Name for abelian category in which every short exact sequence splits Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Does the Grothendieck group depend on the embedding?Is there an additive functor between abelian categories which isn't exact in the middle?Tame abelian tensor categoriesTerminology - subcategories of Abelian categoriesIs there a standard name for a 2-category which has an object z such that, for every object x, the category Hom(x,z) has a terminal object?Rigid monoidal abelian category without an exact tensor functor to VectWhy are the left exact functors from an abelian category to abelian groups cocomplete and have a injective generator?Can one characterize the category of finite-dimensional vector spaces?Picard-Brauer exact sequence for infinity categoriesIs the category of left exact functors abelian?
Name for abelian category in which every short exact sequence splits
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Does the Grothendieck group depend on the embedding?Is there an additive functor between abelian categories which isn't exact in the middle?Tame abelian tensor categoriesTerminology - subcategories of Abelian categoriesIs there a standard name for a 2-category which has an object z such that, for every object x, the category Hom(x,z) has a terminal object?Rigid monoidal abelian category without an exact tensor functor to VectWhy are the left exact functors from an abelian category to abelian groups cocomplete and have a injective generator?Can one characterize the category of finite-dimensional vector spaces?Picard-Brauer exact sequence for infinity categoriesIs the category of left exact functors abelian?
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What is the name of the class of abelian categories defined by the following property: every short exact sequence splits?
ct.category-theory terminology abelian-categories
edited Apr 12 at 15:30
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What is the name of the class of abelian categories defined by the following property: every short exact sequence splits?
ct.category-theory terminology abelian-categories
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I would call it semisimple.
$endgroup$
– Donu Arapura
Apr 12 at 16:25
add a comment |
$begingroup$
What is the name of the class of abelian categories defined by the following property: every short exact sequence splits?
ct.category-theory terminology abelian-categories
edited Apr 12 at 15:30
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What is the name of the class of abelian categories defined by the following property: every short exact sequence splits?
ct.category-theory terminology abelian-categories
ct.category-theory terminology abelian-categories
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13
$begingroup$
I would call it semisimple.
$endgroup$
– Donu Arapura
Apr 12 at 16:25
add a comment |
13
$begingroup$
I would call it semisimple.
$endgroup$
– Donu Arapura
Apr 12 at 16:25
13
13
$begingroup$
I would call it semisimple.
$endgroup$
– Donu Arapura
Apr 12 at 16:25
$begingroup$
I would call it semisimple.
$endgroup$
– Donu Arapura
Apr 12 at 16:25
add a comment |
2 Answers
2
active
oldest
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$begingroup$
The abelian categories in which all short exact sequences split I would call "split abelian categories", reserving the term "semisimple abelian category" for a more restrictive condition. Roughly, perhaps an abelian category should be called "semisimple" if all objects in it are coproducts of simple objects (a nonzero object is called simple if it has no nonzero proper subobjects).
A related question: is every semisimple abelian category split? (Answer: yes, it is; see Jeremy Rickard's comment.) The converse implication is certainly not true.
Grothendieck abelian categories in which all short exact sequences split are known as "spectral categories". The word "spectral" here refers roughly to the spectral theory of operators in Hilbert spaces etc. (in functional analysis), where there is a fundamental opposition between the "discrete" and "continuous" spectrum.
A spectral category in which every object is a coproduct of simple objects is called discrete. A spectral category having no simple objects is called continuous. Every spectral category has a natural, unique decomposition into the Cartesian product of a discrete spectral category and a continuous spectral category.
It is a remarkable and unexpected fact that nondiscrete spectral categories (and in particular, nonzero continuous spectral categories) exist.
Spectral categories $mathcal A$ with a chosen generator $G$ are in bijective correspondence with right self-injective von Neumann regular associative rings $R$. To a spectral category $mathcal A$ with a generator $G$ one simply assigns the ring $R=operatornameHom_mathcal A(G,G)$.
