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This was a statement I came across in a paper on what are called Transnormal curves by Bernd Wegner.

Two immersions $f_1$ and $f_2$ of a smooth manifold $M$ into $\mathbb{R}^n$ such that at each point $p \in M$, both immersions are connected by a parallel section of their respective normal bundles iff the normal planes to $f_1(M)$ and $f_2(M)$ coincide $\forall p \in M$ at $f_1(p)$ and $f_2(p)$
How can one prove this fact??

Further the author claims that in the special case of $M $ being $\mathbb{R}$ or $\mathbb{S}$, given the curves $\alpha = f_1$ and $\beta = f_2$, this boils down to saying

$$\beta(t)= \alpha(t)+ \lambda N(t)$$ and $$\text{prn}(\nabla_{\alpha'(t)}\bf{N}) = 0$$ where $\bf{N}$ denotes the principal normal vector field along $\alpha$ and $prn$ denotes the orthogonal projection to the corresponding normal (vector) space of $\alpha$ and $\alpha'(t)$ is the tangent vector field, with $\lambda$ being a constant.

What does the above condition tell me exactly??The first one is quite clearly saying that $\beta$ is a translation of $\alpha$ along the normal to $\alpha$, but what about the second condition about the projection of the covariant derivative??

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I found the result you quoted, also quoted in

Bernd Wegner: Parallel and transnormal curves on surfaces. New developments in differential geometry (Debrecen, 1994), 423–434, Math. Appl., 350, Kluwer Acad. Publ., Dordrecht, 1996.

There a reference for the proof is given:

B. WEGNER: Some remarks on parallel immersions, Coll. Math. Soc. J. Bolyai 56 (1989 /1991), 707-717.

It seems a little hard to find though.

As to the curves. I think your first condition is not correct, Wegner does not require $\mathbf{N}$ to be the principal normal: any normal vector field $\mathbf{N}$ satisfying the conditions will suffice. The second condition says precisely that $\mathbf{N}$ is parallel along $\alpha$. That is the covariant derivative of $\mathbf{N}$ along $\alpha$ is zero. Recall that the covariant derivative along $\alpha$ is defined to by taking a Euclidean derivative in directions tangent to $\alpha$ (which means $\alpha'$ for curves) and projecting it onto the tangent space $\alpha$.

Unfortunately, Wegner's phrasing seems a little imprecise. For example, take $M = \mathbb{R}$, $$ \alpha(t) = (0, 0, t) $$ the $z$-axis in $\mathbb{R}^3$, and $$ \beta(t) = (\cos t, \sin t, t) $$ a helix. Let $N(t) = (\cos t, \sin t, 0)$ along $\alpha$. Then $N$ is normal, and parallel with $$ \beta(t) = (0, 0, t) + (\cos t, \sin t, 0) = \alpha(t) + \mathbf{N}(t) $$ but the normal planes do not agree.

I think the correct statement would require $\beta'(t) \cdot \mathbf{N}(t) = 0$. That is $\mathbf{N}$ meets $\beta$ normally and I guess also that $\mathbf{N}(t)$ is parallel along $\beta$.

In general, it's not clear to me whether the normal section joining $f_1$ to $f_2$ is supposed to be the same section from the definition, but the example above seems to suggest this must be so. Matters are complicated further when Wegner states (in the first ref above)

But it should be pointed out, that in general a copy of a regular curve $\alpha$, obtained from $\alpha$ by some parallel transfer of the ambient Euclidean space, must not be parallel to $\alpha$ in the sense defined above.

Perhaps "must" should be changed to "may", since parallel curves do seem to satisfy the requirements provided the correct parametrisation is chosen, but not for every parametrisation.

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  • $\begingroup$ Thanks...your answer clears up some of my confusion. The paper cited in the references doesnt seem to be available anywhere... atleast where I am from. Any hints for a proof? Also, the curve you took as an example is just the axis of the helix, right? $\endgroup$ – Vishesh Aug 27 '16 at 0:55
  • $\begingroup$ Right, $\alpha$ is the axis of the helix $\beta$ and $\alpha(t) - \beta(t)$ is not normal to $\beta$ so you can't write $\alpha(t) = \beta(t) + \lambda \mathbf{N}(t)$. $\endgroup$ – Paul Bryan Aug 27 '16 at 12:58
  • $\begingroup$ As to the proof, I assume the following interpretation of the problem: $\mathbf{N}$ is a normal section along $f_1$, $f_2(p) = f_1(p) + \mathbf{N}(p)$ with $\mathbf{N}(p)$ meeting $f_2$ normally so we can think of $\mathbf{N}$ also as a section of the normal bundle of $f_2$. Then in the hypersurface case, you're done. For the higher co-dimension case, presumably one uses the parallel assumption. $\endgroup$ – Paul Bryan Aug 27 '16 at 13:05
  • $\begingroup$ Thanks! if you invert your example and take $\alpha(t)$ as the helix and then take $\beta(t) = \alpha(t) + \lambda N(t)$. $N$ here being the principal normal(a special case) to the helix, then assume that $\beta$'s binormal coincides with $\alpha$'s normal, you will get $\beta$ as the axis of the helix $\endgroup$ – Vishesh Aug 27 '16 at 13:35
  • $\begingroup$ Indeed. I think this is what Wegner's comment quote above is about. In this parametrisation of the helix you get that relation, but in the one I gave above you do not, which is why I suggest "may" instead of "must". $\endgroup$ – Paul Bryan Aug 27 '16 at 13:45

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