Mixing
Derived functor of derived functor
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This is a dumb question.
Q1: Is there a reason to consider doubly/triply derived functors(i.e. derived functor of derived functors? Say given projective resolution and a right exact functor. One can obtain the left derived functor. Suppose to some degree left derived functor is left exact. Then I can further derive this left exact functor. Does this procedure have to stop at some time point? Of course, right derived functor of right derived functor might be trivial like Ext.
Q2: Does the derived functors contain most of information on the complex except homotopy information? Any further derived functors do not yield extra information?
abstract-algebra commutative-algebra homology-cohomology homological-algebra
asked Dec 7 '17 at 15:52
user45765user45765
2,6702724
$endgroup$
add a comment |
$begingroup$
This is a dumb question.
Q1: Is there a reason to consider doubly/triply derived functors(i.e. derived functor of derived functors? Say given projective resolution and a right exact functor. One can obtain the left derived functor. Suppose to some degree left derived functor is left exact. Then I can further derive this left exact functor. Does this procedure have to stop at some time point? Of course, right derived functor of right derived functor might be trivial like Ext.
Q2: Does the derived functors contain most of information on the complex except homotopy information? Any further derived functors do not yield extra information?
abstract-algebra commutative-algebra homology-cohomology homological-algebra
asked Dec 7 '17 at 15:52
user45765user45765
2,6702724
$endgroup$
$begingroup$
The derived functor is really a sequence ${R^iF}$, isn't it? Do you wish to further derive only one of them, or the whole sequence somehow?
$endgroup$
– Randall
Dec 7 '17 at 17:34
$begingroup$
@Randall I wish to derive one of them if it happens to be either left exact of right exact. If I can derive one of them, then I probably could see whether further derive all of them yields anything useful. However, I do not see any reason why not to further derive the derived functors.
$endgroup$
– user45765
Dec 7 '17 at 18:06
add a comment |
$begingroup$
This is a dumb question.
Q1: Is there a reason to consider doubly/triply derived functors(i.e. derived functor of derived functors? Say given projective resolution and a right exact functor. One can obtain the left derived functor. Suppose to some degree left derived functor is left exact. Then I can further derive this left exact functor. Does this procedure have to stop at some time point? Of course, right derived functor of right derived functor might be trivial like Ext.
Q2: Does the derived functors contain most of information on the complex except homotopy information? Any further derived functors do not yield extra information?
abstract-algebra commutative-algebra homology-cohomology homological-algebra
asked Dec 7 '17 at 15:52
user45765user45765
2,6702724
$endgroup$
This is a dumb question.
Q1: Is there a reason to consider doubly/triply derived functors(i.e. derived functor of derived functors? Say given projective resolution and a right exact functor. One can obtain the left derived functor. Suppose to some degree left derived functor is left exact. Then I can further derive this left exact functor. Does this procedure have to stop at some time point? Of course, right derived functor of right derived functor might be trivial like Ext.
Q2: Does the derived functors contain most of information on the complex except homotopy information? Any further derived functors do not yield extra information?
abstract-algebra commutative-algebra homology-cohomology homological-algebra
abstract-algebra commutative-algebra homology-cohomology homological-algebra
asked Dec 7 '17 at 15:52
user45765user45765
2,6702724
asked Dec 7 '17 at 15:52
user45765user45765
2,6702724
asked Dec 7 '17 at 15:52
user45765user45765
2,6702724
asked Dec 7 '17 at 15:52
user45765user45765
2,6702724
asked Dec 7 '17 at 15:52
user45765user45765
2,6702724
2,6702724
$begingroup$
The derived functor is really a sequence ${R^iF}$, isn't it? Do you wish to further derive only one of them, or the whole sequence somehow?
$endgroup$
– Randall
Dec 7 '17 at 17:34
$begingroup$
@Randall I wish to derive one of them if it happens to be either left exact of right exact. If I can derive one of them, then I probably could see whether further derive all of them yields anything useful. However, I do not see any reason why not to further derive the derived functors.
$endgroup$
– user45765
Dec 7 '17 at 18:06
add a comment |
$begingroup$
The derived functor is really a sequence ${R^iF}$, isn't it? Do you wish to further derive only one of them, or the whole sequence somehow?
$endgroup$
– Randall
Dec 7 '17 at 17:34
$begingroup$
@Randall I wish to derive one of them if it happens to be either left exact of right exact. If I can derive one of them, then I probably could see whether further derive all of them yields anything useful. However, I do not see any reason why not to further derive the derived functors.
$endgroup$
– user45765
Dec 7 '17 at 18:06
$begingroup$
The derived functor is really a sequence ${R^iF}$, isn't it? Do you wish to further derive only one of them, or the whole sequence somehow?
$endgroup$
– Randall
Dec 7 '17 at 17:34
$begingroup$
The derived functor is really a sequence ${R^iF}$, isn't it? Do you wish to further derive only one of them, or the whole sequence somehow?
$endgroup$
– Randall
Dec 7 '17 at 17:34
$begingroup$
@Randall I wish to derive one of them if it happens to be either left exact of right exact. If I can derive one of them, then I probably could see whether further derive all of them yields anything useful. However, I do not see any reason why not to further derive the derived functors.
$endgroup$
– user45765
Dec 7 '17 at 18:06
$begingroup$
@Randall I wish to derive one of them if it happens to be either left exact of right exact. If I can derive one of them, then I probably could see whether further derive all of them yields anything useful. However, I do not see any reason why not to further derive the derived functors.
$endgroup$
– user45765
Dec 7 '17 at 18:06
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I think it is rare for a left-derived functor to be right exact again. This is because given any exact sequence
$$0 to A to B to C to 0$$
we get a long exact sequence, a piece of which looks like
$$L^{i+1}F(C) to L^iF(A) to L^iF(B) to L^iF(C) to L^{i-1}F(A)$$
so the functor $L^iF$ will be right-exact if and only if for all injections $0 to A to B$ one has that the map $L^{i-1}F(A) to L^{i-1}F(B)$ is injective -- i.e. that $L^{i-1}F$ is left exact.
For example, $L^1F$ is right exact if and only if $F$ is exact, which implies in fact that $L^1F$ is zero.
answered Dec 7 '17 at 20:19
equinequin
52339
$endgroup$
$begingroup$
For left derived functor, consider $otimes_ZD$. It is clear that $Tor^1_Z(-,D)$ is left exact. So you can derive right derived functors again.
$endgroup$
– user45765
Dec 7 '17 at 23:47
$begingroup$
Sorry, what are $Z$ and $D$ here?
$endgroup$
– equin
Dec 8 '17 at 0:33
$begingroup$
Z is the integer ring. $D$ is any torsion $Z-$ module.
