Is the concept of a “numerable” fiber bundle really useful or an empty generalization? The Next CEO of Stack OverflowNon trivial vector bundle over non-paracompact contractible spaceExample of fiber bundle that is not a fibrationGlobalising fibrations by schedulesFiber bundle = principal bundle + fiber?Numerable covers from the point of view of Grothendieck topologiesGlobal sections for torus fiber bundleAre there analogs of smooth partitions of unity and good open covers for PL-manifolds?Two natural maps asssociated with the nerve of a coverDescent theory, fibrations, and bundlesIn which sense are Euler-Lagrange PDE's on fiber bundles quasi-linear?What is the local structure of a fibration?Complete proof of Homotopy invariance of a numerable fiber bundle based on CHPLocally trivial fibration over a suspension

Is the concept of a “numerable” fiber bundle really useful or an empty generalization?



The Next CEO of Stack OverflowNon trivial vector bundle over non-paracompact contractible spaceExample of fiber bundle that is not a fibrationGlobalising fibrations by schedulesFiber bundle = principal bundle + fiber?Numerable covers from the point of view of Grothendieck topologiesGlobal sections for torus fiber bundleAre there analogs of smooth partitions of unity and good open covers for PL-manifolds?Two natural maps asssociated with the nerve of a coverDescent theory, fibrations, and bundlesIn which sense are Euler-Lagrange PDE's on fiber bundles quasi-linear?What is the local structure of a fibration?Complete proof of Homotopy invariance of a numerable fiber bundle based on CHPLocally trivial fibration over a suspension










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Numerable fiber bundles are defined by Dold (DOLD 1962 - Partitions of Unity in theory of Fibrations) as a generalization of fiber bundles over a paracompact space : the trivialization cover of the base admit a subordinate partition of unity (locally finite). He proves in this paper that almost all important theorems for fiber bundles over paracompact spaces are also valid for numerable bundles.



But is it really an interesting generalization ? Are there examples of "natural" or "useful" numerable fiber bundles that are not paracompact?










share|cite|improve this question











$endgroup$







  • 7




    $begingroup$
    I am not a homotopy theorist but thought the point is that "numerable" is the right notion which makes the theory of fiber bundles work. Paracompactness of the base is not relevant, and it may be unavailable or hard to check.
    $endgroup$
    – Igor Belegradek
    2 days ago










  • $begingroup$
    I don’t know of an example, but maybe this is useful when you want to do differential geometry on the bundle!
    $endgroup$
    – user51223
    yesterday















13












$begingroup$


Numerable fiber bundles are defined by Dold (DOLD 1962 - Partitions of Unity in theory of Fibrations) as a generalization of fiber bundles over a paracompact space : the trivialization cover of the base admit a subordinate partition of unity (locally finite). He proves in this paper that almost all important theorems for fiber bundles over paracompact spaces are also valid for numerable bundles.



But is it really an interesting generalization ? Are there examples of "natural" or "useful" numerable fiber bundles that are not paracompact?










share|cite|improve this question











$endgroup$







  • 7




    $begingroup$
    I am not a homotopy theorist but thought the point is that "numerable" is the right notion which makes the theory of fiber bundles work. Paracompactness of the base is not relevant, and it may be unavailable or hard to check.
    $endgroup$
    – Igor Belegradek
    2 days ago










  • $begingroup$
    I don’t know of an example, but maybe this is useful when you want to do differential geometry on the bundle!
    $endgroup$
    – user51223
    yesterday













13












13








13


4



$begingroup$


Numerable fiber bundles are defined by Dold (DOLD 1962 - Partitions of Unity in theory of Fibrations) as a generalization of fiber bundles over a paracompact space : the trivialization cover of the base admit a subordinate partition of unity (locally finite). He proves in this paper that almost all important theorems for fiber bundles over paracompact spaces are also valid for numerable bundles.



But is it really an interesting generalization ? Are there examples of "natural" or "useful" numerable fiber bundles that are not paracompact?










share|cite|improve this question











$endgroup$




Numerable fiber bundles are defined by Dold (DOLD 1962 - Partitions of Unity in theory of Fibrations) as a generalization of fiber bundles over a paracompact space : the trivialization cover of the base admit a subordinate partition of unity (locally finite). He proves in this paper that almost all important theorems for fiber bundles over paracompact spaces are also valid for numerable bundles.



But is it really an interesting generalization ? Are there examples of "natural" or "useful" numerable fiber bundles that are not paracompact?







at.algebraic-topology fibre-bundles






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









David White

13k462104




13k462104










asked 2 days ago









ychemamaychemama

531210




531210







  • 7




    $begingroup$
    I am not a homotopy theorist but thought the point is that "numerable" is the right notion which makes the theory of fiber bundles work. Paracompactness of the base is not relevant, and it may be unavailable or hard to check.
    $endgroup$
    – Igor Belegradek
    2 days ago










  • $begingroup$
    I don’t know of an example, but maybe this is useful when you want to do differential geometry on the bundle!
    $endgroup$
    – user51223
    yesterday












  • 7




    $begingroup$
    I am not a homotopy theorist but thought the point is that "numerable" is the right notion which makes the theory of fiber bundles work. Paracompactness of the base is not relevant, and it may be unavailable or hard to check.
    $endgroup$
    – Igor Belegradek
    2 days ago










  • $begingroup$
    I don’t know of an example, but maybe this is useful when you want to do differential geometry on the bundle!
    $endgroup$
    – user51223
    yesterday







7




7




$begingroup$
I am not a homotopy theorist but thought the point is that "numerable" is the right notion which makes the theory of fiber bundles work. Paracompactness of the base is not relevant, and it may be unavailable or hard to check.
$endgroup$
– Igor Belegradek
2 days ago




$begingroup$
I am not a homotopy theorist but thought the point is that "numerable" is the right notion which makes the theory of fiber bundles work. Paracompactness of the base is not relevant, and it may be unavailable or hard to check.
$endgroup$
– Igor Belegradek
2 days ago












$begingroup$
I don’t know of an example, but maybe this is useful when you want to do differential geometry on the bundle!
$endgroup$
– user51223
yesterday




$begingroup$
I don’t know of an example, but maybe this is useful when you want to do differential geometry on the bundle!
$endgroup$
– user51223
yesterday










5 Answers
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active

oldest

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In algebraic topology, it is often more convenient to know that a map is a fibration (has the homotopy lifting property with respect to all spaces) than a fibre bundle, because then calculational tools such as long exact sequences of homotopy groups and Serre spectral sequences of (co)homology groups become available.



