Mixing
Explanation of $Theta$ semi stable points (or representation)
$begingroup$
I try to understand the precise meaning of a $Theta$-semi-stable set of points.
Say in the context of quiver representations: Let $Q = (Q_0, Q_1)$ be a quiver $(V_i, alpha_{i,j})$ a representation of $Q$ with dimension vector $d = (dim V_0,... dim V_n)$.
And $Theta$ a weight vector s.t. $Theta * d = 0$
and for every sub-representation with weight vector $d'$ we have $Theta * d' <= 0$.
Technically i understand whats going on, and I understand the intended behavior to only consider 'nice' representations (points). But the 'how' is unclear to me.
I know there are answers about what $Theta$ is in terms of line bundles, but what I am interested in is a more 'down to earth' answer.
- Why does it make sense to look at the dimension vector? I know why it make sense in terms of vectors or the dot-product. but what is the intuition behind this?
- What does $Theta * d = 0$ mean? Does it make sense think about it in terms of orthogonal in any way?
- Does it make any sense to think about $Theta * d'< 0$ as $d'$ laying on one side of a hyperplane defined by $Theta$?
- Finally: why does this construction provide us with the 'good' representations and does not restrict too much?
representation-theory quiver
asked Jan 2 at 9:47
JohannesJohannes
1158
$endgroup$
add a comment |
$begingroup$
I try to understand the precise meaning of a $Theta$-semi-stable set of points.
Say in the context of quiver representations: Let $Q = (Q_0, Q_1)$ be a quiver $(V_i, alpha_{i,j})$ a representation of $Q$ with dimension vector $d = (dim V_0,... dim V_n)$.
And $Theta$ a weight vector s.t. $Theta * d = 0$
and for every sub-representation with weight vector $d'$ we have $Theta * d' <= 0$.
Technically i understand whats going on, and I understand the intended behavior to only consider 'nice' representations (points). But the 'how' is unclear to me.
I know there are answers about what $Theta$ is in terms of line bundles, but what I am interested in is a more 'down to earth' answer.
- Why does it make sense to look at the dimension vector? I know why it make sense in terms of vectors or the dot-product. but what is the intuition behind this?
- What does $Theta * d = 0$ mean? Does it make sense think about it in terms of orthogonal in any way?
- Does it make any sense to think about $Theta * d'< 0$ as $d'$ laying on one side of a hyperplane defined by $Theta$?
- Finally: why does this construction provide us with the 'good' representations and does not restrict too much?
representation-theory quiver
asked Jan 2 at 9:47
JohannesJohannes
1158
$endgroup$
add a comment |
$begingroup$
I try to understand the precise meaning of a $Theta$-semi-stable set of points.
Say in the context of quiver representations: Let $Q = (Q_0, Q_1)$ be a quiver $(V_i, alpha_{i,j})$ a representation of $Q$ with dimension vector $d = (dim V_0,... dim V_n)$.
And $Theta$ a weight vector s.t. $Theta * d = 0$
and for every sub-representation with weight vector $d'$ we have $Theta * d' <= 0$.
Technically i understand whats going on, and I understand the intended behavior to only consider 'nice' representations (points). But the 'how' is unclear to me.
I know there are answers about what $Theta$ is in terms of line bundles, but what I am interested in is a more 'down to earth' answer.
- Why does it make sense to look at the dimension vector? I know why it make sense in terms of vectors or the dot-product. but what is the intuition behind this?
- What does $Theta * d = 0$ mean? Does it make sense think about it in terms of orthogonal in any way?
- Does it make any sense to think about $Theta * d'< 0$ as $d'$ laying on one side of a hyperplane defined by $Theta$?
- Finally: why does this construction provide us with the 'good' representations and does not restrict too much?
representation-theory quiver
asked Jan 2 at 9:47
JohannesJohannes
1158
$endgroup$
I try to understand the precise meaning of a $Theta$-semi-stable set of points.
Say in the context of quiver representations: Let $Q = (Q_0, Q_1)$ be a quiver $(V_i, alpha_{i,j})$ a representation of $Q$ with dimension vector $d = (dim V_0,... dim V_n)$.
