Return to Answer

Remark: There may be similar arguments that can be made for $S^n$ ($n>3$) using an appropriate symmetry reduction via a group action of cohomogeneity $2$, but not all of the induced metrics on the orbit spaces have a continuous symmetry, and hence the Clairaut integral does not always exist, which was essential in the above argument (so that an appeal could be made to Liouville's Theorem).

Remark 1: There may be similar arguments that can be made for $S^n$ ($n>3$) using an appropriate symmetry reduction via a group action of cohomogeneity $2$, but not all of the induced metrics on the orbit spaces have a continuous symmetry, and hence the Clairaut integral does not always exist, which was essential in the above argument (so that an appeal could be made to Liouville's Theorem).

Remark 2: There is an interesting example associated to this question: If, instead of the standard metric on $S^3$ of constant sectional curvature $+1$, one considers the 'squashed' (but still smooth and positive definite) metric $$ G = \mathrm{d}z\,\mathrm{d}\bar z + \mathrm{d}w\,\mathrm{d}\bar w - (\mathrm{d}f)^2 $$ where $f = \tfrac14\sqrt{1+8z\bar z}$ (a smooth function on $S^3$) and sets $(p,q)=(3,1)$, then the metric on the (orbifold) orbit space whose geodesics describe minimal surfaces in this metric that are invariant under the circle action turns out to be the metric on the Tannery pear orbifold (up to a constant multiple). In particular, not only are all the geodesics closed in Tannery's pear, but they are all, in addition, algebraic. Hence, the corresponding minimal tori of revolution in this metric are algebraic as well (using the standard algebraic structure on $S^3$).

The answer to this question is 'no' for minimal surfaces of revolution in the $3$-sphere:

Consider surfaces in $S^3 = \{\,(z,w)\in\mathbb{C}^2\,|\,|z|^2+|w|^2=1\,\}$ that are invariant under the circle action $$ \mathrm{e}^{i\theta}\cdot(z,w) = \bigl(\mathrm{e}^{ip\theta}\,z,\,\mathrm{e}^{iq\theta}\,w\,\bigr) $$ where $p\ge q\ge 0$ are relatively prime integers. A smooth surface invariant under this action that avoids the circles $z=0$ and $w=0$ can be parametrized in the form $$ \bigl(z(\theta,s),w(\theta,s)\bigr) = \left( \mathrm{e}^{i(p\theta-q^*\phi(s))}\left(\tfrac12(1+u(s)\right)^{1/2},\, \mathrm{e}^{i(q\theta+p^*\phi(s))}\left(\tfrac12(1-u(s)\right)^{1/2} \right) $$ for some functions $\phi(s)$, defined modulo $2\pi$, and $u(s)$, satisfying $|u(s)|<1$. In this formula, $q^* = q/(p^2+q^2$ and $p^*=p/(p^2+q^2)$.

Such a surface can be shown to be minimal if and only if the curve $\bigl(u(s),\phi(s)\bigr)$ is, up to reparametrization, a geodesic for the metric $$ g = \frac{\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{2(1-u^2)}\,\mathrm{d}u^2 + (1-u^2)\,\mathrm{d}\phi^2. $$ Moreover, the parametrized surface will be an algebraic surface if and only if $u(s)$ and $v(s) = \tan\phi(s)$ are algebraically related.

When $p\ge q>0$, the metric $g$ is a smooth metric on an orbifold of rotation (the 'poles' of the rotation are the orbifold points, of orders $p$ and $q$ respectively), with Gauss curvature $$ K = \frac{2p^2+2q^2+(p^2{-}q^2)\,u}{(\,p^2+q^2+(p^2{-}q^2)\,u\,)^2} > 0. $$ (In the special case $(p,q)=(1,0)$, the singularity at the 'pole' $u=-1$ is not an orbifold point, but the metric is, of course, smooth on the disk $u>-1$.)

Making use of the Clairaut first integral, one finds that a geodesic that does not have $\phi$ constant (and hence, does not pass through the orbifold 'poles') and does not have $u$ constant must satisfy a relation of the form $$ \frac{m}{1-u^2}\sqrt{\frac{2\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{\bigl(1-u^2-4m^2\bigr)}}\,\mathrm{d}u -\frac{\mathrm{d}v}{1+v^2}\, = 0\tag1 $$ for some constant $m$ with $0<|m|<\tfrac12$.

Now, it is not hard to show, using Liouville's Theorem, that the first term in $(1)$ is not the differential of any elementary function of $u$ when $p^2{-}q^2>0$. (Obviously, it is such a differential when $p^2{-}q^2 = 0$, but, otherwise, this is an elliptic integral.) Hence, $u$ and $v = \tan\phi$ cannot be algebraically related, since expressing $v$, even locally, as an algebraic function of $u$ would express the first term in $(1)$ as the differential of $\tan^{-1}(v)$, which would be an elementary function of $u$. (For those unfamiliar with Liouville's Theorem and the theory of integration in terms of elementary functions, these notes by Brian Conrad give a very useful and clear introduction.)

