Why is $Y_\delta = \{(x,v): x \in Y, |v| < \delta\}$ open in $TM$?

Question

Why is $Y_\delta = \{(x,v): x \in Y, |v| < \delta\}$ open in $TM$? $Y$ is an open subset of $M$, a Riemannian manifold. The norm is given by the Riemannian metric.

To prove it, given $(x,v) \in Y_\delta$ , want to find an open set around it that is contained in $Y_\delta$. To start, choose a coordinate chart $(x^i, v^i)$ of $TM$ around $(x,v)$. To make it easier, I might want to choose a normal coordinate of $M$ around $x$, then use the corresponding chart of $TM$. However, it only simplifies things for $v \in T_xM$. What should I do?

Answer 1

Pick any coordinate chart $U \subset Y \subset M$ with local coordinates $x^i$. The Riemannian metric is smooth, which means that is can be represented by smooth functions $g_{ij}(x)$ in this local chart. (This is the definition of the Riemannian metric being smooth.)

Now consider the coordinate chart $U \times \mathbb R^n$ for $TM$, parametrised by coordinates $(x^i, v^i)$, where the $v^i$ are tangent coordinates. The norm function is $$ (x, v) \mapsto \sum_{i, j} g_{ij}(x) v^i v^j.$$ Since the norm function is a sum of products of continuous functions, it is itself continuous. Hence, by a continuity argument, your $Y_\delta$ is open.

Answer 2

The norm is a continuous function on $TM$. It follows that the set $U=\{(p,v)\in TM|\,|v|<\delta\}$ is open. In addition, the projection $\pi:TM\to M$ is continuous, and so, the set $W=\{(p,v)\in TM|p\in Y\}$ is also open. The intersection of two open sets is open.