Proof. Let (2.1) be the expression of $P$ as the sum of homogeneous polynomials. Writing a point in $\mathbb {R}^{n+1}$ as $x=(x_1,\,\ldots ,\,x_{n+1})$, let $U^{+}=\{x\in \mathbb {R}^{n+1}:P(x)>0\}$ and $U^{-}=\{x\in \mathbb {R}^{n+1}:P(x)<0\}$. Since the mean curvature $H$ of $M$ is a non-zero constant by hypothesis, changing $P$ by $-P$ if necessary, we can assume that $H=c>0$. This means that the mean curvature vector $\overrightarrow {H}$ of $M^{n}$ points in the direction of $U^{+}$.

By (2.1), for any $t\in \mathbb {R}$ and any vector $v$ in the unit sphere $S^{n}_1$ we have

(2.2)\begin{equation} P(tv)=P_m(v)t^{m}+P_{m-1}(v)t^{m-1}+\cdots+P_1(v)t+P_0, \end{equation}

and so

(2.3)\begin{equation} \frac{P(tv)}{t^{m}}-P_m(v)=\sum_{i=0}^{m-1}\frac{P_i(v)}{t^{m-i}},\quad v\in S^{n}_1,\ t\neq 0. \end{equation}

Using (2.3) and the compactness of $S^{n}_1$, one easily sees that $t^{-m}P(tv)\to P_m(v)$ uniformly on $S^{n}_1$ when $t\to \infty$.

Suppose, by contradiction, that $P_m$ changes sign. Then, since $P_m$ is homogeneous, there exists a vector $w$ in the unit sphere $S^{n}_1\subset \mathbb {R}^{n+1}$ such that $P_m(w)>0$. Hence, by continuity, there is a closed disk $W$ around $w$ in $S^{n}_1$ such that $P_m(v)>P_m(w)/2$ for all $v\in W$. Since $t^{-m}P(tv)\to P_m(v)$ uniformly on $S^{n}_1$, there exists $t_0>0$ such that $|t^{-m}P(tv)-P_m(v)|< P_m(w)/4$ for all $t>t_0$ and $v\in S^{n}_1$. Combining these informations, one obtains

(2.4)\begin{align} t^{{-}m}P(tv)&\geq{-}|t^{{-}m}P(tv)-P_m(v)|+P_m(v)\nonumber\\ &>{-}\frac{1}{4}P_m(w)+\frac{1}{2}P_m(w)=\frac{1}{4}P_m(w)>0,\quad t>t_0,\ v\in W, \end{align}

and so

(2.5)\begin{equation} \{tv:t>t_0,\;v\in W\}\subset U^{+}. \end{equation}

Let $R$ be an arbitrary positive number. By (2.5), there exists a ball $B$ of radius $R$ in $\mathbb {R}^{n+1}$ such that $B\subset U^{+}$. Let $x_0$ be the centre of $B$ and $f:M\to \mathbb {R}$ the function defined by $f(x)=||x-x_0||^{2}$. Since $M$ is a closed subset of $\mathbb {R}^{n+1}$, there is a point $p\in M$ such that $f(p)=\inf _M f$. Let $v$ be an arbitrary unit vector of $T_pM$ and $\gamma :I\to M$ a smooth curve such that $\gamma (0)=p$ and $\gamma '(0)=v$. Since the function $f(\gamma (t))$ attains a minimum at $t=0$, one has

(2.6)\begin{equation} 0=\frac{\textrm{d}}{\textrm{d}t}\Big|_{t=0}f(\gamma(t))=\frac{\textrm{d}}{\textrm{d}t}\Big|_{t=0}\langle\gamma(t)-x_0,\gamma(t)-x_0\rangle=2\langle v,p-x_0\rangle \end{equation}

and

(2.7)\begin{equation} 0\leq\frac{\textrm{d}^{2}}{\textrm{d}t^{2}}\Big|_{t=0}f(\gamma(t))=2\langle\gamma''(0),p-x_0\rangle+2. \end{equation}

From (2.6) one obtains that $p-x_0$ is orthogonal to $T_pM$. Since $N$ points inward $U^{+}$ and $x_0\in U^{+}$, it follows that $N(p)=(x_0-p)/||x_0-p||$. Using this information in (2.7), one obtains

\[ \langle Av,v\rangle={-}\langle(N\circ\gamma)'(0),\gamma'(0)\rangle=\langle N(p),\gamma''(0)\rangle\leq\frac{1}{||x_0-p||}, \]

for any unit vector $v\in T_pM$, where $A$ is the shape operator of $M$ with respect to $N$. Taking the trace in the above inequality and using the fact that $p\not \in B$, one concludes that

\[ H=H(p)\leq\frac{1}{||x_0-p||}\leq\frac{1}{R}, \]

for every $R>0$, contradicting $H=c>0$. This contradiction shows that $P_m$ does not change sign, and the theorem is proved.

Proof. We will prove the lemma in the case where $m=2d,\,\;d\geq 1$. The proof in the case that the degree of $P(x)$ is odd is entirely analogous and will be omitted.

