Motives and representations
$begingroup$
Assuming standard conjectures, the category of motives is equivalent to the category of representations of a certain group over $mathbb Q$, but I don't understand the abstract construction.
Now, if we start with a nonsingular projective variety $X$ we get a motive $(X, Delta_X, 0)$, so we should also get a representation $rho$. I think it would help me to understand concretely what representation arises this way.
As far as I understood, the space of the representation should be $bigoplus_{i=0}^{2 dim(X)} H^i(X)$, and it should probably factor through some simpler group. I would be grateful to anyone explainig me how to explicitely build/ describe $rho$ from $X$.
algebraic-geometry category-theory representation-theory
$endgroup$
add a comment |
$begingroup$
Assuming standard conjectures, the category of motives is equivalent to the category of representations of a certain group over $mathbb Q$, but I don't understand the abstract construction.
Now, if we start with a nonsingular projective variety $X$ we get a motive $(X, Delta_X, 0)$, so we should also get a representation $rho$. I think it would help me to understand concretely what representation arises this way.
As far as I understood, the space of the representation should be $bigoplus_{i=0}^{2 dim(X)} H^i(X)$, and it should probably factor through some simpler group. I would be grateful to anyone explainig me how to explicitely build/ describe $rho$ from $X$.
algebraic-geometry category-theory representation-theory
$endgroup$
add a comment |
$begingroup$
Assuming standard conjectures, the category of motives is equivalent to the category of representations of a certain group over $mathbb Q$, but I don't understand the abstract construction.
Now, if we start with a nonsingular projective variety $X$ we get a motive $(X, Delta_X, 0)$, so we should also get a representation $rho$. I think it would help me to understand concretely what representation arises this way.
As far as I understood, the space of the representation should be $bigoplus_{i=0}^{2 dim(X)} H^i(X)$, and it should probably factor through some simpler group. I would be grateful to anyone explainig me how to explicitely build/ describe $rho$ from $X$.
algebraic-geometry category-theory representation-theory
$endgroup$
Assuming standard conjectures, the category of motives is equivalent to the category of representations of a certain group over $mathbb Q$, but I don't understand the abstract construction.
Now, if we start with a nonsingular projective variety $X$ we get a motive $(X, Delta_X, 0)$, so we should also get a representation $rho$. I think it would help me to understand concretely what representation arises this way.
As far as I understood, the space of the representation should be $bigoplus_{i=0}^{2 dim(X)} H^i(X)$, and it should probably factor through some simpler group. I would be grateful to anyone explainig me how to explicitely build/ describe $rho$ from $X$.
algebraic-geometry category-theory representation-theory
algebraic-geometry category-theory representation-theory
asked Jan 8 at 18:50
RégisRégis
678310
678310
add a comment |
add a comment |
1 Answer
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$begingroup$
(Here we assume that $ksubsetmathbb{C}$).
Ok, so we assume the standard conjectures, in particular that the numerical equivalence coincide with the homological one. Then the category of motives $M(k)$ is abelian semi-simple and we have a functor, the Betti-realization : $H_B:M(k)to mathbb{Q}-Vect$ which commutes with $otimes$, is exact and faithful.
The Betti-realization satisfy indeed $H_B((X,Delta_X,0))=bigoplus_{i=0}^{2dim X}H^i(X,mathbb{Q})$.
Now obviously, the group of automorphism of the functor $H_B$ will acts on $H_B(M)$. This is purely abstract non-sense, and does not give much informations. Of course, if we take only the subgroup of $otimes$-automorphism of $H_B$, it will still acts on $H_B(M)$. Denote this group by $G_{mot}$ (the so-called motivic Galois group). So you have that, for any algebraic variety $X$, its cohomology $H^*(X)$ is a $G_{mot}$-representation.
So far, we only said trivialities (even if we assumed deep conjectures...), and we have absolutely no idea what $G_{mot}$ is !
But we have that the functor $H_B$ factor through the category of $G_{mot}$ representations. It happens that this is an equivalence of categories ! Now, this is still not a deep fact (but not a triviality neither). This is purely categorical : it does not use anything from algebraic geometry, nor what the category of motives is, nor what $H_B$ is...
So we can still say that we know nothing about $G_{mot}$ and its representation on $H^*(X)$. And we can't say much more...
