Intuition behind exact sequences as extensions in algebraic geometry
$begingroup$
Introduction: There are certain categories of group schemes or group varieties in which one can define exact sequences. For instance, the categories of commutative group schemes of finite type over a field $k$ and of commutative affine group schemes over $k$ are abelian (see for instance here).
So let's say that we have a short exact sequence
$$ 0 to A to B to C to 0$$
in such a suitable category (bonus question: examples of abelian categories of group varieties?). There is, a priori, no reason why the sequence should be split. Yet people call such a $B$ (together with the maps) fitting in the above exact sequence an extension of $C$ by $A$. Note that this is true also for other categories and does not have to do with algebraic geometry specifically, but I am most interested in this setting.
My question: What kind of information can one deduce, in general, about $B$, knowing only $A$ and $C$? Does one know something about the underlying topological space of $B$ or of $B(k)$ up to homotopy, or even up to homeomorphism? Is it any better if we consider maps up to isogeny? Feel free to restrict to some specific category where there are better results.
I am asking this because the terminology seems to suggest, and it seems to me that people often assume, that $B$ is some kind of direct sum between $A$ and $C$, even though there is a priori no reason for this to be the case. The isomorphism theorem does imply that $B/A cong C$, but this does not imply that we know how to obtain $B$ knowing $A$ and $C$, since there might be many possibilities for doing so.
Approximate paraphrase 1: A paraphrase of my question is: what do all possible $B$'s that fit in the above sequence have in common, topologically?
Approximate paraphrase 2: Is there some exact functor $F$ from some "geometric" category $mathcal{I}$ to some "topological" category $mathcal{J}$ such that in $mathcal{J}$ exact sequences are always split or $F(B)$ is at least somehow isomorphic to $F(A) oplus F(C)$ or something similar?
Note: I am aware of the fact that, at least for modules, the set of the $B$'s fitting in the above exact sequence, up to some equivalence and with some sum operation, is isomorphic to $Ext^1(C,A)$, but I think that the term extension gave rise to the notation $Ext$ rather than the converse, so this fact does not provide any useful insight.
algebraic-geometry soft-question algebraic-groups exact-sequence abelian-categories
asked Jan 14 at 15:36
57Jimmy57Jimmy
3,384422
$endgroup$
add a comment |
$begingroup$
Introduction: There are certain categories of group schemes or group varieties in which one can define exact sequences. For instance, the categories of commutative group schemes of finite type over a field $k$ and of commutative affine group schemes over $k$ are abelian (see for instance here).
So let's say that we have a short exact sequence
$$ 0 to A to B to C to 0$$
in such a suitable category (bonus question: examples of abelian categories of group varieties?). There is, a priori, no reason why the sequence should be split. Yet people call such a $B$ (together with the maps) fitting in the above exact sequence an extension of $C$ by $A$. Note that this is true also for other categories and does not have to do with algebraic geometry specifically, but I am most interested in this setting.
My question: What kind of information can one deduce, in general, about $B$, knowing only $A$ and $C$? Does one know something about the underlying topological space of $B$ or of $B(k)$ up to homotopy, or even up to homeomorphism? Is it any better if we consider maps up to isogeny? Feel free to restrict to some specific category where there are better results.
I am asking this because the terminology seems to suggest, and it seems to me that people often assume, that $B$ is some kind of direct sum between $A$ and $C$, even though there is a priori no reason for this to be the case. The isomorphism theorem does imply that $B/A cong C$, but this does not imply that we know how to obtain $B$ knowing $A$ and $C$, since there might be many possibilities for doing so.
Approximate paraphrase 1: A paraphrase of my question is: what do all possible $B$'s that fit in the above sequence have in common, topologically?
Approximate paraphrase 2: Is there some exact functor $F$ from some "geometric" category $mathcal{I}$ to some "topological" category $mathcal{J}$ such that in $mathcal{J}$ exact sequences are always split or $F(B)$ is at least somehow isomorphic to $F(A) oplus F(C)$ or something similar?
Note: I am aware of the fact that, at least for modules, the set of the $B$'s fitting in the above exact sequence, up to some equivalence and with some sum operation, is isomorphic to $Ext^1(C,A)$, but I think that the term extension gave rise to the notation $Ext$ rather than the converse, so this fact does not provide any useful insight.
algebraic-geometry soft-question algebraic-groups exact-sequence abelian-categories
asked Jan 14 at 15:36
57Jimmy57Jimmy
3,384422
$endgroup$
add a comment |
$begingroup$
Introduction: There are certain categories of group schemes or group varieties in which one can define exact sequences. For instance, the categories of commutative group schemes of finite type over a field $k$ and of commutative affine group schemes over $k$ are abelian (see for instance here).
