On a scale of 1 to 10 how far is this manifold from being a normal bundle?

Question

(DIS)-CLAIMER: All the manifolds considered in this post are completely humble and have no additional structure beyond the smooth structure.

Let $Y$ be a submanifold of $M$ and let $(-)^0$ denote the annihilator.

I claim (and about to prove hopefully) that $NY = \{(m,\omega)\in T^*M : m \in Y, \omega \in (T_m Y)^0 \}$ can be given a natural structure as a submanifold of $T^*M$.

(So before the storm of notation comes...) Here's my question :

On a scale of 1 to ten how far is $NY$ from being a normal bundle?

Furtheremore, i'd like to know how come this manifold is canonical with a unique smooth structure (depending on the original atlas of course) whereas the usual Normal bundle depends on a choice of inner product?

The proof (I am so very terribly sorry about the length):

Let $y \in Y$ and let $(\phi,U)$ be an adapted chart at point the $y$. Meaning $\phi(U \cap Y)=\phi(U) \cap\mathbb{R}^k \times\{0\}^{n-k}$ and $\phi(y)=0$ where $n$ and $k$ are the dimensions of $M$ and $Y$ respectively.

We denote the tangent map of $\phi$ at $m$ by $T_m \phi$ and the cotangent bundle of $M$ by $T^*M$. Define $T^*\phi: T^*M \to \mathbb{R}^{2n}$ to be the map:

$$T^*\phi :(m,\omega) \mapsto (\phi(m), (T_m\phi^*)^{-1} \omega)$$

We look at the image of the set $NY = \{(m,\omega)\in T^*M : m \in Y, \omega \in (T_m Y)^0 \}$ where $(-)^0$ is the annihilator.

$$T^*\phi(NY)= \{(\phi(m),(T_m\phi^*)^{-1}\omega)\in T^*M : m \in Y, \omega \in (T_m Y)^0 \}$$

Let $\pi : \mathbb{R}^k \times \mathbb{R}^{n-k} \to \mathbb{R}^{n-k}$ be the natural projection. Clearly $\ker T_m (\pi o \phi)=T_mY$.

The annihilator of the kernel is the image of the dual so: $\omega \in (T_mY)^0 \implies \omega = T_m(\pi o \phi)^* \alpha$ for some $\alpha \in (\mathbb{R}^{n-k})^*$ substituting we get:

$$(T_m\phi^*)^{-1}\omega=(T_m\phi^*)^{-1}T_m(\pi o \phi)^* \alpha = (T_m\pi)^*\alpha$$

Now $Im (T_m \pi^*)=(\ker T_m \pi)^0= \{0\}^k \times\mathbb{R}^{n-k}$. So basically what we proved is that $T^*\phi$ is in fact an adapted chart of $NY$. Why? Because:

$$T^*\phi(U \cap NY)= T^*\phi(U) \cap \mathbb{R}^k \times\{0\}^{n-k} \times \{0\}^k\times \mathbb{R}^{n-k}$$

And the image of the right side under the right projections is $\mathbb{R}^n$. So $dim(NY)=n$ which is exactly what we want for a normal bundle!

Answer 1

The bundle you're denoting by $NY$ is called the conormal bundle of $Y$, more commonly denoted by $N^*Y$. It is indeed a smooth subbundle of $T^*M$, defined independently of any choice of Riemannian metric. As @Mariano observed, if you put a Riemannian metric on $M$, you obtain a normal bundle $N^gY$ (the orthogonal complement of $TY$ in $TM|_Y$), which is naturally isomorphic to $N^*Y$.

With respect to the natural symplectic structure on $T^*M$, the conormal bundle is a Lagrangian submanifold. It plays an extremely important role in PDE theory and microlocal analysis. For example, if $M$ is a manifold with boundary and $Y=\partial M$, the class of distributions on $M$ whose wave front set lies in $N^*M$, called conormal distributions, is important for studying singular boundary value problems.

Answer 2

If you fix a Riemannian metric on $M$, then you have an isomorphism $TM\cong T^*M$, and under that isomorphism your $NY$, which is the subbundle of $(T^*M)|_Y$ which annihilates $TY$, is the same thing as the orthogonal complement of $TY$ in $(TM)|_Y$. This last bundle is, precisely, the normal bundle of $Y$ in $M$.