$ and $y=\gamma(\tfrac{\pi}{2})$. The argument in our sphere theorem implies that $|x-y|=1$. The rigidity case in the bow lemma implies that $\gamma$ is a half-circle of curvature $2$. Since any two points in $M$ can be connected by a geodesic, we get the following.

• $\diamond$

  The diameter of $M$ is 1.

• $\diamond$

  The intrinsic diameter and injectivity radius of $M$ are equal to $\tfrac{\pi}{2}$.

• $\diamond$

  All geodesics in $M$ are circles of curvature 2 in $\mathbb{R}^{d}$.

Furthermore, for $x$ and $y$ as above, there is another point $z\in M$ such that $|x-z|=|y-z|=1$. If not, then again, the argument in the sphere theorem would imply that $M$ is a sphere. But since $x,y,z\in\mathbb{B}^{d}$, the equalities $|x-y|=|y-z|=|x-z|=1$ imply that $x\in\partial\mathbb{B}^{d}$.

The choice of $x\in M$ was arbitrary. Therefore, $M$ lies in the sphere $\partial\mathbb{B}^{d}$ of radius $r=1/\sqrt{3}$. This sphere has sectional curvature $1/r^{2}=3$; all normal curvatures of $M$ in the sphere are $\kappa=\sqrt{2^{2}-1/r^{2}}=1$. By the Gauss formula [9, Lemma 5], the sectional curvatures of $M$ are at least $3-2{\hskip 0.5pt\cdot\hskip 0.5pt}\kappa^{2}=1$.

By the Gromoll–Grove–Wilking theorem, the universal cover $\tilde{M}$ of $M$ is isometric to a rank-one symmetric space. Taking into account the injectivity radius and curvature of $M$, we get that $\tilde{M}$ must be isometric to one of the following spaces $\tfrac{1}{2}{\hskip 0.5pt\cdot\hskip 0.5pt}\mathbb{S}^{n}$, $\mathbb{S}^{n}$, $\mathbb{C}\mathrm{P}^{n}$, $\mathbb{H}\mathrm{P}^{n}$ for some $n$, or $\mathbb{O}\mathrm{P}^{2}$. Note that the points $x,y,z\in M$ constructed above lie at an intrinsic distance $\tfrac{\pi}{2}$ from each other. It forbids $\tfrac{1}{2}{\hskip 0.5pt\cdot\hskip 0.5pt}\mathbb{S}^{n}$ for every $n$. Furthermore, if $n\geqslant 3$, then each space $\mathbb{S}^{n}$, $\mathbb{C}\mathrm{P}^{n}$ and $\mathbb{H}\mathrm{P}^{n}$ contain 4 points at a distance $\tfrac{\pi}{2}$ from each other. Since the injectivity radius of $M$ is $\tfrac{\pi}{2}$, their projections in $M$ must lie at a distance $\tfrac{\pi}{2}$ from each other as well. It follows that $\mathbb{B}^{d}$ must contain 4 points at a distance 1 from each other, which is impossible.

Hence, $\tilde{M}$ must be isometric to one of the following spaces $\mathbb{S}^{2}$, $\mathbb{C}\mathrm{P}^{2}$, $\mathbb{H}\mathrm{P}^{2}$, or $\mathbb{O}\mathrm{P}^{2}$. Since the injectivity radius of $M$ is $\tfrac{\pi}{2}$, it has to be isometric to $\mathbb{R}\mathrm{P}^{2}$, $\mathbb{C}\mathrm{P}^{2}$, $\mathbb{H}\mathrm{P}^{2}$, or $\mathbb{O}\mathrm{P}^{2}$.

Denote by $M^{\prime}\subset\mathbb{R}^{d}$ the image of the corresponding Veronese embedding provided by the proposition in § 3. Without loss of generality, we can assume that $d$ is large, so $M^{\prime}$ exists. Applying the lemma in § 3, we get that $M$ is congruent $M^{\prime}$ — hence the result. ∎

Recall that the Veronese embeddings map $\mathbb{R}\mathrm{P}^{n}$, $\mathbb{C}\mathrm{P}^{n}$, and $\mathbb{H}\mathrm{P}^{n}$ into balls of radius $r_{n}=\sqrt{n/(2{\hskip 0.5pt\cdot\hskip 0.5pt}n+2)}$, which is the circumradius of a regular $n$-simplex with edge length $1$. This note is motivated by the following question [10].

The same question can be asked about $\mathbb{C}\mathrm{P}^{n}$ and $\mathbb{H}\mathrm{P}^{n}$. A keen reader might have noticed that the case $n=2$ is already solved.

I suspect that the answer is yes. If, in addition, $M$ lies on the boundary of the $r$-ball, then by the Gauss formula [9, Lemma 5], $M$ has strictly quarter-pinched curvature; so, it has to be diffeomorphic to a standard sphere [1].

I want to thank Alexander Lytchak for help. This work was partially supported by the National Science Foundation, grant DMS-2005279.