$endgroup$
– user45765
Dec 8 '17 at 0:39
$begingroup$
@Hurkyl If I remember correctly from what I was told, it is related to Ext rather than $-otimes_Z D$. Could you clarify how you deduced this reasoning. This also states that $R^i=0$ for $igeq 2$ as injective resolution length maximal 1 from projective resolution maximal length 1.
$endgroup$
– user45765
Dec 8 '17 at 23:00
$begingroup$
@user45765: Ah, nevermind, I overlooked the requirement that the right derived functors vanish on injective objects.
$endgroup$
– Hurkyl
Dec 9 '17 at 3:28
|
show 1 more comment
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Based on their comments, it seems the OP was specifically interested in the computation of right derived functors for $mathrm{Tor}_1^A(M,square )$ when $A$ is a PID. I have not been able to find this computation anywhere on StackExchange (or the internet, for that matter). I think the result is interesting, if only as a kind of "derived functor trivia", and will therefore give an outline of my approach here. The proof is not difficult.
(I apologize for the French notation, but it is less confusing when dealing with right-derived functors. I also apologize for the abhorrent diagram, but I'm not used to the AMScd package.)
Let $tM$ denote the torsion submodule of $M$.
Proposition 1. Let $A$ be a PID, and let $M$ be a module over $A$. For every $R$-module $N$, we then have
$$
(R^nmathrm{Tor}_1^A(M,square ))(N)congbegin{cases} mathrm{Tor}_1^A(M,N) quadtext{when}, n=0,\tMotimes_ANqquad,,text{when}, n = 1,\0qquadqquadqquad text{otherwise}.end{cases}
$$
Getting the right idea. Since $A$ is a PID, the only interesting right-derived functor is in degree 1; that is, we immediately have
$$
(R^0mathrm{Tor}_1^A(M,square )(N)cong mathrm{Tor}_1^A(M,N) quadtext{and}quad (R^nmathrm{Tor_1}^A(M,square))(N)cong 0quadtext{for}, n > 1.
$$
From now on we abuse notation wherever possible (e.g. by dropping superscripts from $mathrm{Tor}_1^A$) to make the proof no more painful than it has to be. Given an SES $0rightarrow N'rightarrow Nrightarrow N''rightarrow 0$, the right derived functor will appear in an LES of the form
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'')rightarrow R^1mathrm{Tor}_1(M,N')rightarrow R^1mathrm{Tor}_1(M,N)rightarrow R^1mathrm{Tor}_1(M,N'') rightarrow 0.
$$
But thinking back to the LES of $mathrm{Tor}$, we also have the LES:
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'') rightarrow Motimes N'rightarrow Motimes Nrightarrow Motimes N'' rightarrow 0.
$$
These long exact sequences share the first three terms. Based on this, a naive guess might be: $R^1mathrm{Tor}_1(M,N) cong Motimes N$. However, this is obviously not the case since a right derived functor must vanish on injectives. Note that if $E$ is injective, then $Motimes E$ does vanish when $M = tM$ is torsion (since $A$ is a PID, we have injective $Leftrightarrow$ divisible), so it seems torsion modules might be relevant. This leads us to the hypothesis in the proposition.
Proof (sketch). Consider first the case where $M = tM$ is torsion. Let $N$ be an arbitrary $R$-module and choose an injective resolution
$$
0 rightarrow Nrightarrow E_0 rightarrow E_1rightarrow 0.
$$
(It is not important here that $E_1$ is injective. We have already used the global injective dimension of $A$ above, and will not be using it for the rest of the argument.) Compare the following segments of the long exact sequences described above:
$$
require{AMScd}
begin{CD}
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>delta >> R^1mathrm{Tor}_1(M,N) @>>> R^1mathrm{Tor}_1(M,E_0) \
@| @| \
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>partial >> Motimes N @>>> Motimes E_0
end{CD}
$$
Both terms on the far right cancel. Specifically, $R^1mathrm{Tor}_1(M,E_0)$ cancels since $E_0$ is injective and $Motimes E_0$ cancels since $E_0$ is divisible (remind yourself that $A$ is a PID). But then we have an isomorphism from $R^1mathrm{Tor}_1(M,N)$ to $Motimes N$.
The general case follows from the torsion case by considering the SES
$$
0 rightarrow tM rightarrow M rightarrow M/tM rightarrow 0,
$$
where in particular $tM$ is torsion and $M/tM$ is torsionfree (hence flat). Q.E.D.
Applying the same methods to, say, covariant $mathrm{Ext}^1(M,square )$ returns
Proposition 2. Let $A$ be a PID, and let $M$ and $N$ be $A$-modules. Then
$$
(L_nmathrm{Ext}_A^1(M,square )) (N) cong begin{cases} mathrm{Ext}_A^1(M,N)quadtext{when}, n = 0,\
mathrm{Hom}_A(tM,N):text{when}, n = 1,\
0qquadqquadqquadtext{otherwise.}
end{cases}
$$
Some conclusions. The above computations answer some of the original questions, albeit only for the classical derived functors $mathrm{Tor}$ and $mathrm{Ext}$. For instance, we see that repeatedly deriving these functors (in the appropriate semi-exact term) quickly becomes repetitive.
I don't know if the computations are meaningful. They are a classic exercise (e.g. in Rotman), but this is probably because they can be solved by neat LES arguments like the one above. The $mathrm{Tor}$-computation gives us, as a special case, the derived functors of the torsion functor $tcong mathrm{Tor}_1^A(mathrm{Frac}(A)/A ,square )$, so that's something.
answered Jan 21 at 18:02
o.h.o.h.
4917
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The total derived functor is already exact, but in the relevant homotopical sense.
In fact, you can say more; if $A to B to C$ is a sequence of complexes that has a long exact sequence of homology groups, then there should a quasi-isomorphic complex $A' to B' to C'$ that is short exact in the ordinary sense.
(by a "quasi-isomorphism" of complexes, I mean a map that is a quasi-isomorphism on each object)
(I don't know if you can arrange for the total derived functor itself to be exact in the ordinary sense)
The collection of homology group functors don't remember everything, but they are enough to detect equivalences; a map of complexes is a quasi-isomorphism iff it induces an isomorphism on homology, and two maps of complexes are equivalent iff they induce the same map on homology.
answered Dec 7 '17 at 20:40
HurkylHurkyl
112k9120262
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rather, the derived functor ($i$-th component, or the total one as a graded module) is never left/right exact. otherwise we have every long exact sequence splitted, that is equivalent to being the initial functor exact, and in this case the derived functor should be zero. i can write this in details if needed
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– Andrey Ryabichev
Dec 7 '17 at 21:49
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@AndreyRyabichev: By the "total" one I mean the one that takes values in the ($infty$-) category of complexes, not merely the amalgamation of the homology groups.