Ok so before I say more, lets have a a new look at Galois theory. Consider only the motives of $0$-dimensional varieties. This will be a subcategory $AM(k)subset M(k)$ (the category of so called Artin motives). A motive will essentially be the data of a finite reduced $k$-algebra $A$ and its realization $H_B(A)$ is the $mathbb{Q}$-vector space spanned by the $Hom_{k-alg}(A,mathbb{C})$. Because $A$ is a finite over $k$, a morphism $Atomathbb{C}$ factor through $overline{k}subsetmathbb{C}$. So $Hom_{k-alg}(A,mathbb{C})=Hom_{k-alg}(A,overline{k})$. Obviously, $Gal(overline{k}/k)$ acts on this set (by composition), and so we have a representation of $Gal(overline{k}/k)$ on $H_B(A)=mathbb{Q}.Hom_{k-alg}(A,overline{k})$.
This is not hard to see that this action is natural in $A$. In other words, $Gal(overline{k}/k)$ really acts on the functor $H_B$ (and by $otimes$-automorphism). So we have a map $Gal(overline{k}/k)to Aut_otimes(H_B)$.
It turns out that this is an isomorphism !
By Tannakian reconstruction, the category of Artin motives is equivalent to the category of representation of the Galois group.
Now what can we say if we consider the whole category of motives. Well not much. Though, every natural action of a group on every $H^*(X)$ will contributes to the motivic Galois group. For example, we can let $mathbb{Q}^times$ acts on $H^*(X)$ by $lambda.x=lambda^{deg x} x$. This will be a natural action because every morphism $Xto Y$ preserve the degree of cohomology classes. We thus have a morphism $mathbb{G}_mto G_{mot}$ (this is called the weight-cocharacter).
The Galois group does not define a natural action anymore, this is because an element of $Gal(overline{k}/k)$ does not induce a continuous map $X(mathbb{C})to X(mathbb{C})$ and so does not induce a map in cohomology. I believe however that $Gal(overline{k}/k)$ is supposed to be $pi_0(G_{mot})=G_{mot}/G_{mot}^0$ (the group of connected components) but I am not sure about that.
In conclusion : I like to think of a motive as the finest structure we can put on the cohomology of algebraic varieties. These structures can be encoded in the action of a group (like the cohomological degree can be encoded by the action of $mathbb{Q}^times$ as I did previously). Every structure will contribute to the motivic Galois group (for example, we can put an Hodge structure on $H^*$, and so we will have a map from the Mumford-Tate group to the motivic Galois group.)
So asking what is the representation of the motivic Galois group, is not really possible without a much much much.... deeper understanding of the cohomology of algebraic varieties (and will probably require a huge amount of time).
$endgroup$
$begingroup$
Thank you for your answer. I thought that all the categorical abstract nonsense would become more transparent if we consider some special cases, but you seem to say this is not the case?
$endgroup$
– Régis
Jan 9 at 12:43
$begingroup$
@Régis You said you want to understand the representation $rho$ on $H^*(X)$. But the representation of which group ? As I said, this group has a very abstract definition (which requires deep conjectures). So we know (almost) nothing about $G$. So we can't say much about the representation itself. As I said, the representation is the finest structure we can put on the cohomology of an algebraic variety. By finest, I mean "all of them". So for example on the cohomology $H^*(E)$ on an elliptic curve, what kind of structure there is ? There is the graduation by degree. Are there others ? Yes...
$endgroup$
– Roland
Jan 9 at 12:56
$begingroup$
I understand your point. I hoped (maybe very naively) that in the same way that in some cases representations of the absolute Galois group factor through the Galois group of a finite extension, which as a group is much simpler, something similar would happen with motives. Since it's not the case, I guess I have to understand the construction of the motivic Galois group, and as I said, I don't understand the abstract construction, partly because I miss some definitions and I don't know where to look for them.
$endgroup$
– Régis
Jan 9 at 13:13
$begingroup$
To begin with, I don't know exactly what the Betti realization is, nor $otimes$-automorphism functors.
$endgroup$
– Régis
Jan 9 at 13:14
$begingroup$
Yes, the motivic Galois group acts on $H^*(X)$ through a quotient, even a small one ! This does not mean that we know more about it. The Betti realization is just the singular cohomology of complex points. A $otimes$-automorphism of a monoidal functor $F$ is an natural automorphism $Fto F$ which is compatible with the natural isomorphism $F(Aotimes B)simeq F(A)otimes F(B)$.
$endgroup$
– Roland
Jan 9 at 13:29
|
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$begingroup$
(Here we assume that $ksubsetmathbb{C}$).
Ok, so we assume the standard conjectures, in particular that the numerical equivalence coincide with the homological one. Then the category of motives $M(k)$ is abelian semi-simple and we have a functor, the Betti-realization : $H_B:M(k)to mathbb{Q}-Vect$ which commutes with $otimes$, is exact and faithful.