Assume first $a=0$ and $r=1$. Write $P$ as

(2.8)\begin{equation} P=\sum_{i=0}^{d}P_{2i}+\sum_{i=0}^{d-1}P_{2i+1}, \end{equation}

where $P_j,\, j=0,\,\ldots ,\,2d,$ is a homogeneous polynomial of degree $j$. By hypothesis, for every $v\in S^{k}_1(0)\times \mathbb {R}^{n-k}$ one has

\begin{align*} 0&=P(v) =\sum_{i=0}^{d}P_{2i}(v)+\sum_{i=0}^{d-1}P_{2i+1}(v),\\ 0&=P({-}v)=\sum_{i=0}^{d}P_{2i}(v)-\sum_{i=0}^{d-1}P_{2i+1}(v). \end{align*}

From the two equalities above one obtains $\sum_{i=0}^{d}P_{2i}(v)=0=\sum_{i=0}^{d-1}P_{2i+1}(v)$, which implies

(2.9)\begin{equation} P_{2d}(v)={-}\sum_{i=0}^{d-1}P_{2i}(v),\quad P_{2d-1}(v)={-}\sum_{i=0}^{d-2}P_{2i+1}(v),\ \forall v\in S^{k}_1(0)\times\mathbb{R}^{n-k}. \end{equation}

Let $\{e_1,\,\ldots ,\,e_{n+1}\}$ be the canonical basis of $\mathbb {R}^{n+1}$. For any $x=(x_1,\,\ldots ,\,x_{n+1})\in \mathbb {R}^{n+1}$ such that $y:=\sum_{i=1}^{k+1}x_ie_i\neq 0$ it holds that $x/|y|\in S^{k}_1(0)\times \mathbb {R}^{n-k}$. By (2.8) and (2.9), for all such $x$ one has

\begin{align*} P(x)&=\sum_{i=0}^{d-1}P_{2i}(x)+P_{2d}(x)+\sum_{i=0}^{d-2}P_{2i+1}(x)+P_{2d-1}(x)\nonumber\\ &=\sum_{i=0}^{d-1}P_{2i}(x)+|y|^{2d}P_{2d}(x/|y|)+\sum_{i=0}^{d-2}P_{2i+1}(x)+|y|^{2d-1}P_{2d-1}(x/|y|)\nonumber\\ &=\sum_{i=0}^{d-1}P_{2i}(x)-|y|^{2d}\sum_{i=0}^{d-1}P_{2i}(x/|y|)+\sum_{i=0}^{d-2}P_{2i+1}(x)-|y|^{2d-1}\sum_{i=0}^{d-2}P_{2i+1}(x/|y|)\nonumber\\ &=\sum_{i=0}^{d-1}(1-|y|^{2(d-i)})P_{2i}(x)+\sum_{i=0}^{d-2}(1-|y|^{2(d-i-1)})P_{2i+1}(x).\nonumber \end{align*}

Notice that the second term on the right-hand side of the above equality vanishes when $d=1$. Since $(1-|y|^{2k})$ is divisible by $1-|y|^{2}$ for any integer $k\geq 1$, it follows from the above equality that

(2.10)\begin{equation} P(x)=(|y|^{2}-1)R(x), \end{equation}

for some polynomial $R(x)$ of degree $2(d-1)$. This proves the lemma in the case $a=0$ and $r=1$, since $|y|^{2}=\sum_{i=1}^{k+1}x_i^{2}$. To obtain the lemma in the general case, it suffices to apply (2.10) for the polynomial $\overline {P}(x):=P(rx+a)$.

Proof. Let (2.1) be the expression of $P$ as the sum of its homogeneous factors. Since $P_m$ is definite by hypothesis, changing $P$ by $-P$ if necessary we can assume that $P_m(v)>0$ for all $v\in S^{n}_1$. Let $\alpha =\inf \{P_m(v):v\in S^{n}_1\}>0$. By the proof of theorem 2.1,

\[ t^{{-}m}P(tv)\overset{unif}{\longrightarrow} P_m(v) \text{ on } S^{n}_1 \text{ as }t\to\infty. \]

Then there exists $t_0>0$ such that $|t^{-m}P(tv)-P_m(v)|<\alpha /2$, for all $t>t_0$ and $v\in S^{n}_1$, and so

\[ t^{{-}m}P(tv)\geq{-}|t^{{-}m}P(tv)-P_m(v)|+P_m(v)>{-}\frac{\alpha}{2}+\alpha=\frac{\alpha}{2}>0, \]

for all $t>t_0$ and $v\in S^{n}_1$. Hence, $P(tv)>0$ for all $t>t_0$ and $v\in S^{n}_1$, which implies that the set $U^{-}=\{x\in \mathbb {R}^{n+1}:P(x)<0\}$ is bounded. Since $M=\partial (U^{-})$, one concludes that $M$ is compact.

Assume now that $M^{n}$ has constant mean curvature. Since $M^{n}$ is embedded, by a well known theorem of Alexandrov [Reference Aleksandrov1] each connected component of $M^{n}$ is a hypersphere of $\mathbb {R}^{n+1}$.