It is easy to cook up examples of fibrations which are not fibre bundles (the projection of a $2$-simplex onto one of its edges being the easiest example I know). It is somewhat harder to find examples of fibre bundles which are not fibrations, but they do exist; see here and here.



Numerability is precisely the extra condition on fibre bundles which makes them into fibrations. Of course this means that any fibre bundle over a paracompact base is a fibration.



The homotopy lifting property is used extensively when proving the homotopy classification of principal $G$-bundles, i.e. that isomorphism classes of principal $G$-bundles a given base space $B$ are in one-to-one correspondence with homotopy classes $[B,BG]$. To obtain such a result for arbitrary base spaces $B$ you had better therefore restrict to numerable bundles.



This doesn't really answer your question, in that I haven't given you a natural example of a numerable bundle over a non-paracompact base. But hopefully it indicates why numerable bundles are a useful concept in homotopy theory.






share|cite|improve this answer











$endgroup$








  • 5




    $begingroup$
    This is a bit disingenuous, because for those tools to be available one only needs a Serre fibration (homotopy lifting property with respect to discs), which the most naive definition of a fibre bundle is.
    $endgroup$
    – Oscar Randal-Williams
    2 days ago










  • $begingroup$
    @OscarRandal-Williams: no disingenuity on my part, just ignorance - I hadn't realised that every fibre bundle is a Serre fibration (for reference, a proof is given as Theorem 9 in these notes math.ru.nl/~mgroth/teaching/htpy13/Section06.pdf). My last point about classifying bundles over all spaces still stands though, I believe.
    $endgroup$
    – Mark Grant
    yesterday






  • 2




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    @MarkGrant Another way of seeing that fiber bundles are Serre fibrations, is to notice that a map $p:E→B$ is a Serre fibration iff $E×_B D^n→D^n$ is a Serre fibration for every continuous map $D^n→B$ (this is because you can pull back every lifting diagram). Then you can use that $D^n$ is paracompact and that the pullback of a fiber bundle is a fiber bundle.
    $endgroup$
    – Denis Nardin
    yesterday


















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Dold's condition in on the bundle, while paracompactness is a condition on the base space. These are two very different sorts of concepts, and the former is more categorical. Note that the property of being a numerable bundle is preserved by pullbacks. So e.g., if one pulled back a bundle over a paracompact space to a bundle that was no longer paracompact, one would still automatically know that one has a bundle for which Dold's results hold.






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    To expand and correct my comment...



    Let $Top$ be the category of all topological spaces, $PHTop$ the category paracompact Hausdorff spaces. Let $open$ denote the Grothendieck topology of open covers on either of these categories and $num$ the Grothendieck topology of open covers for which there exists a subordinate partition of unity. Then there are morphisms of sites $Top_num to Top_open$ and $PHTop_num to PHTop_open$. By definition, the latter is an equivalence. But the former is not: there are genuinely more general open covers.



    One can repeat this story for the small sites associated to a fixed space.



    Now the stack $Bun_G$ of principal $G$-bundles on any of these sites is presented by the one-object topological groupoid $mathbfBG = Grightrightarrows *$. Any principal bundle $Pto X$ gives a map of stacks $Xto Bun_G$. If $P$ trivialises over $U to X$, $U = coprod_i U_i$ an open cover, then there is a span of topological groupoids $X leftarrow checkC(U) to mathbfBG$. Here $checkC(U)$ is the topological groupoid with object space $U$ and arrow space $Utimes_X U$. None of this is particularly specific to the case at hand. Since we are dealing with topological groupoids, though, we can geometrically realise and get a span of spaces $X leftarrow BcheckC(U) to BG$, where now $BG$ (plain $B$!) is Segal's construction of the classifying space of $G$.



    However, for $U$ an open cover with a subordinary partition of unity, the map $X leftarrow BcheckC(U)$ is a homotopy equivalence (a result of Segal), whereas in general it is a weak homotopy equivalence. So for numerable bundles, where local triviality is with respect to covers with partitions of unity, you get a map $Xto BG$ and this is well-defined up to homotopy. For arbitrary bundles on non-paracompact Hausdorff spaces, there is only a morphism in the homotopy category, a span with backwards-pointing leg a weak homotopy equivalence.



    Additional to this, the universal $G$-bundle on $BG$ is already numerable, so any bundle gotten by pulling it back is numerable, so this construction also goes in the opposite direction: one cannot get a non-numerable bundle via pulling back along any $Xto BG$. The same goes for Milnor's construction of $BG$ as an infinite join.



    So if you want the topological/homotopy version of classifying theory of principal bundles (using homotopy classes of maps) to agree with the stack-theoretic version, you'd better only use numerable bundles. Which is to say, you need to use the site $Top_num$. When restricted to paracompact Hausdorff spaces this the same as arbitrary open covers, but in general this is a real restriction that affects the theorem about classification of bundles.



    As an additional comment, I think an example of a non-numerable bundle is the frame bundle $F(L)$ (or, equivalently, the tangent bundle $TL$) of the long line $L$. You can think of $L$ as a manifold in the sense of being locally Euclidean and Hausdorff, but isn't metrizable, hence not paracompact. It is connected, path connected and simply-connected. As a result, $F(L)$ is not trivial, despite all this (otherwise $L$ would be metrizable).



    Added Ok, so $TL$ (and so $F(L)$) can't be numerable, otherwise one could define a Riemannian metric on $TL$, and hence a metric on $L$ giving its topology.



    And so this gives another good point why you want numerable bundles in general: if you want connections to exist, or metrics, or indeed any type of geometric information that can be patched together using a partition of unity, then bundles need to trivialise over a numerable cover. A numerable bundle over a non-paracompact Hausdorff space does always admit connections; a numerable vector bundle admits metrics and so on.






    share|cite|improve this answer











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      It is a standard and useful practice to give precise conditions under which a theorem is known to be true, especially if these bring out methods used in the proof. Eldon Dyer suggested to me in the late 1960s that Dold's paper was the first full proof of the globalisation result, and this realisation led to the Dyer-Eilenberg paper referred to in this mathoverflow question.






      share|cite|improve this answer











      $endgroup$








      • 1




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        There is an extra // between the first : and the https:// in the link in your post (i.e. [1]://https://" should be "[1]: https://") and this breaks the link. Unfortunately MathOverflow will not let me edit to repair this because edits need to be at least 6 characters.
        $endgroup$
        – Robert Furber
        yesterday



















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      Question: "... but is it really an interesting generalization?"



