Why is this normal bundle a subset of the tangent bundle

differential-topologygeneral-topologymanifoldssoft-question

Let $M \subset V$ be a submanifold, $\delta M = \delta V = \emptyset$. Let $v = (p,E,M)$ be the normal bundle of $M$ in $V$. Then:

Then we have:

$v \subset T_M V \subset T_M \mathbb{R}^n = M \times \mathbb{R}^n$

I'm a little bit confused on why the normal bundle of $M$ in $V$ is containd in $T_M V$. A part of me wants to think they are disjoint, because one is the tangent bundle and the other is the normal bundle… W hat am I not understanding here? Thanks!

Best Answer

To put together a few things from the comments: If you put a Riemannian metric on $V$ (for example, think of $V\subset\Bbb R^N$ for some $N$ by Whitney), then $\upsilon_p \cong (T_p M)^\perp\subset T_p V$.

What you were reading wrong is that $\upsilon$ is a subbundle of $T_M V$, which is the restriction of $TV$ to $M$. Indeed, $\upsilon \cong (TM)^\perp \subset T_M V$.

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