Identification tangent bundle over affine space with product bundle

[cianfa72]

TL;DR

About the canonical/natural identification of tangent bundle over an affine space with product bundle

Hi,
as in this thread Newton Galilean spacetime as fiber bundle I'd like to clarify some point about tangent bundle for an Affine space.

As said there, I believe the tangent space $T_pE$ at every point $p$ on the affine space manifold $E$ is canonically/naturally identified with the "translation" vector space $V$ of the affine structure $(E,V)$ (i.e. such identification does not require any arbitrary choice).
$$ T_pE \cong V$$ From the above it should follow that $\tau(E)$, the tangent bundle over $E$, is canonically identified with the product bundle $E\times V$.

Is the above correct ? Thanks.

 

[martinbn]

TL;DR Summary: About the canonical/natural identification of tangent bundle over an affine space with product bundle

Hi,
as in this thread Newton Galilean spacetime as fiber bundle I'd like to clarify some point about tangent bundle for an Affine space.

As said there, I believe the tangent space $T_pE$ at every point $p$ on the affine space manifold $E$ is canonically/naturally identified with the "translation" vector space $V$ of the affine structure $(E,V)$ (i.e. such identification does not require any arbitrary choice).
$$ T_pE \cong V$$ From the above it should follow that $\tau(E)$, the tangent bundle over $E$, is canonically identified with the product bundle $E\times V$.

Is the above correct ? Thanks.

Yes.

 

[cianfa72]

My doubt arose since in a linked paper in that thread, namely https://trautman.fuw.edu.pl/publications/Papers-in-pdf/25.pdf
there is a claim that for Galilean spacetime $E$ its principal fiber bundle $(P(E),E,\pi)$ is a product bundle.

I'd say that Galilean spacetime $E$ is an affine space and that its tangent bundle $(\tau(E), E, \pi)$ is a product bundle since it is naturally identified with the product bundle $E \times V$.

Does it makes sense ?

 

[martinbn]

The principle bundle and the tangent bundle are different things. The principle bundle in question is the bundle of frames.

 

[cianfa72]

The principle bundle and the tangent bundle are different things. The principle bundle in question is the bundle of frames.

Sorry, given a smooth manifold $E$ (in this case the spacetime manifold) it should be possible that its principle fiber bundle is (or identified with) a product bundle $E \times X$ while its tangent bundle is not ?

 

[martinbn]

Sorry, given a smooth manifold $E$ (in this case the spacetime manifold) it should be possible that its principle fiber bundle is (or identified with) a product bundle $E \times X$ while its tangent bundle is not ?

No, it is not possible.

 

[cianfa72]

No, it is not possible.

Ok, so what Trautman says in that paper implies that for Galilean spacetime $E$ both its tangent bundle and principal fiber bundle (i.e. bundle of frames) are actually product bundles.

At the same time Galilean spacetime $E$ is both fiber bundle and affine space (actually the fiber bundle structure single out preferred affine charts for the affine structure, i.e. the inertial affine charts).

 

[martinbn]

Ok, so what Trautman says in that paper implies that for Galilean spacetime $E$ both its tangent bundle and principal fiber bundle (i.e. bundle of frames) are actually product bundles.

I didn't look at it, but yes.

At the same time Galilean spacetime $E$ is both fiber bundle and affine space (actually the fiber bundle structure single out preferred affine charts for the affine structure, i.e. the inertial affine charts).

Yes. I don't know why you keep saying!

 

[cianfa72]

Another point related to this. To prove that the map $$v \mapsto \left . \frac {d} {dt} \right |_{t=0} f(a + tv) = D_vf(a), v \in V$$ is isomorphism I believe we cannot avoid to introduce an arbitrary chart for the affine space $(E,V)$.

Indeed $(E,V)$ and $V$ are abstract affine and vector spaces respectively. In order to use the chain rule to evaluate the above derivative, we need a chart $\varphi$ for the affine space $(E,V)$. Then we get
$$(f \circ {\varphi}^{-1}) \circ (\varphi \circ \gamma (t))$$ where $\varphi \circ \gamma(t)$ is the repesentative in the chart $\varphi$ of the curve $\gamma(t) = a + vt$. Now we're in position to use the chain rule to get $$D_v|_a f = D_vf(a) = \left . v^i \frac {\partial f} {\partial x^i} \right |_a$$ Where $v^i$ are the components of the derivatives w.r.t. $t$ of the curve representative $\varphi \circ \gamma(t)$.

Does exist a way to intrinsically define it avoiding a such process?

 

[jbergman]

Sorry, given a smooth manifold $E$ (in this case the spacetime manifold) it should be possible that its principle fiber bundle is (or identified with) a product bundle $E \times X$ while its tangent bundle is not ?

The tangent bundle is a vector bundle. A principal bundle is different. Typically a manifold has both. For instance a Frame bundle is an associated principle bundle to a vector bundle like a tangent bundle that contains all possible basis for the tangent bundle and whose structure group is GL(n).

A good reference for all of this is Hamilton's Mathematical Gauge Theory, chapter 4. You could also read the Wikipedia articles on associated vector bundles.

 

[jbergman]

https://en.m.wikipedia.org/wiki/Associated_bundle
https://en.m.wikipedia.org/wiki/Frame_bundle

 

[jbergman]

I am not that familiar with affine spaces so I would have to think about it a bit more to say anything. I think I would want to express the affine conditions as a Principal Bundle over a base manifold that is a quotient of the group of symmetries. I am guessing E(n)/SO(n). Then the rest would follow naturally.

 

[cianfa72]

Indeed $(E,V)$ and $V$ are abstract affine and vector spaces respectively. In order to use the chain rule to evaluate the above derivative, we need a chart $\varphi$ for the affine space $(E,V)$. Then we get
$$(f \circ {\varphi}^{-1}) \circ (\varphi \circ \gamma (t))$$ where $\varphi \circ \gamma(t)$ is the repesentative in the chart $\varphi$ of the curve $\gamma(t) = a + vt$. Now we're in position to use the chain rule to get $$D_v|_a f = D_vf(a) = \left . v^i \frac {\partial f} {\partial x^i} \right |_a$$ Where $v^i$ are the components of the derivatives w.r.t. $t$ of the curve representative $\varphi \circ \gamma(t)$.

Does exist a way to intrinsically define it avoiding a such process?

Re-thinking about this, I believe $$v \mapsto \left . \frac {d} {dt} \right |_{t=0} f(a + tv) := D_vf(a), v \in V$$ is actually a canonical/natural defined map since no choice is involved and $a + tv$ is well-defined from affine space axioms as the function $f(a + tv)$.

On the other hand, to prove it is a linear map that is isomorphism we must introduce an arbitrary chart as in the quoted post. Therefore, proved it is an isomorphism this way, it follows that it is actually a canonical isomorphism, i.e. $V\cong T_pE$.

 

[cianfa72]

See also Canonical isomorphism between tangent space and translation space of affine space.