      Answer: No. I do not believe that this is a really interesting generalization.
      It is a technical condition which appears when one attempts to formulate results at their a maximal level of generality.



      Let me illustrate this by an example:

      For example, it would be a desirable result to have that given a topological space $X$ and vector bundle $V$ over $Xtimes [0,1]$ then $V|_Xtimes0$ is isomorphic to $V|_Xtimes1$. Unfortunately, that's not true (counterexample: Let $L$ be the long line; equip it with a smooth structure, and let $TL$ be its tangent bundle. Then we the various real powers $(TL)^otimes r$ for $rin [0,1]$ assemble to a line bundle over $L times [0,1]$. That line bundle is trivial over $Ltimes0$, and non-trivial over $Ltimes1$). But if you put the "numerable" condition, then the result becomes true.



      Personally, I would much rather restrict my all my spaces to be paracompact (or any condition which implies paracompact), and then I wouldn't ever need to say the word "numerable".






      share|cite|improve this answer









      $endgroup$








      • 1




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        What are the real tensor powers? Are they the line bundles whose transition functions are real powers of the usual ones for $TL$?
        $endgroup$
        – David Roberts
        2 days ago










      • $begingroup$
        Correct. They can only be defined because TL is orientable (a non-orientable real line bundle does not have a notion of real power).
        $endgroup$
        – André Henriques
        yesterday











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      5 Answers
      5






      active

      oldest

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      5 Answers
      5






      active

      oldest

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      active

      oldest

      votes






      active

      oldest

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      13












      $begingroup$

      In algebraic topology, it is often more convenient to know that a map is a fibration (has the homotopy lifting property with respect to all spaces) than a fibre bundle, because then calculational tools such as long exact sequences of homotopy groups and Serre spectral sequences of (co)homology groups become available.



      It is easy to cook up examples of fibrations which are not fibre bundles (the projection of a $2$-simplex onto one of its edges being the easiest example I know). It is somewhat harder to find examples of fibre bundles which are not fibrations, but they do exist; see here and here.



      Numerability is precisely the extra condition on fibre bundles which makes them into fibrations. Of course this means that any fibre bundle over a paracompact base is a fibration.



      The homotopy lifting property is used extensively when proving the homotopy classification of principal $G$-bundles, i.e. that isomorphism classes of principal $G$-bundles a given base space $B$ are in one-to-one correspondence with homotopy classes $[B,BG]$. To obtain such a result for arbitrary base spaces $B$ you had better therefore restrict to numerable bundles.



      This doesn't really answer your question, in that I haven't given you a natural example of a numerable bundle over a non-paracompact base. But hopefully it indicates why numerable bundles are a useful concept in homotopy theory.






      share|cite|improve this answer











      $endgroup$








      • 5




        $begingroup$
        This is a bit disingenuous, because for those tools to be available one only needs a Serre fibration (homotopy lifting property with respect to discs), which the most naive definition of a fibre bundle is.
        $endgroup$
        – Oscar Randal-Williams
        2 days ago










      • $begingroup$
        @OscarRandal-Williams: no disingenuity on my part, just ignorance - I hadn't realised that every fibre bundle is a Serre fibration (for reference, a proof is given as Theorem 9 in these notes math.ru.nl/~mgroth/teaching/htpy13/Section06.pdf). My last point about classifying bundles over all spaces still stands though, I believe.
        $endgroup$
        – Mark Grant
        yesterday






      • 2




        $begingroup$
        @MarkGrant Another way of seeing that fiber bundles are Serre fibrations, is to notice that a map $p:E→B$ is a Serre fibration iff $E×_B D^n→D^n$ is a Serre fibration for every continuous map $D^n→B$ (this is because you can pull back every lifting diagram). Then you can use that $D^n$ is paracompact and that the pullback of a fiber bundle is a fiber bundle.
        $endgroup$
        – Denis Nardin
        yesterday















      13












      $begingroup$

      In algebraic topology, it is often more convenient to know that a map is a fibration (has the homotopy lifting property with respect to all spaces) than a fibre bundle, because then calculational tools such as long exact sequences of homotopy groups and Serre spectral sequences of (co)homology groups become available.



      It is easy to cook up examples of fibrations which are not fibre bundles (the projection of a $2$-simplex onto one of its edges being the easiest example I know). It is somewhat harder to find examples of fibre bundles which are not fibrations, but they do exist; see here and here.



      Numerability is precisely the extra condition on fibre bundles which makes them into fibrations. Of course this means that any fibre bundle over a paracompact base is a fibration.



      The homotopy lifting property is used extensively when proving the homotopy classification of principal $G$-bundles, i.e. that isomorphism classes of principal $G$-bundles a given base space $B$ are in one-to-one correspondence with homotopy classes $[B,BG]$. To obtain such a result for arbitrary base spaces $B$ you had better therefore restrict to numerable bundles.



      This doesn't really answer your question, in that I haven't given you a natural example of a numerable bundle over a non-paracompact base. But hopefully it indicates why numerable bundles are a useful concept in homotopy theory.






      share|cite|improve this answer











      $endgroup$








      • 5




        $begingroup$
        This is a bit disingenuous, because for those tools to be available one only needs a Serre fibration (homotopy lifting property with respect to discs), which the most naive definition of a fibre bundle is.
        $endgroup$
        – Oscar Randal-Williams
        2 days ago










      • $begingroup$
        @OscarRandal-Williams: no disingenuity on my part, just ignorance - I hadn't realised that every fibre bundle is a Serre fibration (for reference, a proof is given as Theorem 9 in these notes math.ru.nl/~mgroth/teaching/htpy13/Section06.pdf). My last point about classifying bundles over all spaces still stands though, I believe.
        $endgroup$
        – Mark Grant
        yesterday






      • 2




        $begingroup$
        @MarkGrant Another way of seeing that fiber bundles are Serre fibrations, is to notice that a map $p:E→B$ is a Serre fibration iff $E×_B D^n→D^n$ is a Serre fibration for every continuous map $D^n→B$ (this is because you can pull back every lifting diagram). Then you can use that $D^n$ is paracompact and that the pullback of a fiber bundle is a fiber bundle.
        $endgroup$
        – Denis Nardin
        yesterday













      13












      13








      13





      $begingroup$

      In algebraic topology, it is often more convenient to know that a map is a fibration (has the homotopy lifting property with respect to all spaces) than a fibre bundle, because then calculational tools such as long exact sequences of homotopy groups and Serre spectral sequences of (co)homology groups become available.



