Defining Quotient Bundles

This is an extremely elementary question but I just can't seem to get things to work out. What I am looking for is a natural definition of the quotient bundle of a subbundle $E'\subset E$ of $\mathbb R$ (say) vector bundles over a fixed base space $B$. Every source I find on this essentially leaves the construction to the reader. I would like to glue together sets of the form $U\times E_x/E'_x$ where $x\in U$ is a locally trivial neighborhood by some sort of transition function derived from those corresponding to $E$ and $E'$, but this doesn't actually make sense in any meaningful way.

While I am tagging this as differential geometry, I would like a construction that works in the topological category (i.e., does not invoke Riemannian metrics) and avoids passing to the category of locally free sheafs.

Sorry if this is a repost (I'm sure it is, but I can't seem to find anything) and thanks in advance.

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This is a good challenge for someone learning the ins and outs of vector bundles. Just take it slow and struggle with it for a while. It is important to do things that are not canonical or functorial when trivializing the quotient bundle. I suggest trying to work with local frames of sections of the three bundles in question and build trivializations to $U \times \mathbb{R}^k$ (with different values of $k$ for each bundle of course). –  Deane Yang Jun 8 '10 at 3:44

(I was going to leave this as a comment but decided that it's a bit long for that)

A couple of remarks:

1. You express an aversion to Riemannian metrics because you want to be able to apply this in the topological category. That's fine, except for two things: firstly, Riemannian metrics would not be explicitly involved in this construction as it is a general construction that applies to all vector bundles, not just tangent bundles. Secondly, having inner products on the fibres of a vector bundle is not something that is special to the smooth category. Using a partition of unity argument (assuming you're working over a sufficiently nice space, or your vector bundle is a pull-back of a universal one - look up "numerable cover" for more on this - but note that all the answers to this question tacitly assume this), any finite dimensional vector bundle admits a continuous choice of inner product on its fibres. So the standard argument: "choose an orthogonal structure and take the orthogonal complement" works equally well in the continuous category as the smooth one.

2. What is really going on here is a reduction of structure group. The structure group of the big bundle is $Gl(n)$. The inclusion of the subbundle implies that it reduces to the subgroup that preserves $\mathbb{R}^k \subseteq \mathbb{R}^n$. At this point, you should work out what this subgroup consists of - think in terms of matrices if you don't see it immediately. General Nonsense (although for this case, the more junior Lieutenant Nonsense will do) implies that this subgroup is homotopy equivalent to $Gl(k) \times Gl(n-k)$. A reduction to this defines an isomorphism $E \cong E' \oplus E''$, where $E''$ is the required quotient bundle. The two previous answers can be viewed as constructing this reduction. The "standard orthogonality" argument rests on the observation that $Gl(m) \cong O(m)$ for any $m$ so we can reduce everything to the corresponding orthogonal group. Thus we start with $O(n)$, reduce to the subgroup that preserves $\mathbb{R}^k$ and then ... but there is no "and then" because this subgroup is already $O(k) \times O(n-k)$. So either way, you are doing one "reduction of structure group", the difference between the two methods is simply a choice of doing $Gl(n) \to O(n)$ at the start, or doing $Gl(n;k) \to Gl(k) \times Gl(n-k)$ in the middle.

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Andrew, I agree. Isn't that what is going on in my construction, in guise? I construct $\bar \psi$ first as a vector bundle in which $E'$ sits in as a direct sum. –  David Carchedi Jun 8 '10 at 11:23
Absolutely! As I tried to make clear, this is a comment on the question and the two answers. As this is (as the questioner admits) a fairly basic question, the educator in me could not resist adding a little of the background story to the explicit constructions (and the phrase about "Riemannian metrics" needed clarifying!). –  Loop Space Jun 8 '10 at 11:55

I agree this isn't completely obvious. Here's a slightly different take on it. Our intended vector bundle is

$E/E' :=\coprod E_x/E'_x$,

the disjoint union of the quotient vector spaces of fibres. We just need to specify the topology on it. We do this by describing a family of maps which we intend to be continuous local trivializations for the bundle.

So, take a point $p\in B$, and a nhd $U$ of $p$ on which we have a frame $(e_1, \cdots e_{n+k})$ for $E$. Choose $1\leq i_1<\ldots< i_k\leq n+k$ such that on the fibre $p$,

$E_p=E_p' + span(e_{i_1},\ldots e_{i_k})$.

By the continuity of the determinant function, in fact there's a neighbourhood of $p$ on which this is true; that is, there's a (perhaps smaller) nhd $V$ of $p$ such that for all $x\in V$,

$E_x=E_x' + span(e_{i_1},\ldots e_{i_k})$.

