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Equivariant homotopy theory, topos theory and intuitionistic algebraic topology
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Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Is there a category whose internal logic is paraconsistent?Book for Algebraic Topology- Spanier vs Tom DieckModuli spaces, stacks and homotopy theoryApplications of algebraic topology?The Classifying Topos of Local Rings and Algebraic GeometryReference for "topos obtained by adjoining an indeterminate set' theoremAlgebraic topology for non-nice spacesLogic for physicsHaving problem with tom Dieck's algebraic topology textResearch in the intersection of mathematical logic and algebraic topology
$begingroup$
This might be a very naive question, but I don't really see what would go wrong, so I'm wondering if this has already been done.
The idea is the following : equivariant homotpy theory as far as I can understand (though my knowledge is very restricted in that topic) is about homotopy theory in $G-mathbf{Set}$ for some group $G$ (that we may want to vary); which happens to be a topos, and hence has an internal logic (which is intuitionistic).
Therefore whatever theorem I can prove intuitionistically is valid in $G-mathbf{Set}$ and has an external interpretation that we can find in terms of $G$-sets : here I'm not inventing anything, for instance this idea has already been applied to toposes/topoi of sheaves to get information about algebraic geometry by proving intuitionistic theorems of commutative algebra (see for instance Ingo Blechschmidt's work).
But now it's reasonable to think that "intuitionistic algebraic topology interpreted in $G-mathbf{Set}$" has some relation with equivariant algebraic topology.
Hence my question is the following : does it ? Is there anything interesting there ? Can we study equivariant homotopy theory by studying intuitionistic algebraic topology and interpreting it in $G-mathbf{Set}$ ?
This is a soft question, in the sense that I'm not necessarily looking for technical answers, unless it's necessary (for instance "the answer is no but for technical reasons"); I would gladly hear motivated answers/see references, surveys about this if they exist, or analogies with other similar ideas that failed/succeeded and why it indicates failure/success here, etc.etc.
EDIT : as noted in the comments, this actually has very little to do with intuitionistic logic, as the topos $mathbf{Set}^G$ is boolean. However, it does not satisfy choice and is not extensional (when $Gneq 1$), so its internal set theory is still different from $mathbf{Set}$. With that in mind, all of the above uses of the word "intuitionist" should be interpreted as meaning "nonclassical set theory"
algebraic-topology homotopy-theory topos-theory intuitionistic-logic equivariant-topology
asked Mar 25 at 10:06
MaxMax
16.5k11144
$endgroup$
add a comment |
$begingroup$
This might be a very naive question, but I don't really see what would go wrong, so I'm wondering if this has already been done.
The idea is the following : equivariant homotpy theory as far as I can understand (though my knowledge is very restricted in that topic) is about homotopy theory in $G-mathbf{Set}$ for some group $G$ (that we may want to vary); which happens to be a topos, and hence has an internal logic (which is intuitionistic).
Therefore whatever theorem I can prove intuitionistically is valid in $G-mathbf{Set}$ and has an external interpretation that we can find in terms of $G$-sets : here I'm not inventing anything, for instance this idea has already been applied to toposes/topoi of sheaves to get information about algebraic geometry by proving intuitionistic theorems of commutative algebra (see for instance Ingo Blechschmidt's work).
But now it's reasonable to think that "intuitionistic algebraic topology interpreted in $G-mathbf{Set}$" has some relation with equivariant algebraic topology.
Hence my question is the following : does it ? Is there anything interesting there ? Can we study equivariant homotopy theory by studying intuitionistic algebraic topology and interpreting it in $G-mathbf{Set}$ ?
This is a soft question, in the sense that I'm not necessarily looking for technical answers, unless it's necessary (for instance "the answer is no but for technical reasons"); I would gladly hear motivated answers/see references, surveys about this if they exist, or analogies with other similar ideas that failed/succeeded and why it indicates failure/success here, etc.etc.
