I am wondering what a normal bundle looks like given this definition for Riemannian manifolds: Say $M = S^2$. $T_pM$ is the tangent plane ($\mathbb R^2$) at $p \in S^2$. Lets denote the north pole by $0$ and let $S$ be a ball of some radius around $0$. Naturally, I expect the normal vector at $0$ to stick up perpendicularly to $T_0M$. But given the definition we're talking of normal to all vectors in $T_0S$. I don't see how this is possible. Could you explain to me what these normal vectors look like on $S^2$?
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$\begingroup$ We're not talking about all vectors in $T_0M$, were talking about all vectors perpendicular to $T_0M$, namely those which stick straight up. What gave you the first idea? $\endgroup$– Alex BeckerJun 27, 2013 at 16:07
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Perhaps you are confusing a few notions: The normal bundle of a submanifold only makes sense for points in that submanifold, so if you are taking $S$ to be an open geodesic ball around the north pole, then the normal bundle at $0$ is trivial. Normals are taken to be with respect to the manifold the submanifold lives in; if you take a $1$-dimensional submanifold of $S^2$, you'll get a $1$-dimensional normal bundle (here's a crappy drawing of what the normal space at a point might look like), while if you take a $2$-dimensional submanifold, all directions will be tangent directions so the only possibility left for a normal direction is the zero one.
To get the case you are talking about, where the normal points upwards (or downwards), you could view $S^2$ as a submanifold of $\mathbb{R}^3$, embedded in the usual way. Then indeed the normal space $N_0 S^2 \subseteq T_0\mathbb{R}^3 \cong \mathbb{R}^3$ consists of the vectors in the vertical direction.
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$\begingroup$ Why is the north pole not contained in a ball around it? $\endgroup$– studentJun 28, 2013 at 8:10
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$\begingroup$ I've edited my answer a bit; perhaps it's clearer now. $\endgroup$– fugledeJun 28, 2013 at 8:20
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1$\begingroup$ @user84127: The tangent space is always the same dimension as the smooth manifold itself, so it's $3$-dimensional. Now, I don't know how you've gone about defining the tangent space but the most geometric definition is that it consists of all possible tangents of all possible curves through the point. In particular, for a point $(x,y,z) \in \mathbb{R}^3$, $T_{(x,y,z)}\mathbb{R^3}$ spanned by the tangents of three straight lines through the point $(x,y,z)$, each of them parallel to one of your $3$ standard axes. $\endgroup$– fugledeJun 28, 2013 at 12:35