To a right self-injective von Neumann regular ring $R$, one assigns the full subcategory $mathcal A$ in the category of right $R$-modules $Mod-R$ consisting of all the direct summands of products of copies of the (injective) right $R$-module $R$. It may be better to think of $mathcal A$ as a quotient category of $Mod-R$, as in the Gabriel-Popescu theorem; so there is an exact, colimit-preserving localization functor $Mod-Rlongrightarrowmathcal A$ which has a fully faithful right adjoint $mathcal Alongrightarrow Mod-R$. The image of this fully faithful functor is described above.
To produce an example of a nondiscrete spectral category, one has to come up with a "nontrivial enough" example of right self-injective von Neumann regular ring. E.g., complete Boolean algebras with no atoms correspond to some continuous spectral categories.
References:
P. Gabriel, U. Oberst. Spektralkategorien und reguläre Ringe im von-Neumannschen Sinn. Math. Zeitschrift 92, #5, p.389-395, 1966.
B. Stenström. Rings of quotients. An introduction to methods of ring theory.
Springer, 1975. Sections V.6-7 and XII.1-3.
answered Apr 13 at 0:28
Leonid PositselskiLeonid Positselski
11.1k13977
$endgroup$
$begingroup$
OK, good to know. I withdraw my earlier suggestion.
$endgroup$
– Donu Arapura
Apr 13 at 1:14
1
$begingroup$
Regarding “Is every semisimple abelian category split?”: If every object is a coproduct of simples, can you not directly construct a splitting of an epimorphism $f:Xto Y$ by decomposing $Y$ as a coproduct of simples $S$ and then for each $S$ decomposing $f^-1(S)$ as a coproduct of simples?
$endgroup$
– Jeremy Rickard
Apr 13 at 7:59
1
$begingroup$
@JeremyRickard Thank you, yes, you are right. So every semisimple abelian category is split.
$endgroup$
– Leonid Positselski
Apr 13 at 11:21
$begingroup$
A semisimple Grothendieck abelian category is another name for a discrete spectral category. It would be interesting to know whether there exist semisimple abelian categories with coproducts and a generator that are not Grothendieck.
$endgroup$
– Leonid Positselski
Apr 13 at 11:24
add a comment |
$begingroup$
In section III.2.3 in the book "Methods of Homological Algebra" by Gelfand and Manin such an abelian category is indeed called semisimple as Donu Arapura suggested in the comment. One might also call them "abelian categories of global dimension 0", since every object is projective.
answered Apr 13 at 11:43
MareMare
3,55531336
$endgroup$
1
$begingroup$
It seems to me that the proper references are sections III.2.3 and III.5.6, and Exercise IV.1.1 in the book of Gelfand and Manin.
$endgroup$
– Leonid Positselski
Apr 13 at 12:04
2
$begingroup$
Yes, "abelian categories of global/homological dimension 0" is a good terminology for abelian categories in which all short exact sequences split. Existence of projectives or injectives is not needed for that (as one can always define the Ext functor in an abelian category using Yoneda's construction).
$endgroup$
– Leonid Positselski
Apr 13 at 12:07
1
$begingroup$
I think that a good reason for avoiding the term “semisimple” for abelian categories, at least without explanation, is that there are at least three different uses that I have seen: (i) every short exact sequence splits, (ii) every object is a coproduct of simples, or (iii) every object is a finite coproduct of simples. Sometimes we can have clearly correct opinions on what the terminology should be ... but we’re too late.
$endgroup$
– Jeremy Rickard
Apr 13 at 12:15
2
$begingroup$
I almost regret making my original suggestion, but if I were given a choice between "semisimple" and "abelian category with global dimension 0", I'd pick the first. At least it's semisimpler.
$endgroup$
– Donu Arapura
Apr 13 at 12:55
add a comment |
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2 Answers
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$begingroup$
The abelian categories in which all short exact sequences split I would call "split abelian categories", reserving the term "semisimple abelian category" for a more restrictive condition. Roughly, perhaps an abelian category should be called "semisimple" if all objects in it are coproducts of simple objects (a nonzero object is called simple if it has no nonzero proper subobjects).
A related question: is every semisimple abelian category split? (Answer: yes, it is; see Jeremy Rickard's comment.) The converse implication is certainly not true.
Grothendieck abelian categories in which all short exact sequences split are known as "spectral categories". The word "spectral" here refers roughly to the spectral theory of operators in Hilbert spaces etc. (in functional analysis), where there is a fundamental opposition between the "discrete" and "continuous" spectrum.