Meanwhile, the set of values of $m$ satisfying $0<|m| < \tfrac12$ for which the geodesic described by the above relation closes smoothly is dense in the interval $(-\tfrac12,\tfrac12)$, since the integral of the first term in $(1)$ over the interval $u^2<1{-}4m^2$ is not constant in $m$, as it is an odd function of $m$ that is not equal to zero when $m$ is not zero.

I would expect that a similar argument can be made for $S^n$ ($n> 3$) using an appropriate symmetry reduction via a group action of cohomogeneity $2$.

The answer to this question is 'no' for most minimal surfaces of revolution in the $3$-sphere.

Consider surfaces in $S^3 = \{\,(z,w)\in\mathbb{C}^2\,|\,|z|^2+|w|^2=1\,\}$ that are invariant under the circle action $$ \mathrm{e}^{i\theta}\cdot(z,w) = \bigl(\mathrm{e}^{ip\theta}\,z,\,\mathrm{e}^{iq\theta}\,w\,\bigr) $$ where $p\ge q\ge 0$ are relatively prime integers. The ring of polynomials invariant under this circle action is generated by $z\bar z$ and $w\bar w$ (of degree $2$) and the real and imaginary parts of $\zeta = {\bar z}^q\,w^p$ (of degree $p{+}q$). These are subject to a single relation, $\zeta\bar\zeta = (z\bar z)^q(w\bar w)^p$ of degree $2(p{+}q)$. Setting $u = z\bar z-w\bar w$, the mapping $(u,\zeta):S^3\to \mathbb{R}\times\mathbb{C}$ embeds the space of orbits in $S^3$ (where $z\bar z+ w\bar w = 1$) as the algebraic (orbifold) variety $(1{+}u)^q(1{-}u)^p-2^{p+q}\,\zeta\bar\zeta=0$.

A smooth surface in $S^3$ that is invariant under this action and that avoids the circles $z=0$ and $w=0$ can be parametrized in the form $$ \bigl(z(\theta,s),w(\theta,s)\bigr) = \left( \mathrm{e}^{i(p\theta-q^*\phi(s))} \left(\tfrac12(1+u(s)\right)^{1/2}, \, \mathrm{e}^{i(q\theta+p^*\phi(s))} \left(\tfrac12(1-u(s)\right)^{1/2} \right) $$ for some functions $\phi(s)$, defined modulo $2\pi$, and $u(s)$, satisfying $|u(s)|<1$. In this formula, $q^* = q/(p^2{+}q^2)$ and $p^*=p/(p^2{+}q^2)$.

Calculation shows that such a surface is minimal if and only if the curve $\bigl(u(s),\phi(s)\bigr)$ is, up to reparametrization, a geodesic for the metric $$ g = \frac{\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{2(1-u^2)}\,\mathrm{d}u^2 + (1-u^2)\,\mathrm{d}\phi^2. $$ Moreover, the parametrized surface will be an algebraic surface if and only if $u(s)$ and $v(s) = \tan\phi(s)$ are algebraically related, since, on such a surface, we will have $$ \zeta(s) = \mathrm{e}^{i\phi(s)} \left(\tfrac12(1{+}u(s)\right)^{q/2} \left(\tfrac12(1{-}u(s)\right)^{p/2} = \frac{1{+}i\,v(s)}{\sqrt{1{+}v(s)^2}} \left(\tfrac12(1{+}u(s)\right)^{q/2} \left(\tfrac12(1{-}u(s)\right)^{p/2} $$

When $p\ge q>0$, the metric $g$ is a smooth metric on an orbifold of rotation (the 'poles' of the rotation are the orbifold points, of orders $p$ and $q$ respectively), with Gauss curvature $$ K = \frac{2p^2+2q^2+(p^2{-}q^2)\,u}{(\,p^2+q^2+(p^2{-}q^2)\,u\,)^2} > 0. $$ (In the special case $(p,q)=(1,0)$, the singularity at the 'pole' $u=-1$ is not an orbifold point, but the metric is, of course, still smooth on the disk $u>-1$.)

Making use of the Clairaut first integral for metrics of revolution, one finds that a geodesic that does not have $\phi$ constant (and hence, does not pass through the orbifold 'poles') and does not have $u$ constant must satisfy a relation of the form $$ \frac{m}{1-u^2}\sqrt{\frac{2\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{\bigl(1-u^2-4m^2\bigr)}}\,\mathrm{d}u -\frac{\mathrm{d}v}{1+v^2}\, = 0\tag1 $$ for some constant $m$ with $0<|m|<\tfrac12$.

