Inverse of the tangent map for tangent bundles

196 Views Asked by At

(Physicist here trying to understand some more involved differential geometry). In the context of differential geometry we define the tangent bundle as

$ TM = \bigsqcup_{p \in M} T_p(M) $ where a general element is given by the cartesian product $(p,v_P)$ for $p \in M$ and $v_p \in T_p(M)$.

We have the natural map $\pi: TM \rightarrow M$, explicitly $\pi(p,v_p)=p$, called the tangent map.

However, we then define the inverse tangent map as $\pi^{-1}: M \rightarrow TM$ but I dont understand how we can do this. The tangent map is surjective and so its inverse does not uniquely map onto an element in the tangent bundle.

1

There are 1 best solutions below

2
On BEST ANSWER

As Thorgott said we cannot have a true inverse function from $M$ to $TM$ because $\pi : TM \to M$ is not injective.

However, vector fields $X$ (i.e. sections of $TM$) are a "half inverse" to $\pi$ in the sense that they give you a map from $M$ back up to $TM$. Specifically, for each $p \in M$ the vector field $X$ gives a vector $X_p \in T_pM \subset TM$. Formally, vector fields satisfy the equation $\pi \circ X = \text{id}_M$, so they are right inverses to $\pi$.