A spectral category in which every object is a coproduct of simple objects is called discrete. A spectral category having no simple objects is called continuous. Every spectral category has a natural, unique decomposition into the Cartesian product of a discrete spectral category and a continuous spectral category.
It is a remarkable and unexpected fact that nondiscrete spectral categories (and in particular, nonzero continuous spectral categories) exist.
Spectral categories $mathcal A$ with a chosen generator $G$ are in bijective correspondence with right self-injective von Neumann regular associative rings $R$. To a spectral category $mathcal A$ with a generator $G$ one simply assigns the ring $R=operatornameHom_mathcal A(G,G)$.
To a right self-injective von Neumann regular ring $R$, one assigns the full subcategory $mathcal A$ in the category of right $R$-modules $Mod-R$ consisting of all the direct summands of products of copies of the (injective) right $R$-module $R$. It may be better to think of $mathcal A$ as a quotient category of $Mod-R$, as in the Gabriel-Popescu theorem; so there is an exact, colimit-preserving localization functor $Mod-Rlongrightarrowmathcal A$ which has a fully faithful right adjoint $mathcal Alongrightarrow Mod-R$. The image of this fully faithful functor is described above.
To produce an example of a nondiscrete spectral category, one has to come up with a "nontrivial enough" example of right self-injective von Neumann regular ring. E.g., complete Boolean algebras with no atoms correspond to some continuous spectral categories.
References:
P. Gabriel, U. Oberst. Spektralkategorien und reguläre Ringe im von-Neumannschen Sinn. Math. Zeitschrift 92, #5, p.389-395, 1966.
B. Stenström. Rings of quotients. An introduction to methods of ring theory.
Springer, 1975. Sections V.6-7 and XII.1-3.
answered Apr 13 at 0:28
Leonid PositselskiLeonid Positselski
11.1k13977
$endgroup$
$begingroup$
OK, good to know. I withdraw my earlier suggestion.
$endgroup$
– Donu Arapura
Apr 13 at 1:14
1
$begingroup$
Regarding “Is every semisimple abelian category split?”: If every object is a coproduct of simples, can you not directly construct a splitting of an epimorphism $f:Xto Y$ by decomposing $Y$ as a coproduct of simples $S$ and then for each $S$ decomposing $f^-1(S)$ as a coproduct of simples?
$endgroup$
– Jeremy Rickard
Apr 13 at 7:59
1
$begingroup$
@JeremyRickard Thank you, yes, you are right. So every semisimple abelian category is split.
$endgroup$
– Leonid Positselski
Apr 13 at 11:21
$begingroup$
A semisimple Grothendieck abelian category is another name for a discrete spectral category. It would be interesting to know whether there exist semisimple abelian categories with coproducts and a generator that are not Grothendieck.
$endgroup$
– Leonid Positselski
Apr 13 at 11:24
add a comment |
$begingroup$
The abelian categories in which all short exact sequences split I would call "split abelian categories", reserving the term "semisimple abelian category" for a more restrictive condition. Roughly, perhaps an abelian category should be called "semisimple" if all objects in it are coproducts of simple objects (a nonzero object is called simple if it has no nonzero proper subobjects).
A related question: is every semisimple abelian category split? (Answer: yes, it is; see Jeremy Rickard's comment.) The converse implication is certainly not true.
Grothendieck abelian categories in which all short exact sequences split are known as "spectral categories". The word "spectral" here refers roughly to the spectral theory of operators in Hilbert spaces etc. (in functional analysis), where there is a fundamental opposition between the "discrete" and "continuous" spectrum.
A spectral category in which every object is a coproduct of simple objects is called discrete. A spectral category having no simple objects is called continuous. Every spectral category has a natural, unique decomposition into the Cartesian product of a discrete spectral category and a continuous spectral category.
It is a remarkable and unexpected fact that nondiscrete spectral categories (and in particular, nonzero continuous spectral categories) exist.
