Is the concept of a “numerable” fiber bundle really useful or an empty generalization?
$begingroup$
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
$endgroup$
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
$endgroup$
5
$begingroup$
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
14 hours ago
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 12 hours ago
David White
12.9k462104
12.9k462104
asked 16 hours ago
ychemamaychemama
516210
516210
5
$begingroup$
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
14 hours ago
add a comment |
5
$begingroup$
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
14 hours ago
5
5
$begingroup$
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
14 hours 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
14 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
$endgroup$
3
$begingroup$
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
3 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 $mathbf{B}G = Grightrightarrows *$. 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 $X leftarrow check{C}(U) to mathbf{B}G$. Here $check{C}(U)$ 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 $X leftarrow Bcheck{C}(U) to BG$, 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 $X leftarrow Bcheck{C}(U)$ 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 |
$begingroup$
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 $V|_{Xtimes{0}}$ is isomorphic to $V|_{Xtimes{1}}$. 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 $Ltimes{0}$, and non-trivial over $Ltimes{1}$). 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".
$endgroup$
$begingroup$
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
7 mins ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "504"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326565%2fis-the-concept-of-a-numerable-fiber-bundle-really-useful-or-an-empty-generaliz%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
3
$begingroup$
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
3 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.
$endgroup$
3
$begingroup$
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
3 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 11 hours ago
answered 11 hours ago
Mark GrantMark Grant
22.2k657135
22.2k657135
3
$begingroup$
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
3 hours ago
add a comment |
3
$begingroup$
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
3 hours ago
3
3
$begingroup$
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
3 hours 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
3 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 $mathbf{B}G = Grightrightarrows *$. 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 $X leftarrow check{C}(U) to mathbf{B}G$. Here $check{C}(U)$ 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 $X leftarrow Bcheck{C}(U) to BG$, 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 $X leftarrow Bcheck{C}(U)$ 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 $mathbf{B}G = Grightrightarrows *$. 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 $X leftarrow check{C}(U) to mathbf{B}G$. Here $check{C}(U)$ 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 $X leftarrow Bcheck{C}(U) to BG$, 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 $X leftarrow Bcheck{C}(U)$ 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 $mathbf{B}G = Grightrightarrows *$. 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 $X leftarrow check{C}(U) to mathbf{B}G$. Here $check{C}(U)$ 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 $X leftarrow Bcheck{C}(U) to BG$, 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 $X leftarrow Bcheck{C}(U)$ 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 $mathbf{B}G = Grightrightarrows *$. 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 $X leftarrow check{C}(U) to mathbf{B}G$. Here $check{C}(U)$ 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 $X leftarrow Bcheck{C}(U) to BG$, 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 $X leftarrow Bcheck{C}(U)$ 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 2 hours ago
answered 3 hours ago
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 $V|_{Xtimes{0}}$ is isomorphic to $V|_{Xtimes{1}}$. 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 $Ltimes{0}$, and non-trivial over $Ltimes{1}$). 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".
$endgroup$
$begingroup$
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
7 mins 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 $V|_{Xtimes{0}}$ is isomorphic to $V|_{Xtimes{1}}$. 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 $Ltimes{0}$, and non-trivial over $Ltimes{1}$). 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".
$endgroup$
$begingroup$
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
7 mins 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 $V|_{Xtimes{0}}$ is isomorphic to $V|_{Xtimes{1}}$. 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 $Ltimes{0}$, and non-trivial over $Ltimes{1}$). 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 $V|_{Xtimes{0}}$ is isomorphic to $V|_{Xtimes{1}}$. 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 $Ltimes{0}$, and non-trivial over $Ltimes{1}$). 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 1 hour ago
André HenriquesAndré Henriques
27.9k484214
27.9k484214
$begingroup$
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
7 mins ago
add a comment |
$begingroup$
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
7 mins 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
7 mins 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
7 mins ago
add a comment |
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f326565%2fis-the-concept-of-a-numerable-fiber-bundle-really-useful-or-an-empty-generaliz%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
5
$begingroup$
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
14 hours ago