- #71
garrett
Gold Member
- 413
- 47
OK, spiffy. I'm liking your rolling sphere example more and more, and looking forward (OK, maybe impatiently) to seeing it come together. We can describe the rolling of the sphere locally with a connection. The "without slipping or twisting" restriction is what GR aficionados think of as "parallel transport." If a point on the sphere is labeled r, and written in good old cartesian 3-space, relative to the sphere's center, as a column vector, then any rotation of the sphere moves this to the new point,
[tex]
r' = g r
[/tex]
with g an element of the rotation group, G=SO(3), represented as a matrix. (phew, I got the O right that time) Of course, we could also do a rotation by writing r as a row vector, and operating with g from the right, or by writing r as a quaternion and operating with g as [itex]g r g^-[/itex] (but I don't think it will matter). Adopting the left acting representation, the connection is a 1-form over the surface, M, valued in the Lie algebra of G,
[tex]
\underrightarrow{w} \in \underrightarrow{Lie(G)} = \underrightarrow{so(3)}
[/tex]
Now, there are two ways clear to me to build this into a fiber bundle picture. The first is to use the sphere, [itex]S^2 = SO(3)/SO(2)[/itex], as the typical fiber for a bundle that is locally [itex]M \times S^2 [/itex]. The "covariant derivative" of a "[itex]S^2[/itex] valued field" is
[tex]
\underrightarrow{\nabla} r(x) = \underrightarrow{\partial} r + \underrightarrow{w} r
[/tex]
and a sphere position is parallel transported (rolled without slipping or twisting) along a path with velocity [itex]\vec{v}[/itex] over the surface iff
[tex]
0 = \vec{v}\underrightarrow{\nabla} r = \frac{d}{d t} r + \vec{v} \underrightarrow{w} r
[/tex]
The second way to choose a fiber bundle is to see that we can choose some arbitrary starting position on the sphere, and write any position using a SO(3) group element,
[tex]
r(x) = g(x) r_0
[/tex]
This way, parallel transport along a path happens iff
[tex]
\frac{d}{d t} g(x(t)) = - \vec{v} \underrightarrow{w} g
[/tex]
with the SAME connection, [itex]\underrightarrow{w}[/itex]. This is called the principle bundle associated to [itex]M \times S^2[/itex], locally [itex]Q = M \times SO(3)[/itex] -- "associated" because it has the same connection (technically, the same structure group, G), and "principle" because the structure group and fiber are the same.
I went ahead and worked out what this connection should be for the sphere rolling over a flat 2D surface:
[tex]
\underrightarrow{w} = \underrightarrow{dx^1} \frac{1}{R} T_2 - \underrightarrow{dx^2} \frac{1}{R} T_1
[/tex]
And computed its curvature.
Now, John Baez is promising to somehow relate this to a Cartan connection, [itex]\underrightarrow{\omega}[/itex], which is a 1-form over the ENTIRE space of a different bundle, locally [itex]P = M \times SO(2)[/itex], and still valued in the so(3) Lie algebra. In terms of the various basis, this will be able to be written as:
[tex]
\underrightarrow{\omega} = \underrightarrow{dx^1} \omega_1{}^A T_A + \underrightarrow{dx^2} \omega_2{}^A T_A + \underrightarrow{d\theta} \omega_3{}^A T_A
[/tex]
And have to satisfy some restrictions.
My pedantic self will be especially happy when I see exactly how [itex]\underrightarrow{\omega}[/itex] relates to [itex]\underrightarrow{w}[/itex].
And I can see another reason why this Cartan connection construction is especially pretty. When one plays around with the geometry of a Lie group, G, one of the best things to build and work with is Cartan's 1-form,
[tex]
\underrightarrow{W} = g^- \underrightarrow{\partial} g \in \underrightarrow{Lie(G)}
[/tex]
which is a 1-form over the entire group manifold. From the definition, its curvature vanishes. But what if this 1-form over the group manifold were a little different, and some of its curvature wasn't necessarily zero? Say, if the part of the curvature valued in the Lie algebra of some subgroup, H, could be nonzero, while the rest of the curvature was zero. I think this would then be a Cartan connection. And this point of view may tie together how [itex]\underrightarrow{\omega}[/itex] relates to [itex]\underrightarrow{w}[/itex]. But, I'm not sure.
