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Hi,

in bordism-theory and many bordering areas one has the following construction: Given a manifold M (say closed for the purposes of this discussion and k-dimensional), we embed it into some $\mathbb R^n$ for n large and look at the normal bundle of that embedding. This pulls back to give an n-k-dim vectorbundle over M, and we consider the homotopy class $M \rightarrow BGL(n-k) \rightarrow BGL$, where the first map is the classifying map for that bundle and the second one is induced by the obvious inclusion. One now finds that the homotopy class of this composition does not depend on the particular embedding chosen. Since $BGL$ classifies principal-$GL$-bundles we have thus constructed an isomorphism class of such bundles and from what I gather this is what is called the stable normal bundle.

Now my question is:
Is there a sufficiently nice construction of an actual $GL$-bundle representing this isomorphism class?

There certainly seems to be none for the individual normal bundles (for they of course DO depend on the embedding for small n), but for the infinite one there just might be, right? By 'construction' I mean construction out of intrinsic data of the manifold and not one along the lines of 'embed M into $\mathbb R^\infty$ and look at the frames of the arising normal bundle'.

If a construction can be found at all then there are probably many, so there won't be a canonical one, which is why I don't really want to specify what 'nice' is supposed to mean.

Thank you for any answers

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    $\begingroup$ The tangent bundle is classified by a map $t:M\to BO(k)$, which composed with $BO(k)\to BO$, gives a map $\tau:M\to BO$ which has an homotopy inverse $\nu:M\to BO$. This "is" the stable normal bundle, no? $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 26, 2011 at 14:04
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    $\begingroup$ The homotopy class of the inverse is relatively easy to describe, too. If $M$ is compact (for the sake of simplicity), Mariano's map $t : M \to BO(k) = Gr_{\infty,k}$ is a map into $Gr_{m,k}$ for some large $m$. There is an isometry / diffeomorphism $Gr_{m,k} \equiv Gr_{m,m-k}$ given by taking the orthogonal complement. The composite $t : M \to Gr_{m,k} \equiv Gr_{m,m-k} \to BO(m-k)$ is the classifying map for the stable normal bundle. $\endgroup$
    – Ryan Budney
    Commented Oct 26, 2011 at 14:49
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    $\begingroup$ I think your question is too vague. It would be more of an actual question if you tell us precisely what kind of bundle construction you're looking for. Or if you can't tell us that, perhaps you can tell us some kind of functorial or categorical setting you need this construction for. $\endgroup$
    – Ryan Budney
    Commented Oct 26, 2011 at 16:34
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    $\begingroup$ Maybe this point is made elsewhere, but the notion of a normal bundle implicitly requires an embedding. What is a normal vector supposed to be orthogonal to? $\endgroup$
    – Sean Tilson
    Commented Oct 26, 2011 at 19:19
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    $\begingroup$ @Sean Tilson: The point is that the stable normal bundle is intrinsic to the manifold, despite the fact that all the constructions that have been bounced around depend on an embedding. So we might hope to read it off from the manifold without talking about any ambient anything. $\endgroup$
    – Aaron Mazel-Gee
    Commented Nov 1, 2011 at 18:21

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Perhaps I'm confused about what you are looking for, but haven't you already constructed an "actual" $GL$-bundle in your question?

What I mean is the following. The usual definition of $GL$ is a direct limit of $GL(m)$'s. So an element of $GL$ is just an element of $GL(m)$ for some $m$. Similarly, if $M$ is compact then a $GL$ bundle over $M$ is just a $GL(m)$ bundle over $M$, for some $m$. As you note in your question, one can construct such bundles by embedding $M$ in $\mathbb{R}^{k+m}$, taking the frame bundle of the normal bundle, and then interpreting this as a $GL$-bundle rather than a $GL(m)$-bundle. Any two such embeddings of $M$ give isomorphic $GL$-bundles.

In response to pudin's comment below, here's a second construction. Embed $M$ into $\mathbb{R}^\infty$. Define a bundle $F$ over $M$ whose fiber at $x$ is frames of the normal bundle of the embedding at $x$ which eventually coincide with the standard framing of $\mathbb{R}^\infty$. (This is possible because the image of the embedding will lie in some $\mathbb{R}^n \subset \mathbb{R}^\infty$ if $M$ is compact.) $F$ is a principal $GL$ bundle, where in this case we take $GL$ to be invertible linear maps $\mathbb{R}^\infty \to \mathbb{R}^\infty$ which differ from the identity only on a finite subspace.

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    $\begingroup$ i like to think of GL as invertible linear maps $\mathbb R^\infty \rightarrow \mathbb R^\infty$ that differ from the identity only on a finite subspace, but that does not really matter. But what is not true is that a $GL$-bundle IS a $GL(m)$-bundle for some m. True, over a compact space every $GL$-bundle admits a reduction to a $GL(m)$-bundle along $GL(m) \rightarrow GL$ for some m; but as you state you end up with a lot of choices giving isomorphic $GL$-bundles, which is what we started with. The question is for an actual bundle, not just an isomorphism class, just as for the tangentbundle. $\endgroup$
    – old account
    Commented Oct 26, 2011 at 14:40
  • $\begingroup$ I disagree that this is not an "actual" $GL$ bundle. If we describe a $GL$ bundle in terms of charts and transition functions, then these transition functions will take values in some $GL(M)\subset GL$. So every $GL$ bundle is also a $GL(m)$ bundle (for some $m$); the data for a $GL$ bundle is precisely the data for a $GL(m)$ bundle. (I'm assuming $M$ is compact, of course.) Yes, the construction depends on some choices, but each of those choices yields a concrete, "actual" $GL$ bundle. $\endgroup$
    – Kevin Walker
    Commented Oct 26, 2011 at 15:21
  • $\begingroup$ The tangent bundle is also an isomorphism class, if you really insist... $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 26, 2011 at 18:13
  • $\begingroup$ @Kevin Walker: the second construction you gave now is the one i tried to mention in the lines immediatly after my question (except i forgot to pass to frames, which i have edited now). The bundle ypu end up with will depend on the embedding! Granted two choices of embeddings certainly give isomorphic bundles, but what i'm looking for is a construction that does not depend on such choices at all. For the tangent bundle there are such constructions, eg germs of curves mod some relation on derivatives or derivations on the rings of germs of functions to R $\endgroup$
    – old account
    Commented Nov 1, 2011 at 10:24
  • $\begingroup$ Ah. I misinterpreted the original version of your question to be asking for a concrete ("actual") construction of the stable normal bundle, as opposed to a finite-dimensional bundle which determined the stable normal bundle up to isomorphism. Now I understand that you really want a construction that does not involve making arbitrary choices. $\endgroup$
    – Kevin Walker
    Commented Nov 1, 2011 at 13:06

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