A special parallelization of $S^3$

Consider the set $S^{3} \subset \mathbb{R}^4$, as known $S^3$ is a Lie Group and, therefore, is parallelizable.

If we define the sets $T_1$, $T_2$ $\subset$ $S^3$ as

$$T_1 = \{(x_1, x_2,x_3,x_4) \in S^3; x_1^2 + x_2^2 \leq 1/2\}, $$ $$T_2 = \{(x_1, x_2,x_3,x_4) \in S^3; x_1^2 + x_2^2 \geq 1/2\}. $$

Then, we found a nice way to decompose $S^3$, once $T_1 \cong D^2 \times S^1$, $T_2 \cong S^1 \times D^2$, $\partial T_1 = \partial T_2 = T_1 \cap T_2 \cong S^{1} \times S^{1}$ and $S^3 = T_1 \cup T_2$.

Note that $N := T_1 \cap T_2$ is also parallelizable (because $S^1 \times S^1$ is a Lie Group).

My Question: Is there a special parallelization of $S^3$, $H: TS^3 \rightarrow S^3\times \mathbb{R}^3 $ such that $$H(T N ) = N \times \mathbb{R}^2 \times \{0\}?$$

Unfortunately, everything that I have tried has led me to nowhere, and I don't have the faintest idea of how to approach this problem.

Can anyone help me?

I give below an argument to show that such a trivialization does exist. The argument is based on 'elementary' homotopy theory (after all, if such an extension were not possible, the obstruction would probably be of topological nature). It is not especially constructive, though it suggests that a constructive demonstration should be possible in principle. However there are many subtle problems to avoid which, I expect, hint to the practical difficulty to come up with an explicit, constructive, calculatory proof of this result.

Step 1. Identify general constraints on the trivialization of $TS^3$:

Since a trivialization of a rank $n$ real vector bundle $p : E \to B$ is usually defined as a map $T : E \to B \times \mathbb{R}^n$ such that $\mathrm{id}_B \circ p = \mathrm{proj}_1 \circ T$, a trivialization $T$ is completely encoded in the map $\mathrm{proj}_2 \circ T : E \to \mathbb{R}^n$. Based on this observation, we shall simply describe trivializations of $TS^3$ as maps $TS^3 \to \mathbb{R}^3$.

Let $\pi : TS^3 \to S^3$ denote the canonical bundle map. Since $S^3$ is parallelizable, there exists a trivialization $P : TS^3 \to \mathbb{R}^3$ which we shall fix throughout the argument. Given any trivialization $P' : TS^3 \to \mathbb{R}^3$, for each $b \in S^3$ we get a map $P'_b : T_bS^3 \to \mathbb{R}^3$ and thus a map $T_{P'}(b) := P'_b \circ (P_b)^{-1} \in Gl(3)$. (Observe that this map $T_{P'}(b)$ makes sense as long as $P'_b$ exists, regardless of whether $P'$ is defined on the whole of $TS^3$ or not.) Post-composing if necessary $P'$ with the map $\mathbb{R}^3 \to \mathbb{R}^3 : (x,y,z) \mapsto (x,y,-z)$, we can restrict our attention to those $P'$ such that $T_{P'}(b)$ as positive determinant as some (and hence at all) $b \in S^3$. Notice also that $T_P = Id$ for all $b$.

A key observation is that given a continuous circle $c : S^1 \to S^3$, the map $S^1 \to Gl_+(3) : t \mapsto T_{P'}(c(t))$ has to be contractible since $c$ is contractible in $S^3$. This put some constraints on the possible trivialization $P'$ you seek.

Step 2. Construct a restriction to $T_NS^3$ of a possible trivialization as we seek:

The 2-torus $N \subset S^3$ is a smooth submanifold, since it is the preimage of the regular value $1/2$ of the smooth function $f(x_1, x_2, x_3, x_4) = x_1^2 + x_2^2$ on $S^3$. Given any Riemaniann metric on $S^3$, the gradient vector field $\nabla f$ is nonvanishing on $N$ and everywhere transverse to $TN$ along $N$. We thus get a splitting $T_NS^3 \cong TN \oplus \mathbb{R}\langle \nabla f \rangle$. Consider some smooth parametrization $g : S^1 \times S^1 \to N$; the inverse of its differential $Dg$ induces a trivialization $\tau : TN \to \mathbb{R}^2$. Hence we obtain a trivialization $Q : TN \oplus \mathbb{R}\langle \nabla f \rangle \to \mathbb{R}^2 \oplus \mathbb{R}$ in the obvious way.

Given a circle $c : S^1 \to N$, there is a priori no reason for the map $t \in S^1 \mapsto T_{Q}(c(t)) \in Gl_+(3)$ (we assume it has positive determinant) to be contractible. However, since $Gl_+(3) \simeq SO(3)$, for any fixed matrix $\ast \in Gl_+(3)$ we have $\pi_1(Gl_+(3), \ast) = \mathbb{Z}/2\mathbb{Z}$ with the generator represented by any full turn rotation around the first column of the matrix $\ast$.