The Betti-realization satisfy indeed $H_B((X,Delta_X,0))=bigoplus_{i=0}^{2dim X}H^i(X,mathbb{Q})$.
Now obviously, the group of automorphism of the functor $H_B$ will acts on $H_B(M)$. This is purely abstract non-sense, and does not give much informations. Of course, if we take only the subgroup of $otimes$-automorphism of $H_B$, it will still acts on $H_B(M)$. Denote this group by $G_{mot}$ (the so-called motivic Galois group). So you have that, for any algebraic variety $X$, its cohomology $H^*(X)$ is a $G_{mot}$-representation.
So far, we only said trivialities (even if we assumed deep conjectures...), and we have absolutely no idea what $G_{mot}$ is !
But we have that the functor $H_B$ factor through the category of $G_{mot}$ representations. It happens that this is an equivalence of categories ! Now, this is still not a deep fact (but not a triviality neither). This is purely categorical : it does not use anything from algebraic geometry, nor what the category of motives is, nor what $H_B$ is...
So we can still say that we know nothing about $G_{mot}$ and its representation on $H^*(X)$. And we can't say much more...
Ok so before I say more, lets have a a new look at Galois theory. Consider only the motives of $0$-dimensional varieties. This will be a subcategory $AM(k)subset M(k)$ (the category of so called Artin motives). A motive will essentially be the data of a finite reduced $k$-algebra $A$ and its realization $H_B(A)$ is the $mathbb{Q}$-vector space spanned by the $Hom_{k-alg}(A,mathbb{C})$. Because $A$ is a finite over $k$, a morphism $Atomathbb{C}$ factor through $overline{k}subsetmathbb{C}$. So $Hom_{k-alg}(A,mathbb{C})=Hom_{k-alg}(A,overline{k})$. Obviously, $Gal(overline{k}/k)$ acts on this set (by composition), and so we have a representation of $Gal(overline{k}/k)$ on $H_B(A)=mathbb{Q}.Hom_{k-alg}(A,overline{k})$.
This is not hard to see that this action is natural in $A$. In other words, $Gal(overline{k}/k)$ really acts on the functor $H_B$ (and by $otimes$-automorphism). So we have a map $Gal(overline{k}/k)to Aut_otimes(H_B)$.
It turns out that this is an isomorphism !
By Tannakian reconstruction, the category of Artin motives is equivalent to the category of representation of the Galois group.
Now what can we say if we consider the whole category of motives. Well not much. Though, every natural action of a group on every $H^*(X)$ will contributes to the motivic Galois group. For example, we can let $mathbb{Q}^times$ acts on $H^*(X)$ by $lambda.x=lambda^{deg x} x$. This will be a natural action because every morphism $Xto Y$ preserve the degree of cohomology classes. We thus have a morphism $mathbb{G}_mto G_{mot}$ (this is called the weight-cocharacter).
The Galois group does not define a natural action anymore, this is because an element of $Gal(overline{k}/k)$ does not induce a continuous map $X(mathbb{C})to X(mathbb{C})$ and so does not induce a map in cohomology. I believe however that $Gal(overline{k}/k)$ is supposed to be $pi_0(G_{mot})=G_{mot}/G_{mot}^0$ (the group of connected components) but I am not sure about that.
In conclusion : I like to think of a motive as the finest structure we can put on the cohomology of algebraic varieties. These structures can be encoded in the action of a group (like the cohomological degree can be encoded by the action of $mathbb{Q}^times$ as I did previously). Every structure will contribute to the motivic Galois group (for example, we can put an Hodge structure on $H^*$, and so we will have a map from the Mumford-Tate group to the motivic Galois group.)
So asking what is the representation of the motivic Galois group, is not really possible without a much much much.... deeper understanding of the cohomology of algebraic varieties (and will probably require a huge amount of time).
$endgroup$
$begingroup$
Thank you for your answer. I thought that all the categorical abstract nonsense would become more transparent if we consider some special cases, but you seem to say this is not the case?
$endgroup$
– Régis
Jan 9 at 12:43
$begingroup$
@Régis You said you want to understand the representation $rho$ on $H^*(X)$. But the representation of which group ? As I said, this group has a very abstract definition (which requires deep conjectures). So we know (almost) nothing about $G$. So we can't say much about the representation itself. As I said, the representation is the finest structure we can put on the cohomology of an algebraic variety. By finest, I mean "all of them". So for example on the cohomology $H^*(E)$ on an elliptic curve, what kind of structure there is ? There is the graduation by degree. Are there others ? Yes...