So let's say that we have a short exact sequence
$$ 0 to A to B to C to 0$$
in such a suitable category (bonus question: examples of abelian categories of group varieties?). There is, a priori, no reason why the sequence should be split. Yet people call such a $B$ (together with the maps) fitting in the above exact sequence an extension of $C$ by $A$. Note that this is true also for other categories and does not have to do with algebraic geometry specifically, but I am most interested in this setting.
My question: What kind of information can one deduce, in general, about $B$, knowing only $A$ and $C$? Does one know something about the underlying topological space of $B$ or of $B(k)$ up to homotopy, or even up to homeomorphism? Is it any better if we consider maps up to isogeny? Feel free to restrict to some specific category where there are better results.
I am asking this because the terminology seems to suggest, and it seems to me that people often assume, that $B$ is some kind of direct sum between $A$ and $C$, even though there is a priori no reason for this to be the case. The isomorphism theorem does imply that $B/A cong C$, but this does not imply that we know how to obtain $B$ knowing $A$ and $C$, since there might be many possibilities for doing so.
Approximate paraphrase 1: A paraphrase of my question is: what do all possible $B$'s that fit in the above sequence have in common, topologically?
Approximate paraphrase 2: Is there some exact functor $F$ from some "geometric" category $mathcal{I}$ to some "topological" category $mathcal{J}$ such that in $mathcal{J}$ exact sequences are always split or $F(B)$ is at least somehow isomorphic to $F(A) oplus F(C)$ or something similar?
Note: I am aware of the fact that, at least for modules, the set of the $B$'s fitting in the above exact sequence, up to some equivalence and with some sum operation, is isomorphic to $Ext^1(C,A)$, but I think that the term extension gave rise to the notation $Ext$ rather than the converse, so this fact does not provide any useful insight.
algebraic-geometry soft-question algebraic-groups exact-sequence abelian-categories
asked Jan 14 at 15:36
57Jimmy57Jimmy
3,384422
$endgroup$
Introduction: There are certain categories of group schemes or group varieties in which one can define exact sequences. For instance, the categories of commutative group schemes of finite type over a field $k$ and of commutative affine group schemes over $k$ are abelian (see for instance here).
So let's say that we have a short exact sequence
$$ 0 to A to B to C to 0$$
in such a suitable category (bonus question: examples of abelian categories of group varieties?). There is, a priori, no reason why the sequence should be split. Yet people call such a $B$ (together with the maps) fitting in the above exact sequence an extension of $C$ by $A$. Note that this is true also for other categories and does not have to do with algebraic geometry specifically, but I am most interested in this setting.
My question: What kind of information can one deduce, in general, about $B$, knowing only $A$ and $C$? Does one know something about the underlying topological space of $B$ or of $B(k)$ up to homotopy, or even up to homeomorphism? Is it any better if we consider maps up to isogeny? Feel free to restrict to some specific category where there are better results.
I am asking this because the terminology seems to suggest, and it seems to me that people often assume, that $B$ is some kind of direct sum between $A$ and $C$, even though there is a priori no reason for this to be the case. The isomorphism theorem does imply that $B/A cong C$, but this does not imply that we know how to obtain $B$ knowing $A$ and $C$, since there might be many possibilities for doing so.
Approximate paraphrase 1: A paraphrase of my question is: what do all possible $B$'s that fit in the above sequence have in common, topologically?
Approximate paraphrase 2: Is there some exact functor $F$ from some "geometric" category $mathcal{I}$ to some "topological" category $mathcal{J}$ such that in $mathcal{J}$ exact sequences are always split or $F(B)$ is at least somehow isomorphic to $F(A) oplus F(C)$ or something similar?
Note: I am aware of the fact that, at least for modules, the set of the $B$'s fitting in the above exact sequence, up to some equivalence and with some sum operation, is isomorphic to $Ext^1(C,A)$, but I think that the term extension gave rise to the notation $Ext$ rather than the converse, so this fact does not provide any useful insight.
algebraic-geometry soft-question algebraic-groups exact-sequence abelian-categories
algebraic-geometry soft-question algebraic-groups exact-sequence abelian-categories
asked Jan 14 at 15:36
57Jimmy57Jimmy
3,384422
asked Jan 14 at 15:36
57Jimmy57Jimmy
3,384422
asked Jan 14 at 15:36
57Jimmy57Jimmy
3,384422
asked Jan 14 at 15:36
57Jimmy57Jimmy
3,384422
asked Jan 14 at 15:36
57Jimmy57Jimmy
3,384422
3,384422
add a comment |
add a comment |
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