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– Hurkyl
Dec 8 '17 at 3:50
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but such a category is not abelian, how to define a derived functor in that sence?
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– Andrey Ryabichev
Dec 8 '17 at 10:38
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@Andrey: The usual way; take a projective resolution and apply your functor termwise. (And the category of chain complexes is abelian. And the $infty$-category it presents is a stable one)
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– Hurkyl
Dec 8 '17 at 20:15
add a comment |
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3 Answers
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3 Answers
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$begingroup$
I think it is rare for a left-derived functor to be right exact again. This is because given any exact sequence
$$0 to A to B to C to 0$$
we get a long exact sequence, a piece of which looks like
$$L^{i+1}F(C) to L^iF(A) to L^iF(B) to L^iF(C) to L^{i-1}F(A)$$
so the functor $L^iF$ will be right-exact if and only if for all injections $0 to A to B$ one has that the map $L^{i-1}F(A) to L^{i-1}F(B)$ is injective -- i.e. that $L^{i-1}F$ is left exact.
For example, $L^1F$ is right exact if and only if $F$ is exact, which implies in fact that $L^1F$ is zero.
answered Dec 7 '17 at 20:19
equinequin
52339
$endgroup$
$begingroup$
For left derived functor, consider $otimes_ZD$. It is clear that $Tor^1_Z(-,D)$ is left exact. So you can derive right derived functors again.
$endgroup$
– user45765
Dec 7 '17 at 23:47
$begingroup$
Sorry, what are $Z$ and $D$ here?
$endgroup$
– equin
Dec 8 '17 at 0:33
$begingroup$
Z is the integer ring. $D$ is any torsion $Z-$ module.
$endgroup$
– user45765
Dec 8 '17 at 0:39
$begingroup$
@Hurkyl If I remember correctly from what I was told, it is related to Ext rather than $-otimes_Z D$. Could you clarify how you deduced this reasoning. This also states that $R^i=0$ for $igeq 2$ as injective resolution length maximal 1 from projective resolution maximal length 1.
$endgroup$
– user45765
Dec 8 '17 at 23:00
$begingroup$
@user45765: Ah, nevermind, I overlooked the requirement that the right derived functors vanish on injective objects.
$endgroup$
– Hurkyl
Dec 9 '17 at 3:28
|
show 1 more comment
$begingroup$
I think it is rare for a left-derived functor to be right exact again. This is because given any exact sequence
$$0 to A to B to C to 0$$
we get a long exact sequence, a piece of which looks like
$$L^{i+1}F(C) to L^iF(A) to L^iF(B) to L^iF(C) to L^{i-1}F(A)$$
so the functor $L^iF$ will be right-exact if and only if for all injections $0 to A to B$ one has that the map $L^{i-1}F(A) to L^{i-1}F(B)$ is injective -- i.e. that $L^{i-1}F$ is left exact.
For example, $L^1F$ is right exact if and only if $F$ is exact, which implies in fact that $L^1F$ is zero.
answered Dec 7 '17 at 20:19
equinequin
52339
$endgroup$
$begingroup$
For left derived functor, consider $otimes_ZD$. It is clear that $Tor^1_Z(-,D)$ is left exact. So you can derive right derived functors again.
$endgroup$
– user45765
Dec 7 '17 at 23:47
$begingroup$
Sorry, what are $Z$ and $D$ here?
$endgroup$
– equin
Dec 8 '17 at 0:33
$begingroup$
Z is the integer ring. $D$ is any torsion $Z-$ module.
$endgroup$
– user45765
Dec 8 '17 at 0:39
$begingroup$
@Hurkyl If I remember correctly from what I was told, it is related to Ext rather than $-otimes_Z D$. Could you clarify how you deduced this reasoning. This also states that $R^i=0$ for $igeq 2$ as injective resolution length maximal 1 from projective resolution maximal length 1.
$endgroup$
– user45765
Dec 8 '17 at 23:00
$begingroup$
@user45765: Ah, nevermind, I overlooked the requirement that the right derived functors vanish on injective objects.
$endgroup$
– Hurkyl
Dec 9 '17 at 3:28
|
show 1 more comment
$begingroup$
I think it is rare for a left-derived functor to be right exact again. This is because given any exact sequence
$$0 to A to B to C to 0$$
we get a long exact sequence, a piece of which looks like
$$L^{i+1}F(C) to L^iF(A) to L^iF(B) to L^iF(C) to L^{i-1}F(A)$$
so the functor $L^iF$ will be right-exact if and only if for all injections $0 to A to B$ one has that the map $L^{i-1}F(A) to L^{i-1}F(B)$ is injective -- i.e. that $L^{i-1}F$ is left exact.
For example, $L^1F$ is right exact if and only if $F$ is exact, which implies in fact that $L^1F$ is zero.
answered Dec 7 '17 at 20:19
equinequin
52339
$endgroup$
I think it is rare for a left-derived functor to be right exact again. This is because given any exact sequence
$$0 to A to B to C to 0$$
we get a long exact sequence, a piece of which looks like
$$L^{i+1}F(C) to L^iF(A) to L^iF(B) to L^iF(C) to L^{i-1}F(A)$$
so the functor $L^iF$ will be right-exact if and only if for all injections $0 to A to B$ one has that the map $L^{i-1}F(A) to L^{i-1}F(B)$ is injective -- i.e. that $L^{i-1}F$ is left exact.
For example, $L^1F$ is right exact if and only if $F$ is exact, which implies in fact that $L^1F$ is zero.
answered Dec 7 '17 at 20:19
equinequin
52339
answered Dec 7 '17 at 20:19
equinequin
52339
answered Dec 7 '17 at 20:19
equinequin
52339
answered Dec 7 '17 at 20:19
equinequin
52339
52339
$begingroup$
For left derived functor, consider $otimes_ZD$. It is clear that $Tor^1_Z(-,D)$ is left exact. So you can derive right derived functors again.
$endgroup$
– user45765
Dec 7 '17 at 23:47
$begingroup$
Sorry, what are $Z$ and $D$ here?
$endgroup$
– equin
Dec 8 '17 at 0:33
$begingroup$
Z is the integer ring. $D$ is any torsion $Z-$ module.
$endgroup$
– user45765
Dec 8 '17 at 0:39
$begingroup$
@Hurkyl If I remember correctly from what I was told, it is related to Ext rather than $-otimes_Z D$. Could you clarify how you deduced this reasoning. This also states that $R^i=0$ for $igeq 2$ as injective resolution length maximal 1 from projective resolution maximal length 1.