$endgroup$
– Roland
Jan 9 at 12:56
$begingroup$
I understand your point. I hoped (maybe very naively) that in the same way that in some cases representations of the absolute Galois group factor through the Galois group of a finite extension, which as a group is much simpler, something similar would happen with motives. Since it's not the case, I guess I have to understand the construction of the motivic Galois group, and as I said, I don't understand the abstract construction, partly because I miss some definitions and I don't know where to look for them.
$endgroup$
– Régis
Jan 9 at 13:13
$begingroup$
To begin with, I don't know exactly what the Betti realization is, nor $otimes$-automorphism functors.
$endgroup$
– Régis
Jan 9 at 13:14
$begingroup$
Yes, the motivic Galois group acts on $H^*(X)$ through a quotient, even a small one ! This does not mean that we know more about it. The Betti realization is just the singular cohomology of complex points. A $otimes$-automorphism of a monoidal functor $F$ is an natural automorphism $Fto F$ which is compatible with the natural isomorphism $F(Aotimes B)simeq F(A)otimes F(B)$.
$endgroup$
– Roland
Jan 9 at 13:29
|
show 3 more comments
$begingroup$
(Here we assume that $ksubsetmathbb{C}$).
Ok, so we assume the standard conjectures, in particular that the numerical equivalence coincide with the homological one. Then the category of motives $M(k)$ is abelian semi-simple and we have a functor, the Betti-realization : $H_B:M(k)to mathbb{Q}-Vect$ which commutes with $otimes$, is exact and faithful.
The Betti-realization satisfy indeed $H_B((X,Delta_X,0))=bigoplus_{i=0}^{2dim X}H^i(X,mathbb{Q})$.
Now obviously, the group of automorphism of the functor $H_B$ will acts on $H_B(M)$. This is purely abstract non-sense, and does not give much informations. Of course, if we take only the subgroup of $otimes$-automorphism of $H_B$, it will still acts on $H_B(M)$. Denote this group by $G_{mot}$ (the so-called motivic Galois group). So you have that, for any algebraic variety $X$, its cohomology $H^*(X)$ is a $G_{mot}$-representation.
So far, we only said trivialities (even if we assumed deep conjectures...), and we have absolutely no idea what $G_{mot}$ is !
But we have that the functor $H_B$ factor through the category of $G_{mot}$ representations. It happens that this is an equivalence of categories ! Now, this is still not a deep fact (but not a triviality neither). This is purely categorical : it does not use anything from algebraic geometry, nor what the category of motives is, nor what $H_B$ is...
So we can still say that we know nothing about $G_{mot}$ and its representation on $H^*(X)$. And we can't say much more...
Ok so before I say more, lets have a a new look at Galois theory. Consider only the motives of $0$-dimensional varieties. This will be a subcategory $AM(k)subset M(k)$ (the category of so called Artin motives). A motive will essentially be the data of a finite reduced $k$-algebra $A$ and its realization $H_B(A)$ is the $mathbb{Q}$-vector space spanned by the $Hom_{k-alg}(A,mathbb{C})$. Because $A$ is a finite over $k$, a morphism $Atomathbb{C}$ factor through $overline{k}subsetmathbb{C}$. So $Hom_{k-alg}(A,mathbb{C})=Hom_{k-alg}(A,overline{k})$. Obviously, $Gal(overline{k}/k)$ acts on this set (by composition), and so we have a representation of $Gal(overline{k}/k)$ on $H_B(A)=mathbb{Q}.Hom_{k-alg}(A,overline{k})$.
This is not hard to see that this action is natural in $A$. In other words, $Gal(overline{k}/k)$ really acts on the functor $H_B$ (and by $otimes$-automorphism). So we have a map $Gal(overline{k}/k)to Aut_otimes(H_B)$.
It turns out that this is an isomorphism !
By Tannakian reconstruction, the category of Artin motives is equivalent to the category of representation of the Galois group.
Now what can we say if we consider the whole category of motives. Well not much. Though, every natural action of a group on every $H^*(X)$ will contributes to the motivic Galois group. For example, we can let $mathbb{Q}^times$ acts on $H^*(X)$ by $lambda.x=lambda^{deg x} x$. This will be a natural action because every morphism $Xto Y$ preserve the degree of cohomology classes. We thus have a morphism $mathbb{G}_mto G_{mot}$ (this is called the weight-cocharacter).
The Galois group does not define a natural action anymore, this is because an element of $Gal(overline{k}/k)$ does not induce a continuous map $X(mathbb{C})to X(mathbb{C})$ and so does not induce a map in cohomology. I believe however that $Gal(overline{k}/k)$ is supposed to be $pi_0(G_{mot})=G_{mot}/G_{mot}^0$ (the group of connected components) but I am not sure about that.
