# Thursday February 27th

Setup: $M^m, N^n \subset V^{m+n}$ closed submanifolds, $M\transverse N$, $M$ oriented (i.e. an orientation of $TM)$ and $N$ co-oriented (i.e. an orientation $\nu_N = TM/TV$).

Each $p\in M\intersect N$ has a sign $\eps(p) \in \theset{-1, 1}$.
If $p, q \in M\intersect N$ with $\eps(p) = 1, \eps(1) = -1$, we would like an isotopy $(h_t)_{0 \leq t \leq 1}$ of $V$ such that $h_0 = \id$ and $h(M) \intersect N = (M\intersect N) \setminus\theset{p, q}$.

Idea: we want to push $M$ off of $N$:

![Image](figures/2020-02-27-11:16.png)\

From last week, assume $\dim V \geq 5$ and $\dim N \geq 3$ and $\pi_1(V\setminus N) = \pi(V) = 0$.

> Why? In the 2-dimensional model above, we want the disc in the middle to be contractible.

![Image](figures/2020-02-27-11:21.png)\

![Image](figures/2020-02-27-11:25.png)\

Then there is a smooth embedding $\phi_3: U \to V$, 2-dimensional to $n+1$ dimensional, sending $U \intersect C_i$ to $C_i$ and $U(C_0 \union C_0')$ to $V\setminus (M\union N)$.

Goal for today: under the same hypotheses, $\phi_3$ extends to an embedding $\phi: U \cross \RR^{m-1} \cross \RR^{n-1} \to V$ such that $\phi\inv(M) = (U\intersect C_0) \cross \RR^{m-1} \cross \theset{0}$ and $\phi\inv(N) = (U \intersect C_0') \cross \theset{0} \cross \RR^{n-1}$.

Let $U' = \phi_3(U) \subset V$:

![Image](figures/2020-02-27-11:26.png)\

Lemma
:   There exist vector fields along $U'$, $\xi_1, \cdots, \xi_{m-1}, \eta_1, \cdots, \eta_{n-1}$, such that

    1. These are orthonormal to each other and orthogonal to $U'$.

        > Note that we'll need a Riemannian metric to make sense of this, and particularly one such that $M, N$ are totally geodesic, and $T_p M \perp T_p N$ and $T_q M \perp T_q N$.

    2. $\xi_1, \cdots, \xi_{m-1}$ are tangent to $M$ along $C$

    3. $\eta_1, \cdots, \eta_{n-1}$ are tangent to $N$ along $C_1'$.

Given this, we have $\phi(u, \vector x, \vector y) = \exp_{\psi_3(u)} \qty{ \sum x_i \xi_i(\phi_3(u)) + \sum y_j \eta_j(\phi_r(u))  }$, where the exponential maps is evaluating a geodesic path at time 1.


### Proof of Lemma
Let $\tau$ be the unit tangent vector field along $C$, oriented from $p$ to $q$:

![Image](figures/2020-02-27-11:42.png)\

Let $\nu'$ be the unit vector field along $C'$ normal to $C_1'$ pointing toward the interior of $U'$.
Thus $\nu'(p) = \tau(p)$ and $\nu'(q) = -\tau(q)$.

First, complete tangents to an orthonomal basis: choose $\xi_i(p)$ such that $\theset{\tau(p), \xi_1(p), \cdots}$ is an oriented orthonormal basis for $T_p M$.
Riemannian metrics induce a unique notion of parallel transport, extend $\xi_i$ to all of $C$ by parallel transport.
This preserves inner products, and in particular we obtain an orthonormal basis for $T_q M$.

We can use this to obtain bases for the orthogonal complements, and thus for the normal bundles.
Since $\eps(p) = 1$, an orientation of $T_p M$ yields an orientation of $(\nu_N)_p$.
Thus $\theset{\nu'(p), \xi_i(p)}$ is an oreinted basis, and similarly by flipping the sign of the first term, since $\nu'(q) = -\tau(q)$, $\theset{\nu'(q), \xi_i(q)}$ is an oriented basis for $(\nu_N)_q$.

Consider the bundle over $C'$ with fibers equal to orthonormal bases $\theset{w_1, \cdots, w_{n-1}} \in (T_x V)^{n-1}$ with each $w_i$ orthonormal and orthogonal to $\nu'(x)$.
This has fiber $O(n-1)$, and since the base $C'$ is contractible, this is a trivial bundle.

We have elements in the fiber over $p$ and $q$ inducing the same orientation, so are related by an element of $SO(n-1)$, which is connected and thus path-connected.
This gives a path in the frame bundle connecting $\theset{ \xi_1(p), \xi_{m-1}(p)}$ to $\theset{\xi_1(q), \cdots, \xi_{m-1}(q)}$.

So we extend the $\xi'$ over all of $C'$, remaining orthogonal to $N$ and $U'$.
So we have vector fields $\xi_1, \cdots, x_m$ along $C \union C'$.
We want to show that these can be extended over all of $U_1'$ remaining orthogonal to $U'$.

Consider the bundle over $U'$ whose fibers are $(m-1)\dash$tuples of vectors in $T_x V$ that are orthonormal to $T_x U'$.
Since $U'$ is contractible, this bundle is trivial, and the fiber is orthonormal $(m-1)\dash$frames in an $(m+n-2)\dash$dimensional vector space, the orthogonal complement of $T_x U'$.

Since any orthonormal basis of size $m+n-2$ will send $m-1$ frames to other $m-1$ frames, with some redundancy if the upper-left block is the identity.
Thus the fibers are isomorphic to $O(m + n - 2)/O(n-1)$.

The construction of $\xi_1, \cdots, \xi_n$ over all of $U'$ is now reduced to extending the loop on $O(m+n-2) / O(n-1)$ determined by $\xi_i$ on $C \union C'$ to a disk, i.e. $U'$.

In fact, $\pi_1$ of this space is 0, so this can be done.
Once we have $\xi_i$, just take $\nu_i$ to be any orthonormal over $U'$ such that $\xi$ is orthogonal to $TN$ along $C'$.

To see why this is, consider the fibrations $O(n-1) \to O(n-m-2) \to Q$ the quotient above and take the LES in homotopy, also consider $O(n) \to O(n+1) \to S^k$.