# In search of a new isometric twisting invariant $T= \tau_1.\tau_2$

A curved line in $$\mathbb R^3$$ has properties of curvature and torsion, and, on an $$\mathbb R^2$$ possesses surface scalar properties of normal curvature and geodesic torsion $$(\kappa_n, \tau_g).$$

In isometric mappings product of extreme $$\kappa_n's$$ remains constant by Egregium theorem.

Euler normal curvatures are from tangent plane $$(\dfrac{dT}{ds}= \kappa_n N ):$$

$$\kappa_n= \kappa_1 \cos^2 \psi+\kappa_2 \sin^2 \psi$$

What about $$\tau_g ?$$ Is there any theorem corresponding to curvature dealing with torsions as well? Like in normal plane ( since $$\dfrac{dB}{ds}= -\tau_g N ):$$

$$\tau_g = \tau_1 \cos^2\phi + \tau_2 \sin ^2\phi ?$$

where $$\phi$$ is angle between principal binormal direction and reference binormal direction.

And what about the product of extreme $$\tau_g's ?$$ Does a corresponding scalar $$"T "$$ also remain constant in isometry?

The question is about conditions for existence of an isometric twisting invariant $$"T"$$ gleaned from this geometrical analogue just as Gauss curvature $$K$$ is a bending invariant.

The statement needs to be improved to remove vagueness of expressing torsion. I request your help to refine the question and point to an answer to the extent it makes sense in this context. And please suggest references also.

I have tried to represent ( a rough sketch.. shall improve it) the two parameters in bipolar coordinates together. Motivation for bipolar plot representation of curvature/torsion tensor view is.. to work on a prospect to build premises that if $$K$$ invariance is true then $$T$$ invariance must consequently be also true...and a twist invariant $$T$$ that rank with $$K$$ would be found/identified.

EDIT 1:

I mean we should be able to formulate for a surface of revolution of constant torsion product T.

T=K in magnitude as geometric mean of segment lengths as shown in rough sketch.. but its definition is drastically different. Can we have a formulation (and views if possible) of any example of a surface of constant P with parametrization obtained by integration of an ODE with arbitrary constants in two or three twisted stages where individual principal geodesic torsions multiply to same product P? The answer to such question for constant K is well known.

A metric in normal $$(B-N)$$ plane has to be found.

I wonder if topic of Spinor Bundles is relavant here.

• I am looking at curvature & torsion together. Developable surfaces give K zero, away from generality. Also Apollonian curvature circles and the concurrent circles rank together in isometry.for curvature and torsion respectively. $\phi$ is expected to be bounded as shown in the $\tau_g$ circles. – Narasimham Dec 15 '18 at 16:36
• @Ben McKay In that situation $K=0,T=0$ for both circle sets tangential to $\kappa_n, \tau_g$ axes at origin.If formulation of latter is easy, please indicate how. – Narasimham Dec 16 '18 at 5:49

Any nonconstant scalar invariant $$T$$ depending on the values of $$\tau_g$$ in the various directions at a point is not invariant under isometry. Compute that $$\tau_g=(k_1-k_2)\sin \psi \cos \psi$$, so maxima and minima are from directions perpendicular to the principal directions. If nonconstant as a function of $$k_1,k_2$$, any invariant $$T=T(k_1,k_2)$$ will determine (along with the isometry invariant Gauss curvature) the squared mean curvature, as these are the elementary symmetric functions of $$k_1,k_2$$ (up to constant multiples). But the squared mean curvature is not isometry invariant. In fact, because all real analytic metrics on surface are locally embeddable isometrically in Euclidean 3-space, any isometry invariants can only depend on the intrinsic geometry. If an invariant is rational in highest derivatives it is expressible (by a theorem of Weyl, I think) in the Gauss curvature and its covariant derivatives. I don't know a good reference, though, so I hope someone will help with that. Anyway, we aren't missing any invariants in surface theory.

• Thanks. Was thinking if we may be able to derive a relation for $T= \tau_1.\tau_2$ starting from first/ second/third fundamental form coefficients, their derivatives and Christoffel symbols in a way similar to the derivation of $K$. – Narasimham Dec 16 '18 at 13:17
• Mc Kay I am not referring to extreme values $\tau_g = \pm (k1-k2)/2$. This is the case when still referring to red $\kappa$ Apollonian circles. To be clear I have marked the extreme values $\tau_1,\tau_2$ as top and bottom points of new "Mohr" circles shown in black. I place in quotes as they are $not$ Mohr Circles representing curvature,stress/strain moment of inertia etc. among other tensors. – Narasimham Dec 16 '18 at 13:51
• It doesn't matter how you define $T$, you still cannot produce a new isometry invariant as a function of $k_1$ and $k_2$. The values of $k_1$ and $k_2$ can vary under isometric deformation arbitrarily, subject only to their product remaining invariant. You don't get any other invariant. And no matter how you define $T$ in terms of $\tau_g$, since $\tau_g$ is a quadratic form with $k_1,k_2$ as coefficients, defined in terms of the surface 2-jet, there is no more data to work from than the isometry invariants of the 2-jet of the surface, which is just given by Gauss curvature. – Ben McKay Dec 16 '18 at 15:11
• As far I understood your answer and comments relate to the curvature set but not to the twist set choice between two possibilities of the bipolar coordinate scheme. – Narasimham Dec 20 '18 at 18:56