      It is easy to cook up examples of fibrations which are not fibre bundles (the projection of a $2$-simplex onto one of its edges being the easiest example I know). It is somewhat harder to find examples of fibre bundles which are not fibrations, but they do exist; see here and here.



      Numerability is precisely the extra condition on fibre bundles which makes them into fibrations. Of course this means that any fibre bundle over a paracompact base is a fibration.



      The homotopy lifting property is used extensively when proving the homotopy classification of principal $G$-bundles, i.e. that isomorphism classes of principal $G$-bundles a given base space $B$ are in one-to-one correspondence with homotopy classes $[B,BG]$. To obtain such a result for arbitrary base spaces $B$ you had better therefore restrict to numerable bundles.



      This doesn't really answer your question, in that I haven't given you a natural example of a numerable bundle over a non-paracompact base. But hopefully it indicates why numerable bundles are a useful concept in homotopy theory.






      share|cite|improve this answer











      $endgroup$



      In algebraic topology, it is often more convenient to know that a map is a fibration (has the homotopy lifting property with respect to all spaces) than a fibre bundle, because then calculational tools such as long exact sequences of homotopy groups and Serre spectral sequences of (co)homology groups become available.



      It is easy to cook up examples of fibrations which are not fibre bundles (the projection of a $2$-simplex onto one of its edges being the easiest example I know). It is somewhat harder to find examples of fibre bundles which are not fibrations, but they do exist; see here and here.



      Numerability is precisely the extra condition on fibre bundles which makes them into fibrations. Of course this means that any fibre bundle over a paracompact base is a fibration.



      The homotopy lifting property is used extensively when proving the homotopy classification of principal $G$-bundles, i.e. that isomorphism classes of principal $G$-bundles a given base space $B$ are in one-to-one correspondence with homotopy classes $[B,BG]$. To obtain such a result for arbitrary base spaces $B$ you had better therefore restrict to numerable bundles.



      This doesn't really answer your question, in that I haven't given you a natural example of a numerable bundle over a non-paracompact base. But hopefully it indicates why numerable bundles are a useful concept in homotopy theory.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited 2 days ago

























      answered 2 days ago









      Mark GrantMark Grant

      22.2k657135




      22.2k657135







      • 5




        $begingroup$
        This is a bit disingenuous, because for those tools to be available one only needs a Serre fibration (homotopy lifting property with respect to discs), which the most naive definition of a fibre bundle is.
        $endgroup$
        – Oscar Randal-Williams
        2 days ago










      • $begingroup$
        @OscarRandal-Williams: no disingenuity on my part, just ignorance - I hadn't realised that every fibre bundle is a Serre fibration (for reference, a proof is given as Theorem 9 in these notes math.ru.nl/~mgroth/teaching/htpy13/Section06.pdf). My last point about classifying bundles over all spaces still stands though, I believe.
        $endgroup$
        – Mark Grant
        yesterday






      • 2




        $begingroup$
        @MarkGrant Another way of seeing that fiber bundles are Serre fibrations, is to notice that a map $p:E→B$ is a Serre fibration iff $E×_B D^n→D^n$ is a Serre fibration for every continuous map $D^n→B$ (this is because you can pull back every lifting diagram). Then you can use that $D^n$ is paracompact and that the pullback of a fiber bundle is a fiber bundle.
        $endgroup$
        – Denis Nardin
        yesterday












      • 5




        $begingroup$
        This is a bit disingenuous, because for those tools to be available one only needs a Serre fibration (homotopy lifting property with respect to discs), which the most naive definition of a fibre bundle is.
        $endgroup$
        – Oscar Randal-Williams
        2 days ago










      • $begingroup$
        @OscarRandal-Williams: no disingenuity on my part, just ignorance - I hadn't realised that every fibre bundle is a Serre fibration (for reference, a proof is given as Theorem 9 in these notes math.ru.nl/~mgroth/teaching/htpy13/Section06.pdf). My last point about classifying bundles over all spaces still stands though, I believe.
        $endgroup$
        – Mark Grant
        yesterday






      • 2




        $begingroup$
        @MarkGrant Another way of seeing that fiber bundles are Serre fibrations, is to notice that a map $p:E→B$ is a Serre fibration iff $E×_B D^n→D^n$ is a Serre fibration for every continuous map $D^n→B$ (this is because you can pull back every lifting diagram). Then you can use that $D^n$ is paracompact and that the pullback of a fiber bundle is a fiber bundle.
        $endgroup$
        – Denis Nardin
        yesterday







      5




      5




      $begingroup$
      This is a bit disingenuous, because for those tools to be available one only needs a Serre fibration (homotopy lifting property with respect to discs), which the most naive definition of a fibre bundle is.
      $endgroup$
      – Oscar Randal-Williams
      2 days ago




      $begingroup$
      This is a bit disingenuous, because for those tools to be available one only needs a Serre fibration (homotopy lifting property with respect to discs), which the most naive definition of a fibre bundle is.
      $endgroup$
      – Oscar Randal-Williams
      2 days ago












      $begingroup$
      @OscarRandal-Williams: no disingenuity on my part, just ignorance - I hadn't realised that every fibre bundle is a Serre fibration (for reference, a proof is given as Theorem 9 in these notes math.ru.nl/~mgroth/teaching/htpy13/Section06.pdf). My last point about classifying bundles over all spaces still stands though, I believe.
      $endgroup$
      – Mark Grant
      yesterday




      $begingroup$
      @OscarRandal-Williams: no disingenuity on my part, just ignorance - I hadn't realised that every fibre bundle is a Serre fibration (for reference, a proof is given as Theorem 9 in these notes math.ru.nl/~mgroth/teaching/htpy13/Section06.pdf). My last point about classifying bundles over all spaces still stands though, I believe.
      $endgroup$
      – Mark Grant
      yesterday




      2




      2




      $begingroup$
      @MarkGrant Another way of seeing that fiber bundles are Serre fibrations, is to notice that a map $p:E→B$ is a Serre fibration iff $E×_B D^n→D^n$ is a Serre fibration for every continuous map $D^n→B$ (this is because you can pull back every lifting diagram). Then you can use that $D^n$ is paracompact and that the pullback of a fiber bundle is a fiber bundle.
      $endgroup$
      – Denis Nardin
      yesterday




      $begingroup$
      @MarkGrant Another way of seeing that fiber bundles are Serre fibrations, is to notice that a map $p:E→B$ is a Serre fibration iff $E×_B D^n→D^n$ is a Serre fibration for every continuous map $D^n→B$ (this is because you can pull back every lifting diagram). Then you can use that $D^n$ is paracompact and that the pullback of a fiber bundle is a fiber bundle.
      $endgroup$
      – Denis Nardin
      yesterday