$endgroup$
– user45765
Dec 8 '17 at 23:00
$begingroup$
@user45765: Ah, nevermind, I overlooked the requirement that the right derived functors vanish on injective objects.
$endgroup$
– Hurkyl
Dec 9 '17 at 3:28
|
show 1 more comment
$begingroup$
For left derived functor, consider $otimes_ZD$. It is clear that $Tor^1_Z(-,D)$ is left exact. So you can derive right derived functors again.
$endgroup$
– user45765
Dec 7 '17 at 23:47
$begingroup$
Sorry, what are $Z$ and $D$ here?
$endgroup$
– equin
Dec 8 '17 at 0:33
$begingroup$
Z is the integer ring. $D$ is any torsion $Z-$ module.
$endgroup$
– user45765
Dec 8 '17 at 0:39
$begingroup$
@Hurkyl If I remember correctly from what I was told, it is related to Ext rather than $-otimes_Z D$. Could you clarify how you deduced this reasoning. This also states that $R^i=0$ for $igeq 2$ as injective resolution length maximal 1 from projective resolution maximal length 1.
$endgroup$
– user45765
Dec 8 '17 at 23:00
$begingroup$
@user45765: Ah, nevermind, I overlooked the requirement that the right derived functors vanish on injective objects.
$endgroup$
– Hurkyl
Dec 9 '17 at 3:28
$begingroup$
For left derived functor, consider $otimes_ZD$. It is clear that $Tor^1_Z(-,D)$ is left exact. So you can derive right derived functors again.
$endgroup$
– user45765
Dec 7 '17 at 23:47
$begingroup$
For left derived functor, consider $otimes_ZD$. It is clear that $Tor^1_Z(-,D)$ is left exact. So you can derive right derived functors again.
$endgroup$
– user45765
Dec 7 '17 at 23:47
$begingroup$
Sorry, what are $Z$ and $D$ here?
$endgroup$
– equin
Dec 8 '17 at 0:33
$begingroup$
Sorry, what are $Z$ and $D$ here?
$endgroup$
– equin
Dec 8 '17 at 0:33
$begingroup$
Z is the integer ring. $D$ is any torsion $Z-$ module.
$endgroup$
– user45765
Dec 8 '17 at 0:39
$begingroup$
Z is the integer ring. $D$ is any torsion $Z-$ module.
$endgroup$
– user45765
Dec 8 '17 at 0:39
$begingroup$
@Hurkyl If I remember correctly from what I was told, it is related to Ext rather than $-otimes_Z D$. Could you clarify how you deduced this reasoning. This also states that $R^i=0$ for $igeq 2$ as injective resolution length maximal 1 from projective resolution maximal length 1.
$endgroup$
– user45765
Dec 8 '17 at 23:00
$begingroup$
@Hurkyl If I remember correctly from what I was told, it is related to Ext rather than $-otimes_Z D$. Could you clarify how you deduced this reasoning. This also states that $R^i=0$ for $igeq 2$ as injective resolution length maximal 1 from projective resolution maximal length 1.
$endgroup$
– user45765
Dec 8 '17 at 23:00
$begingroup$
@user45765: Ah, nevermind, I overlooked the requirement that the right derived functors vanish on injective objects.
$endgroup$
– Hurkyl
Dec 9 '17 at 3:28
$begingroup$
@user45765: Ah, nevermind, I overlooked the requirement that the right derived functors vanish on injective objects.
$endgroup$
– Hurkyl
Dec 9 '17 at 3:28
|
show 1 more comment
$begingroup$
Based on their comments, it seems the OP was specifically interested in the computation of right derived functors for $mathrm{Tor}_1^A(M,square )$ when $A$ is a PID. I have not been able to find this computation anywhere on StackExchange (or the internet, for that matter). I think the result is interesting, if only as a kind of "derived functor trivia", and will therefore give an outline of my approach here. The proof is not difficult.
(I apologize for the French notation, but it is less confusing when dealing with right-derived functors. I also apologize for the abhorrent diagram, but I'm not used to the AMScd package.)
Let $tM$ denote the torsion submodule of $M$.
Proposition 1. Let $A$ be a PID, and let $M$ be a module over $A$. For every $R$-module $N$, we then have
$$
(R^nmathrm{Tor}_1^A(M,square ))(N)congbegin{cases} mathrm{Tor}_1^A(M,N) quadtext{when}, n=0,\tMotimes_ANqquad,,text{when}, n = 1,\0qquadqquadqquad text{otherwise}.end{cases}
$$
Getting the right idea. Since $A$ is a PID, the only interesting right-derived functor is in degree 1; that is, we immediately have
$$
(R^0mathrm{Tor}_1^A(M,square )(N)cong mathrm{Tor}_1^A(M,N) quadtext{and}quad (R^nmathrm{Tor_1}^A(M,square))(N)cong 0quadtext{for}, n > 1.
$$
From now on we abuse notation wherever possible (e.g. by dropping superscripts from $mathrm{Tor}_1^A$) to make the proof no more painful than it has to be. Given an SES $0rightarrow N'rightarrow Nrightarrow N''rightarrow 0$, the right derived functor will appear in an LES of the form
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'')rightarrow R^1mathrm{Tor}_1(M,N')rightarrow R^1mathrm{Tor}_1(M,N)rightarrow R^1mathrm{Tor}_1(M,N'') rightarrow 0.
$$
But thinking back to the LES of $mathrm{Tor}$, we also have the LES:
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'') rightarrow Motimes N'rightarrow Motimes Nrightarrow Motimes N'' rightarrow 0.
$$
These long exact sequences share the first three terms. Based on this, a naive guess might be: $R^1mathrm{Tor}_1(M,N) cong Motimes N$. However, this is obviously not the case since a right derived functor must vanish on injectives. Note that if $E$ is injective, then $Motimes E$ does vanish when $M = tM$ is torsion (since $A$ is a PID, we have injective $Leftrightarrow$ divisible), so it seems torsion modules might be relevant. This leads us to the hypothesis in the proposition.
Proof (sketch). Consider first the case where $M = tM$ is torsion. Let $N$ be an arbitrary $R$-module and choose an injective resolution
$$
0 rightarrow Nrightarrow E_0 rightarrow E_1rightarrow 0.
$$
(It is not important here that $E_1$ is injective. We have already used the global injective dimension of $A$ above, and will not be using it for the rest of the argument.) Compare the following segments of the long exact sequences described above:
$$
require{AMScd}
begin{CD}
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>delta >> R^1mathrm{Tor}_1(M,N) @>>> R^1mathrm{Tor}_1(M,E_0) \
@| @| \
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>partial >> Motimes N @>>> Motimes E_0
end{CD}
$$
Both terms on the far right cancel. Specifically, $R^1mathrm{Tor}_1(M,E_0)$ cancels since $E_0$ is injective and $Motimes E_0$ cancels since $E_0$ is divisible (remind yourself that $A$ is a PID). But then we have an isomorphism from $R^1mathrm{Tor}_1(M,N)$ to $Motimes N$.