So at each $x\in V$, we have a basis

$(e_{i_1}+E_x',\ \ldots \ e_{i_k}+E_x')$

for $E_x/E'_x$. We demand that this collection of bases give a (continuous) frame for $E/E'$ over $V$. It's an easy check that the transition functions between two thus-constructed local trivializations are continuous, as required.

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Let $h:E' \hookrightarrow E$ denote the inclusion of vector bundles. Let $p:Coker(h) \to B$ be defined in the obvious way (fiber at $x$ is $E_x/h(E'_x)$). It suffices to construct local trivializations of it. Let $E'$ have rank $n$ and $E$ rank $n+k$. Let $U$ be an open subset of $B$ over which both bundles admit a trivialization. Consider the induced map $h:\pi'^{-1}(U) \to \pi^{-1}(U)$ where $\pi'$ and $\pi$ are the associated projections. Let $\phi:\pi'^{-1}(U) \to U \times \mathbb{R}^{n}$ and $\psi:\pi^{-1}(U) \to U \times \mathbb{R}^{n+k}$ be the associated trivializations. Let $\sigma=\psi \circ h \circ \phi^{-1}:U \times \mathbf{R}^{n} \to U \times \mathbb{R}^{n+k}$. Note, $\sigma$ must be fiber-wise linear since we have a vector-bundle. So we have an identification $R^{n+k} \cong Im(\sigma_x) \oplus Coker(\sigma_x)$ for each x. So we get an iso $U \times R^{n+k} \cong U \times Im(\sigma) \oplus Coker(\sigma)$. Composing this with $\sigma$ yields a morphism $U \times \mathbb{R}^{n} \to U \times Im(\sigma) \oplus Coker(\sigma)$. Call this composition $g$. Let $f$ be the map $U \times \mathbb{R}^{n} \oplus \mathbb{R}^{k} \to U \times \mathbb{R}^{n+k}$ which is identify on $U$ and $g \oplus I_k$ on $\mathbb{R}^{n} \oplus \mathbb{R}^{k}$, where $I_k$ is the $k \times k$-identity matrix. Then $\bar \psi:=f^{-1} \circ \psi:\pi'^{-1}(U) \to U \times \mathbb{R}^{n} \oplus \mathbb{R}^{k}$ is a bundle-chart. Note that $Ker(proj_2 \circ \bar \psi)=Im(h)$. So, $\bar \psi$ induces a diffeomorphism $p^{-1}(U) \to U \times \mathbb{R}^{k}$ via $(x,w+h(E'_x)) \mapsto (x,(proj_2 \circ \bar \psi)(w))$. This will serve for the trivialization.

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Hi David, I'm a little worried about this. First, do you mean "coker" instead of "ker" throughout? Secondly and more importantly, I think as you've set this up the identification of $Im(\sigma_x)\oplus Coker(\sigma_x)$ with $\mathbb{R}^{n+k}$ isn't canonical. This is a problem because it then isn't clear that the identification on each fibre can be chosen so as to "vary continuously from fibre to fibre" (which is the key point for the local trivialization). –  macbeth Jun 8 '10 at 2:23
(Reading Deane's comment, I notice my remarks might be ambiguous: by "canonical" I mean "uniquely specified from the data (including trivializations) so far given," not "independent of the choice of trivializations.") –  macbeth Jun 8 '10 at 8:00
No, no. Every short-exact sequence of vector spaces SPLITS, so I mean kernel, not cokernel. Now, to define the splitting in a natural way, you need to pick a basis this is true- but then this boils down to the same comment you made about the continuity of the determinant function. Everything is fine.. –  David Carchedi Jun 8 '10 at 10:23
Maybe this is just a misunderstanding of notation: as I've read your answer, $\sigma$ denotes the inclusion of fibres of $E'$ into fibres of $E$, and in particular has zero kernel. Sure, the ses $0\to E'_x\to E_x\to E_x/E_x'=coker(\sigma)\to 0$ splits, and I'm happy to believe that essentially our answers say the same thing. –  macbeth Jun 8 '10 at 11:43
(The style of the question is that of a beginner to DG so I'm being more pedantic than usual. I was sure you knew what was going on but couldn't be sure the questioner did.) You should be careful to distinguish between a (local) section of the vector bundle and a (local) section of the associated principal bundle. The latter is, of course, equivalent to a trivialisation of the vector bundle (by definition, if you set things up correctly) but the former is most definitely not. As originally phrased, your answer read as if you just needed a local section of the vector bundle. –  Loop Space Jun 8 '10 at 12:26