      4












      $begingroup$

      Dold's condition in on the bundle, while paracompactness is a condition on the base space. These are two very different sorts of concepts, and the former is more categorical. Note that the property of being a numerable bundle is preserved by pullbacks. So e.g., if one pulled back a bundle over a paracompact space to a bundle that was no longer paracompact, one would still automatically know that one has a bundle for which Dold's results hold.






      share|cite|improve this answer









      $endgroup$

















        4












        $begingroup$

        Dold's condition in on the bundle, while paracompactness is a condition on the base space. These are two very different sorts of concepts, and the former is more categorical. Note that the property of being a numerable bundle is preserved by pullbacks. So e.g., if one pulled back a bundle over a paracompact space to a bundle that was no longer paracompact, one would still automatically know that one has a bundle for which Dold's results hold.






        share|cite|improve this answer









        $endgroup$















          4












          4








          4





          $begingroup$

          Dold's condition in on the bundle, while paracompactness is a condition on the base space. These are two very different sorts of concepts, and the former is more categorical. Note that the property of being a numerable bundle is preserved by pullbacks. So e.g., if one pulled back a bundle over a paracompact space to a bundle that was no longer paracompact, one would still automatically know that one has a bundle for which Dold's results hold.






          share|cite|improve this answer









          $endgroup$



          Dold's condition in on the bundle, while paracompactness is a condition on the base space. These are two very different sorts of concepts, and the former is more categorical. Note that the property of being a numerable bundle is preserved by pullbacks. So e.g., if one pulled back a bundle over a paracompact space to a bundle that was no longer paracompact, one would still automatically know that one has a bundle for which Dold's results hold.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          Nicholas KuhnNicholas Kuhn

          3,5751221




          3,5751221





















              3












              $begingroup$

              To expand and correct my comment...



              Let $Top$ be the category of all topological spaces, $PHTop$ the category paracompact Hausdorff spaces. Let $open$ denote the Grothendieck topology of open covers on either of these categories and $num$ the Grothendieck topology of open covers for which there exists a subordinate partition of unity. Then there are morphisms of sites $Top_num to Top_open$ and $PHTop_num to PHTop_open$. By definition, the latter is an equivalence. But the former is not: there are genuinely more general open covers.



              One can repeat this story for the small sites associated to a fixed space.



              Now the stack $Bun_G$ of principal $G$-bundles on any of these sites is presented by the one-object topological groupoid $mathbfBG = Grightrightarrows *$. Any principal bundle $Pto X$ gives a map of stacks $Xto Bun_G$. If $P$ trivialises over $U to X$, $U = coprod_i U_i$ an open cover, then there is a span of topological groupoids $X leftarrow checkC(U) to mathbfBG$. Here $checkC(U)$ is the topological groupoid with object space $U$ and arrow space $Utimes_X U$. None of this is particularly specific to the case at hand. Since we are dealing with topological groupoids, though, we can geometrically realise and get a span of spaces $X leftarrow BcheckC(U) to BG$, where now $BG$ (plain $B$!) is Segal's construction of the classifying space of $G$.



              However, for $U$ an open cover with a subordinary partition of unity, the map $X leftarrow BcheckC(U)$ is a homotopy equivalence (a result of Segal), whereas in general it is a weak homotopy equivalence. So for numerable bundles, where local triviality is with respect to covers with partitions of unity, you get a map $Xto BG$ and this is well-defined up to homotopy. For arbitrary bundles on non-paracompact Hausdorff spaces, there is only a morphism in the homotopy category, a span with backwards-pointing leg a weak homotopy equivalence.



              Additional to this, the universal $G$-bundle on $BG$ is already numerable, so any bundle gotten by pulling it back is numerable, so this construction also goes in the opposite direction: one cannot get a non-numerable bundle via pulling back along any $Xto BG$. The same goes for Milnor's construction of $BG$ as an infinite join.



              So if you want the topological/homotopy version of classifying theory of principal bundles (using homotopy classes of maps) to agree with the stack-theoretic version, you'd better only use numerable bundles. Which is to say, you need to use the site $Top_num$. When restricted to paracompact Hausdorff spaces this the same as arbitrary open covers, but in general this is a real restriction that affects the theorem about classification of bundles.



              As an additional comment, I think an example of a non-numerable bundle is the frame bundle $F(L)$ (or, equivalently, the tangent bundle $TL$) of the long line $L$. You can think of $L$ as a manifold in the sense of being locally Euclidean and Hausdorff, but isn't metrizable, hence not paracompact. It is connected, path connected and simply-connected. As a result, $F(L)$ is not trivial, despite all this (otherwise $L$ would be metrizable).



              Added Ok, so $TL$ (and so $F(L)$) can't be numerable, otherwise one could define a Riemannian metric on $TL$, and hence a metric on $L$ giving its topology.



              And so this gives another good point why you want numerable bundles in general: if you want connections to exist, or metrics, or indeed any type of geometric information that can be patched together using a partition of unity, then bundles need to trivialise over a numerable cover. A numerable bundle over a non-paracompact Hausdorff space does always admit connections; a numerable vector bundle admits metrics and so on.






              share|cite|improve this answer











              $endgroup$

















                3












                $begingroup$

                To expand and correct my comment...



                Let $Top$ be the category of all topological spaces, $PHTop$ the category paracompact Hausdorff spaces. Let $open$ denote the Grothendieck topology of open covers on either of these categories and $num$ the Grothendieck topology of open covers for which there exists a subordinate partition of unity. Then there are morphisms of sites $Top_num to Top_open$ and $PHTop_num to PHTop_open$. By definition, the latter is an equivalence. But the former is not: there are genuinely more general open covers.



                One can repeat this story for the small sites associated to a fixed space.



                Now the stack $Bun_G$ of principal $G$-bundles on any of these sites is presented by the one-object topological groupoid $mathbfBG = Grightrightarrows *$. Any principal bundle $Pto X$ gives a map of stacks $Xto Bun_G$. If $P$ trivialises over $U to X$, $U = coprod_i U_i$ an open cover, then there is a span of topological groupoids $X leftarrow checkC(U) to mathbfBG$. Here $checkC(U)$ is the topological groupoid with object space $U$ and arrow space $Utimes_X U$. None of this is particularly specific to the case at hand. Since we are dealing with topological groupoids, though, we can geometrically realise and get a span of spaces $X leftarrow BcheckC(U) to BG$, where now $BG$ (plain $B$!) is Segal's construction of the classifying space of $G$.