Spectral categories $mathcal A$ with a chosen generator $G$ are in bijective correspondence with right self-injective von Neumann regular associative rings $R$. To a spectral category $mathcal A$ with a generator $G$ one simply assigns the ring $R=operatornameHom_mathcal A(G,G)$.
To a right self-injective von Neumann regular ring $R$, one assigns the full subcategory $mathcal A$ in the category of right $R$-modules $Mod-R$ consisting of all the direct summands of products of copies of the (injective) right $R$-module $R$. It may be better to think of $mathcal A$ as a quotient category of $Mod-R$, as in the Gabriel-Popescu theorem; so there is an exact, colimit-preserving localization functor $Mod-Rlongrightarrowmathcal A$ which has a fully faithful right adjoint $mathcal Alongrightarrow Mod-R$. The image of this fully faithful functor is described above.
To produce an example of a nondiscrete spectral category, one has to come up with a "nontrivial enough" example of right self-injective von Neumann regular ring. E.g., complete Boolean algebras with no atoms correspond to some continuous spectral categories.
References:
P. Gabriel, U. Oberst. Spektralkategorien und reguläre Ringe im von-Neumannschen Sinn. Math. Zeitschrift 92, #5, p.389-395, 1966.
B. Stenström. Rings of quotients. An introduction to methods of ring theory.
Springer, 1975. Sections V.6-7 and XII.1-3.
answered Apr 13 at 0:28
Leonid PositselskiLeonid Positselski
11.1k13977
$endgroup$
$begingroup$
OK, good to know. I withdraw my earlier suggestion.
$endgroup$
– Donu Arapura
Apr 13 at 1:14
1
$begingroup$
Regarding “Is every semisimple abelian category split?”: If every object is a coproduct of simples, can you not directly construct a splitting of an epimorphism $f:Xto Y$ by decomposing $Y$ as a coproduct of simples $S$ and then for each $S$ decomposing $f^-1(S)$ as a coproduct of simples?
$endgroup$
– Jeremy Rickard
Apr 13 at 7:59
1
$begingroup$
@JeremyRickard Thank you, yes, you are right. So every semisimple abelian category is split.
$endgroup$
– Leonid Positselski
Apr 13 at 11:21
$begingroup$
A semisimple Grothendieck abelian category is another name for a discrete spectral category. It would be interesting to know whether there exist semisimple abelian categories with coproducts and a generator that are not Grothendieck.
$endgroup$
– Leonid Positselski
Apr 13 at 11:24
add a comment |
$begingroup$
The abelian categories in which all short exact sequences split I would call "split abelian categories", reserving the term "semisimple abelian category" for a more restrictive condition. Roughly, perhaps an abelian category should be called "semisimple" if all objects in it are coproducts of simple objects (a nonzero object is called simple if it has no nonzero proper subobjects).
A related question: is every semisimple abelian category split? (Answer: yes, it is; see Jeremy Rickard's comment.) The converse implication is certainly not true.
Grothendieck abelian categories in which all short exact sequences split are known as "spectral categories". The word "spectral" here refers roughly to the spectral theory of operators in Hilbert spaces etc. (in functional analysis), where there is a fundamental opposition between the "discrete" and "continuous" spectrum.
A spectral category in which every object is a coproduct of simple objects is called discrete. A spectral category having no simple objects is called continuous. Every spectral category has a natural, unique decomposition into the Cartesian product of a discrete spectral category and a continuous spectral category.
It is a remarkable and unexpected fact that nondiscrete spectral categories (and in particular, nonzero continuous spectral categories) exist.
Spectral categories $mathcal A$ with a chosen generator $G$ are in bijective correspondence with right self-injective von Neumann regular associative rings $R$. To a spectral category $mathcal A$ with a generator $G$ one simply assigns the ring $R=operatornameHom_mathcal A(G,G)$.
To a right self-injective von Neumann regular ring $R$, one assigns the full subcategory $mathcal A$ in the category of right $R$-modules $Mod-R$ consisting of all the direct summands of products of copies of the (injective) right $R$-module $R$. It may be better to think of $mathcal A$ as a quotient category of $Mod-R$, as in the Gabriel-Popescu theorem; so there is an exact, colimit-preserving localization functor $Mod-Rlongrightarrowmathcal A$ which has a fully faithful right adjoint $mathcal Alongrightarrow Mod-R$. The image of this fully faithful functor is described above.