In conclusion : I like to think of a motive as the finest structure we can put on the cohomology of algebraic varieties. These structures can be encoded in the action of a group (like the cohomological degree can be encoded by the action of $mathbb{Q}^times$ as I did previously). Every structure will contribute to the motivic Galois group (for example, we can put an Hodge structure on $H^*$, and so we will have a map from the Mumford-Tate group to the motivic Galois group.)
So asking what is the representation of the motivic Galois group, is not really possible without a much much much.... deeper understanding of the cohomology of algebraic varieties (and will probably require a huge amount of time).
$endgroup$
$begingroup$
Thank you for your answer. I thought that all the categorical abstract nonsense would become more transparent if we consider some special cases, but you seem to say this is not the case?
$endgroup$
– Régis
Jan 9 at 12:43
$begingroup$
@Régis You said you want to understand the representation $rho$ on $H^*(X)$. But the representation of which group ? As I said, this group has a very abstract definition (which requires deep conjectures). So we know (almost) nothing about $G$. So we can't say much about the representation itself. As I said, the representation is the finest structure we can put on the cohomology of an algebraic variety. By finest, I mean "all of them". So for example on the cohomology $H^*(E)$ on an elliptic curve, what kind of structure there is ? There is the graduation by degree. Are there others ? Yes...
$endgroup$
– Roland
Jan 9 at 12:56
$begingroup$
I understand your point. I hoped (maybe very naively) that in the same way that in some cases representations of the absolute Galois group factor through the Galois group of a finite extension, which as a group is much simpler, something similar would happen with motives. Since it's not the case, I guess I have to understand the construction of the motivic Galois group, and as I said, I don't understand the abstract construction, partly because I miss some definitions and I don't know where to look for them.
$endgroup$
– Régis
Jan 9 at 13:13
$begingroup$
To begin with, I don't know exactly what the Betti realization is, nor $otimes$-automorphism functors.
$endgroup$
– Régis
Jan 9 at 13:14
$begingroup$
Yes, the motivic Galois group acts on $H^*(X)$ through a quotient, even a small one ! This does not mean that we know more about it. The Betti realization is just the singular cohomology of complex points. A $otimes$-automorphism of a monoidal functor $F$ is an natural automorphism $Fto F$ which is compatible with the natural isomorphism $F(Aotimes B)simeq F(A)otimes F(B)$.
$endgroup$
– Roland
Jan 9 at 13:29
|
show 3 more comments
$begingroup$
(Here we assume that $ksubsetmathbb{C}$).
Ok, so we assume the standard conjectures, in particular that the numerical equivalence coincide with the homological one. Then the category of motives $M(k)$ is abelian semi-simple and we have a functor, the Betti-realization : $H_B:M(k)to mathbb{Q}-Vect$ which commutes with $otimes$, is exact and faithful.
The Betti-realization satisfy indeed $H_B((X,Delta_X,0))=bigoplus_{i=0}^{2dim X}H^i(X,mathbb{Q})$.
Now obviously, the group of automorphism of the functor $H_B$ will acts on $H_B(M)$. This is purely abstract non-sense, and does not give much informations. Of course, if we take only the subgroup of $otimes$-automorphism of $H_B$, it will still acts on $H_B(M)$. Denote this group by $G_{mot}$ (the so-called motivic Galois group). So you have that, for any algebraic variety $X$, its cohomology $H^*(X)$ is a $G_{mot}$-representation.
So far, we only said trivialities (even if we assumed deep conjectures...), and we have absolutely no idea what $G_{mot}$ is !
But we have that the functor $H_B$ factor through the category of $G_{mot}$ representations. It happens that this is an equivalence of categories ! Now, this is still not a deep fact (but not a triviality neither). This is purely categorical : it does not use anything from algebraic geometry, nor what the category of motives is, nor what $H_B$ is...
So we can still say that we know nothing about $G_{mot}$ and its representation on $H^*(X)$. And we can't say much more...
Ok so before I say more, lets have a a new look at Galois theory. Consider only the motives of $0$-dimensional varieties. This will be a subcategory $AM(k)subset M(k)$ (the category of so called Artin motives). A motive will essentially be the data of a finite reduced $k$-algebra $A$ and its realization $H_B(A)$ is the $mathbb{Q}$-vector space spanned by the $Hom_{k-alg}(A,mathbb{C})$. Because $A$ is a finite over $k$, a morphism $Atomathbb{C}$ factor through $overline{k}subsetmathbb{C}$. So $Hom_{k-alg}(A,mathbb{C})=Hom_{k-alg}(A,overline{k})$. Obviously, $Gal(overline{k}/k)$ acts on this set (by composition), and so we have a representation of $Gal(overline{k}/k)$ on $H_B(A)=mathbb{Q}.Hom_{k-alg}(A,overline{k})$.