EDIT : as noted in the comments, this actually has very little to do with intuitionistic logic, as the topos $mathbf{Set}^G$ is boolean. However, it does not satisfy choice and is not extensional (when $Gneq 1$), so its internal set theory is still different from $mathbf{Set}$. With that in mind, all of the above uses of the word "intuitionist" should be interpreted as meaning "nonclassical set theory"
algebraic-topology homotopy-theory topos-theory intuitionistic-logic equivariant-topology
asked Mar 25 at 10:06
MaxMax
16.5k11144
$endgroup$
1
$begingroup$
I think it's worth noting that the internal logic of a category of $G$-sets is classical, so there will be nothing novel about said logic.
$endgroup$
– Malice Vidrine
Mar 25 at 12:38
$begingroup$
@MaliceVidrine : oh right, $mathbf{Set}^G$ is boolean, I had forgotten that ! But it definitely doesn't satisfy choice unless $G$ is trivial, and it's not extensional; so it's not the logic that would change but the "set theory", right ?
$endgroup$
– Max
Mar 25 at 12:45
$begingroup$
Correct, its "higher order" features are definitely unlike $mathbf{Set}$, in just the ways you mention.
$endgroup$
– Malice Vidrine
Mar 25 at 12:54
2
$begingroup$
Although the topos of $G$-sets doesn't satisfy the axiom of choice (i.e., epimorphisms need not split), it does satisfy the internal axiom of choice. That is, sentences of the form $(forall xin A)(exists yin B),R(x,y)to(exists fin B^A)(forall xin A),R(x,f(x))$ are internally valid. (And this is enough to imply Boolean logic.)
$endgroup$
– Andreas Blass
Mar 25 at 14:47
$begingroup$
@AndreasBlass that is good to know, altough I knew that internal and external AC weren't equivalent I would've bet that in this case both would fail
$endgroup$
– Max
Mar 25 at 15:16
add a comment |
$begingroup$
This might be a very naive question, but I don't really see what would go wrong, so I'm wondering if this has already been done.
The idea is the following : equivariant homotpy theory as far as I can understand (though my knowledge is very restricted in that topic) is about homotopy theory in $G-mathbf{Set}$ for some group $G$ (that we may want to vary); which happens to be a topos, and hence has an internal logic (which is intuitionistic).
Therefore whatever theorem I can prove intuitionistically is valid in $G-mathbf{Set}$ and has an external interpretation that we can find in terms of $G$-sets : here I'm not inventing anything, for instance this idea has already been applied to toposes/topoi of sheaves to get information about algebraic geometry by proving intuitionistic theorems of commutative algebra (see for instance Ingo Blechschmidt's work).
But now it's reasonable to think that "intuitionistic algebraic topology interpreted in $G-mathbf{Set}$" has some relation with equivariant algebraic topology.
Hence my question is the following : does it ? Is there anything interesting there ? Can we study equivariant homotopy theory by studying intuitionistic algebraic topology and interpreting it in $G-mathbf{Set}$ ?
This is a soft question, in the sense that I'm not necessarily looking for technical answers, unless it's necessary (for instance "the answer is no but for technical reasons"); I would gladly hear motivated answers/see references, surveys about this if they exist, or analogies with other similar ideas that failed/succeeded and why it indicates failure/success here, etc.etc.
EDIT : as noted in the comments, this actually has very little to do with intuitionistic logic, as the topos $mathbf{Set}^G$ is boolean. However, it does not satisfy choice and is not extensional (when $Gneq 1$), so its internal set theory is still different from $mathbf{Set}$. With that in mind, all of the above uses of the word "intuitionist" should be interpreted as meaning "nonclassical set theory"
algebraic-topology homotopy-theory topos-theory intuitionistic-logic equivariant-topology
asked Mar 25 at 10:06
MaxMax
16.5k11144
$endgroup$
This might be a very naive question, but I don't really see what would go wrong, so I'm wondering if this has already been done.
The idea is the following : equivariant homotpy theory as far as I can understand (though my knowledge is very restricted in that topic) is about homotopy theory in $G-mathbf{Set}$ for some group $G$ (that we may want to vary); which happens to be a topos, and hence has an internal logic (which is intuitionistic).
Therefore whatever theorem I can prove intuitionistically is valid in $G-mathbf{Set}$ and has an external interpretation that we can find in terms of $G$-sets : here I'm not inventing anything, for instance this idea has already been applied to toposes/topoi of sheaves to get information about algebraic geometry by proving intuitionistic theorems of commutative algebra (see for instance Ingo Blechschmidt's work).