And $Theta$ a weight vector s.t. $Theta * d = 0$
and for every sub-representation with weight vector $d'$ we have $Theta * d' <= 0$.
Technically i understand whats going on, and I understand the intended behavior to only consider 'nice' representations (points). But the 'how' is unclear to me.
I know there are answers about what $Theta$ is in terms of line bundles, but what I am interested in is a more 'down to earth' answer.
- Why does it make sense to look at the dimension vector? I know why it make sense in terms of vectors or the dot-product. but what is the intuition behind this?
- What does $Theta * d = 0$ mean? Does it make sense think about it in terms of orthogonal in any way?
- Does it make any sense to think about $Theta * d'< 0$ as $d'$ laying on one side of a hyperplane defined by $Theta$?
- Finally: why does this construction provide us with the 'good' representations and does not restrict too much?
representation-theory quiver
representation-theory quiver
asked Jan 2 at 9:47
JohannesJohannes
1158
asked Jan 2 at 9:47
JohannesJohannes
1158
asked Jan 2 at 9:47
JohannesJohannes
1158
asked Jan 2 at 9:47
JohannesJohannes
1158
asked Jan 2 at 9:47
JohannesJohannes
1158
1158
add a comment |
add a comment |
1 Answer
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$begingroup$
One sense in which the semistable representations with respect to $Theta$ are a "good" set of representations is that they form an exact abelian extension-closed subcategory of the representations of the quiver. (This is fairly easy to deduce directly from the definition you have stated.)
In general, it is not the case that the semistable representations do not "restrict too much" --- if $Theta$ is chosen at random, it is quite possible for there to be no non-zero $Theta$-semistable representations.
A geometric explanation of the significance of this version of semistability in terms of geometric invariant theory is provided by Alastair King's paper Moduli of representations of finite-dimensional algebras.
I am not sure how to address your questions about what $Thetaast d$ means, and what it means to have $Thetaast d' < 0$. King's paper provides a GIT explanation, but on a down-to-earth level they are exactly conditions that $d$ lie on the hyperplane defined by $Theta$, and that $d'$ lie on one side of the hyperplane defined by $Theta$. (In general it is probably a good idea to think of $d$ and $Theta$ is living, not in the same $mathbb Z^n$, but rather in two dual copies of $mathbb Z^n$, so that instead of the dot product between them, you have the canonical pairing, but concretely this is not going to make a difference.)
edited Jan 5 at 2:28
Martin Sleziak
44.7k8117272
answered Jan 2 at 19:35
Hugh ThomasHugh Thomas
1,06868
$endgroup$
add a comment |
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1 Answer
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oldest
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$begingroup$
One sense in which the semistable representations with respect to $Theta$ are a "good" set of representations is that they form an exact abelian extension-closed subcategory of the representations of the quiver. (This is fairly easy to deduce directly from the definition you have stated.)
In general, it is not the case that the semistable representations do not "restrict too much" --- if $Theta$ is chosen at random, it is quite possible for there to be no non-zero $Theta$-semistable representations.
A geometric explanation of the significance of this version of semistability in terms of geometric invariant theory is provided by Alastair King's paper Moduli of representations of finite-dimensional algebras.
I am not sure how to address your questions about what $Thetaast d$ means, and what it means to have $Thetaast d' < 0$. King's paper provides a GIT explanation, but on a down-to-earth level they are exactly conditions that $d$ lie on the hyperplane defined by $Theta$, and that $d'$ lie on one side of the hyperplane defined by $Theta$. (In general it is probably a good idea to think of $d$ and $Theta$ is living, not in the same $mathbb Z^n$, but rather in two dual copies of $mathbb Z^n$, so that instead of the dot product between them, you have the canonical pairing, but concretely this is not going to make a difference.)
edited Jan 5 at 2:28
Martin Sleziak
44.7k8117272
answered Jan 2 at 19:35
Hugh ThomasHugh Thomas
1,06868
$endgroup$
add a comment |
$begingroup$
One sense in which the semistable representations with respect to $Theta$ are a "good" set of representations is that they form an exact abelian extension-closed subcategory of the representations of the quiver. (This is fairly easy to deduce directly from the definition you have stated.)