To produce an example of a nondiscrete spectral category, one has to come up with a "nontrivial enough" example of right self-injective von Neumann regular ring. E.g., complete Boolean algebras with no atoms correspond to some continuous spectral categories.
References:
P. Gabriel, U. Oberst. Spektralkategorien und reguläre Ringe im von-Neumannschen Sinn. Math. Zeitschrift 92, #5, p.389-395, 1966.
B. Stenström. Rings of quotients. An introduction to methods of ring theory.
Springer, 1975. Sections V.6-7 and XII.1-3.
answered Apr 13 at 0:28
Leonid PositselskiLeonid Positselski
11.1k13977
$endgroup$
The abelian categories in which all short exact sequences split I would call "split abelian categories", reserving the term "semisimple abelian category" for a more restrictive condition. Roughly, perhaps an abelian category should be called "semisimple" if all objects in it are coproducts of simple objects (a nonzero object is called simple if it has no nonzero proper subobjects).
A related question: is every semisimple abelian category split? (Answer: yes, it is; see Jeremy Rickard's comment.) The converse implication is certainly not true.
Grothendieck abelian categories in which all short exact sequences split are known as "spectral categories". The word "spectral" here refers roughly to the spectral theory of operators in Hilbert spaces etc. (in functional analysis), where there is a fundamental opposition between the "discrete" and "continuous" spectrum.
A spectral category in which every object is a coproduct of simple objects is called discrete. A spectral category having no simple objects is called continuous. Every spectral category has a natural, unique decomposition into the Cartesian product of a discrete spectral category and a continuous spectral category.
It is a remarkable and unexpected fact that nondiscrete spectral categories (and in particular, nonzero continuous spectral categories) exist.
Spectral categories $mathcal A$ with a chosen generator $G$ are in bijective correspondence with right self-injective von Neumann regular associative rings $R$. To a spectral category $mathcal A$ with a generator $G$ one simply assigns the ring $R=operatornameHom_mathcal A(G,G)$.
To a right self-injective von Neumann regular ring $R$, one assigns the full subcategory $mathcal A$ in the category of right $R$-modules $Mod-R$ consisting of all the direct summands of products of copies of the (injective) right $R$-module $R$. It may be better to think of $mathcal A$ as a quotient category of $Mod-R$, as in the Gabriel-Popescu theorem; so there is an exact, colimit-preserving localization functor $Mod-Rlongrightarrowmathcal A$ which has a fully faithful right adjoint $mathcal Alongrightarrow Mod-R$. The image of this fully faithful functor is described above.
To produce an example of a nondiscrete spectral category, one has to come up with a "nontrivial enough" example of right self-injective von Neumann regular ring. E.g., complete Boolean algebras with no atoms correspond to some continuous spectral categories.
References:
P. Gabriel, U. Oberst. Spektralkategorien und reguläre Ringe im von-Neumannschen Sinn. Math. Zeitschrift 92, #5, p.389-395, 1966.
B. Stenström. Rings of quotients. An introduction to methods of ring theory.
Springer, 1975. Sections V.6-7 and XII.1-3.
answered Apr 13 at 0:28
Leonid PositselskiLeonid Positselski
11.1k13977
edited Apr 13 at 11:31
answered Apr 13 at 0:28
Leonid PositselskiLeonid Positselski
11.1k13977
answered Apr 13 at 0:28
Leonid PositselskiLeonid Positselski
11.1k13977
answered Apr 13 at 0:28
Leonid PositselskiLeonid Positselski
11.1k13977
11.1k13977
$begingroup$
OK, good to know. I withdraw my earlier suggestion.
$endgroup$
– Donu Arapura
Apr 13 at 1:14
1
$begingroup$
Regarding “Is every semisimple abelian category split?”: If every object is a coproduct of simples, can you not directly construct a splitting of an epimorphism $f:Xto Y$ by decomposing $Y$ as a coproduct of simples $S$ and then for each $S$ decomposing $f^-1(S)$ as a coproduct of simples?