But now it's reasonable to think that "intuitionistic algebraic topology interpreted in $G-mathbf{Set}$" has some relation with equivariant algebraic topology.
Hence my question is the following : does it ? Is there anything interesting there ? Can we study equivariant homotopy theory by studying intuitionistic algebraic topology and interpreting it in $G-mathbf{Set}$ ?
This is a soft question, in the sense that I'm not necessarily looking for technical answers, unless it's necessary (for instance "the answer is no but for technical reasons"); I would gladly hear motivated answers/see references, surveys about this if they exist, or analogies with other similar ideas that failed/succeeded and why it indicates failure/success here, etc.etc.
EDIT : as noted in the comments, this actually has very little to do with intuitionistic logic, as the topos $mathbf{Set}^G$ is boolean. However, it does not satisfy choice and is not extensional (when $Gneq 1$), so its internal set theory is still different from $mathbf{Set}$. With that in mind, all of the above uses of the word "intuitionist" should be interpreted as meaning "nonclassical set theory"
algebraic-topology homotopy-theory topos-theory intuitionistic-logic equivariant-topology
algebraic-topology homotopy-theory topos-theory intuitionistic-logic equivariant-topology
asked Mar 25 at 10:06
MaxMax
16.5k11144
asked Mar 25 at 10:06
MaxMax
16.5k11144
edited Mar 25 at 12:57
Max
asked Mar 25 at 10:06
MaxMax
16.5k11144
asked Mar 25 at 10:06
MaxMax
16.5k11144
asked Mar 25 at 10:06
MaxMax
16.5k11144
16.5k11144
1
$begingroup$
I think it's worth noting that the internal logic of a category of $G$-sets is classical, so there will be nothing novel about said logic.
$endgroup$
– Malice Vidrine
Mar 25 at 12:38
$begingroup$
@MaliceVidrine : oh right, $mathbf{Set}^G$ is boolean, I had forgotten that ! But it definitely doesn't satisfy choice unless $G$ is trivial, and it's not extensional; so it's not the logic that would change but the "set theory", right ?
$endgroup$
– Max
Mar 25 at 12:45
$begingroup$
Correct, its "higher order" features are definitely unlike $mathbf{Set}$, in just the ways you mention.
$endgroup$
– Malice Vidrine
Mar 25 at 12:54
2
$begingroup$
Although the topos of $G$-sets doesn't satisfy the axiom of choice (i.e., epimorphisms need not split), it does satisfy the internal axiom of choice. That is, sentences of the form $(forall xin A)(exists yin B),R(x,y)to(exists fin B^A)(forall xin A),R(x,f(x))$ are internally valid. (And this is enough to imply Boolean logic.)
$endgroup$
– Andreas Blass
Mar 25 at 14:47
$begingroup$
@AndreasBlass that is good to know, altough I knew that internal and external AC weren't equivalent I would've bet that in this case both would fail
$endgroup$
– Max
Mar 25 at 15:16
add a comment |
1
$begingroup$
I think it's worth noting that the internal logic of a category of $G$-sets is classical, so there will be nothing novel about said logic.
$endgroup$
– Malice Vidrine
Mar 25 at 12:38
$begingroup$
@MaliceVidrine : oh right, $mathbf{Set}^G$ is boolean, I had forgotten that ! But it definitely doesn't satisfy choice unless $G$ is trivial, and it's not extensional; so it's not the logic that would change but the "set theory", right ?
$endgroup$
– Max
Mar 25 at 12:45
$begingroup$
Correct, its "higher order" features are definitely unlike $mathbf{Set}$, in just the ways you mention.
$endgroup$
– Malice Vidrine
Mar 25 at 12:54
2
$begingroup$
Although the topos of $G$-sets doesn't satisfy the axiom of choice (i.e., epimorphisms need not split), it does satisfy the internal axiom of choice. That is, sentences of the form $(forall xin A)(exists yin B),R(x,y)to(exists fin B^A)(forall xin A),R(x,f(x))$ are internally valid. (And this is enough to imply Boolean logic.)
