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request for counterexamples
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Reid Barton
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Let me first try to answer a simpler question:

Why are long exact sequences so ubiquitous?

Almost anything that is written as a capital letter, followed by a subscript i or superscript -i, i an integer, and finally some stuff in parentheses, can be interpreted as πi of some spectrum (or sometimes space, as in nonabelian group cohomology, or maybe a sheaf of spectra or spaces...). And almost any long exact sequence which involves three similar terms in a cycle, with i decreasing by 1 every three terms, comes from the long exact sequence of homotopy groups of a fiber sequence of spectra or spaces. For instance, the long exact sequence for the cohomology of a group G with coefficients in a short exact sequence of G-modules A → B → C corresponds to (HA)hG → (HB)hG → (HC)hG, since H-i(G, M) = πi((HG)hM) (which is nonzero only for nonpositive i), which is a fiber sequence because HA → HB → HC is one (since A → B → C is a SES) and (–)hG is a homotopy limit, so it preserves fiber sequences. Or, I could draw a square with these three terms in it and 0 in the lower left, which is both a pullback and pushout square (since spectra form a stable (∞,1)-category).

Long exact sequences are actually a special case of spectral sequences—those whose E1 page has only two columns. If you have never done this before, you should check for yourself that the d1 and the extension problems exactly tell you that there is a long exact sequence formed by the two columns and whatever the spectral sequence converges to. This suggests that we could try to generalize the picture with a pushout/pullback square of spectra to find something to which a more general spectral sequence corresponds. Two possibilites are: we could extend the top row of the square to a directed sequence of spectra, and take the homotopy cofiber of each map; or we could extend the right column of the square to an inverse sequence of spectra and take the fiber of each map. These are the homotopy version of filtered and cofiltered objects, respectively; however, there is no condition on the maps (the notion of "inclusion" does not make much sense in the homotopy-theoretic world). Associated to each is a spectral sequence, though there are convergence issues when the sequence of spectra is infinite. It could be that most spectral sequences encountered in practice can be viewed as arising from an underlying sequence of spectra, though I have not attempted to convince myself of this fact.

Edit: Clark Barwick suggests that one may indeed view all "natural" spectral sequences as arising from filtered spectra. He and I would like to know whether there are any convincing counterexamples, so please let me know if you have any! Note however that 1 and 2 from VA's answer are not counterexamples.

Let me first try to answer a simpler question:

Why are long exact sequences so ubiquitous?

Almost anything that is written as a capital letter, followed by a subscript i or superscript -i, i an integer, and finally some stuff in parentheses, can be interpreted as πi of some spectrum (or sometimes space, as in nonabelian group cohomology, or maybe a sheaf of spectra or spaces...). And almost any long exact sequence which involves three similar terms in a cycle, with i decreasing by 1 every three terms, comes from the long exact sequence of homotopy groups of a fiber sequence of spectra or spaces. For instance, the long exact sequence for the cohomology of a group G with coefficients in a short exact sequence of G-modules A → B → C corresponds to (HA)hG → (HB)hG → (HC)hG, since H-i(G, M) = πi((HG)hM) (which is nonzero only for nonpositive i), which is a fiber sequence because HA → HB → HC is one (since A → B → C is a SES) and (–)hG is a homotopy limit, so it preserves fiber sequences. Or, I could draw a square with these three terms in it and 0 in the lower left, which is both a pullback and pushout square (since spectra form a stable (∞,1)-category).

Long exact sequences are actually a special case of spectral sequences—those whose E1 page has only two columns. If you have never done this before, you should check for yourself that the d1 and the extension problems exactly tell you that there is a long exact sequence formed by the two columns and whatever the spectral sequence converges to. This suggests that we could try to generalize the picture with a pushout/pullback square of spectra to find something to which a more general spectral sequence corresponds. Two possibilites are: we could extend the top row of the square to a directed sequence of spectra, and take the homotopy cofiber of each map; or we could extend the right column of the square to an inverse sequence of spectra and take the fiber of each map. These are the homotopy version of filtered and cofiltered objects, respectively; however, there is no condition on the maps (the notion of "inclusion" does not make much sense in the homotopy-theoretic world). Associated to each is a spectral sequence, though there are convergence issues when the sequence of spectra is infinite. It could be that most spectral sequences encountered in practice can be viewed as arising from an underlying sequence of spectra, though I have not attempted to convince myself of this fact.

Let me first try to answer a simpler question:

Why are long exact sequences so ubiquitous?

Almost anything that is written as a capital letter, followed by a subscript i or superscript -i, i an integer, and finally some stuff in parentheses, can be interpreted as πi of some spectrum (or sometimes space, as in nonabelian group cohomology, or maybe a sheaf of spectra or spaces...). And almost any long exact sequence which involves three similar terms in a cycle, with i decreasing by 1 every three terms, comes from the long exact sequence of homotopy groups of a fiber sequence of spectra or spaces. For instance, the long exact sequence for the cohomology of a group G with coefficients in a short exact sequence of G-modules A → B → C corresponds to (HA)hG → (HB)hG → (HC)hG, since H-i(G, M) = πi((HG)hM) (which is nonzero only for nonpositive i), which is a fiber sequence because HA → HB → HC is one (since A → B → C is a SES) and (–)hG is a homotopy limit, so it preserves fiber sequences. Or, I could draw a square with these three terms in it and 0 in the lower left, which is both a pullback and pushout square (since spectra form a stable (∞,1)-category).