This is not hard to see that this action is natural in $A$. In other words, $Gal(overline{k}/k)$ really acts on the functor $H_B$ (and by $otimes$-automorphism). So we have a map $Gal(overline{k}/k)to Aut_otimes(H_B)$.
It turns out that this is an isomorphism !
By Tannakian reconstruction, the category of Artin motives is equivalent to the category of representation of the Galois group.
Now what can we say if we consider the whole category of motives. Well not much. Though, every natural action of a group on every $H^*(X)$ will contributes to the motivic Galois group. For example, we can let $mathbb{Q}^times$ acts on $H^*(X)$ by $lambda.x=lambda^{deg x} x$. This will be a natural action because every morphism $Xto Y$ preserve the degree of cohomology classes. We thus have a morphism $mathbb{G}_mto G_{mot}$ (this is called the weight-cocharacter).
The Galois group does not define a natural action anymore, this is because an element of $Gal(overline{k}/k)$ does not induce a continuous map $X(mathbb{C})to X(mathbb{C})$ and so does not induce a map in cohomology. I believe however that $Gal(overline{k}/k)$ is supposed to be $pi_0(G_{mot})=G_{mot}/G_{mot}^0$ (the group of connected components) but I am not sure about that.
In conclusion : I like to think of a motive as the finest structure we can put on the cohomology of algebraic varieties. These structures can be encoded in the action of a group (like the cohomological degree can be encoded by the action of $mathbb{Q}^times$ as I did previously). Every structure will contribute to the motivic Galois group (for example, we can put an Hodge structure on $H^*$, and so we will have a map from the Mumford-Tate group to the motivic Galois group.)
So asking what is the representation of the motivic Galois group, is not really possible without a much much much.... deeper understanding of the cohomology of algebraic varieties (and will probably require a huge amount of time).
$endgroup$
(Here we assume that $ksubsetmathbb{C}$).
Ok, so we assume the standard conjectures, in particular that the numerical equivalence coincide with the homological one. Then the category of motives $M(k)$ is abelian semi-simple and we have a functor, the Betti-realization : $H_B:M(k)to mathbb{Q}-Vect$ which commutes with $otimes$, is exact and faithful.
The Betti-realization satisfy indeed $H_B((X,Delta_X,0))=bigoplus_{i=0}^{2dim X}H^i(X,mathbb{Q})$.
Now obviously, the group of automorphism of the functor $H_B$ will acts on $H_B(M)$. This is purely abstract non-sense, and does not give much informations. Of course, if we take only the subgroup of $otimes$-automorphism of $H_B$, it will still acts on $H_B(M)$. Denote this group by $G_{mot}$ (the so-called motivic Galois group). So you have that, for any algebraic variety $X$, its cohomology $H^*(X)$ is a $G_{mot}$-representation.
So far, we only said trivialities (even if we assumed deep conjectures...), and we have absolutely no idea what $G_{mot}$ is !
But we have that the functor $H_B$ factor through the category of $G_{mot}$ representations. It happens that this is an equivalence of categories ! Now, this is still not a deep fact (but not a triviality neither). This is purely categorical : it does not use anything from algebraic geometry, nor what the category of motives is, nor what $H_B$ is...
So we can still say that we know nothing about $G_{mot}$ and its representation on $H^*(X)$. And we can't say much more...
Ok so before I say more, lets have a a new look at Galois theory. Consider only the motives of $0$-dimensional varieties. This will be a subcategory $AM(k)subset M(k)$ (the category of so called Artin motives). A motive will essentially be the data of a finite reduced $k$-algebra $A$ and its realization $H_B(A)$ is the $mathbb{Q}$-vector space spanned by the $Hom_{k-alg}(A,mathbb{C})$. Because $A$ is a finite over $k$, a morphism $Atomathbb{C}$ factor through $overline{k}subsetmathbb{C}$. So $Hom_{k-alg}(A,mathbb{C})=Hom_{k-alg}(A,overline{k})$. Obviously, $Gal(overline{k}/k)$ acts on this set (by composition), and so we have a representation of $Gal(overline{k}/k)$ on $H_B(A)=mathbb{Q}.Hom_{k-alg}(A,overline{k})$.