I really am eager to see how John fits all this together! Hmm, maybe I should promise to pay him twice as much... Wait, of course, I can get him to work for me if I can figure out his True Name.
[tex]
r' = g r
[/tex]
with g an element of the rotation group, G=SO(3), represented as a matrix. (phew, I got the O right that time) Of course, we could also do a rotation by writing r as a row vector, and operating with g from the right, or by writing r as a quaternion and operating with g as [itex]g r g^-[/itex] (but I don't think it will matter). Adopting the left acting representation, the connection is a 1-form over the surface, M, valued in the Lie algebra of G,
[tex]
\underrightarrow{w} \in \underrightarrow{Lie(G)} = \underrightarrow{so(3)}
[/tex]
Now, there are two ways clear to me to build this into a fiber bundle picture. The first is to use the sphere, [itex]S^2 = SO(3)/SO(2)[/itex], as the typical fiber for a bundle that is locally [itex]M \times S^2 [/itex]. The "covariant derivative" of a "[itex]S^2[/itex] valued field" is
[tex]
\underrightarrow{\nabla} r(x) = \underrightarrow{\partial} r + \underrightarrow{w} r
[/tex]
and a sphere position is parallel transported (rolled without slipping or twisting) along a path with velocity [itex]\vec{v}[/itex] over the surface iff
[tex]
0 = \vec{v}\underrightarrow{\nabla} r = \frac{d}{d t} r + \vec{v} \underrightarrow{w} r
[/tex]
The second way to choose a fiber bundle is to see that we can choose some arbitrary starting position on the sphere, and write any position using a SO(3) group element,
[tex]
r(x) = g(x) r_0
[/tex]
This way, parallel transport along a path happens iff
[tex]
\frac{d}{d t} g(x(t)) = - \vec{v} \underrightarrow{w} g
[/tex]
with the SAME connection, [itex]\underrightarrow{w}[/itex]. This is called the principle bundle associated to [itex]M \times S^2[/itex], locally [itex]Q = M \times SO(3)[/itex] -- "associated" because it has the same connection (technically, the same structure group, G), and "principle" because the structure group and fiber are the same.
I went ahead and worked out what this connection should be for the sphere rolling over a flat 2D surface:
[tex]
\underrightarrow{w} = \underrightarrow{dx^1} \frac{1}{R} T_2 - \underrightarrow{dx^2} \frac{1}{R} T_1
[/tex]
And computed its curvature.
Now, John Baez is promising to somehow relate this to a Cartan connection, [itex]\underrightarrow{\omega}[/itex], which is a 1-form over the ENTIRE space of a different bundle, locally [itex]P = M \times SO(2)[/itex], and still valued in the so(3) Lie algebra. In terms of the various basis, this will be able to be written as:
[tex]
\underrightarrow{\omega} = \underrightarrow{dx^1} \omega_1{}^A T_A + \underrightarrow{dx^2} \omega_2{}^A T_A + \underrightarrow{d\theta} \omega_3{}^A T_A
[/tex]
And have to satisfy some restrictions.
My pedantic self will be especially happy when I see exactly how [itex]\underrightarrow{\omega}[/itex] relates to [itex]\underrightarrow{w}[/itex].
And I can see another reason why this Cartan connection construction is especially pretty. When one plays around with the geometry of a Lie group, G, one of the best things to build and work with is Cartan's 1-form,
[tex]
\underrightarrow{W} = g^- \underrightarrow{\partial} g \in \underrightarrow{Lie(G)}
[/tex]
which is a 1-form over the entire group manifold. From the definition, its curvature vanishes. But what if this 1-form over the group manifold were a little different, and some of its curvature wasn't necessarily zero? Say, if the part of the curvature valued in the Lie algebra of some subgroup, H, could be nonzero, while the rest of the curvature was zero. I think this would then be a Cartan connection. And this point of view may tie together how [itex]\underrightarrow{\omega}[/itex] relates to [itex]\underrightarrow{w}[/itex]. But, I'm not sure.
I really am eager to see how John fits all this together! Hmm, maybe I should promise to pay him twice as much... Wait, of course, I can get him to work for me if I can figure out his True Name.
Last edited: