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Let $\tilde V, \tilde W$ be orientable real vector bundles of rank $k$ over a smooth manifold $M$.

Suppose there exist isomorphic subbundles $V \subseteq \tilde V,W \subseteq \tilde W$ of rank $k-1$, i.e $V \cong W$.

Then we have the following claim:

Proposition:

$\tilde V,\tilde W$ are isomorphic as vector bundles, and so are the quotients $\tilde V/V \cong \tilde W/W$.

Question: I would like to find a proof of the proposition which does not involve introducing a metric.

Here is my proof (using metrics):

Fix orientations on $\tilde V,\tilde W$. Let $\Phi:V \to W$ be an isomorphism. Put a (smooth) metric on $W$, and pull it back by $\Phi$ to $V$, thus making $\Phi$ as isometry. Now extend the metrics to $\tilde V,\tilde W$.

Since the "codimension" of $V$ in $\tilde V$ is $1$ (and so for $W, \tilde W$), there exist a unique isometric bundle isomorphism $\Phi^{\perp}:V^{\perp} \to W^{\perp}$ making $\Phi \oplus \Phi^{\perp}:\tilde V \to \tilde W$ an isometric orientation preserving bundle isomorphism.

Note that this $\Phi^{\perp}$ is smooth.

In particular, we showed that $\tilde V \cong \tilde W$, and $\tilde V/V \cong V^{\perp} \cong W^{\perp} \cong \tilde W/W$.

Edit: (elaboration on the construction of $\Phi^{\perp}$)

Here is an explicit construction of $\Phi^{\perp}:V^{\perp} \to W^{\perp}$:

Let $v_1,\dots,v_{k-1},v_k $ be a local positive orthonormal frame for $\tilde V$ such that $\text{span}\{v_1,\dots,v_{k-1}\}=V$ and $\text{span}(v_k)=V^{\perp}$. (such a frame can always be chosen locally, see below).

Now, define $$\Phi^{\perp}(v_k)=\Phi^{\perp}\big(\star_{\tilde V}^{k-1}(v_1 \wedge \dots \wedge v_{k-1})\big):=\star_{\tilde W}^{k-1}\big(\Phi(v_1) \wedge \dots \wedge \Phi(v_{k-1})\big), \tag{1}$$ where $\star_{\tilde V}^{k-1}:\Lambda_{k-1}(\tilde V) \to \Lambda_{1}(\tilde V)=\tilde V$ is the Hodge dual of $\tilde V$ which is defined via the metric and orientation (same for $\tilde W$).

It is immediate from the definitions of the Hodge dual, that $$\Phi^{\perp}(v_k) \in \big(\text{span}\{\Phi(v_1),\dots,\Phi(v_{k-1})\}\big)^{\perp}=W|_{U}^{\perp},$$

where the last equality follows from the assumed surjectivity of $\Phi:V \to W$. ($U \subseteq M$ is the small set over which we have our local frame).

This makes $\Phi^{\perp}$ an isometric isomorphism $V^{\perp} \to W^{\perp}$. Moreover, this makes $\Phi \oplus \Phi^{\perp}:\tilde V \to \tilde W$ an isometric orientation preserving isomorphism (indeed, it takes the positive orthonormal frame $v_1,\dots,v_k$ of $\tilde V$ to a positive orhtonormal frame in $\tilde W$).

In particular this shows $\Phi^{\perp}:V^{\perp} \to W^{\perp}$ is well defined, i.e it does not depend on the frame we chose. The reason is that $V^{\perp},W^{\perp}$ are of rank $1$, and there are only two isometric maps between $1$-dim inner product spaces. Here we know how to "choose the right one" at each fiber, due to the requirement that the "big" combined map $\Phi \oplus \Phi^{\perp}$ will be orientation-preserving.

Moreover, the continuity of the orientations, implies this choice is "smooth" along changing fibers.

It is here that we use the fact $\tilde V,\tilde W$ were orientable.

An extended comment regarding the uniqueness (and smoothness) of $\Phi^{\perp}$:

We didn't really had a choice in constructing $\Phi^{\perp}$. The uniqueness is a "pointwise-linear algebra" phenomenon. Indeed, suppose $\Phi \oplus \Phi^{\perp}$ is an isometric orientation preserving isomorphism. Then, since orientation preserving isometries commute with Hodge duals we must have $$ \Phi^{\perp}(v_k)=\big(\Phi \oplus \Phi^{\perp}\big)(v_k)=\big(\Phi \oplus \Phi^{\perp}\big)\big(\star_{\tilde V}^{k-1}(v_1 \wedge \dots \wedge v_{k-1})\big)=$$

$$ \star_{\tilde W}^{k-1}\big( \bigwedge^{k-1}\big(\Phi \oplus \Phi^{\perp}\big)(v_1 \wedge \dots \wedge v_{k-1})\big)=\star_{\tilde W}^{k-1}\big( \big(\Phi \oplus \Phi^{\perp}\big)(v_1) \wedge \dots \wedge \big(\Phi \oplus \Phi^{\perp}\big)(v_{k-1})\big)=$$

$$ \star_{\tilde W}^{k-1}\big( \Phi (v_1) \wedge \dots \wedge \Phi (v_{k-1})\big).$$

This is exactly our definition in equation $(1)$.

In particular this uniqueness implies $\Phi^{\perp}$ is smooth, since it takes a smooth frame to a smooth frame.

A comment regarding the existence of a suitable frame $v_1,\dots,v_{k-1},v_k$:

We can always choose a local orthonormal frame $v_1,\dots,v_{k-1}$ for $V$. Now, we can also choose an orthonormal frame $v_k$ for $V^{\perp}$. If we restrict everything to be over a connected subset $U \subseteq M$, then we must have that the resulting frame $v_1,\dots,v_{k-1},v_k$ for $\tilde V$ is positively oriented or negatively oriented. So, if need be, we just replace $v_k$ by $-v_k$ and we are done.

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  • $\begingroup$ How do you derive the bundle isomorphism $\Phi^\perp$? $\endgroup$ – Amitai Yuval Nov 16 '17 at 11:55
  • $\begingroup$ I mean, are you suggesting that two line bundles are always isomorphic to one another? (This is not true in general). $\endgroup$ – Amitai Yuval Nov 16 '17 at 11:56
  • $\begingroup$ @AmitaiYuval No, I am not suggesting this. What I am saying is, that every two line bundles which are obtained as quotients of two orientable rank-$k$ bundles by isomorphic rank-$(k-1)$ subbundles, must be themselves isomorphic. (And the large $k$-bundles were also in fact isomorphic). I have added many details on the construction, uniqueness and smoothness of the map $\Phi^{\perp}$. The point is that I would like to see a proof which does not involve metrics, since there are none in the statement (it's purely topological). $\endgroup$ – Asaf Shachar Nov 16 '17 at 13:13
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Essentially, it seems to me that your proof is the right one. The fact that it uses Riemannian metrics shows that the big isomorphism $\Phi\oplus\Phi^\perp$ is not canonical. Indeed, it makes sense that one would have to make arbitrary choices in order to extend an isomorphism of subbundles to the whole bundles, and as it happens so frequently in differential topology, all those choices can be packed in the construction of a Riemannian metric. However, it can be interesting to avoid using a metric and see where it is exactly that a choice is needed. This is what this answer will try and do.

Step 1: Let $\widetilde{V}$ and $\widetilde{W}$ be two oriented real vector spaces, let $V\subset\widetilde{V}$ and $W\subset\widetilde{W}$ be two subspaces, and let $\Phi:V\to W$ be an isomorphism. Then there is a natural bijection $$\mathcal{F}:\mathrm{or}\left(\widetilde{V}/V\right)\to\mathrm{or}\left(\widetilde{W}/W\right),$$where $\mathrm{or}(\cdot)$ denotes the set of orientations.

The proof of Step $1$ is basically your argument using the Hodge star operator, only here, one does not need any metric since we only care about the orientations. Note also that this step holds in any codimension.

Step 2: In the question's setting, there is a natural isomorphism between the orientation bundles of $\widetilde{V}/V$ and $\widetilde{W}/W$.

This follows immediately from Step $1$. Up until here, no choices were made. This is about to change drastically.

Step 3: Let $E$ and $E'$ be two real line bundles over a manifold $M$ with isomorphic orientation bundles. Then $E$ and $E'$ are isomorphic line bundles.

Of course, Step 3 can be proved easily using a Riemannian metric. If one prefers to avoid using a metric, then one needs to imitate the construction of a metric and construct a bundle isomorphism using a partition-of-unity argument. This way or another, one cannot avoid taking a massive choice here (and this makes sense).

(Question: How come Step $3$ doesn't work when higher rank bundles are involved?).

Step 4: Let $\widetilde{V}$ and $\widetilde{W}$ be vector bundles over a manifold $M$. Let $V\subset\widetilde{V}$ and $W\subset\widetilde{W}$ be subbundles. Let $\Phi:V\to W$ and $\Phi^\perp:\widetilde{V}/V\to\widetilde{W}/W$ be isomorphisms of vector bundles. Then the bundles $\widetilde{V}$ and $\widetilde{W}$ are isomorphic to one another.

Once again, Step $4$ is easy to prove with the aid of Riemannian metrics. Furthermore, there is no natural way to construct an isomorphism. Indeed, different Riemannian metrics yield different isomorphisms. However, as this answer tries to minimize the extent of the choices being made, it may be interesting to note that we only need splittings of $\widetilde{V}$ and $\widetilde{W}$. That is, we only need subbundles $V^\perp\subset\widetilde{V}$ and $W^\perp\subset\widetilde{W}$ that are complementary to $V$ and $W$. Such splittings may appear even when there is no Riemannian metric in the picture.

Edit: Let us go into a proof of Step 3. To do that, first we need to understand what it means for two line bundles to have isomorphic orientation bundles. Given a real line $L$, an orientation on $L$ is just a choice of a connected component of $L\setminus\{0\}$. So, if $L$ and $L'$ are two line bundles over a manifold $M$ and $\varphi:\mathrm{or}(L)\to\mathrm{or}(L')$ is an isomorphism of fiber bundles, this means that we have a global correspondence between connected components of $L_p\setminus\{0_p\}$ and $L'_p\setminus\{0'_p\}$ for every $p\in M$. Now we can construct local isomorphisms between $L$ and $L'$, that respect this correspondence (this involves choices), and then patch these local isomorphisms to a global one using a partition of unity (more choices). The reason we can do this, is that for every $p\in M$, the set of correspondence-respecting isomorphisms between $L_p$ and $L'_p$ is convex.

Another edit: Here is another, perhaps nicer, way to see Step 3. In the setting of the previous edit, let us examine the homomorphism bundle $\hom(L,L')$. Of course, this is a line bundle. The isomorphism $\varphi:\mathrm{or}(L)\to\mathrm{or}(L')$ induces an orientation on $\hom(L,L')$. But every orientable real line bundle is trivial, and this means that the $\hom$ bundle has a non-vanishing section.

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  • $\begingroup$ Thanks, great answer. Can you please elaborate on step 3? i.e why two real line bundles with isomorphic orientation bundles, are isomorphic? I see why two orientable line bundles are always isomorphic (since the orinetation form gives you a global non-zero section on both bundles). However, I do not see why the more general claim is correct. $\endgroup$ – Asaf Shachar Nov 18 '17 at 6:41
  • $\begingroup$ @AsafShachar Check the edit please. $\endgroup$ – Amitai Yuval Nov 18 '17 at 9:48
  • $\begingroup$ @AsafShachar I added another edit. I think you may like it more. $\endgroup$ – Amitai Yuval Nov 18 '17 at 10:00
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This is an elaboration on "Step I" in Amitay Yuval's answer:

Claim (Step 1): Let $\widetilde{V}$ and $\widetilde{W}$ be two oriented real vector spaces, let $V\subset\widetilde{V}$ and $W\subset\widetilde{W}$ be two subspaces, and let $\Phi:V\to W$ be an isomorphism. Then there is a natural bijection $$\mathcal{F}:\mathrm{or}\left(\widetilde{V}/V\right)\to\mathrm{or}\left(\widetilde{W}/W\right),$$where $\mathrm{or}(\cdot)$ denotes the set of orientations.

A proof of the claim::

We start with the following

Lemma 1: Let $W$ be an oriented vector space, $V \subseteq W$ a subspace. There exist a bijection $$ \mathcal{F}_{W/V \to V}:\mathrm{or}\left(W/V\right)\to\mathrm{or}\left(V \right).$$

We denote its inverse by $$ \mathcal{F}_{V \to W/V}:=(\mathcal{F}_{W/V \to V})^{-1}:\mathrm{or}\left(V \right) \to \mathrm{or}\left(W/V\right).$$

Given lemma 1, we proceed as follows:

Note that $\Phi$ induces a bijective map $\mathcal{F}_{\Phi}:\mathrm{or}\left(V\right) \to \mathrm{or}\left( W \right)$. (given an orientation on $V$, we define an orientation on $W$ by requiring $\Phi$ to be orientation-preserving).

Now, we construct $$\mathcal{F}:\mathrm{or}\left(\widetilde{V}/V\right)\to\mathrm{or}\left(\widetilde{W}/W\right),$$ by setting

$$ \mathcal{F}=\mathcal{F}_{W \to \tilde W/W} \circ \mathcal{F}_{\Phi} \circ \mathcal{F}_{\tilde V/V \to V} \,\,.$$


Proof of lemma 1:

Suppose $\dim V=k,\dim W=r+k$, so $\dim\left(W/V\right)=r$. Let $O_{W/V} \in \mathrm{or}\left(W/V\right)$. We show how $O_{W/V}$ induces an orientation on $V$.

Given a base $(v_1, \ldots, v_k)$ of $V$, choose some base of $W/V$, say $(w_1 + V, \ldots, w_r + V)$, and lift it to $W$, i.e consider $(w_1, \ldots, w_r)$. Note $(w_1, \ldots, w_r, v_1, \ldots, v_k)$ is base for $W$. Now define $O(v_1, \ldots, v_k) $ by requiring $$ O_{W/V}(w_1 + V, \ldots, w_r + V) = O_W(w_1, \ldots, w_r, v_1, \ldots, v_k)O(v_1, \ldots, v_k) $$

Let's check this is well-defined, i.e does not depend on the choice of the representatives.

Let $\bar w_i$ be another choice of representatives. We need to show that $$O_W(w_1, \ldots, w_r, v_1, \ldots, v_k)=O_W(\bar w_1, \ldots, \bar w_r, v_1, \ldots, v_k).$$ Since $w_i-\bar w_i \in V,$ we can write $w_i-\bar w_i=a_i^jv_j.$ Then the transition matrix from $(w_1, \ldots, w_r, v_1, \ldots, v_k)$ to $(\bar w_1, \ldots, \bar w_r, v_1, \ldots, v_k)$ is a block matrix of the form

$$ B=\left(\begin{array}{cc} \text{Id}_{r \times r}&A\\ 0&\text{Id}_{k \times k} \end{array}\right),$$ where $A=a_i^j$. Since $\det B=1$, we are done.

It is easy to see the map $\mathcal{F}_{W/V \to V}:\mathrm{or}\left(W/V\right)\to\mathrm{or}\left(V \right)$ we built in injective, hence bijective.

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