$endgroup$
– Jeremy Rickard
Apr 13 at 7:59
1
$begingroup$
@JeremyRickard Thank you, yes, you are right. So every semisimple abelian category is split.
$endgroup$
– Leonid Positselski
Apr 13 at 11:21
$begingroup$
A semisimple Grothendieck abelian category is another name for a discrete spectral category. It would be interesting to know whether there exist semisimple abelian categories with coproducts and a generator that are not Grothendieck.
$endgroup$
– Leonid Positselski
Apr 13 at 11:24
add a comment |
$begingroup$
OK, good to know. I withdraw my earlier suggestion.
$endgroup$
– Donu Arapura
Apr 13 at 1:14
1
$begingroup$
Regarding “Is every semisimple abelian category split?”: If every object is a coproduct of simples, can you not directly construct a splitting of an epimorphism $f:Xto Y$ by decomposing $Y$ as a coproduct of simples $S$ and then for each $S$ decomposing $f^-1(S)$ as a coproduct of simples?
$endgroup$
– Jeremy Rickard
Apr 13 at 7:59
1
$begingroup$
@JeremyRickard Thank you, yes, you are right. So every semisimple abelian category is split.
$endgroup$
– Leonid Positselski
Apr 13 at 11:21
$begingroup$
A semisimple Grothendieck abelian category is another name for a discrete spectral category. It would be interesting to know whether there exist semisimple abelian categories with coproducts and a generator that are not Grothendieck.
$endgroup$
– Leonid Positselski
Apr 13 at 11:24
$begingroup$
OK, good to know. I withdraw my earlier suggestion.
$endgroup$
– Donu Arapura
Apr 13 at 1:14
$begingroup$
OK, good to know. I withdraw my earlier suggestion.
$endgroup$
– Donu Arapura
Apr 13 at 1:14
1
1
$begingroup$
Regarding “Is every semisimple abelian category split?”: If every object is a coproduct of simples, can you not directly construct a splitting of an epimorphism $f:Xto Y$ by decomposing $Y$ as a coproduct of simples $S$ and then for each $S$ decomposing $f^-1(S)$ as a coproduct of simples?
$endgroup$
– Jeremy Rickard
Apr 13 at 7:59
$begingroup$
Regarding “Is every semisimple abelian category split?”: If every object is a coproduct of simples, can you not directly construct a splitting of an epimorphism $f:Xto Y$ by decomposing $Y$ as a coproduct of simples $S$ and then for each $S$ decomposing $f^-1(S)$ as a coproduct of simples?
$endgroup$
– Jeremy Rickard
Apr 13 at 7:59
1
1
$begingroup$
@JeremyRickard Thank you, yes, you are right. So every semisimple abelian category is split.
$endgroup$
– Leonid Positselski
Apr 13 at 11:21
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@JeremyRickard Thank you, yes, you are right. So every semisimple abelian category is split.
$endgroup$
– Leonid Positselski
Apr 13 at 11:21
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A semisimple Grothendieck abelian category is another name for a discrete spectral category. It would be interesting to know whether there exist semisimple abelian categories with coproducts and a generator that are not Grothendieck.
$endgroup$
– Leonid Positselski
Apr 13 at 11:24
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A semisimple Grothendieck abelian category is another name for a discrete spectral category. It would be interesting to know whether there exist semisimple abelian categories with coproducts and a generator that are not Grothendieck.
$endgroup$
– Leonid Positselski
Apr 13 at 11:24
add a comment |
$begingroup$
In section III.2.3 in the book "Methods of Homological Algebra" by Gelfand and Manin such an abelian category is indeed called semisimple as Donu Arapura suggested in the comment. One might also call them "abelian categories of global dimension 0", since every object is projective.
answered Apr 13 at 11:43
MareMare
3,55531336
$endgroup$
1
$begingroup$
It seems to me that the proper references are sections III.2.3 and III.5.6, and Exercise IV.1.1 in the book of Gelfand and Manin.
$endgroup$
– Leonid Positselski
Apr 13 at 12:04
2
$begingroup$
Yes, "abelian categories of global/homological dimension 0" is a good terminology for abelian categories in which all short exact sequences split. Existence of projectives or injectives is not needed for that (as one can always define the Ext functor in an abelian category using Yoneda's construction).
