Understanding why the Adams Spectral Sequence works
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I am trying to learn about the Adams Spectral Sequence and my question is basically summed up in the title.
More precisely, let $X$, $Y$, and $E$ be spectra. We have a homomorphism $[X,Y] to Hom_{E^∗E}(E^∗Y,E^∗X)$, the latter giving the first page of the Adams Spectral Sequence. But why does $Ext^∗_{E^∗E}(E^∗Y,E^∗X)$ give a "better approximation" to $[X,Y]$ than $Hom_{E^∗E}(E^∗Y,E^∗X)$?
And more generally, it seems like magic that we can get homotopy from successively taking homology. Any intuition about this?
algebraic-topology homology-cohomology homotopy-theory spectral-sequences
asked Dec 3 '18 at 1:13
math1234567math1234567
8661919
$endgroup$
add a comment |
$begingroup$
I am trying to learn about the Adams Spectral Sequence and my question is basically summed up in the title.
More precisely, let $X$, $Y$, and $E$ be spectra. We have a homomorphism $[X,Y] to Hom_{E^∗E}(E^∗Y,E^∗X)$, the latter giving the first page of the Adams Spectral Sequence. But why does $Ext^∗_{E^∗E}(E^∗Y,E^∗X)$ give a "better approximation" to $[X,Y]$ than $Hom_{E^∗E}(E^∗Y,E^∗X)$?
And more generally, it seems like magic that we can get homotopy from successively taking homology. Any intuition about this?
algebraic-topology homology-cohomology homotopy-theory spectral-sequences
asked Dec 3 '18 at 1:13
math1234567math1234567
8661919
$endgroup$
3
$begingroup$
Pretend that you understand homotopy classes of maps in one case: if the codomain is $E$, then $[X,E] = E^*X$. Then if you want to compute $[X,Y]$ for a general $Y$, you should try to approximate $Y$ by copies of $E$ somehow. This leads to the $E$-Adams tower and then to the Adams SS, which is built up from a sort of resolution of $Y$ by copies of $E$. This at least makes it plausible that Ext is relevant, rather than just Hom, or at least that homological algebra could enter into things.
$endgroup$
– John Palmieri
Dec 3 '18 at 6:46
add a comment |
$begingroup$
I am trying to learn about the Adams Spectral Sequence and my question is basically summed up in the title.
More precisely, let $X$, $Y$, and $E$ be spectra. We have a homomorphism $[X,Y] to Hom_{E^∗E}(E^∗Y,E^∗X)$, the latter giving the first page of the Adams Spectral Sequence. But why does $Ext^∗_{E^∗E}(E^∗Y,E^∗X)$ give a "better approximation" to $[X,Y]$ than $Hom_{E^∗E}(E^∗Y,E^∗X)$?
And more generally, it seems like magic that we can get homotopy from successively taking homology. Any intuition about this?
algebraic-topology homology-cohomology homotopy-theory spectral-sequences
asked Dec 3 '18 at 1:13
math1234567math1234567
8661919
$endgroup$
I am trying to learn about the Adams Spectral Sequence and my question is basically summed up in the title.
More precisely, let $X$, $Y$, and $E$ be spectra. We have a homomorphism $[X,Y] to Hom_{E^∗E}(E^∗Y,E^∗X)$, the latter giving the first page of the Adams Spectral Sequence. But why does $Ext^∗_{E^∗E}(E^∗Y,E^∗X)$ give a "better approximation" to $[X,Y]$ than $Hom_{E^∗E}(E^∗Y,E^∗X)$?
And more generally, it seems like magic that we can get homotopy from successively taking homology. Any intuition about this?
algebraic-topology homology-cohomology homotopy-theory spectral-sequences
algebraic-topology homology-cohomology homotopy-theory spectral-sequences
asked Dec 3 '18 at 1:13
math1234567math1234567
8661919
asked Dec 3 '18 at 1:13
math1234567math1234567
8661919
asked Dec 3 '18 at 1:13
math1234567math1234567
8661919
asked Dec 3 '18 at 1:13
math1234567math1234567
8661919
asked Dec 3 '18 at 1:13
math1234567math1234567
8661919
8661919
3
$begingroup$
Pretend that you understand homotopy classes of maps in one case: if the codomain is $E$, then $[X,E] = E^*X$. Then if you want to compute $[X,Y]$ for a general $Y$, you should try to approximate $Y$ by copies of $E$ somehow. This leads to the $E$-Adams tower and then to the Adams SS, which is built up from a sort of resolution of $Y$ by copies of $E$. This at least makes it plausible that Ext is relevant, rather than just Hom, or at least that homological algebra could enter into things.
$endgroup$
– John Palmieri
Dec 3 '18 at 6:46
add a comment |
3
$begingroup$
Pretend that you understand homotopy classes of maps in one case: if the codomain is $E$, then $[X,E] = E^*X$. Then if you want to compute $[X,Y]$ for a general $Y$, you should try to approximate $Y$ by copies of $E$ somehow. This leads to the $E$-Adams tower and then to the Adams SS, which is built up from a sort of resolution of $Y$ by copies of $E$. This at least makes it plausible that Ext is relevant, rather than just Hom, or at least that homological algebra could enter into things.
$endgroup$
– John Palmieri
Dec 3 '18 at 6:46
3
3
$begingroup$
Pretend that you understand homotopy classes of maps in one case: if the codomain is $E$, then $[X,E] = E^*X$. Then if you want to compute $[X,Y]$ for a general $Y$, you should try to approximate $Y$ by copies of $E$ somehow. This leads to the $E$-Adams tower and then to the Adams SS, which is built up from a sort of resolution of $Y$ by copies of $E$. This at least makes it plausible that Ext is relevant, rather than just Hom, or at least that homological algebra could enter into things.
$endgroup$
– John Palmieri
Dec 3 '18 at 6:46
$begingroup$
Pretend that you understand homotopy classes of maps in one case: if the codomain is $E$, then $[X,E] = E^*X$. Then if you want to compute $[X,Y]$ for a general $Y$, you should try to approximate $Y$ by copies of $E$ somehow. This leads to the $E$-Adams tower and then to the Adams SS, which is built up from a sort of resolution of $Y$ by copies of $E$. This at least makes it plausible that Ext is relevant, rather than just Hom, or at least that homological algebra could enter into things.
$endgroup$
– John Palmieri
Dec 3 '18 at 6:46
add a comment |
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$begingroup$
Pretend that you understand homotopy classes of maps in one case: if the codomain is $E$, then $[X,E] = E^*X$. Then if you want to compute $[X,Y]$ for a general $Y$, you should try to approximate $Y$ by copies of $E$ somehow. This leads to the $E$-Adams tower and then to the Adams SS, which is built up from a sort of resolution of $Y$ by copies of $E$. This at least makes it plausible that Ext is relevant, rather than just Hom, or at least that homological algebra could enter into things.
$endgroup$
– John Palmieri
Dec 3 '18 at 6:46