Now, it is not hard to show, using Liouville's Theorem, that the first term in $(1)$ is not the differential of any elementary function of $u$ when $p^2{-}q^2>0$. (Obviously, it is the differential of an elementary function when $p^2{-}q^2 = 0$, but, otherwise, it is the differential of a nontrivial linear combination of incomplete elliptic integrals of the third kind in Legendre's terminology.) Hence, $u$ and $v = \tan\phi$ cannot be algebraically related, since expressing $v$, even locally, as an algebraic function of $u$ would express the first term in $(1)$ as the differential of $\tan^{-1}(v)$, which would be an elementary function of $u$. (For those unfamiliar with Liouville's Theorem and the theory of integration in terms of elementary functions, these notes by Brian Conrad give a very useful and clear introduction.)

Meanwhile, the set of values of $m$ satisfying $0< |m| < \tfrac12$ for which the geodesic described by the above relation closes smoothly is dense in the interval $(-\tfrac12,\tfrac12)$, since the integral of the first term in $(1)$ over the interval $u^2<1{-}4m^2$ is not constant in $m$, as it is an odd function of $m$ that is not equal to zero when $m$ is not zero.

Remark: There may be similar arguments that can be made for $S^n$ ($n>3$) using an appropriate symmetry reduction via a group action of cohomogeneity $2$, but not all of the induced metrics on the orbit spaces have a continuous symmetry, and hence the Clairaut integral does not always exist, which was essential in the above argument (so that an appeal could be made to Liouville's Theorem).

The answer to this question is 'no' for minimal surfaces of revolution in the $3$-sphere:

Consider surfaces in $S^3 = \{\,(z,w)\in\mathbb{C}^2\,|\,|z|^2+|w|^2=1\,\}$ that are invariant under the circle action $$ \mathrm{e}^{i\theta}\cdot(z,w) = \bigl(\mathrm{e}^{ip\theta}\,z,\,\mathrm{e}^{iq\theta}\,w\,\bigr) $$ where $p\ge q\ge 0$ are relatively prime integers. A smooth surface invariant under this action that avoids the circles $z=0$ and $w=0$ can be parametrized in the form $$ \bigl(z(\theta,s),w(\theta,s)\bigr) = \left( \mathrm{e}^{i(p\theta-q\phi(s))}\left(\tfrac12(1+u(s)\right)^{1/2},\, \mathrm{e}^{i(q\theta+p\phi(s))}\left(\tfrac12(1-u(s)\right)^{1/2} \right) $$ for some functions $\phi(s)$ and $u(s)$ with $u(s)^2<1$.

Such a surface can be shown to be minimal if and only if the curve $\bigl(u(s),\phi(s)\bigr)$ is, up to reparametrization, a geodesic for the metric $$ g = \frac{\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{8(1-u^2)}\,\mathrm{d}u^2 + \tfrac14(p^2{+}q^2)^2(1-u^2)\,\mathrm{d}\phi^2. $$ Moreover, the parametrized surface will be an algebraic surface if and only if $u(s)$ and $v(s) = \tan\phi(s)$ are algebraically related.

Now, the metric $g$ can be seen as a smooth metric on an orbifold of rotation (the 'poles' of the rotation are the orbifold points, of orders $p$ and $q$ respectively). Making use of the Clairaut first integral, one finds that a geodesic that does not have $\phi$ constant (and hence, does not pass through the orbifold 'poles') and does not have $u$ constant must satisfy a relation of the form $$ \frac{m}{1-u^2}\sqrt{\frac{2\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{\bigl(1-u^2-4m^2\bigr)}}\,\mathrm{d}u -\frac{(p^2{+}q^2)}{1+v^2}\,\mathrm{d}v = 0\tag1 $$ for some constant $m\not=0$ with $|m|<\tfrac12$.

Now, it is not hard to show, using Liouville's Theorem, that the first term is not the differential of any elementary function of $u$ when $p^2-q^2>0$. (Obviously, it is such a differential when $p^2 -q^2 = 0$, but, otherwise, this is clearly an elliptic integral.) Hence, $u$ and $v = \tan\phi$ cannot be algebraically related, since expressing $v$, even locally, as an algebraic function of $u$ would express the first term in $(1)$ as the differential of $\tan^{-1}(v)$, which would be an elementary function of $u$. (For those unfamiliar with Liouville's Theorem and the theory of integration in terms of elementary functions, these notes by Brian Conrad give a very useful and clear introduction.)

