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?
at.algebraic-topology fibre-bundles
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add a comment |
$begingroup$
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?
at.algebraic-topology fibre-bundles
$endgroup$
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
yesterday
add a comment |
$begingroup$
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?
at.algebraic-topology fibre-bundles
$endgroup$
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
at.algebraic-topology fibre-bundles
edited yesterday
David White
13k462104
13k462104
asked yesterday
ychemamaychemama
526210
526210
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
yesterday
add a comment |
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
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
yesterday
$begingroup$
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
yesterday
add a comment |
5 Answers
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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.
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5
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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
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
15 hours ago
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.
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– Denis Nardin
8 hours ago
add a comment |
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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.
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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".
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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 $TL$?
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– David Roberts
21 hours ago
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Correct. They can only be defined because TL is orientable (a non-orientable real line bundle does not have a notion of real power).
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– André Henriques
6 hours ago
add a comment |
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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.
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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.
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– Robert Furber
10 hours ago
add a comment |
$begingroup$
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.
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5 Answers
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5 Answers
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$begingroup$
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.
$endgroup$
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
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
15 hours ago
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
8 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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
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
15 hours ago
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
8 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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.
edited yesterday
answered yesterday
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
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
15 hours ago
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
8 hours ago
add a comment |
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
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
15 hours ago
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
8 hours ago
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
yesterday
$begingroup$
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
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
15 hours 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
15 hours ago
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
8 hours ago
$begingroup$
@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
8 hours ago
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 23 hours ago
answered yesterday
David RobertsDavid Roberts
17.5k463177
17.5k463177
add a comment |
add a comment |
$begingroup$
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".
$endgroup$
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
21 hours 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
6 hours ago
add a comment |
$begingroup$
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".
$endgroup$
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
21 hours 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
6 hours ago
add a comment |
$begingroup$
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".
$endgroup$
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".
answered 22 hours 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
21 hours 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
6 hours ago
add a comment |
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
21 hours 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
6 hours ago
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
21 hours 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
21 hours 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
6 hours 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
6 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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
10 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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
10 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 4 hours ago
Danny Ruberman
11k3657
11k3657
answered 11 hours ago
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
10 hours ago
add a comment |
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
10 hours ago
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
10 hours ago
$begingroup$
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
10 hours ago
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 3 hours ago
Nicholas KuhnNicholas Kuhn
3,5351220
3,5351220
add a comment |
add a comment |
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$begingroup$
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
yesterday