Now, any curve that lies on a sphere has positive curvature $\kappa$, but it need not have nonvanishing torsion $\tau$, of course. Thus, the above criterion does not make sense for all nondegenerate curves, i.e., space curves for which $\kappa$ is positive (which are the curves for which the classical Frenet frame is well-defined). The following argument shows that one cannot hope to have a necessary and sufficient condition expressed in terms of local conditions for spherical curves that works for all nondegenerate space curves: Consider two distinct spheres $S_1$ and $S_2$ that intersect along a circle $C$. It is easy to construct a smooth curve $x(s)$ with positive curvature $\kappa$ that starts out on $S_1$ (but not on $S_2$), runs along $C$ for some interval, and then continues on $S_2$ (after leaving $S_1$). This curve is locally spherical, but not globally spherical, so no local condition can be necessary and sufficient for all nondegenerate space curves.

Meanwhile, one can get an expression that makes sense for all nondegenerate curves by considering the identity $$ \frac{d\ }{ds}\left(\frac{(\kappa')^2 + \kappa^2\tau^2}{\kappa^4\tau^2}\right) = \frac{2\kappa'\bigl((\kappa\kappa''-2\kappa'^2-\kappa^2\tau^2)\,\tau - \kappa\kappa'\tau'\bigr)}{\kappa^5\tau^3}, $$ which shows that any nondegenerate space curve that satisfies the nondegeneracy condition that $\kappa$, $\kappa'$, and $\tau$ be nonvanishing while $$ P(\kappa,\kappa',\kappa'',\tau,\tau') = (\kappa\kappa''-2\kappa'^2-\kappa^2\tau^2)\,\tau - \kappa\kappa'\tau' = 0 $$ must necessarily lie on a sphere (of some unspecified radius). This is, in some sense, the correct statement of the classical result.

Now, any curve that lies on a sphere has positive curvature $\kappa$, but it need not have nonvanishing torsion $\tau$. Thus, the above criterion does not make sense for all nondegenerate curves, i.e., space curves for which $\kappa$ is positive (which are the curves for which the classical Frenet frame is well-defined). The following argument shows that one cannot hope to have a necessary and sufficient condition expressed in terms of local conditions for spherical curves that works for all nondegenerate space curves: Consider two distinct spheres $S_1$ and $S_2$ that intersect along a circle $C$. It is easy to construct a smooth curve $x(s)$ with positive curvature $\kappa$ that starts out on $S_1$ (but not on $S_2$), runs along $C$ for some interval, and then continues on $S_2$ (after leaving $S_1$). This curve is locally spherical, but not globally spherical, so no local condition can be necessary and sufficient for all nondegenerate space curves.

Meanwhile, one can get an expression that makes sense and works for all nondegenerate curves by considering the identity $$ \frac{d\ }{ds}\left(\frac{(\kappa')^2 + \kappa^2\tau^2}{\kappa^4\tau^2}\right) = \frac{2\kappa'\bigl((\kappa\kappa''-2\kappa'^2-\kappa^2\tau^2)\,\tau - \kappa\kappa'\tau'\bigr)}{\kappa^5\tau^3}. $$ In fact, any space curve that satisfies the nondegeneracy condition that $\kappa\tau$ be nonvanishing while $$ P(\kappa,\kappa',\kappa'',\tau,\tau') = (\kappa\kappa''-2\kappa'^2-\kappa^2\tau^2)\,\tau - \kappa\kappa'\tau' = 0 $$ must necessarily lie on a sphere of radius $r$ where $r^2 =\bigl((\kappa')^2 + \kappa^2\tau^2\bigr)/(\kappa^4\tau^2)$ . This is, in some sense, the correct statement of the classical result. (Proof: If the above equation holds and $\kappa\tau$ is nonvanishing, then the corresponding space curve $X:(a,b)\to\mathbb{R}^3$ satisfies the condition that $\bigl((\kappa')^2 + \kappa^2\tau^2\bigr)/(\kappa^4\tau^2)$ be constant. Moreover, if $T$, $N$, and $B$ are the Frenet frame of $X$, so that $X' = T$, $T' = \kappa N$, $N' = -\kappa T+\tau B$ and $B' = -\tau N$, where the prime denotes differentiation with respect to arclength, then the curve $Y = X + (1/\kappa)N - (\kappa'/(\kappa^2\tau)) B$ satisfies $Y' = 0$.)

When a space curve is fully nondegenerate, i.e., its curvature and torsion are nowhere vanishing, the necessary and sufficient condition for the curve to lie on a sphere is that the expression $((\kappa')^2 + \kappa^2\tau^2)/(\kappa^4\tau^2)$ should be constant. Indeed, when it is constant, this expression is just the square of the radius of the sphere on which the curve lies. Now, any curve that lies on a sphere has positive curvature $\kappa$, but it need not have nonvanishing torsion $\tau$, of course. Thus, the above criterion does not make sense for all nondegenerate curves, i.e., space curves for which $\kappa$ is positive (which are the curves for which the classical Frenet frame is well-defined). The following argument shows that one cannot hope to have a necessary and sufficient condition expressed in terms of local conditions for spherical curves that works for all nondegenerate space curves: Consider two distinct spheres $S_1$ and $S_2$ that intersect along a circle $C$. It is easy to construct a smooth curve $x(s)$ with positive curvature $\kappa$ that starts out on $S_1$ (but not on $S_2$), runs along $C$ for some interval, and then continues on $S_2$ (after leaving $S_1$). This curve is locally spherical, but not globally spherical, so no local condition can be necessary and sufficient for all nondegenerate space curves.