This is not hard to see that this action is natural in $A$. In other words, $Gal(overline{k}/k)$ really acts on the functor $H_B$ (and by $otimes$-automorphism). So we have a map $Gal(overline{k}/k)to Aut_otimes(H_B)$.
It turns out that this is an isomorphism !
By Tannakian reconstruction, the category of Artin motives is equivalent to the category of representation of the Galois group.
Now what can we say if we consider the whole category of motives. Well not much. Though, every natural action of a group on every $H^*(X)$ will contributes to the motivic Galois group. For example, we can let $mathbb{Q}^times$ acts on $H^*(X)$ by $lambda.x=lambda^{deg x} x$. This will be a natural action because every morphism $Xto Y$ preserve the degree of cohomology classes. We thus have a morphism $mathbb{G}_mto G_{mot}$ (this is called the weight-cocharacter).
The Galois group does not define a natural action anymore, this is because an element of $Gal(overline{k}/k)$ does not induce a continuous map $X(mathbb{C})to X(mathbb{C})$ and so does not induce a map in cohomology. I believe however that $Gal(overline{k}/k)$ is supposed to be $pi_0(G_{mot})=G_{mot}/G_{mot}^0$ (the group of connected components) but I am not sure about that.
In conclusion : I like to think of a motive as the finest structure we can put on the cohomology of algebraic varieties. These structures can be encoded in the action of a group (like the cohomological degree can be encoded by the action of $mathbb{Q}^times$ as I did previously). Every structure will contribute to the motivic Galois group (for example, we can put an Hodge structure on $H^*$, and so we will have a map from the Mumford-Tate group to the motivic Galois group.)
So asking what is the representation of the motivic Galois group, is not really possible without a much much much.... deeper understanding of the cohomology of algebraic varieties (and will probably require a huge amount of time).
answered Jan 8 at 20:22
RolandRoland
7,45911015
7,45911015
$begingroup$
Thank you for your answer. I thought that all the categorical abstract nonsense would become more transparent if we consider some special cases, but you seem to say this is not the case?
$endgroup$
– Régis
Jan 9 at 12:43
$begingroup$
@Régis You said you want to understand the representation $rho$ on $H^*(X)$. But the representation of which group ? As I said, this group has a very abstract definition (which requires deep conjectures). So we know (almost) nothing about $G$. So we can't say much about the representation itself. As I said, the representation is the finest structure we can put on the cohomology of an algebraic variety. By finest, I mean "all of them". So for example on the cohomology $H^*(E)$ on an elliptic curve, what kind of structure there is ? There is the graduation by degree. Are there others ? Yes...
$endgroup$
– Roland
Jan 9 at 12:56
$begingroup$
I understand your point. I hoped (maybe very naively) that in the same way that in some cases representations of the absolute Galois group factor through the Galois group of a finite extension, which as a group is much simpler, something similar would happen with motives. Since it's not the case, I guess I have to understand the construction of the motivic Galois group, and as I said, I don't understand the abstract construction, partly because I miss some definitions and I don't know where to look for them.
$endgroup$
– Régis
Jan 9 at 13:13
$begingroup$
To begin with, I don't know exactly what the Betti realization is, nor $otimes$-automorphism functors.
$endgroup$
– Régis
Jan 9 at 13:14
$begingroup$
Yes, the motivic Galois group acts on $H^*(X)$ through a quotient, even a small one ! This does not mean that we know more about it. The Betti realization is just the singular cohomology of complex points. A $otimes$-automorphism of a monoidal functor $F$ is an natural automorphism $Fto F$ which is compatible with the natural isomorphism $F(Aotimes B)simeq F(A)otimes F(B)$.
$endgroup$
– Roland
Jan 9 at 13:29
|
show 3 more comments
$begingroup$
Thank you for your answer. I thought that all the categorical abstract nonsense would become more transparent if we consider some special cases, but you seem to say this is not the case?
$endgroup$
– Régis
Jan 9 at 12:43
$begingroup$
@Régis You said you want to understand the representation $rho$ on $H^*(X)$. But the representation of which group ? As I said, this group has a very abstract definition (which requires deep conjectures). So we know (almost) nothing about $G$. So we can't say much about the representation itself. As I said, the representation is the finest structure we can put on the cohomology of an algebraic variety. By finest, I mean "all of them". So for example on the cohomology $H^*(E)$ on an elliptic curve, what kind of structure there is ? There is the graduation by degree. Are there others ? Yes...