Long exact sequences are actually a special case of spectral sequences—those whose E1 page has only two columns. If you have never done this before, you should check for yourself that the d1 and the extension problems exactly tell you that there is a long exact sequence formed by the two columns and whatever the spectral sequence converges to. This suggests that we could try to generalize the picture with a pushout/pullback square of spectra to find something to which a more general spectral sequence corresponds. Two possibilites are: we could extend the top row of the square to a directed sequence of spectra, and take the homotopy cofiber of each map; or we could extend the right column of the square to an inverse sequence of spectra and take the fiber of each map. These are the homotopy version of filtered and cofiltered objects, respectively; however, there is no condition on the maps (the notion of "inclusion" does not make much sense in the homotopy-theoretic world). Associated to each is a spectral sequence, though there are convergence issues when the sequence of spectra is infinite. It could be that most spectral sequences encountered in practice can be viewed as arising from an underlying sequence of spectra, though I have not attempted to convince myself of this fact.

Edit: Clark Barwick suggests that one may indeed view all "natural" spectral sequences as arising from filtered spectra. He and I would like to know whether there are any convincing counterexamples, so please let me know if you have any! Note however that 1 and 2 from VA's answer are not counterexamples.

deleted 19 characters in body
Source Link
Reid Barton
  • 25.2k
  • 1
  • 76
  • 133

Let me first try to answer a simpler question:

Why are long exact sequences so ubiquitous?

Almost anything that is written as a capital letter, followed by a subscript i or superscript -i, i an integer, and finally some stuff in parentheses, can be interpreted as πi of some spectrum (or sometimes space, as in nonabelian group cohomology, or maybe a sheaf of spectra or spaces...). And almost any long exact sequence which involves three similar terms in a cycle, with i decreasing by 1 every three terms, comes from the long exact sequence of homotopy groups of a fiber sequence of spectra or spaces. For instance, the long exact sequence for the cohomology of a group G with coefficients in a short exact sequence of G-modules A → B → C corresponds to (HA)hG → (HB)hG → (HC)hG, since H-i(G, M) = πi((HG)hM) (which is nonzero only for nonpositive i), which is a fiber sequence because HA → HB → HC is one (since A → B → C is a SES) and (–)hG is a homotopy limit, so it preserves fiber sequences. Or, I could draw a square with these three terms in it and 0 in the lower left, which is both a pullback and pushout square (since spectra form a stable (∞,1)-category).

Long exact sequences are actually a special case of spectral sequences—those whose E21 page has only two columns (or only two rows). If you have never done this before, you should check for yourself that the d21 and the extension problems exactly tell you that there is a long exact sequence formed by the two columns and whatever the spectral sequence converges to. This suggests that we could try to generalize the picture with a pushout/pullback square of spectra to find something to which a more general spectral sequence corresponds. Two possibilites are: we could extend the top row of the square to a directed sequence of spectra, and take the homotopy cofiber of each map; or we could extend the right column of the square to an inverse sequence of spectra and take the fiber of each map. These are the homotopy version of filtered and cofiltered objects, respectively; however, there is no condition on the maps (the notion of "inclusion" does not make much sense in the homotopy-theoretic world). Associated to each is a spectral sequence, though there are convergence issues when the sequence of spectra is infinite. It could be that most spectral sequences encountered in practice can be viewed as arising from an underlying sequence of spectra, though I have not attempted to convince myself of this fact.

Let me first try to answer a simpler question:

Why are long exact sequences so ubiquitous?

Almost anything that is written as a capital letter, followed by a subscript i or superscript -i, i an integer, and finally some stuff in parentheses, can be interpreted as πi of some spectrum (or sometimes space, as in nonabelian group cohomology, or maybe a sheaf of spectra or spaces...). And almost any long exact sequence which involves three similar terms in a cycle, with i decreasing by 1 every three terms, comes from the long exact sequence of homotopy groups of a fiber sequence of spectra or spaces. For instance, the long exact sequence for the cohomology of a group G with coefficients in a short exact sequence of G-modules A → B → C corresponds to (HA)hG → (HB)hG → (HC)hG, since H-i(G, M) = πi((HG)hM) (which is nonzero only for nonpositive i), which is a fiber sequence because HA → HB → HC is one (since A → B → C is a SES) and (–)hG is a homotopy limit, so it preserves fiber sequences. Or, I could draw a square with these three terms in it and 0 in the lower left, which is both a pullback and pushout square (since spectra form a stable (∞,1)-category).