Fix $pt = g(1, 1) \in N$ and let $c_1, c_2 : (S^1, 1) \to (N, pt)$ be the two usual generators of $\pi_1(N, pt)$, namely thinking of $S^1$ as the unit circle in $\mathbb{C}$, $c_1(e^{it}) = g(e^{it}, 1)$ and $c_2(e^{it}) = g(1, e^{it})$ for $e^{it} \in S^1$. Take $\ast = T_Q(pt) = T_Q(c_1(1)) = T_Q(c_2(1))$. If $T_{Q}(c_i(-)) \in \pi_1(Gl_+(3), \ast)$ is the nontrivial element, it is possible to change the trivialization $Q$ by pre-composing the differential $Dg : T(S^1 \times S^1) \to TN$ by a bundle endomorphism

$$h : T(S^1 \times S^1) = (S^1 \times S^1) \times \mathbb{R}^2 \to (S^1 \times S^1) \times \mathbb{R}^2 : (b, X) \mapsto (b, R(b)(X))$$ where $R(b)$ is a rotation of $\mathbb{R}^2$ such that $R(pt) = Id$ and the map $S^1 \to SO(2) \simeq S^1 : e^{it} \mapsto R(c_i(e^{it}))$ makes an odd number of full turns if and only if the element $T_{Q}(c_i(-))$ is nontrivial. For instance, identifying $SO(2) \cong U(1)$, we can take $R(g(e^{ir}, e^{is})) = e^{i(mr+ns)}$ with $m,n \in \mathbb{Z}$; then in particular $R(g(e^{ir}, 1)) = e^{imr}$ makes $m$ turns when $e^{ir}$ makes one turn, so we only have to choose the parity of $m,n$ according to the above prescription.

The new trivialization $Q' : T_NS^3 \to \mathbb{R}^3$ where $Dg \circ h$ is used in replacement of $Dg$ is such that the elements $T_{Q'}(c_i(-))$ are trivial.

It turns out that since $c_1$ and $c_2$ generates $\pi_1(N, pt)$, then $T_{Q'}(c(-))$ is trivial for any continuous circle $c : S^1 \to N$, but this will follow from our forthcoming considerations.

Step 3. Construct a homotopy over $N$ between $T_{Q'}$ and $T_{P}$:

Since each map $T_{Q'}(c_i(-)) : (S^1, 1) \to (Gl_+(3), \ast)$ is contractible, it extends to a map $T_i : (S^1 \times [0,1], \{1\} \times [0,1]) \to (Gl_+(3), \ast)$ such that $T_i(t,0) = T_{Q'}(c_i(t))$ and $T_i(t,1) = \ast$.

Our previous considerations yield a continuous map $F : \partial([0, 2\pi]^2 \times [0,1]) \to Gl_+(3)$ as follows:

• for $(r,s,t) \in [0,2\pi]^2 \times \{0\}$, $F(r,s,t) = T_{Q'}(e^{ir}, e^{is})$;
• for $(r,s,t) \in [0,2\pi] \times \{0, 2\pi\} \times [0,1]$, $F(r,s,t) = T_1(e^{ir}, t)$;
• for $(r,s,t) \in \{0, 2\pi\} \times [0, 2\pi] \times [0,1]$, $F(r,s,t) = T_2(e^{is}, t)$;
• for $(r,s,t) \in [0, 2\pi]^2 \times \{1\}$, $F(r,s,t) = \ast$.

Since $Gl_+(3) \simeq SO(3)$ and since $SO(3)$ is a compact Lie group, we have $\pi_2(Gl_+(3), \ast) = 0$. Since $[0, 2\pi]^2 \times [0,1] \cong D^3$, it follows that the map $F$ extends into a continuous map $F : [0, 2\pi]^2 \times [0,1] \to Gl_+(3)$.

We then define a map $F' : (S^1 \times S^1) \times [0,1] \to Gl_+(3) : ((e^{ir}, e^{is}), t) \mapsto F(r,s,t)$. In other words, we obtained a homotopy of map $F'_t : (N, pt) \to (Gl_+(3), \ast)$ such that $F'_0 = T_{Q'}$ and $F'_1 = \ast$. Since $\ast$ is path-connected to $Id \in Gl_+(3)$, concatenation with such a path yields a homotopy over $N$ from the map $T_{Q'}$ to the constant map $Id = \left. T_{P'} \right|_{T_N S^3}$.

Step 4. Existence of a trivialization $P'$ as we seek:

We have obtained a continuous map $H : (S^3 \times \{0\}) \cup (N \times [0,1]) \to Gl_+(3)$ such that $H(b, 0) = Id = T_{P}(b)$ for $b \in S^3$ and such that $H(b, 1) = T_{Q'}(b)$ for $b \in N$. Since $N$ is a submanifold of the manifold $S^3$, the pair $(S^3, N)$ has the homotopy extension property, i.e. there exists an extension $H' : S^3 \times [0,1] \to Gl_+(3)$ to the above map $H$.

Set $T_{P'}(b) = H'(b, 1)$ for $b \in S^3$ and $P'_b := T_{P'}(b) \circ P_b$. The map $P' : TS^3 \to \mathbb{R}^3$ is a trivialization such that $P' = Q'$ on $T_N S^3$, as we seek.

Step 5. Smoothing:

The above is a continuous trivialization. However, essentially by Whitney's approximation, it is possible to homotopically perturb any continuous function into a smooth one; the best is probably not to smoothen $P'$ (since its domain is noncompact), but rather the map $T_{P'} : S^3 \to Gl_+(3)$.