$endgroup$
– Leonid Positselski
Apr 13 at 12:07
1
$begingroup$
I think that a good reason for avoiding the term “semisimple” for abelian categories, at least without explanation, is that there are at least three different uses that I have seen: (i) every short exact sequence splits, (ii) every object is a coproduct of simples, or (iii) every object is a finite coproduct of simples. Sometimes we can have clearly correct opinions on what the terminology should be ... but we’re too late.
$endgroup$
– Jeremy Rickard
Apr 13 at 12:15
2
$begingroup$
I almost regret making my original suggestion, but if I were given a choice between "semisimple" and "abelian category with global dimension 0", I'd pick the first. At least it's semisimpler.
$endgroup$
– Donu Arapura
Apr 13 at 12:55
add a comment |
$begingroup$
In section III.2.3 in the book "Methods of Homological Algebra" by Gelfand and Manin such an abelian category is indeed called semisimple as Donu Arapura suggested in the comment. One might also call them "abelian categories of global dimension 0", since every object is projective.
answered Apr 13 at 11:43
MareMare
3,55531336
$endgroup$
1
$begingroup$
It seems to me that the proper references are sections III.2.3 and III.5.6, and Exercise IV.1.1 in the book of Gelfand and Manin.
$endgroup$
– Leonid Positselski
Apr 13 at 12:04
2
$begingroup$
Yes, "abelian categories of global/homological dimension 0" is a good terminology for abelian categories in which all short exact sequences split. Existence of projectives or injectives is not needed for that (as one can always define the Ext functor in an abelian category using Yoneda's construction).
$endgroup$
– Leonid Positselski
Apr 13 at 12:07
1
$begingroup$
I think that a good reason for avoiding the term “semisimple” for abelian categories, at least without explanation, is that there are at least three different uses that I have seen: (i) every short exact sequence splits, (ii) every object is a coproduct of simples, or (iii) every object is a finite coproduct of simples. Sometimes we can have clearly correct opinions on what the terminology should be ... but we’re too late.
$endgroup$
– Jeremy Rickard
Apr 13 at 12:15
2
$begingroup$
I almost regret making my original suggestion, but if I were given a choice between "semisimple" and "abelian category with global dimension 0", I'd pick the first. At least it's semisimpler.
$endgroup$
– Donu Arapura
Apr 13 at 12:55
add a comment |
$begingroup$
In section III.2.3 in the book "Methods of Homological Algebra" by Gelfand and Manin such an abelian category is indeed called semisimple as Donu Arapura suggested in the comment. One might also call them "abelian categories of global dimension 0", since every object is projective.
answered Apr 13 at 11:43
MareMare
3,55531336
$endgroup$
In section III.2.3 in the book "Methods of Homological Algebra" by Gelfand and Manin such an abelian category is indeed called semisimple as Donu Arapura suggested in the comment. One might also call them "abelian categories of global dimension 0", since every object is projective.
answered Apr 13 at 11:43
MareMare
3,55531336
edited Apr 13 at 12:56
answered Apr 13 at 11:43
MareMare
3,55531336
answered Apr 13 at 11:43
MareMare
3,55531336
answered Apr 13 at 11:43
MareMare
3,55531336
3,55531336
1
$begingroup$
It seems to me that the proper references are sections III.2.3 and III.5.6, and Exercise IV.1.1 in the book of Gelfand and Manin.
$endgroup$
– Leonid Positselski
Apr 13 at 12:04
2
$begingroup$
Yes, "abelian categories of global/homological dimension 0" is a good terminology for abelian categories in which all short exact sequences split. Existence of projectives or injectives is not needed for that (as one can always define the Ext functor in an abelian category using Yoneda's construction).
$endgroup$
– Leonid Positselski
Apr 13 at 12:07
1
$begingroup$
I think that a good reason for avoiding the term “semisimple” for abelian categories, at least without explanation, is that there are at least three different uses that I have seen: (i) every short exact sequence splits, (ii) every object is a coproduct of simples, or (iii) every object is a finite coproduct of simples. Sometimes we can have clearly correct opinions on what the terminology should be ... but we’re too late.