$endgroup$
– Andreas Blass
Mar 25 at 14:47
$begingroup$
@AndreasBlass that is good to know, altough I knew that internal and external AC weren't equivalent I would've bet that in this case both would fail
$endgroup$
– Max
Mar 25 at 15:16
1
1
$begingroup$
I think it's worth noting that the internal logic of a category of $G$-sets is classical, so there will be nothing novel about said logic.
$endgroup$
– Malice Vidrine
Mar 25 at 12:38
$begingroup$
I think it's worth noting that the internal logic of a category of $G$-sets is classical, so there will be nothing novel about said logic.
$endgroup$
– Malice Vidrine
Mar 25 at 12:38
$begingroup$
@MaliceVidrine : oh right, $mathbf{Set}^G$ is boolean, I had forgotten that ! But it definitely doesn't satisfy choice unless $G$ is trivial, and it's not extensional; so it's not the logic that would change but the "set theory", right ?
$endgroup$
– Max
Mar 25 at 12:45
$begingroup$
@MaliceVidrine : oh right, $mathbf{Set}^G$ is boolean, I had forgotten that ! But it definitely doesn't satisfy choice unless $G$ is trivial, and it's not extensional; so it's not the logic that would change but the "set theory", right ?
$endgroup$
– Max
Mar 25 at 12:45
$begingroup$
Correct, its "higher order" features are definitely unlike $mathbf{Set}$, in just the ways you mention.
$endgroup$
– Malice Vidrine
Mar 25 at 12:54
$begingroup$
Correct, its "higher order" features are definitely unlike $mathbf{Set}$, in just the ways you mention.
$endgroup$
– Malice Vidrine
Mar 25 at 12:54
2
2
$begingroup$
Although the topos of $G$-sets doesn't satisfy the axiom of choice (i.e., epimorphisms need not split), it does satisfy the internal axiom of choice. That is, sentences of the form $(forall xin A)(exists yin B),R(x,y)to(exists fin B^A)(forall xin A),R(x,f(x))$ are internally valid. (And this is enough to imply Boolean logic.)
$endgroup$
– Andreas Blass
Mar 25 at 14:47
$begingroup$
Although the topos of $G$-sets doesn't satisfy the axiom of choice (i.e., epimorphisms need not split), it does satisfy the internal axiom of choice. That is, sentences of the form $(forall xin A)(exists yin B),R(x,y)to(exists fin B^A)(forall xin A),R(x,f(x))$ are internally valid. (And this is enough to imply Boolean logic.)
$endgroup$
– Andreas Blass
Mar 25 at 14:47
$begingroup$
@AndreasBlass that is good to know, altough I knew that internal and external AC weren't equivalent I would've bet that in this case both would fail
$endgroup$
– Max
Mar 25 at 15:16
$begingroup$
@AndreasBlass that is good to know, altough I knew that internal and external AC weren't equivalent I would've bet that in this case both would fail
$endgroup$
– Max
Mar 25 at 15:16
add a comment |
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1
$begingroup$
I think it's worth noting that the internal logic of a category of $G$-sets is classical, so there will be nothing novel about said logic.
$endgroup$
– Malice Vidrine
Mar 25 at 12:38
$begingroup$
@MaliceVidrine : oh right, $mathbf{Set}^G$ is boolean, I had forgotten that ! But it definitely doesn't satisfy choice unless $G$ is trivial, and it's not extensional; so it's not the logic that would change but the "set theory", right ?
$endgroup$
– Max
Mar 25 at 12:45
$begingroup$
Correct, its "higher order" features are definitely unlike $mathbf{Set}$, in just the ways you mention.
$endgroup$
– Malice Vidrine
Mar 25 at 12:54
2
$begingroup$
Although the topos of $G$-sets doesn't satisfy the axiom of choice (i.e., epimorphisms need not split), it does satisfy the internal axiom of choice. That is, sentences of the form $(forall xin A)(exists yin B),R(x,y)to(exists fin B^A)(forall xin A),R(x,f(x))$ are internally valid. (And this is enough to imply Boolean logic.)
$endgroup$
– Andreas Blass
Mar 25 at 14:47
$begingroup$
@AndreasBlass that is good to know, altough I knew that internal and external AC weren't equivalent I would've bet that in this case both would fail
$endgroup$
– Max
Mar 25 at 15:16