                However, for $U$ an open cover with a subordinary partition of unity, the map $X leftarrow BcheckC(U)$ is a homotopy equivalence (a result of Segal), whereas in general it is a weak homotopy equivalence. So for numerable bundles, where local triviality is with respect to covers with partitions of unity, you get a map $Xto BG$ and this is well-defined up to homotopy. For arbitrary bundles on non-paracompact Hausdorff spaces, there is only a morphism in the homotopy category, a span with backwards-pointing leg a weak homotopy equivalence.



                Additional to this, the universal $G$-bundle on $BG$ is already numerable, so any bundle gotten by pulling it back is numerable, so this construction also goes in the opposite direction: one cannot get a non-numerable bundle via pulling back along any $Xto BG$. The same goes for Milnor's construction of $BG$ as an infinite join.



                So if you want the topological/homotopy version of classifying theory of principal bundles (using homotopy classes of maps) to agree with the stack-theoretic version, you'd better only use numerable bundles. Which is to say, you need to use the site $Top_num$. When restricted to paracompact Hausdorff spaces this the same as arbitrary open covers, but in general this is a real restriction that affects the theorem about classification of bundles.



                As an additional comment, I think an example of a non-numerable bundle is the frame bundle $F(L)$ (or, equivalently, the tangent bundle $TL$) of the long line $L$. You can think of $L$ as a manifold in the sense of being locally Euclidean and Hausdorff, but isn't metrizable, hence not paracompact. It is connected, path connected and simply-connected. As a result, $F(L)$ is not trivial, despite all this (otherwise $L$ would be metrizable).



                Added Ok, so $TL$ (and so $F(L)$) can't be numerable, otherwise one could define a Riemannian metric on $TL$, and hence a metric on $L$ giving its topology.



                And so this gives another good point why you want numerable bundles in general: if you want connections to exist, or metrics, or indeed any type of geometric information that can be patched together using a partition of unity, then bundles need to trivialise over a numerable cover. A numerable bundle over a non-paracompact Hausdorff space does always admit connections; a numerable vector bundle admits metrics and so on.






                share|cite|improve this answer











                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  To expand and correct my comment...



                  Let $Top$ be the category of all topological spaces, $PHTop$ the category paracompact Hausdorff spaces. Let $open$ denote the Grothendieck topology of open covers on either of these categories and $num$ the Grothendieck topology of open covers for which there exists a subordinate partition of unity. Then there are morphisms of sites $Top_num to Top_open$ and $PHTop_num to PHTop_open$. By definition, the latter is an equivalence. But the former is not: there are genuinely more general open covers.



                  One can repeat this story for the small sites associated to a fixed space.



                  Now the stack $Bun_G$ of principal $G$-bundles on any of these sites is presented by the one-object topological groupoid $mathbfBG = Grightrightarrows *$. Any principal bundle $Pto X$ gives a map of stacks $Xto Bun_G$. If $P$ trivialises over $U to X$, $U = coprod_i U_i$ an open cover, then there is a span of topological groupoids $X leftarrow checkC(U) to mathbfBG$. Here $checkC(U)$ is the topological groupoid with object space $U$ and arrow space $Utimes_X U$. None of this is particularly specific to the case at hand. Since we are dealing with topological groupoids, though, we can geometrically realise and get a span of spaces $X leftarrow BcheckC(U) to BG$, where now $BG$ (plain $B$!) is Segal's construction of the classifying space of $G$.



                  However, for $U$ an open cover with a subordinary partition of unity, the map $X leftarrow BcheckC(U)$ is a homotopy equivalence (a result of Segal), whereas in general it is a weak homotopy equivalence. So for numerable bundles, where local triviality is with respect to covers with partitions of unity, you get a map $Xto BG$ and this is well-defined up to homotopy. For arbitrary bundles on non-paracompact Hausdorff spaces, there is only a morphism in the homotopy category, a span with backwards-pointing leg a weak homotopy equivalence.



                  Additional to this, the universal $G$-bundle on $BG$ is already numerable, so any bundle gotten by pulling it back is numerable, so this construction also goes in the opposite direction: one cannot get a non-numerable bundle via pulling back along any $Xto BG$. The same goes for Milnor's construction of $BG$ as an infinite join.



                  So if you want the topological/homotopy version of classifying theory of principal bundles (using homotopy classes of maps) to agree with the stack-theoretic version, you'd better only use numerable bundles. Which is to say, you need to use the site $Top_num$. When restricted to paracompact Hausdorff spaces this the same as arbitrary open covers, but in general this is a real restriction that affects the theorem about classification of bundles.



                  As an additional comment, I think an example of a non-numerable bundle is the frame bundle $F(L)$ (or, equivalently, the tangent bundle $TL$) of the long line $L$. You can think of $L$ as a manifold in the sense of being locally Euclidean and Hausdorff, but isn't metrizable, hence not paracompact. It is connected, path connected and simply-connected. As a result, $F(L)$ is not trivial, despite all this (otherwise $L$ would be metrizable).



                  Added Ok, so $TL$ (and so $F(L)$) can't be numerable, otherwise one could define a Riemannian metric on $TL$, and hence a metric on $L$ giving its topology.



                  And so this gives another good point why you want numerable bundles in general: if you want connections to exist, or metrics, or indeed any type of geometric information that can be patched together using a partition of unity, then bundles need to trivialise over a numerable cover. A numerable bundle over a non-paracompact Hausdorff space does always admit connections; a numerable vector bundle admits metrics and so on.






                  share|cite|improve this answer











                  $endgroup$



                  To expand and correct my comment...



                  Let $Top$ be the category of all topological spaces, $PHTop$ the category paracompact Hausdorff spaces. Let $open$ denote the Grothendieck topology of open covers on either of these categories and $num$ the Grothendieck topology of open covers for which there exists a subordinate partition of unity. Then there are morphisms of sites $Top_num to Top_open$ and $PHTop_num to PHTop_open$. By definition, the latter is an equivalence. But the former is not: there are genuinely more general open covers.



                  One can repeat this story for the small sites associated to a fixed space.