The general case follows from the torsion case by considering the SES
$$
0 rightarrow tM rightarrow M rightarrow M/tM rightarrow 0,
$$
where in particular $tM$ is torsion and $M/tM$ is torsionfree (hence flat). Q.E.D.
Applying the same methods to, say, covariant $mathrm{Ext}^1(M,square )$ returns
Proposition 2. Let $A$ be a PID, and let $M$ and $N$ be $A$-modules. Then
$$
(L_nmathrm{Ext}_A^1(M,square )) (N) cong begin{cases} mathrm{Ext}_A^1(M,N)quadtext{when}, n = 0,\
mathrm{Hom}_A(tM,N):text{when}, n = 1,\
0qquadqquadqquadtext{otherwise.}
end{cases}
$$
Some conclusions. The above computations answer some of the original questions, albeit only for the classical derived functors $mathrm{Tor}$ and $mathrm{Ext}$. For instance, we see that repeatedly deriving these functors (in the appropriate semi-exact term) quickly becomes repetitive.
I don't know if the computations are meaningful. They are a classic exercise (e.g. in Rotman), but this is probably because they can be solved by neat LES arguments like the one above. The $mathrm{Tor}$-computation gives us, as a special case, the derived functors of the torsion functor $tcong mathrm{Tor}_1^A(mathrm{Frac}(A)/A ,square )$, so that's something.
answered Jan 21 at 18:02
o.h.o.h.
4917
$endgroup$
add a comment |
$begingroup$
Based on their comments, it seems the OP was specifically interested in the computation of right derived functors for $mathrm{Tor}_1^A(M,square )$ when $A$ is a PID. I have not been able to find this computation anywhere on StackExchange (or the internet, for that matter). I think the result is interesting, if only as a kind of "derived functor trivia", and will therefore give an outline of my approach here. The proof is not difficult.
(I apologize for the French notation, but it is less confusing when dealing with right-derived functors. I also apologize for the abhorrent diagram, but I'm not used to the AMScd package.)
Let $tM$ denote the torsion submodule of $M$.
Proposition 1. Let $A$ be a PID, and let $M$ be a module over $A$. For every $R$-module $N$, we then have
$$
(R^nmathrm{Tor}_1^A(M,square ))(N)congbegin{cases} mathrm{Tor}_1^A(M,N) quadtext{when}, n=0,\tMotimes_ANqquad,,text{when}, n = 1,\0qquadqquadqquad text{otherwise}.end{cases}
$$
Getting the right idea. Since $A$ is a PID, the only interesting right-derived functor is in degree 1; that is, we immediately have
$$
(R^0mathrm{Tor}_1^A(M,square )(N)cong mathrm{Tor}_1^A(M,N) quadtext{and}quad (R^nmathrm{Tor_1}^A(M,square))(N)cong 0quadtext{for}, n > 1.
$$
From now on we abuse notation wherever possible (e.g. by dropping superscripts from $mathrm{Tor}_1^A$) to make the proof no more painful than it has to be. Given an SES $0rightarrow N'rightarrow Nrightarrow N''rightarrow 0$, the right derived functor will appear in an LES of the form
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'')rightarrow R^1mathrm{Tor}_1(M,N')rightarrow R^1mathrm{Tor}_1(M,N)rightarrow R^1mathrm{Tor}_1(M,N'') rightarrow 0.
$$
But thinking back to the LES of $mathrm{Tor}$, we also have the LES:
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'') rightarrow Motimes N'rightarrow Motimes Nrightarrow Motimes N'' rightarrow 0.
$$
These long exact sequences share the first three terms. Based on this, a naive guess might be: $R^1mathrm{Tor}_1(M,N) cong Motimes N$. However, this is obviously not the case since a right derived functor must vanish on injectives. Note that if $E$ is injective, then $Motimes E$ does vanish when $M = tM$ is torsion (since $A$ is a PID, we have injective $Leftrightarrow$ divisible), so it seems torsion modules might be relevant. This leads us to the hypothesis in the proposition.
Proof (sketch). Consider first the case where $M = tM$ is torsion. Let $N$ be an arbitrary $R$-module and choose an injective resolution
$$
0 rightarrow Nrightarrow E_0 rightarrow E_1rightarrow 0.
$$
(It is not important here that $E_1$ is injective. We have already used the global injective dimension of $A$ above, and will not be using it for the rest of the argument.) Compare the following segments of the long exact sequences described above:
$$
require{AMScd}
begin{CD}
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>delta >> R^1mathrm{Tor}_1(M,N) @>>> R^1mathrm{Tor}_1(M,E_0) \
@| @| \
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>partial >> Motimes N @>>> Motimes E_0
end{CD}
$$
Both terms on the far right cancel. Specifically, $R^1mathrm{Tor}_1(M,E_0)$ cancels since $E_0$ is injective and $Motimes E_0$ cancels since $E_0$ is divisible (remind yourself that $A$ is a PID). But then we have an isomorphism from $R^1mathrm{Tor}_1(M,N)$ to $Motimes N$.
The general case follows from the torsion case by considering the SES
$$
0 rightarrow tM rightarrow M rightarrow M/tM rightarrow 0,
$$
where in particular $tM$ is torsion and $M/tM$ is torsionfree (hence flat). Q.E.D.
Applying the same methods to, say, covariant $mathrm{Ext}^1(M,square )$ returns
Proposition 2. Let $A$ be a PID, and let $M$ and $N$ be $A$-modules. Then
$$
(L_nmathrm{Ext}_A^1(M,square )) (N) cong begin{cases} mathrm{Ext}_A^1(M,N)quadtext{when}, n = 0,\
mathrm{Hom}_A(tM,N):text{when}, n = 1,\
0qquadqquadqquadtext{otherwise.}
end{cases}
$$
Some conclusions. The above computations answer some of the original questions, albeit only for the classical derived functors $mathrm{Tor}$ and $mathrm{Ext}$. For instance, we see that repeatedly deriving these functors (in the appropriate semi-exact term) quickly becomes repetitive.
I don't know if the computations are meaningful. They are a classic exercise (e.g. in Rotman), but this is probably because they can be solved by neat LES arguments like the one above. The $mathrm{Tor}$-computation gives us, as a special case, the derived functors of the torsion functor $tcong mathrm{Tor}_1^A(mathrm{Frac}(A)/A ,square )$, so that's something.
answered Jan 21 at 18:02
o.h.o.h.
4917
$endgroup$
add a comment |
$begingroup$
Based on their comments, it seems the OP was specifically interested in the computation of right derived functors for $mathrm{Tor}_1^A(M,square )$ when $A$ is a PID. I have not been able to find this computation anywhere on StackExchange (or the internet, for that matter). I think the result is interesting, if only as a kind of "derived functor trivia", and will therefore give an outline of my approach here. The proof is not difficult.
(I apologize for the French notation, but it is less confusing when dealing with right-derived functors. I also apologize for the abhorrent diagram, but I'm not used to the AMScd package.)
Let $tM$ denote the torsion submodule of $M$.
Proposition 1. Let $A$ be a PID, and let $M$ be a module over $A$. For every $R$-module $N$, we then have
$$
(R^nmathrm{Tor}_1^A(M,square ))(N)congbegin{cases} mathrm{Tor}_1^A(M,N) quadtext{when}, n=0,\tMotimes_ANqquad,,text{when}, n = 1,\0qquadqquadqquad text{otherwise}.end{cases}
$$
Getting the right idea. Since $A$ is a PID, the only interesting right-derived functor is in degree 1; that is, we immediately have
$$
(R^0mathrm{Tor}_1^A(M,square )(N)cong mathrm{Tor}_1^A(M,N) quadtext{and}quad (R^nmathrm{Tor_1}^A(M,square))(N)cong 0quadtext{for}, n > 1.
$$
From now on we abuse notation wherever possible (e.g. by dropping superscripts from $mathrm{Tor}_1^A$) to make the proof no more painful than it has to be. Given an SES $0rightarrow N'rightarrow Nrightarrow N''rightarrow 0$, the right derived functor will appear in an LES of the form
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'')rightarrow R^1mathrm{Tor}_1(M,N')rightarrow R^1mathrm{Tor}_1(M,N)rightarrow R^1mathrm{Tor}_1(M,N'') rightarrow 0.
$$
But thinking back to the LES of $mathrm{Tor}$, we also have the LES:
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'') rightarrow Motimes N'rightarrow Motimes Nrightarrow Motimes N'' rightarrow 0.
$$
These long exact sequences share the first three terms. Based on this, a naive guess might be: $R^1mathrm{Tor}_1(M,N) cong Motimes N$. However, this is obviously not the case since a right derived functor must vanish on injectives. Note that if $E$ is injective, then $Motimes E$ does vanish when $M = tM$ is torsion (since $A$ is a PID, we have injective $Leftrightarrow$ divisible), so it seems torsion modules might be relevant. This leads us to the hypothesis in the proposition.
Proof (sketch). Consider first the case where $M = tM$ is torsion. Let $N$ be an arbitrary $R$-module and choose an injective resolution
$$
0 rightarrow Nrightarrow E_0 rightarrow E_1rightarrow 0.
$$
(It is not important here that $E_1$ is injective. We have already used the global injective dimension of $A$ above, and will not be using it for the rest of the argument.) Compare the following segments of the long exact sequences described above:
$$
require{AMScd}
begin{CD}
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>delta >> R^1mathrm{Tor}_1(M,N) @>>> R^1mathrm{Tor}_1(M,E_0) \
@| @| \
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>partial >> Motimes N @>>> Motimes E_0
end{CD}
$$
Both terms on the far right cancel. Specifically, $R^1mathrm{Tor}_1(M,E_0)$ cancels since $E_0$ is injective and $Motimes E_0$ cancels since $E_0$ is divisible (remind yourself that $A$ is a PID). But then we have an isomorphism from $R^1mathrm{Tor}_1(M,N)$ to $Motimes N$.
The general case follows from the torsion case by considering the SES
$$
0 rightarrow tM rightarrow M rightarrow M/tM rightarrow 0,
$$
where in particular $tM$ is torsion and $M/tM$ is torsionfree (hence flat). Q.E.D.
Applying the same methods to, say, covariant $mathrm{Ext}^1(M,square )$ returns
Proposition 2. Let $A$ be a PID, and let $M$ and $N$ be $A$-modules. Then
$$
(L_nmathrm{Ext}_A^1(M,square )) (N) cong begin{cases} mathrm{Ext}_A^1(M,N)quadtext{when}, n = 0,\
mathrm{Hom}_A(tM,N):text{when}, n = 1,\
0qquadqquadqquadtext{otherwise.}
end{cases}
$$
Some conclusions. The above computations answer some of the original questions, albeit only for the classical derived functors $mathrm{Tor}$ and $mathrm{Ext}$. For instance, we see that repeatedly deriving these functors (in the appropriate semi-exact term) quickly becomes repetitive.
I don't know if the computations are meaningful. They are a classic exercise (e.g. in Rotman), but this is probably because they can be solved by neat LES arguments like the one above. The $mathrm{Tor}$-computation gives us, as a special case, the derived functors of the torsion functor $tcong mathrm{Tor}_1^A(mathrm{Frac}(A)/A ,square )$, so that's something.
answered Jan 21 at 18:02
o.h.o.h.
4917
$endgroup$
Based on their comments, it seems the OP was specifically interested in the computation of right derived functors for $mathrm{Tor}_1^A(M,square )$ when $A$ is a PID. I have not been able to find this computation anywhere on StackExchange (or the internet, for that matter). I think the result is interesting, if only as a kind of "derived functor trivia", and will therefore give an outline of my approach here. The proof is not difficult.
(I apologize for the French notation, but it is less confusing when dealing with right-derived functors. I also apologize for the abhorrent diagram, but I'm not used to the AMScd package.)
Let $tM$ denote the torsion submodule of $M$.
Proposition 1. Let $A$ be a PID, and let $M$ be a module over $A$. For every $R$-module $N$, we then have
$$
(R^nmathrm{Tor}_1^A(M,square ))(N)congbegin{cases} mathrm{Tor}_1^A(M,N) quadtext{when}, n=0,\tMotimes_ANqquad,,text{when}, n = 1,\0qquadqquadqquad text{otherwise}.end{cases}
$$
Getting the right idea. Since $A$ is a PID, the only interesting right-derived functor is in degree 1; that is, we immediately have
$$
(R^0mathrm{Tor}_1^A(M,square )(N)cong mathrm{Tor}_1^A(M,N) quadtext{and}quad (R^nmathrm{Tor_1}^A(M,square))(N)cong 0quadtext{for}, n > 1.
$$
From now on we abuse notation wherever possible (e.g. by dropping superscripts from $mathrm{Tor}_1^A$) to make the proof no more painful than it has to be. Given an SES $0rightarrow N'rightarrow Nrightarrow N''rightarrow 0$, the right derived functor will appear in an LES of the form
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'')rightarrow R^1mathrm{Tor}_1(M,N')rightarrow R^1mathrm{Tor}_1(M,N)rightarrow R^1mathrm{Tor}_1(M,N'') rightarrow 0.
$$
But thinking back to the LES of $mathrm{Tor}$, we also have the LES:
$$
0rightarrow mathrm{Tor}_1(M,N') rightarrow mathrm{Tor}_1(M,N)rightarrowmathrm{Tor}_1(M,N'') rightarrow Motimes N'rightarrow Motimes Nrightarrow Motimes N'' rightarrow 0.
$$
These long exact sequences share the first three terms. Based on this, a naive guess might be: $R^1mathrm{Tor}_1(M,N) cong Motimes N$. However, this is obviously not the case since a right derived functor must vanish on injectives. Note that if $E$ is injective, then $Motimes E$ does vanish when $M = tM$ is torsion (since $A$ is a PID, we have injective $Leftrightarrow$ divisible), so it seems torsion modules might be relevant. This leads us to the hypothesis in the proposition.
Proof (sketch). Consider first the case where $M = tM$ is torsion. Let $N$ be an arbitrary $R$-module and choose an injective resolution
$$
0 rightarrow Nrightarrow E_0 rightarrow E_1rightarrow 0.
$$
(It is not important here that $E_1$ is injective. We have already used the global injective dimension of $A$ above, and will not be using it for the rest of the argument.) Compare the following segments of the long exact sequences described above:
$$
require{AMScd}
begin{CD}
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>delta >> R^1mathrm{Tor}_1(M,N) @>>> R^1mathrm{Tor}_1(M,E_0) \
@| @| \
mathrm{Tor}_1(M,E_0) @>>> mathrm{Tor}_1(M,E_1) @>partial >> Motimes N @>>> Motimes E_0
end{CD}
$$
Both terms on the far right cancel. Specifically, $R^1mathrm{Tor}_1(M,E_0)$ cancels since $E_0$ is injective and $Motimes E_0$ cancels since $E_0$ is divisible (remind yourself that $A$ is a PID). But then we have an isomorphism from $R^1mathrm{Tor}_1(M,N)$ to $Motimes N$.
The general case follows from the torsion case by considering the SES
$$
0 rightarrow tM rightarrow M rightarrow M/tM rightarrow 0,
$$
where in particular $tM$ is torsion and $M/tM$ is torsionfree (hence flat). Q.E.D.
Applying the same methods to, say, covariant $mathrm{Ext}^1(M,square )$ returns
Proposition 2. Let $A$ be a PID, and let $M$ and $N$ be $A$-modules. Then
$$
(L_nmathrm{Ext}_A^1(M,square )) (N) cong begin{cases} mathrm{Ext}_A^1(M,N)quadtext{when}, n = 0,\
mathrm{Hom}_A(tM,N):text{when}, n = 1,\
0qquadqquadqquadtext{otherwise.}
end{cases}
$$
Some conclusions. The above computations answer some of the original questions, albeit only for the classical derived functors $mathrm{Tor}$ and $mathrm{Ext}$. For instance, we see that repeatedly deriving these functors (in the appropriate semi-exact term) quickly becomes repetitive.
I don't know if the computations are meaningful. They are a classic exercise (e.g. in Rotman), but this is probably because they can be solved by neat LES arguments like the one above. The $mathrm{Tor}$-computation gives us, as a special case, the derived functors of the torsion functor $tcong mathrm{Tor}_1^A(mathrm{Frac}(A)/A ,square )$, so that's something.
answered Jan 21 at 18:02
o.h.o.h.
4917
edited Mar 3 at 7:43
answered Jan 21 at 18:02
o.h.o.h.
4917
answered Jan 21 at 18:02
o.h.o.h.
4917
answered Jan 21 at 18:02
o.h.o.h.
4917
4917
add a comment |
add a comment |
$begingroup$
The total derived functor is already exact, but in the relevant homotopical sense.
In fact, you can say more; if $A to B to C$ is a sequence of complexes that has a long exact sequence of homology groups, then there should a quasi-isomorphic complex $A' to B' to C'$ that is short exact in the ordinary sense.
(by a "quasi-isomorphism" of complexes, I mean a map that is a quasi-isomorphism on each object)
(I don't know if you can arrange for the total derived functor itself to be exact in the ordinary sense)
The collection of homology group functors don't remember everything, but they are enough to detect equivalences; a map of complexes is a quasi-isomorphism iff it induces an isomorphism on homology, and two maps of complexes are equivalent iff they induce the same map on homology.
answered Dec 7 '17 at 20:40
HurkylHurkyl
112k9120262
$endgroup$
$begingroup$
rather, the derived functor ($i$-th component, or the total one as a graded module) is never left/right exact. otherwise we have every long exact sequence splitted, that is equivalent to being the initial functor exact, and in this case the derived functor should be zero. i can write this in details if needed
$endgroup$
– Andrey Ryabichev
Dec 7 '17 at 21:49
$begingroup$
@AndreyRyabichev: By the "total" one I mean the one that takes values in the ($infty$-) category of complexes, not merely the amalgamation of the homology groups.
$endgroup$
– Hurkyl
Dec 8 '17 at 3:50
$begingroup$
but such a category is not abelian, how to define a derived functor in that sence?
$endgroup$
– Andrey Ryabichev
Dec 8 '17 at 10:38
$begingroup$
@Andrey: The usual way; take a projective resolution and apply your functor termwise. (And the category of chain complexes is abelian. And the $infty$-category it presents is a stable one)
$endgroup$
– Hurkyl
Dec 8 '17 at 20:15
add a comment |
$begingroup$
The total derived functor is already exact, but in the relevant homotopical sense.
In fact, you can say more; if $A to B to C$ is a sequence of complexes that has a long exact sequence of homology groups, then there should a quasi-isomorphic complex $A' to B' to C'$ that is short exact in the ordinary sense.
(by a "quasi-isomorphism" of complexes, I mean a map that is a quasi-isomorphism on each object)
(I don't know if you can arrange for the total derived functor itself to be exact in the ordinary sense)
The collection of homology group functors don't remember everything, but they are enough to detect equivalences; a map of complexes is a quasi-isomorphism iff it induces an isomorphism on homology, and two maps of complexes are equivalent iff they induce the same map on homology.
answered Dec 7 '17 at 20:40
HurkylHurkyl
112k9120262
$endgroup$
$begingroup$
rather, the derived functor ($i$-th component, or the total one as a graded module) is never left/right exact. otherwise we have every long exact sequence splitted, that is equivalent to being the initial functor exact, and in this case the derived functor should be zero. i can write this in details if needed
$endgroup$
– Andrey Ryabichev
Dec 7 '17 at 21:49
$begingroup$
@AndreyRyabichev: By the "total" one I mean the one that takes values in the ($infty$-) category of complexes, not merely the amalgamation of the homology groups.