$endgroup$
– Roland
Jan 9 at 12:56
$begingroup$
I understand your point. I hoped (maybe very naively) that in the same way that in some cases representations of the absolute Galois group factor through the Galois group of a finite extension, which as a group is much simpler, something similar would happen with motives. Since it's not the case, I guess I have to understand the construction of the motivic Galois group, and as I said, I don't understand the abstract construction, partly because I miss some definitions and I don't know where to look for them.
$endgroup$
– Régis
Jan 9 at 13:13
$begingroup$
To begin with, I don't know exactly what the Betti realization is, nor $otimes$-automorphism functors.
$endgroup$
– Régis
Jan 9 at 13:14
$begingroup$
Yes, the motivic Galois group acts on $H^*(X)$ through a quotient, even a small one ! This does not mean that we know more about it. The Betti realization is just the singular cohomology of complex points. A $otimes$-automorphism of a monoidal functor $F$ is an natural automorphism $Fto F$ which is compatible with the natural isomorphism $F(Aotimes B)simeq F(A)otimes F(B)$.
$endgroup$
– Roland
Jan 9 at 13:29
$begingroup$
Thank you for your answer. I thought that all the categorical abstract nonsense would become more transparent if we consider some special cases, but you seem to say this is not the case?
$endgroup$
– Régis
Jan 9 at 12:43
$begingroup$
Thank you for your answer. I thought that all the categorical abstract nonsense would become more transparent if we consider some special cases, but you seem to say this is not the case?
$endgroup$
– Régis
Jan 9 at 12:43
$begingroup$
@Régis You said you want to understand the representation $rho$ on $H^*(X)$. But the representation of which group ? As I said, this group has a very abstract definition (which requires deep conjectures). So we know (almost) nothing about $G$. So we can't say much about the representation itself. As I said, the representation is the finest structure we can put on the cohomology of an algebraic variety. By finest, I mean "all of them". So for example on the cohomology $H^*(E)$ on an elliptic curve, what kind of structure there is ? There is the graduation by degree. Are there others ? Yes...
$endgroup$
– Roland
Jan 9 at 12:56
$begingroup$
@Régis You said you want to understand the representation $rho$ on $H^*(X)$. But the representation of which group ? As I said, this group has a very abstract definition (which requires deep conjectures). So we know (almost) nothing about $G$. So we can't say much about the representation itself. As I said, the representation is the finest structure we can put on the cohomology of an algebraic variety. By finest, I mean "all of them". So for example on the cohomology $H^*(E)$ on an elliptic curve, what kind of structure there is ? There is the graduation by degree. Are there others ? Yes...
$endgroup$
– Roland
Jan 9 at 12:56
$begingroup$
I understand your point. I hoped (maybe very naively) that in the same way that in some cases representations of the absolute Galois group factor through the Galois group of a finite extension, which as a group is much simpler, something similar would happen with motives. Since it's not the case, I guess I have to understand the construction of the motivic Galois group, and as I said, I don't understand the abstract construction, partly because I miss some definitions and I don't know where to look for them.
$endgroup$
– Régis
Jan 9 at 13:13
$begingroup$
I understand your point. I hoped (maybe very naively) that in the same way that in some cases representations of the absolute Galois group factor through the Galois group of a finite extension, which as a group is much simpler, something similar would happen with motives. Since it's not the case, I guess I have to understand the construction of the motivic Galois group, and as I said, I don't understand the abstract construction, partly because I miss some definitions and I don't know where to look for them.
$endgroup$
– Régis
Jan 9 at 13:13
$begingroup$
To begin with, I don't know exactly what the Betti realization is, nor $otimes$-automorphism functors.
$endgroup$
– Régis
Jan 9 at 13:14
$begingroup$
To begin with, I don't know exactly what the Betti realization is, nor $otimes$-automorphism functors.
$endgroup$
– Régis
Jan 9 at 13:14
$begingroup$
Yes, the motivic Galois group acts on $H^*(X)$ through a quotient, even a small one ! This does not mean that we know more about it. The Betti realization is just the singular cohomology of complex points. A $otimes$-automorphism of a monoidal functor $F$ is an natural automorphism $Fto F$ which is compatible with the natural isomorphism $F(Aotimes B)simeq F(A)otimes F(B)$.
$endgroup$
– Roland
Jan 9 at 13:29
$begingroup$
Yes, the motivic Galois group acts on $H^*(X)$ through a quotient, even a small one ! This does not mean that we know more about it. The Betti realization is just the singular cohomology of complex points. A $otimes$-automorphism of a monoidal functor $F$ is an natural automorphism $Fto F$ which is compatible with the natural isomorphism $F(Aotimes B)simeq F(A)otimes F(B)$.
$endgroup$
– Roland
Jan 9 at 13:29
|
show 3 more comments
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