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Let $M$ be a smooth connected oriented $d$-dimensional manifold. Let $S \subseteq M$ be a $k$-dimensional embedded submanifold which is also compact, connected and orientable.

Suppose we are given a smooth embedding $F:S \to \mathbb{R}^d$.

Question:

Suppose that $TM|_S$ is a trivial vector bundle. Is it true that the quotients $TM|_S \big/ TS$ and $S \times \mathbb{R}^d\big/dF(TS)$ are isomorphic as vector bundles?

The triviality of $TM|_S$ is an obvious necessary condition, since if $TM|_S \big/ TS \cong S \times \mathbb{R}^d\big/dF(TS)$, then $$TM|_S \cong TS \oplus TM|_S \big/ TS \cong dF(TS) \oplus S \times \mathbb{R}^d\big/dF(TS) \cong S \times \mathbb{R}^d.$$

Of course, isomorphic bundles and subbundles can induce non-isomorphic quotients, so $TS \cong dF(TS)$ and $TM|_S \cong S \times \mathbb{R}^d$ do not imply the quotients are isomorphic.

My guess is that the answer can be negative in general, but I don't know how to find a counterexample.

Note that when the codimension $d-k=1$, the answer is positive:

The quotients are isomorphic.

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  • $\begingroup$ It seems that you are asking the following question: You have two orthogonal rank $n$ vector bundles $E_1\to S, E_2\to S$, which are isomorphic as vector bundles. Are they isomorphic as orthogonal vector bundles? The answer is positive since $GL(n)/O(n)$ is contractible. $\endgroup$ – Moishe Kohan Nov 13 '17 at 21:00
  • $\begingroup$ I don't understand the Riemannian metric's role in this question. Is it really important that the embedding into Euclidean space is isometric? Your question is topological after all. $\endgroup$ – Amitai Yuval Nov 16 '17 at 15:34
  • $\begingroup$ If $d$ is sufficiently high relative to $k$ (I think, $d\ge 2k+2$ will suffice) then the answer is positive, otherwise, most likely, not. $\endgroup$ – Moishe Kohan Nov 18 '17 at 12:37
  • $\begingroup$ @AmitaiYuval You are absolutely right. I had a specific (Riemannian) application in mind, but after some more thought I am convinced the question is only topological. (e.g all normal bundles are isomorphic, no matter which metric you choose, so a question on the topology of normal bundles is really a question on quotient bundles). I edited the question to remove the irrelevant information. $\endgroup$ – Asaf Shachar Nov 18 '17 at 15:34
  • $\begingroup$ @MoisheCohen This sounds very interesting, I can imagine some intuition for what you are saying (in high dimension), but I have no clue regarding a rigorous argument. Can you elaborate? $\endgroup$ – Asaf Shachar Nov 18 '17 at 15:37
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The answer to your question is no. As a counter example, consider $S=S^2$, the $2$-dimensional sphere, and $M=TS$, the total space of the tangent bundle. The inclusion $S\hookrightarrow M$ we think of is the usual one, that is, we identify $S$ with the zero section.

Why is this a counter example?

Fact $1$: The normal bundle of $S$ in $M$ is non-trivial. Indeed, this normal bundle is just the tangent bundle itself, which is non-trivial.

Fact $2$: The sphere $S$ can be embedded in $\mathbb{R}^4$ with a trivial normal bundle. Indeed, the standard embedding of $S$ in $\mathbb{R}^3$ as the unit sphere has a trivial normal bundle, and so, the composition $$S\hookrightarrow\mathbb{R}^3\hookrightarrow\mathbb{R}^4,$$where the right hand embedding can be any linear one, has a trivial normal bundle too.

The only thing we need now is

Fact $3$: The restricted bundle $TM|_S$ is trivial. To see this, note first that $$TM|_S=TS\oplus TS.$$Now, every rank $4$ bundle over the $2$-sphere is determined by a loop in $SO(4)$ (every such bundle is trivial on both the northern and southern hemispheres, and can thus be described by a transition map along the equator). But the fundamental group of $SO(4)$ is isomorphic to $\mathbb{Z}/2$, and so, every loop in $SO(4)$ of the form $\alpha*\alpha$ is null-homotopic, where $\alpha$ is any other loop in $SO(4)$.

Let $\beta\in\pi_1(SO(2))$ denote the homotopy class corresponding to the tangent bundle $TS$. Then under an isomorphism $\pi_1(SO(2))\to\mathbb{Z}$, the class $\beta$ is mapped to $\pm2$ (this is a known fact), which means that there is an $\alpha\in\pi_1(SO(2))$ such that $\beta=\alpha*\alpha$.

In conclusion, the homotopy class in $\pi_1(SO(4))$ which corresponds to $TS\oplus TS$ is $$\beta\oplus\beta=(\alpha\oplus\alpha)*(\alpha\oplus\alpha)=0,$$verifying Fact $3$.

Edit: In fact, I believe the direct sum $E\oplus E$ is trivial for any rank $2$ bundle over the $2$-sphere. This can be shown using arguments similar to the ones above.

Another edit: I wonder what the answer would be if you required $M$ to be Euclidean too. In other words, we know that in general, a manifold can be embedded in Euclidean space in non-homotopic ways (think of knot theory, for example). But can a manifold be embedded in Euclidean space in two different ways that yield different normal bundles?

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