$endgroup$
– Jeremy Rickard
Apr 13 at 12:15
2
$begingroup$
I almost regret making my original suggestion, but if I were given a choice between "semisimple" and "abelian category with global dimension 0", I'd pick the first. At least it's semisimpler.
$endgroup$
– Donu Arapura
Apr 13 at 12:55
add a comment |
1
$begingroup$
It seems to me that the proper references are sections III.2.3 and III.5.6, and Exercise IV.1.1 in the book of Gelfand and Manin.
$endgroup$
– Leonid Positselski
Apr 13 at 12:04
2
$begingroup$
Yes, "abelian categories of global/homological dimension 0" is a good terminology for abelian categories in which all short exact sequences split. Existence of projectives or injectives is not needed for that (as one can always define the Ext functor in an abelian category using Yoneda's construction).
$endgroup$
– Leonid Positselski
Apr 13 at 12:07
1
$begingroup$
I think that a good reason for avoiding the term “semisimple” for abelian categories, at least without explanation, is that there are at least three different uses that I have seen: (i) every short exact sequence splits, (ii) every object is a coproduct of simples, or (iii) every object is a finite coproduct of simples. Sometimes we can have clearly correct opinions on what the terminology should be ... but we’re too late.
$endgroup$
– Jeremy Rickard
Apr 13 at 12:15
2
$begingroup$
I almost regret making my original suggestion, but if I were given a choice between "semisimple" and "abelian category with global dimension 0", I'd pick the first. At least it's semisimpler.
$endgroup$
– Donu Arapura
Apr 13 at 12:55
1
1
$begingroup$
It seems to me that the proper references are sections III.2.3 and III.5.6, and Exercise IV.1.1 in the book of Gelfand and Manin.
$endgroup$
– Leonid Positselski
Apr 13 at 12:04
$begingroup$
It seems to me that the proper references are sections III.2.3 and III.5.6, and Exercise IV.1.1 in the book of Gelfand and Manin.
$endgroup$
– Leonid Positselski
Apr 13 at 12:04
2
2
$begingroup$
Yes, "abelian categories of global/homological dimension 0" is a good terminology for abelian categories in which all short exact sequences split. Existence of projectives or injectives is not needed for that (as one can always define the Ext functor in an abelian category using Yoneda's construction).
$endgroup$
– Leonid Positselski
Apr 13 at 12:07
$begingroup$
Yes, "abelian categories of global/homological dimension 0" is a good terminology for abelian categories in which all short exact sequences split. Existence of projectives or injectives is not needed for that (as one can always define the Ext functor in an abelian category using Yoneda's construction).
$endgroup$
– Leonid Positselski
Apr 13 at 12:07
1
1
$begingroup$
I think that a good reason for avoiding the term “semisimple” for abelian categories, at least without explanation, is that there are at least three different uses that I have seen: (i) every short exact sequence splits, (ii) every object is a coproduct of simples, or (iii) every object is a finite coproduct of simples. Sometimes we can have clearly correct opinions on what the terminology should be ... but we’re too late.
$endgroup$
– Jeremy Rickard
Apr 13 at 12:15
$begingroup$
I think that a good reason for avoiding the term “semisimple” for abelian categories, at least without explanation, is that there are at least three different uses that I have seen: (i) every short exact sequence splits, (ii) every object is a coproduct of simples, or (iii) every object is a finite coproduct of simples. Sometimes we can have clearly correct opinions on what the terminology should be ... but we’re too late.
$endgroup$
– Jeremy Rickard
Apr 13 at 12:15
2
2
$begingroup$
I almost regret making my original suggestion, but if I were given a choice between "semisimple" and "abelian category with global dimension 0", I'd pick the first. At least it's semisimpler.
$endgroup$
– Donu Arapura
Apr 13 at 12:55
$begingroup$
I almost regret making my original suggestion, but if I were given a choice between "semisimple" and "abelian category with global dimension 0", I'd pick the first. At least it's semisimpler.
$endgroup$
– Donu Arapura
Apr 13 at 12:55
add a comment |
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I would call it semisimple.
$endgroup$
– Donu Arapura
Apr 12 at 16:25