1

Given a Riemannian manifold $(M,\bar{g})$, we can construct the Riemannian cone manifold $(C(M), g )$ as follows. Topologically, $C(M)$ is $M \times \mathbb{R}_{>0}$. We equip this with the cone metric

$g = t^2 \bar{g} + dt^2$

where $t$ is the parameter on $\mathbb{R}_{>0}$.

What can we say about the (Co)tangent bundle of $C(M)$?

I know that at any given point $(m,t) \in C(M)$, we have

$T_{(m,t)}(M \times \mathbb{R}_{>0}) \cong T_m(M) \oplus T_t(\mathbb{R}_{>0})$

Is that enough to conclude $T(M \times \mathbb{R}_{>0}) \cong T(M) \oplus T(\mathbb{R}_{>0})$ ? In that case, do vector fields on the cone decompose into vector fields on the base $M$ plus vector fields along the "cone" direction?

Can we do this with the cotangent bundle?

i.e. Can we decompose differential forms on the cone in terms of forms on the base plus forms along the cone direction?

Community
  • 1
Mark B
  • 2,054

1 Answers1

2

Yes, we can!

If a Riemannian manifold has a metric in the form you have written, then the Levi-Civita connection is split into the two directions, which implies that vectors (and differential forms, just as well) decompose exactly.

It is explained in detail in D.Joyce, Riemannian Holonomy Groups and Calibrated Geometry, if I'm not mistaken.

geodude
  • 8,357