The reader may ask, "What about assuming real-analyticity?". However, real-analyticity is not necessary for a space curve to be spherical, so this does not count as a local criterion that would be be part of a necessary and sufficient condition for any nondegenerate space curve to be spherical. I think that the reasonable thing to do is to simply restrict to the class of fully nondegenerate space curves, for which a necessary and sufficient condition to be spherical is available. (Alternatively, one could restrict to the space of real-analytic nondegenerate space curves, but then one has to make a special exception for the planar circles, since the above criterion does not make sense for them.)

The seriousness of this problem becomes evident in the case of what we might call ellipsoidal space curves, i.e., the space curves that lie on some ellipsoid. Now, even full nondegeneracy is not sufficient to avoid the above difficulty, for one can easily write down a pair of ellipsoids that intersect in a space curve that is not planar and hence contains fully nondegenerate arcs. In particular, one can construct a fully nondegenerate space curve that is locally ellipsoidal but is not globally ellipsoidal. Thus, even the right notion of 'nondegenerate' needs to be specified in order to get anywhere.

When a space curve is sufficiently nondegenerate, i.e., its curvature $\kappa$ and torsion $\tau$ and the first derivative of its curvature with respect to arclength $\kappa'$ are nowhere vanishing, the necessary and sufficient condition for the curve to lie on a sphere is that the expression $\bigl((\kappa')^2 + \kappa^2\tau^2\bigr)/(\kappa^4\tau^2)$ be constant. Indeed, when it is constant under these nondegeneracy hypotheses, this expression is just the square of the radius of the sphere on which the curve lies. Note that the assumption that $\kappa'$ be nonzero is necessary: If $\kappa'$ vanishes identically, then the above expression is constant (because the $\tau$-factors cancel), but not every curve with constant $\kappa$ lies on a sphere (for example, consider the circular helices, which have $\kappa$ and $\tau$ constant and nonzero).

Now, any curve that lies on a sphere has positive curvature $\kappa$, but it need not have nonvanishing torsion $\tau$, of course. Thus, the above criterion does not make sense for all nondegenerate curves, i.e., space curves for which $\kappa$ is positive (which are the curves for which the classical Frenet frame is well-defined). The following argument shows that one cannot hope to have a necessary and sufficient condition expressed in terms of local conditions for spherical curves that works for all nondegenerate space curves: Consider two distinct spheres $S_1$ and $S_2$ that intersect along a circle $C$. It is easy to construct a smooth curve $x(s)$ with positive curvature $\kappa$ that starts out on $S_1$ (but not on $S_2$), runs along $C$ for some interval, and then continues on $S_2$ (after leaving $S_1$). This curve is locally spherical, but not globally spherical, so no local condition can be necessary and sufficient for all nondegenerate space curves.

Meanwhile, one can get an expression that makes sense for all nondegenerate curves by considering the identity $$ \frac{d\ }{ds}\left(\frac{(\kappa')^2 + \kappa^2\tau^2}{\kappa^4\tau^2}\right) = \frac{2\kappa'\bigl((\kappa\kappa''-2\kappa'^2-\kappa^2\tau^2)\,\tau - \kappa\kappa'\tau'\bigr)}{\kappa^5\tau^3}, $$ which shows that any nondegenerate space curve that satisfies the nondegeneracy condition that $\kappa$, $\kappa'$, and $\tau$ be nonvanishing while $$ P(\kappa,\kappa',\kappa'',\tau,\tau') = (\kappa\kappa''-2\kappa'^2-\kappa^2\tau^2)\,\tau - \kappa\kappa'\tau' = 0 $$ must necessarily lie on a sphere (of some unspecified radius). This is, in some sense, the correct statement of the classical result.

The reader may ask, "What about assuming real-analyticity?". However, real-analyticity is not necessary for a space curve to be spherical, so this does not count as a local criterion that would be be part of a necessary and sufficient condition for any nondegenerate space curve to be spherical. I think that the reasonable thing to do is to simply restrict to the appropriate class of sufficiently nondegenerate space curves, for which a necessary and sufficient condition to be spherical is available. (Alternatively, one could restrict to the space of real-analytic nondegenerate space curves, but then one has to make a special exception for the curves with constant $\kappa$, etc.)

The seriousness of this problem becomes evident in the case of what we might call ellipsoidal space curves, i.e., the space curves that lie on some ellipsoid. Now, even the above nondegeneracy ($\kappa$, $\kappa'$, and $\tau$ be nonvanishing) is not sufficient to avoid the above difficulty, for one can easily write down a pair of ellipsoids that intersect in a space curve that is not planar and does not have constant $\kappa$ and hence contains such nondegenerate arcs. In particular, one can construct such a nondegenerate space curve that is locally ellipsoidal but is not globally ellipsoidal. Thus, even the right notion of 'nondegenerate' needs to be specified in order to get anywhere.