                  Now the stack $Bun_G$ of principal $G$-bundles on any of these sites is presented by the one-object topological groupoid $mathbfBG = Grightrightarrows *$. Any principal bundle $Pto X$ gives a map of stacks $Xto Bun_G$. If $P$ trivialises over $U to X$, $U = coprod_i U_i$ an open cover, then there is a span of topological groupoids $X leftarrow checkC(U) to mathbfBG$. Here $checkC(U)$ is the topological groupoid with object space $U$ and arrow space $Utimes_X U$. None of this is particularly specific to the case at hand. Since we are dealing with topological groupoids, though, we can geometrically realise and get a span of spaces $X leftarrow BcheckC(U) to BG$, where now $BG$ (plain $B$!) is Segal's construction of the classifying space of $G$.



                  However, for $U$ an open cover with a subordinary partition of unity, the map $X leftarrow BcheckC(U)$ is a homotopy equivalence (a result of Segal), whereas in general it is a weak homotopy equivalence. So for numerable bundles, where local triviality is with respect to covers with partitions of unity, you get a map $Xto BG$ and this is well-defined up to homotopy. For arbitrary bundles on non-paracompact Hausdorff spaces, there is only a morphism in the homotopy category, a span with backwards-pointing leg a weak homotopy equivalence.



                  Additional to this, the universal $G$-bundle on $BG$ is already numerable, so any bundle gotten by pulling it back is numerable, so this construction also goes in the opposite direction: one cannot get a non-numerable bundle via pulling back along any $Xto BG$. The same goes for Milnor's construction of $BG$ as an infinite join.



                  So if you want the topological/homotopy version of classifying theory of principal bundles (using homotopy classes of maps) to agree with the stack-theoretic version, you'd better only use numerable bundles. Which is to say, you need to use the site $Top_num$. When restricted to paracompact Hausdorff spaces this the same as arbitrary open covers, but in general this is a real restriction that affects the theorem about classification of bundles.



                  As an additional comment, I think an example of a non-numerable bundle is the frame bundle $F(L)$ (or, equivalently, the tangent bundle $TL$) of the long line $L$. You can think of $L$ as a manifold in the sense of being locally Euclidean and Hausdorff, but isn't metrizable, hence not paracompact. It is connected, path connected and simply-connected. As a result, $F(L)$ is not trivial, despite all this (otherwise $L$ would be metrizable).



                  Added Ok, so $TL$ (and so $F(L)$) can't be numerable, otherwise one could define a Riemannian metric on $TL$, and hence a metric on $L$ giving its topology.



                  And so this gives another good point why you want numerable bundles in general: if you want connections to exist, or metrics, or indeed any type of geometric information that can be patched together using a partition of unity, then bundles need to trivialise over a numerable cover. A numerable bundle over a non-paracompact Hausdorff space does always admit connections; a numerable vector bundle admits metrics and so on.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 2 days ago

























                  answered 2 days ago









                  David RobertsDavid Roberts

                  17.5k463177




                  17.5k463177





















                      3












                      $begingroup$

                      It is a standard and useful practice to give precise conditions under which a theorem is known to be true, especially if these bring out methods used in the proof. Eldon Dyer suggested to me in the late 1960s that Dold's paper was the first full proof of the globalisation result, and this realisation led to the Dyer-Eilenberg paper referred to in this mathoverflow question.






                      share|cite|improve this answer











                      $endgroup$








                      • 1




                        $begingroup$
                        There is an extra // between the first : and the https:// in the link in your post (i.e. [1]://https://" should be "[1]: https://") and this breaks the link. Unfortunately MathOverflow will not let me edit to repair this because edits need to be at least 6 characters.
                        $endgroup$
                        – Robert Furber
                        yesterday
















                      3












                      $begingroup$

                      It is a standard and useful practice to give precise conditions under which a theorem is known to be true, especially if these bring out methods used in the proof. Eldon Dyer suggested to me in the late 1960s that Dold's paper was the first full proof of the globalisation result, and this realisation led to the Dyer-Eilenberg paper referred to in this mathoverflow question.






                      share|cite|improve this answer











                      $endgroup$








                      • 1




                        $begingroup$
                        There is an extra // between the first : and the https:// in the link in your post (i.e. [1]://https://" should be "[1]: https://") and this breaks the link. Unfortunately MathOverflow will not let me edit to repair this because edits need to be at least 6 characters.
                        $endgroup$
                        – Robert Furber
                        yesterday














                      3












                      3








                      3





                      $begingroup$

                      It is a standard and useful practice to give precise conditions under which a theorem is known to be true, especially if these bring out methods used in the proof. Eldon Dyer suggested to me in the late 1960s that Dold's paper was the first full proof of the globalisation result, and this realisation led to the Dyer-Eilenberg paper referred to in this mathoverflow question.






                      share|cite|improve this answer











                      $endgroup$



                      It is a standard and useful practice to give precise conditions under which a theorem is known to be true, especially if these bring out methods used in the proof. Eldon Dyer suggested to me in the late 1960s that Dold's paper was the first full proof of the globalisation result, and this realisation led to the Dyer-Eilenberg paper referred to in this mathoverflow question.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited yesterday









                      Danny Ruberman

                      11k3657




                      11k3657










                      answered yesterday









                      Ronnie BrownRonnie Brown

                      10.1k14864




                      10.1k14864







                      • 1




                        $begingroup$
                        There is an extra // between the first : and the https:// in the link in your post (i.e. [1]://https://" should be "[1]: https://") and this breaks the link. Unfortunately MathOverflow will not let me edit to repair this because edits need to be at least 6 characters.
                        $endgroup$
                        – Robert Furber
                        yesterday













                      • 1




                        $begingroup$
                        There is an extra // between the first : and the https:// in the link in your post (i.e. [1]://https://" should be "[1]: https://") and this breaks the link. Unfortunately MathOverflow will not let me edit to repair this because edits need to be at least 6 characters.
                        $endgroup$
                        – Robert Furber
                        yesterday








                      1




                      1




                      $begingroup$
                      There is an extra // between the first : and the https:// in the link in your post (i.e. [1]://https://" should be "[1]: https://") and this breaks the link. Unfortunately MathOverflow will not let me edit to repair this because edits need to be at least 6 characters.
                      $endgroup$
                      – Robert Furber
                      yesterday





                      $begingroup$
                      There is an extra // between the first : and the https:// in the link in your post (i.e. [1]://https://" should be "[1]: https://") and this breaks the link. Unfortunately MathOverflow will not let me edit to repair this because edits need to be at least 6 characters.
                      $endgroup$
                      – Robert Furber
                      yesterday












                      2












                      $begingroup$

                      Question: "... but is it really an interesting generalization?"



                      Answer: No. I do not believe that this is a really interesting generalization.
                      It is a technical condition which appears when one attempts to formulate results at their a maximal level of generality.