$endgroup$
– Hurkyl
Dec 8 '17 at 3:50
$begingroup$
but such a category is not abelian, how to define a derived functor in that sence?
$endgroup$
– Andrey Ryabichev
Dec 8 '17 at 10:38
$begingroup$
@Andrey: The usual way; take a projective resolution and apply your functor termwise. (And the category of chain complexes is abelian. And the $infty$-category it presents is a stable one)
$endgroup$
– Hurkyl
Dec 8 '17 at 20:15
add a comment |
$begingroup$
The total derived functor is already exact, but in the relevant homotopical sense.
In fact, you can say more; if $A to B to C$ is a sequence of complexes that has a long exact sequence of homology groups, then there should a quasi-isomorphic complex $A' to B' to C'$ that is short exact in the ordinary sense.
(by a "quasi-isomorphism" of complexes, I mean a map that is a quasi-isomorphism on each object)
(I don't know if you can arrange for the total derived functor itself to be exact in the ordinary sense)
The collection of homology group functors don't remember everything, but they are enough to detect equivalences; a map of complexes is a quasi-isomorphism iff it induces an isomorphism on homology, and two maps of complexes are equivalent iff they induce the same map on homology.
answered Dec 7 '17 at 20:40
HurkylHurkyl
112k9120262
$endgroup$
The total derived functor is already exact, but in the relevant homotopical sense.
In fact, you can say more; if $A to B to C$ is a sequence of complexes that has a long exact sequence of homology groups, then there should a quasi-isomorphic complex $A' to B' to C'$ that is short exact in the ordinary sense.
(by a "quasi-isomorphism" of complexes, I mean a map that is a quasi-isomorphism on each object)
(I don't know if you can arrange for the total derived functor itself to be exact in the ordinary sense)
The collection of homology group functors don't remember everything, but they are enough to detect equivalences; a map of complexes is a quasi-isomorphism iff it induces an isomorphism on homology, and two maps of complexes are equivalent iff they induce the same map on homology.
answered Dec 7 '17 at 20:40
HurkylHurkyl
112k9120262
answered Dec 7 '17 at 20:40
HurkylHurkyl
112k9120262
answered Dec 7 '17 at 20:40
HurkylHurkyl
112k9120262
answered Dec 7 '17 at 20:40
HurkylHurkyl
112k9120262
112k9120262
$begingroup$
rather, the derived functor ($i$-th component, or the total one as a graded module) is never left/right exact. otherwise we have every long exact sequence splitted, that is equivalent to being the initial functor exact, and in this case the derived functor should be zero. i can write this in details if needed
$endgroup$
– Andrey Ryabichev
Dec 7 '17 at 21:49
$begingroup$
@AndreyRyabichev: By the "total" one I mean the one that takes values in the ($infty$-) category of complexes, not merely the amalgamation of the homology groups.
$endgroup$
– Hurkyl
Dec 8 '17 at 3:50
$begingroup$
but such a category is not abelian, how to define a derived functor in that sence?
$endgroup$
– Andrey Ryabichev
Dec 8 '17 at 10:38
$begingroup$
@Andrey: The usual way; take a projective resolution and apply your functor termwise. (And the category of chain complexes is abelian. And the $infty$-category it presents is a stable one)
$endgroup$
– Hurkyl
Dec 8 '17 at 20:15
add a comment |
$begingroup$
rather, the derived functor ($i$-th component, or the total one as a graded module) is never left/right exact. otherwise we have every long exact sequence splitted, that is equivalent to being the initial functor exact, and in this case the derived functor should be zero. i can write this in details if needed
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– Andrey Ryabichev
Dec 7 '17 at 21:49
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@AndreyRyabichev: By the "total" one I mean the one that takes values in the ($infty$-) category of complexes, not merely the amalgamation of the homology groups.
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– Hurkyl
Dec 8 '17 at 3:50
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but such a category is not abelian, how to define a derived functor in that sence?
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– Andrey Ryabichev
Dec 8 '17 at 10:38
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@Andrey: The usual way; take a projective resolution and apply your functor termwise. (And the category of chain complexes is abelian. And the $infty$-category it presents is a stable one)
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– Hurkyl
Dec 8 '17 at 20:15
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rather, the derived functor ($i$-th component, or the total one as a graded module) is never left/right exact. otherwise we have every long exact sequence splitted, that is equivalent to being the initial functor exact, and in this case the derived functor should be zero. i can write this in details if needed
$endgroup$
– Andrey Ryabichev
Dec 7 '17 at 21:49
$begingroup$
rather, the derived functor ($i$-th component, or the total one as a graded module) is never left/right exact. otherwise we have every long exact sequence splitted, that is equivalent to being the initial functor exact, and in this case the derived functor should be zero. i can write this in details if needed
$endgroup$
– Andrey Ryabichev
Dec 7 '17 at 21:49
$begingroup$
@AndreyRyabichev: By the "total" one I mean the one that takes values in the ($infty$-) category of complexes, not merely the amalgamation of the homology groups.
$endgroup$
– Hurkyl
Dec 8 '17 at 3:50
$begingroup$
@AndreyRyabichev: By the "total" one I mean the one that takes values in the ($infty$-) category of complexes, not merely the amalgamation of the homology groups.
$endgroup$
– Hurkyl
Dec 8 '17 at 3:50
$begingroup$
but such a category is not abelian, how to define a derived functor in that sence?
$endgroup$
– Andrey Ryabichev
Dec 8 '17 at 10:38
$begingroup$
but such a category is not abelian, how to define a derived functor in that sence?
$endgroup$
– Andrey Ryabichev
Dec 8 '17 at 10:38
$begingroup$
@Andrey: The usual way; take a projective resolution and apply your functor termwise. (And the category of chain complexes is abelian. And the $infty$-category it presents is a stable one)
$endgroup$
– Hurkyl
Dec 8 '17 at 20:15
$begingroup$
@Andrey: The usual way; take a projective resolution and apply your functor termwise. (And the category of chain complexes is abelian. And the $infty$-category it presents is a stable one)
$endgroup$
– Hurkyl
Dec 8 '17 at 20:15
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The derived functor is really a sequence ${R^iF}$, isn't it? Do you wish to further derive only one of them, or the whole sequence somehow?
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– Randall
Dec 7 '17 at 17:34
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@Randall I wish to derive one of them if it happens to be either left exact of right exact. If I can derive one of them, then I probably could see whether further derive all of them yields anything useful. However, I do not see any reason why not to further derive the derived functors.
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– user45765
Dec 7 '17 at 18:06