Meanwhile, the set of values of $m$ satisfying $0<4m^2 < 1$ for which the geodesic described by the above relation closes smoothly is dense in the interval $[-1/2,1/2]$, since the integral of the first term in $(1)$ over the interval $u^2<1{-}4m^2$ clearly depends on $m$, as it is an odd function that is clearly not equal to zero when $m$ is not zero.

Hence there exist such $m$ that yield a closed geodesic on the orbifold of rotation, which lifts to a closed minimal surface in $S^3$ that is not algebraic.

I would expect that a similar argument can be made for $S^n$ ($n> 3$) using an appropriate symmetry reduction via a group action of cohomogeneity $2$.

The answer to this question is 'no' for minimal surfaces of revolution in the $3$-sphere:

Consider surfaces in $S^3 = \{\,(z,w)\in\mathbb{C}^2\,|\,|z|^2+|w|^2=1\,\}$ that are invariant under the circle action $$ \mathrm{e}^{i\theta}\cdot(z,w) = \bigl(\mathrm{e}^{ip\theta}\,z,\,\mathrm{e}^{iq\theta}\,w\,\bigr) $$ where $p\ge q\ge 0$ are relatively prime integers. A smooth surface invariant under this action that avoids the circles $z=0$ and $w=0$ can be parametrized in the form $$ \bigl(z(\theta,s),w(\theta,s)\bigr) = \left( \mathrm{e}^{i(p\theta-q^*\phi(s))}\left(\tfrac12(1+u(s)\right)^{1/2},\, \mathrm{e}^{i(q\theta+p^*\phi(s))}\left(\tfrac12(1-u(s)\right)^{1/2} \right) $$ for some functions $\phi(s)$, defined modulo $2\pi$, and $u(s)$, satisfying $|u(s)|<1$. In this formula, $q^* = q/(p^2+q^2$ and $p^*=p/(p^2+q^2)$.

Such a surface can be shown to be minimal if and only if the curve $\bigl(u(s),\phi(s)\bigr)$ is, up to reparametrization, a geodesic for the metric $$ g = \frac{\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{2(1-u^2)}\,\mathrm{d}u^2 + (1-u^2)\,\mathrm{d}\phi^2. $$ Moreover, the parametrized surface will be an algebraic surface if and only if $u(s)$ and $v(s) = \tan\phi(s)$ are algebraically related.

When $p\ge q>0$, the metric $g$ is a smooth metric on an orbifold of rotation (the 'poles' of the rotation are the orbifold points, of orders $p$ and $q$ respectively), with Gauss curvature $$ K = \frac{2p^2+2q^2+(p^2{-}q^2)\,u}{(\,p^2+q^2+(p^2{-}q^2)\,u\,)^2} > 0. $$ (In the special case $(p,q)=(1,0)$, the singularity at the 'pole' $u=-1$ is not an orbifold point, but the metric is, of course, smooth on the disk $u>-1$.)

Making use of the Clairaut first integral, one finds that a geodesic that does not have $\phi$ constant (and hence, does not pass through the orbifold 'poles') and does not have $u$ constant must satisfy a relation of the form $$ \frac{m}{1-u^2}\sqrt{\frac{2\bigl((p^2{+}q^2)+(p^2{-}q^2)u\bigr)}{\bigl(1-u^2-4m^2\bigr)}}\,\mathrm{d}u -\frac{\mathrm{d}v}{1+v^2}\, = 0\tag1 $$ for some constant $m$ with $0<|m|<\tfrac12$.

Now, it is not hard to show, using Liouville's Theorem, that the first term in $(1)$ is not the differential of any elementary function of $u$ when $p^2{-}q^2>0$. (Obviously, it is such a differential when $p^2{-}q^2 = 0$, but, otherwise, this is an elliptic integral.) Hence, $u$ and $v = \tan\phi$ cannot be algebraically related, since expressing $v$, even locally, as an algebraic function of $u$ would express the first term in $(1)$ as the differential of $\tan^{-1}(v)$, which would be an elementary function of $u$. (For those unfamiliar with Liouville's Theorem and the theory of integration in terms of elementary functions, these notes by Brian Conrad give a very useful and clear introduction.)

Meanwhile, the set of values of $m$ satisfying $0<|m| < \tfrac12$ for which the geodesic described by the above relation closes smoothly is dense in the interval $(-\tfrac12,\tfrac12)$, since the integral of the first term in $(1)$ over the interval $u^2<1{-}4m^2$ is not constant in $m$, as it is an odd function of $m$ that is not equal to zero when $m$ is not zero.

Hence there exist such $m$ that yield a closed geodesic on the orbifold of rotation, which lifts to a closed minimal surface in $S^3$ that is not algebraic.

I would expect that a similar argument can be made for $S^n$ ($n> 3$) using an appropriate symmetry reduction via a group action of cohomogeneity $2$.