4

Having finished an introductory course to GR as part of my physics undergrad degree, I decided to look at some differential geometry over the summer, and picked up Schutz's Geometrical methods of mathematical physics. On page 82 (in case you have access to the book), in a section on foliation, the author introduces a helical vector field:

$\frac{\mathrm{dx}}{\mathrm{d}\lambda} = -\sin \lambda, \frac{\mathrm{dy}}{\mathrm{d}\lambda} = \cos \lambda, \frac{\mathrm{dz}}{\mathrm{d}\lambda} = 1 $

The integral curves of this field are helices, which appear to form a congruence over the manifold. Their equations are $x-x_0 = \cos\lambda, y-y_0 = \sin\lambda, z=\lambda$.

Now, the author says, this vector field, together with the x-basis vector field (so I guess that's $\frac{\partial}{\partial x}$ at every point of the manifold), form a family of surfaces, each point in $\mathbb{R}^3$ being on one, so it foliates the manifold.

However, the helical vector field, plus the z-basis vector field, do not together form a submanifold. The author specifically writes: "The plane defined by the two at any point is not tangent to the 'next' spiral curve above or below it."

This has me confused - what does he mean? I thought that I should calculate the Lie brackets for each of the two pairs; they will foliate the manifold if their Lie brackets at all points are linear combinations of the two fields at that point (is that right?). However, I get that for neither pair does this condition hold.

In the first case, I have

$\overline{V} = -\sin\lambda\frac{\partial}{\partial x} + \cos\lambda\frac{\partial}{\partial y} + \frac{\partial}{\partial z},\\ \overline{X} = \frac{\partial}{\partial x}\\ \left[\overline{V},\overline{X}\right] = (\frac{\partial}{\partial x} \sin\lambda)\frac{\partial}{\partial x}-(\frac{\partial}{\partial x}\cos\lambda)\frac{\partial}{\partial y} $

Even without explicitly evaluating $\frac{\partial}{\partial x}\sin\lambda$ etc. I can already tell that the bracket is not tangent to a surface defined by the helix and x-axis vector fields, as the two tangent vectors at any point are $\overline{V}$ and $\overline{X}$, and the Lie bracket is just not a linear combination of them (except for points where the $\frac{\partial}{\partial y}$ coefficient vanishes).

In the second case, the situation is very similar, up to a rescaling factor.

So there are my questions: was I correct in looking at the Lie brackets of the helical vector field and the x-axis (z-axis) vector field to check whether they define a foliation, and if so, how does one get the correct Lie bracket, as I seem to be getting the wrong one!

Many thanks,

Dom

dom_miketa
  • 159
  • It's not clear to me at all what the vector field actually is. He writes an ODE with time-dependent right-hand side ($\lambda$ being the time variable, of course), but vector fields correspond to ODEs with time-independent right-hand sides... – Hans Lundmark Jul 17 '14 at 07:52
  • My understanding is that the vector field defines a vector at every point of the manifold, and, if the field in question is at least once differentiable, it's always possible to construct a curve passing through a fixed point and threading through the manifold, such that at each point the curve's tangent vector is identical to the vector field at that point. $\lambda$ is then better thought of not as time, but as a parameter along just such a curve, and $\overline{V} = \frac{\mathrm{d}}{\mathrm{d}\lambda} = \frac{\mathrm{d}x^i}{\mathrm{d}\lambda} \frac{\partial}{\partial x^i}$. – dom_miketa Jul 17 '14 at 08:01
  • The components of the vector are then the $\frac{\mathrm{d}x^i}{\mathrm{d}\lambda}$, and the basis vectors are $\frac{\partial}{\partial x^i}$. The goal is to find a commutator between $\overline{V}$ and $\overline{X}$ which ought to be a linear combination of the two vector fields, I think. But since that seems to be impossible, I must be wrong about this on some level, and I was hoping someone here could give me a hint. – dom_miketa Jul 17 '14 at 08:13
  • No matter if you call the parameter "time" or something else, it shouldn't appear in the right-hand side of the ODEs. A system of ODEs of the form $dx/dt = F(x)$ corresponds to a vector field (and conversely), namely the one where the vector $F(x)$ is attached to the point $x$. But a system of the form $dx/dt=F(x,t)$ doesn't define a vector field. – Hans Lundmark Jul 17 '14 at 08:27
  • But the solutions are just parametrised integral curves of the vector field. Once you fix $x_0$ and $y_0$ (the centre of the helix), you get a congruence of helices filling out the entire manifold, and a sliding parameter $\lambda$ on each of them, which actually corresponds to the $z$ coordinate, so really the parameter trivially washes out... Am I missing something? – dom_miketa Jul 17 '14 at 08:31
  • OK, I guess one could say that the vector field is $-(\sin z)\partial_x + (\cos z) \partial_y + \partial_z$. That would be unambiguous. But you could also replace $z$ by $z-z_0$ to get another vector field whose flow equally well satisfies the ODE given in the book, so I still think the description given there is unclear. – Hans Lundmark Jul 17 '14 at 09:50
  • I'm not sure if this changes much. Just set $z_0=0$ for the argument's sake. This doesn't make the Lie bracket tangent to the surface defined by $\overline{V}$ and $\overline{X}$. – dom_miketa Jul 17 '14 at 09:57
  • The Lie bracket between $X=\partial_x$ and the vector field $V$ in my previous comment is identically zero. Since $V$ is translationally invariant in the $x$ and $y$ directions, its flow commutes with the flows generated by $\partial_x$ and $\partial_y$. But $V$ is not invariant with respect to translations in the $z$ direction, and in fact $[V,\partial_z]=(\cos z)\partial_x + (\sin z)\partial_y$, which is not in the span of $V$ and $\partial_z$. – Hans Lundmark Jul 17 '14 at 10:27
  • I think the problem with your computation is that $\partial(\sin\lambda)/\partial x$ doesn't really make sense. But $\partial(\sin z)/\partial x$, on the other hand, is simply zero. – Hans Lundmark Jul 17 '14 at 10:31
  • Thanks, Hans, I think that's going to be the right solution. Very appreciated.

    For the record, my $\frac{\partial}{\partial\lambda}\cos\lambda$ etc. were to be calculated by noting that $x-x_0 = \cos\lambda$. I'm still confused why that failed so spectacularly.

     – dom_miketa Jul 17 '14 at 10:35
  • Mmm, I think I have it - $\frac{\partial}{\partial x}$ takes me to the next helix of the congruence along the x axis. There, $\sin\lambda$ at that point is still the same, which is another way to state that the vector field is symmetrical with respect to translation in the x-y plane.

    Thank you again, Hans, this was very helpful.

     – dom_miketa Jul 17 '14 at 12:28
  • You're welcome! – Hans Lundmark Jul 17 '14 at 14:56
  • Related to this: Does anyone understand the figure in the book about this? It is supposed to depict the surfaces of the foliation, but I cannot see this (in fact I have a hard time understanding what I am even looking at). I have posted a figure of this in my separate question: http://math.stackexchange.com/questions/2005207/foliation-from-from-helix-field-and-x-basis-vector – JezuzStardust Nov 21 '16 at 14:14

0 Answers0