In general, it is not the case that the semistable representations do not "restrict too much" --- if $Theta$ is chosen at random, it is quite possible for there to be no non-zero $Theta$-semistable representations.
A geometric explanation of the significance of this version of semistability in terms of geometric invariant theory is provided by Alastair King's paper Moduli of representations of finite-dimensional algebras.
I am not sure how to address your questions about what $Thetaast d$ means, and what it means to have $Thetaast d' < 0$. King's paper provides a GIT explanation, but on a down-to-earth level they are exactly conditions that $d$ lie on the hyperplane defined by $Theta$, and that $d'$ lie on one side of the hyperplane defined by $Theta$. (In general it is probably a good idea to think of $d$ and $Theta$ is living, not in the same $mathbb Z^n$, but rather in two dual copies of $mathbb Z^n$, so that instead of the dot product between them, you have the canonical pairing, but concretely this is not going to make a difference.)
edited Jan 5 at 2:28
Martin Sleziak
44.7k8117272
answered Jan 2 at 19:35
Hugh ThomasHugh Thomas
1,06868
$endgroup$
add a comment |
$begingroup$
One sense in which the semistable representations with respect to $Theta$ are a "good" set of representations is that they form an exact abelian extension-closed subcategory of the representations of the quiver. (This is fairly easy to deduce directly from the definition you have stated.)
In general, it is not the case that the semistable representations do not "restrict too much" --- if $Theta$ is chosen at random, it is quite possible for there to be no non-zero $Theta$-semistable representations.
A geometric explanation of the significance of this version of semistability in terms of geometric invariant theory is provided by Alastair King's paper Moduli of representations of finite-dimensional algebras.
I am not sure how to address your questions about what $Thetaast d$ means, and what it means to have $Thetaast d' < 0$. King's paper provides a GIT explanation, but on a down-to-earth level they are exactly conditions that $d$ lie on the hyperplane defined by $Theta$, and that $d'$ lie on one side of the hyperplane defined by $Theta$. (In general it is probably a good idea to think of $d$ and $Theta$ is living, not in the same $mathbb Z^n$, but rather in two dual copies of $mathbb Z^n$, so that instead of the dot product between them, you have the canonical pairing, but concretely this is not going to make a difference.)
edited Jan 5 at 2:28
Martin Sleziak
44.7k8117272
answered Jan 2 at 19:35
Hugh ThomasHugh Thomas
1,06868
$endgroup$
One sense in which the semistable representations with respect to $Theta$ are a "good" set of representations is that they form an exact abelian extension-closed subcategory of the representations of the quiver. (This is fairly easy to deduce directly from the definition you have stated.)
In general, it is not the case that the semistable representations do not "restrict too much" --- if $Theta$ is chosen at random, it is quite possible for there to be no non-zero $Theta$-semistable representations.
A geometric explanation of the significance of this version of semistability in terms of geometric invariant theory is provided by Alastair King's paper Moduli of representations of finite-dimensional algebras.
I am not sure how to address your questions about what $Thetaast d$ means, and what it means to have $Thetaast d' < 0$. King's paper provides a GIT explanation, but on a down-to-earth level they are exactly conditions that $d$ lie on the hyperplane defined by $Theta$, and that $d'$ lie on one side of the hyperplane defined by $Theta$. (In general it is probably a good idea to think of $d$ and $Theta$ is living, not in the same $mathbb Z^n$, but rather in two dual copies of $mathbb Z^n$, so that instead of the dot product between them, you have the canonical pairing, but concretely this is not going to make a difference.)
edited Jan 5 at 2:28
Martin Sleziak
44.7k8117272
answered Jan 2 at 19:35
Hugh ThomasHugh Thomas
1,06868
edited Jan 5 at 2:28
Martin Sleziak
44.7k8117272
edited Jan 5 at 2:28
Martin Sleziak
44.7k8117272
edited Jan 5 at 2:28
Martin Sleziak
44.7k8117272
44.7k8117272
answered Jan 2 at 19:35
Hugh ThomasHugh Thomas
1,06868
answered Jan 2 at 19:35
Hugh ThomasHugh Thomas
1,06868
answered Jan 2 at 19:35
Hugh ThomasHugh Thomas
1,06868
1,06868
add a comment |
add a comment |
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