## Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorBruce Bartlett
• CommentTimeSep 21st 2013

Added a link to an expository talk I gave on “The geometry of force” giving an elementary explanation of the classical Kaluza-Klein mechanism (i.e. the idea that geodesics on the principal bundle project down to curved trajectories on base space apparently experiencing a “force”). Following the book of Bleecker, Gauge theory and variational principles.

• CommentRowNumber2.
• CommentAuthorBruce Bartlett
• CommentTimeSep 21st 2013

While preparing this talk, and especially after trying to understand the case of the frame bundle of $S^2$ via this approach, I have the following question.

Recall the main idea here (following the book of Bleecker). Let $\pi : P\rightarrow M$ be a principal $G$-bundle with connection $\omega$ over a Riemannian manifold $(M, g)$, and let $k$ be an invariant inner product on the lie algebra of $G$. Then there is an induced “pull-back” metric $h = \pi^*g + k \omega$ on the bundle $P$. The point is that geodesics $\gamma$ on $P$ project down to curved trajectories $\bar{\gamma} := \pi \circ \gamma$ on $M$ which we think are “experiencing a force”. There is a precise equation for these projected geodesics which Bleecker writes down.

Here’s my question. Don’t we also get, in this setup, a new connection $\alpha$ on $M$ itself? Namely, let $\sigma$ be a curve in $M$, and let $F$ be an orthonormal frame at $x = \sigma(0)$. Using the connection $\omega$ on $P$, we can lift $\sigma$ to a curve $\tilde{\sigma}$ in $P$. But now $(P, h)$ is a Riemannian manifold as defined above, so it has a Levi-Civita connection, so we can parallel transport a frame in $P$ along $\tilde{\sigma}$. And a frame in $P$ is just a frame in $M$ plus a frame in the Lie algebra of $G$, and we can choose some fixed one in the beginning. So we can parallel transport the frame in $P$! Then we can project this down to $M$. This should give a moving frame in $M$.

In other words, I have constructed a new connection $\alpha$ on $M$. Is this correct, or am I making a geometric error?

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeSep 21st 2013
• (edited Sep 21st 2013)

Hi Bruce,

I have expanded your line in the References-section

An elementary exposition of the geometry behind the Kaluza-Klein mechanism (the idea that geodesics on the gauge bundle project to curved trajectories on the base manifold) can be found in this talk:

to

An elementary exposition of the geometry behind the Lorentz force in the Kaluza-Klein mechanism (the idea that geodesics on the gauge bundle project to curved trajectories on the base manifold) can be found in this talk:

Okay?

Then I have added to the section The mechanism the following paragraph:

Then one finds that the KK-mechanics indeed not only reproduces gauge fields and their correct dynamics from pure gravity in higher dimensions, but also the forces which they excert on test particles. For instance the rajectory of a charged particle subject to the Lorentz force excerted by an electromagnetic field in $d$-dimensional spacetime is in fact a geodesic in the field of pure gravity of the total space of the corresponding KK-un-compactified circle principal bundle. See (Bartlett 13) for a pedagogical discussion of this effect.

• CommentRowNumber4.
• CommentAuthorBruce Bartlett
• CommentTimeSep 21st 2013

Hi Urs, thanks yes that’s fine. I also added a reference to the book of Bleecker, and his name to the phrase “See (Bartlett13) for a pedagogical…”.

• CommentRowNumber5.
• CommentAuthorBruce Bartlett
• CommentTimeSep 21st 2013

I am still curious about my question. Let me make it more clear: I am suggesting that a connection on a $G$-principal bundle over a Riemannian manifold $(M, g)$ gives rise to a new connection (different from the Levi-Civita connection) on $M$.

For instance, take $M=\mathbb{R}^3$ with the standard metric. And let $P$ be the trivial $U(1)$-bundle over . And let the connection on the $\mathbb{R^U(1)$-bundle $P$ be given by a magnetic field in the $z$-direction (remember, this is a connection on $P$, not on $M$! A connection on $M$ is what I’m about to construct.) . This is a standard setup from 1st year physics. If we gave a particle in $\mathbb{R}^3$ an initial velocity in the $x$ direction, it travels in a circle in the $xy$-plane by the Lorentz force law, $F = q(E + v \times B)$.

I am claiming that we can interpret the magnetic field as giving rise to a new connection on $\mathbb{R}^3$, different from the standard flat connection. To give you this connection, I have to tell you how to parallel transport frames in $\mathbb{R}^3$ along a curve $\sigma$. Answer: each basis vector of the frame rotates as if experiencing the magnetic force appropriate to that basis vector being thought of as a velocity vector. So, if the curve $\sigma$ goes along the $z$-axis, nothing happens to the frame (no force). If $\sigma$ is a circle in the $xy$-plane, the frame rotates around the $z$-axis. For a general curve, a combination of these effects. I believe this connection has torsion, but I haven’t checked. I’m probably making a psychological error here.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeSep 22nd 2013
• (edited Sep 22nd 2013)

Bruce, for the sake of readability, could you dis-ambiguate your uses of “connection on” a bit? There are principal connections on a principal bundle in the game and at the same time affine connections on the tangent bundle over the total space of the principal bundle. When you say “connection on $M$” is is clear that you mean a connection on $T M \to M$, but when you say “connection on $P$” then the reader has to start working out which of the two different meanings you have in mind (the actual principal connection on $P \to M$ or the affine connection on $T P \to P$).

• CommentRowNumber7.
• CommentAuthorBruce Bartlett
• CommentTimeSep 22nd 2013

I am using both those meanings, that’s the point. (Well… I always use connections on principal bundles, and never “affine” connections. But an affine connection on $M$ is the same thing as a connection on the principal frame bundle.) In particular, as you say, there is the connection on $P$ in the sense of $P$ being a principal bundle over $M$ (a $Lie(G)$-valued 1-form on $P$), but there is also the Levi-Civita connection on the frame bundle of $P$, since $P$ is equipped with the bundle metric $h = \pi^* g + k \omega$ where $k$ is an inner product on $Lie(G)$.

When I say “I am suggesting that a connection on a G-principal bundle $P$ over a Riemannian manifold (M,g) gives rise to a new connection (different from the Levi-Civita connection) on M” I mean the following:

“connection on a $G$-principal bundle $P$ over a Riemannian manifold $(M,g)$” means the usual notion of a connection on a principal bundle $P$ (an equivariant distribution of horizontal subspaces of $P$).

” a new connection on $M$” means a new connection on the frame bundle of $M$, in the usual sense of a connection on a frame bundle (i.e. thinking of the frame bundle as a principal bundle).

It can be a bit tricky, I agree.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeSep 22nd 2013
• (edited Sep 22nd 2013)

I don’t think it’s tricky, as we can all distinguish two things at a time. I just think that when you say “connection on $P$” you are being ambiguous and making your readers work harder than necessary to figure out what you want to say.

Another thing: I just noticed the edit of yours in verion 26 of the KK-entry, which is probably based on a misunderstanding of my citation convention. You changed the citation anchor “Bartlett13” to “Bartlett14” and introduced “Bleecker13”. But that number is supposed to be the year of publication. So I changed it (back) to “Bleecker81” and “Bartlett13”.

• CommentRowNumber9.
• CommentAuthorBruce Bartlett
• CommentTimeSep 23rd 2013

Yes, apologies for getting the conventions mixed up. But I guess there are no thoughts on my proposal?