Proof of theorem 1.2. Since, by hypothesis, the degree $m$ of $P$ is less than or equal to $3$ and $M^{n}$ has non-zero constant mean curvature, it follows from theorem 1.1 that $m=2$. Hence, we can express $P$ as

(2.11)\begin{equation} P=P_0+P_1+P_2, \end{equation}

where $P_i,\,\;i=0,\,1,\,2$, is a homogeneous polynomial of degree $i$ in the variables $x_1,\,\ldots ,\,x_{n+1}$. Changing $P$ by $-P$ if necessary, we can assume that $H=c>0$. We will establish the theorem by proving that

(${\dagger}$) Up to a rigid motion of $\mathbb {R}^{n+1}$, $M^{n}=S^{k}\times \mathbb {R}^{n-k}$ for some $k,\,\;1\leq k\leq n$, where $S^{k}$ is a hypersphere of $\mathbb {R}^{k+1}$.

We will prove (${\dagger}$) by induction on $n$. If $n=2$, it follows from the well known classification of quadric surfaces in $\mathbb {R}^{3}$ and from the hypothesis that $M$ has non-zero constant mean curvature that $M$ is a sphere or a right circular cylinder. This shows that (${\dagger}$) holds for $n=2$.

Assume now $n\geq 3$ and that (${\dagger}$) holds for $n-1$. If $P_2$ is definite, then $M^{n}$ is a finite union of hyperspheres of $\mathbb {R}^{n+1}$ by theorem 2.4. Since the degree of $P$ is two, it follows from lemma 2.3 that $M^{n}$ is a single hypersphere of $\mathbb {R}^{n+1}$. Thus (${\dagger}$) holds for $n$ in the case that $P_2$ is definite.

If $P_2$ is indefinite, then there exists $w\in S^{n}_1$ such that $P_2(w)=0$. Then, after a change of coordinates given by an orthogonal transformation that sends $e_{n+1}$ to $w$, the polynomial P can be written as

(2.12)\begin{equation} P(x_1,\ldots,x_n,x_{n+1})=A_1(x_1,\ldots,x_n)x_{n+1}+A_0(x_1,\ldots,x_n), \end{equation}

where $A_1(x_1,\,\ldots ,\,x_n)$ is a polynomial of degree $\leq 1$ and $A_0(x_1,\,\ldots ,\,x_n)$ a polynomial of degree $\leq 2$.

Claim. $A_1=0$.

Assume that the claim is not true. Since the degree of $A_1$ is at most $1$, given $R>0$ there exists a closed ball $B\subset \mathbb {R}^{n}$ of radius $R$ on which $A_1(x_1,\,\ldots , x_n)\neq 0$. If $A_1>0$ on $B$, using (2.12) one easily verifies that $P(x_1,\,\ldots ,\,x_n,\,x_{n+1})>0$ for all $(x_1,\,\ldots ,\,x_n)\in B$ and all $x_{n+1}>\max _B |A_0|/\min _B A_1$. Similarly, if $A_1<0$ on $B$ one has $P(x_1,\,\ldots ,\,x_n,\,x_{n+1})>0$ for all $(x_1,\,\ldots ,\,x_n)\in B$ and all $x_{n+1}<\max _B |A_0|/\max _B A_1$. In either case, one sees that there is a ball of radius R in $\mathbb {R}^{n+1}$ entirely contained in the set $U^{+}=\{x\in \mathbb {R}^{n+1}:P(x)>0\}$. Then, by an argument used in the proof of theorem 2.1, $H\leq 1/R$ for every $R>0$, contradicting $H=c>0$. This proves the claim.

By (2.12) and the claim above, $P(x_1,\,\ldots ,\,x_n,\,x_{n+1})=A_0(x_1,\,\ldots ,\,x_n)$, where $A_0:\mathbb {R}^{n}\to \mathbb {R}$ is a polynomial of degree two in the variables $x_1,\,\ldots ,\,x_n$. Since 0 is a regular value for $P$, so is for $A_0$. Hence $M^{n}=M_0^{n-1}\times \mathbb {R}$, where $M_0^{n-1}=A_0^{-1}(0)$ is a regular algebraic hypersurface of $\mathbb {R}^{n}$. Since $M^{n}$ has non-zero constant mean curvature, the same is true of $M_0^{n-1}$. Then, by the induction hypothesis, up to a rigid motion of $\mathbb {R}^{n}$ one has $M_0^{n-1}=S^{k}\times \mathbb {R}^{n-1-k}$, for some $k,\,\;1\leq k\leq n-1$, where $S^{k}$ is a hypersphere of $\mathbb {R}^{k+1}$. Hence $M^{n}=S^{k}\times \mathbb {R}^{n-k}$, for some $k,\,\;1\leq k\leq n-1$. This shows that (${\dagger}$) holds for $n$ also in the case that $P_2$ is indefinite. Then, by the induction principle, (${\dagger}$) holds for every $n\geq 2$, and the theorem is proved.