                      Let me illustrate this by an example:

                      For example, it would be a desirable result to have that given a topological space $X$ and vector bundle $V$ over $Xtimes [0,1]$ then $V|_Xtimes0$ is isomorphic to $V|_Xtimes1$. Unfortunately, that's not true (counterexample: Let $L$ be the long line; equip it with a smooth structure, and let $TL$ be its tangent bundle. Then we the various real powers $(TL)^otimes r$ for $rin [0,1]$ assemble to a line bundle over $L times [0,1]$. That line bundle is trivial over $Ltimes0$, and non-trivial over $Ltimes1$). But if you put the "numerable" condition, then the result becomes true.



                      Personally, I would much rather restrict my all my spaces to be paracompact (or any condition which implies paracompact), and then I wouldn't ever need to say the word "numerable".






                      share|cite|improve this answer









                      $endgroup$








                      • 1




                        $begingroup$
                        What are the real tensor powers? Are they the line bundles whose transition functions are real powers of the usual ones for $TL$?
                        $endgroup$
                        – David Roberts
                        2 days ago










                      • $begingroup$
                        Correct. They can only be defined because TL is orientable (a non-orientable real line bundle does not have a notion of real power).
                        $endgroup$
                        – André Henriques
                        yesterday















                      2












                      $begingroup$

                      Question: "... but is it really an interesting generalization?"



                      Answer: No. I do not believe that this is a really interesting generalization.
                      It is a technical condition which appears when one attempts to formulate results at their a maximal level of generality.



                      Let me illustrate this by an example:

                      For example, it would be a desirable result to have that given a topological space $X$ and vector bundle $V$ over $Xtimes [0,1]$ then $V|_Xtimes0$ is isomorphic to $V|_Xtimes1$. Unfortunately, that's not true (counterexample: Let $L$ be the long line; equip it with a smooth structure, and let $TL$ be its tangent bundle. Then we the various real powers $(TL)^otimes r$ for $rin [0,1]$ assemble to a line bundle over $L times [0,1]$. That line bundle is trivial over $Ltimes0$, and non-trivial over $Ltimes1$). But if you put the "numerable" condition, then the result becomes true.



                      Personally, I would much rather restrict my all my spaces to be paracompact (or any condition which implies paracompact), and then I wouldn't ever need to say the word "numerable".






                      share|cite|improve this answer









                      $endgroup$








                      • 1




                        $begingroup$
                        What are the real tensor powers? Are they the line bundles whose transition functions are real powers of the usual ones for $TL$?
                        $endgroup$
                        – David Roberts
                        2 days ago










                      • $begingroup$
                        Correct. They can only be defined because TL is orientable (a non-orientable real line bundle does not have a notion of real power).
                        $endgroup$
                        – André Henriques
                        yesterday













                      2












                      2








                      2





                      $begingroup$

                      Question: "... but is it really an interesting generalization?"



                      Answer: No. I do not believe that this is a really interesting generalization.
                      It is a technical condition which appears when one attempts to formulate results at their a maximal level of generality.



                      Let me illustrate this by an example:

                      For example, it would be a desirable result to have that given a topological space $X$ and vector bundle $V$ over $Xtimes [0,1]$ then $V|_Xtimes0$ is isomorphic to $V|_Xtimes1$. Unfortunately, that's not true (counterexample: Let $L$ be the long line; equip it with a smooth structure, and let $TL$ be its tangent bundle. Then we the various real powers $(TL)^otimes r$ for $rin [0,1]$ assemble to a line bundle over $L times [0,1]$. That line bundle is trivial over $Ltimes0$, and non-trivial over $Ltimes1$). But if you put the "numerable" condition, then the result becomes true.



                      Personally, I would much rather restrict my all my spaces to be paracompact (or any condition which implies paracompact), and then I wouldn't ever need to say the word "numerable".






                      share|cite|improve this answer









                      $endgroup$



                      Question: "... but is it really an interesting generalization?"



                      Answer: No. I do not believe that this is a really interesting generalization.
                      It is a technical condition which appears when one attempts to formulate results at their a maximal level of generality.



                      Let me illustrate this by an example:

                      For example, it would be a desirable result to have that given a topological space $X$ and vector bundle $V$ over $Xtimes [0,1]$ then $V|_Xtimes0$ is isomorphic to $V|_Xtimes1$. Unfortunately, that's not true (counterexample: Let $L$ be the long line; equip it with a smooth structure, and let $TL$ be its tangent bundle. Then we the various real powers $(TL)^otimes r$ for $rin [0,1]$ assemble to a line bundle over $L times [0,1]$. That line bundle is trivial over $Ltimes0$, and non-trivial over $Ltimes1$). But if you put the "numerable" condition, then the result becomes true.



                      Personally, I would much rather restrict my all my spaces to be paracompact (or any condition which implies paracompact), and then I wouldn't ever need to say the word "numerable".







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 2 days ago









                      André HenriquesAndré Henriques

                      27.9k484214




                      27.9k484214







                      • 1




                        $begingroup$
                        What are the real tensor powers? Are they the line bundles whose transition functions are real powers of the usual ones for $TL$?
                        $endgroup$
                        – David Roberts
                        2 days ago










                      • $begingroup$
                        Correct. They can only be defined because TL is orientable (a non-orientable real line bundle does not have a notion of real power).
                        $endgroup$
                        – André Henriques
                        yesterday












                      • 1




                        $begingroup$
                        What are the real tensor powers? Are they the line bundles whose transition functions are real powers of the usual ones for $TL$?
                        $endgroup$
                        – David Roberts
                        2 days ago










                      • $begingroup$
                        Correct. They can only be defined because TL is orientable (a non-orientable real line bundle does not have a notion of real power).
                        $endgroup$
                        – André Henriques
                        yesterday







                      1




                      1




                      $begingroup$
                      What are the real tensor powers? Are they the line bundles whose transition functions are real powers of the usual ones for $TL$?
                      $endgroup$
                      – David Roberts
                      2 days ago




                      $begingroup$
                      What are the real tensor powers? Are they the line bundles whose transition functions are real powers of the usual ones for $TL$?
                      $endgroup$
                      – David Roberts
                      2 days ago












                      $begingroup$
                      Correct. They can only be defined because TL is orientable (a non-orientable real line bundle does not have a notion of real power).
                      $endgroup$
                      – André Henriques
                      yesterday




                      $begingroup$
                      Correct. They can only be defined because TL is orientable (a non-orientable real line bundle does not have a notion of real power).
                      $endgroup$
                      – André Henriques
                      yesterday

















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