Long exact sequences are actually a special case of spectral sequences—those whose E2 page has only two columns (or only two rows). If you have never done this before, you should check for yourself that the d2 and the extension problems exactly tell you that there is a long exact sequence formed by the two columns and whatever the spectral sequence converges to. This suggests that we could try to generalize the picture with a pushout/pullback square of spectra to find something to which a more general spectral sequence corresponds. Two possibilites are: we could extend the top row of the square to a directed sequence of spectra, and take the homotopy cofiber of each map; or we could extend the right column of the square to an inverse sequence of spectra and take the fiber of each map. These are the homotopy version of filtered and cofiltered objects, respectively; however, there is no condition on the maps (the notion of "inclusion" does not make much sense in the homotopy-theoretic world). Associated to each is a spectral sequence, though there are convergence issues when the sequence of spectra is infinite. It could be that most spectral sequences encountered in practice can be viewed as arising from an underlying sequence of spectra, though I have not attempted to convince myself of this fact.

Let me first try to answer a simpler question:

Why are long exact sequences so ubiquitous?

Almost anything that is written as a capital letter, followed by a subscript i or superscript -i, i an integer, and finally some stuff in parentheses, can be interpreted as πi of some spectrum (or sometimes space, as in nonabelian group cohomology, or maybe a sheaf of spectra or spaces...). And almost any long exact sequence which involves three similar terms in a cycle, with i decreasing by 1 every three terms, comes from the long exact sequence of homotopy groups of a fiber sequence of spectra or spaces. For instance, the long exact sequence for the cohomology of a group G with coefficients in a short exact sequence of G-modules A → B → C corresponds to (HA)hG → (HB)hG → (HC)hG, since H-i(G, M) = πi((HG)hM) (which is nonzero only for nonpositive i), which is a fiber sequence because HA → HB → HC is one (since A → B → C is a SES) and (–)hG is a homotopy limit, so it preserves fiber sequences. Or, I could draw a square with these three terms in it and 0 in the lower left, which is both a pullback and pushout square (since spectra form a stable (∞,1)-category).

Long exact sequences are actually a special case of spectral sequences—those whose E1 page has only two columns. If you have never done this before, you should check for yourself that the d1 and the extension problems exactly tell you that there is a long exact sequence formed by the two columns and whatever the spectral sequence converges to. This suggests that we could try to generalize the picture with a pushout/pullback square of spectra to find something to which a more general spectral sequence corresponds. Two possibilites are: we could extend the top row of the square to a directed sequence of spectra, and take the homotopy cofiber of each map; or we could extend the right column of the square to an inverse sequence of spectra and take the fiber of each map. These are the homotopy version of filtered and cofiltered objects, respectively; however, there is no condition on the maps (the notion of "inclusion" does not make much sense in the homotopy-theoretic world). Associated to each is a spectral sequence, though there are convergence issues when the sequence of spectra is infinite. It could be that most spectral sequences encountered in practice can be viewed as arising from an underlying sequence of spectra, though I have not attempted to convince myself of this fact.

Source Link
Reid Barton
  • 25.2k
  • 1
  • 76
  • 133

Let me first try to answer a simpler question:

Why are long exact sequences so ubiquitous?

Almost anything that is written as a capital letter, followed by a subscript i or superscript -i, i an integer, and finally some stuff in parentheses, can be interpreted as πi of some spectrum (or sometimes space, as in nonabelian group cohomology, or maybe a sheaf of spectra or spaces...). And almost any long exact sequence which involves three similar terms in a cycle, with i decreasing by 1 every three terms, comes from the long exact sequence of homotopy groups of a fiber sequence of spectra or spaces. For instance, the long exact sequence for the cohomology of a group G with coefficients in a short exact sequence of G-modules A → B → C corresponds to (HA)hG → (HB)hG → (HC)hG, since H-i(G, M) = πi((HG)hM) (which is nonzero only for nonpositive i), which is a fiber sequence because HA → HB → HC is one (since A → B → C is a SES) and (–)hG is a homotopy limit, so it preserves fiber sequences. Or, I could draw a square with these three terms in it and 0 in the lower left, which is both a pullback and pushout square (since spectra form a stable (∞,1)-category).

Long exact sequences are actually a special case of spectral sequences—those whose E2 page has only two columns (or only two rows). If you have never done this before, you should check for yourself that the d2 and the extension problems exactly tell you that there is a long exact sequence formed by the two columns and whatever the spectral sequence converges to. This suggests that we could try to generalize the picture with a pushout/pullback square of spectra to find something to which a more general spectral sequence corresponds. Two possibilites are: we could extend the top row of the square to a directed sequence of spectra, and take the homotopy cofiber of each map; or we could extend the right column of the square to an inverse sequence of spectra and take the fiber of each map. These are the homotopy version of filtered and cofiltered objects, respectively; however, there is no condition on the maps (the notion of "inclusion" does not make much sense in the homotopy-theoretic world). Associated to each is a spectral sequence, though there are convergence issues when the sequence of spectra is infinite. It could be that most spectral sequences encountered in practice can be viewed as arising from an underlying sequence of spectra, though I have not attempted to convince myself of this fact.