Spanning trees of the Tait graph and Khovanov homology.
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I'm reading this paper by Champanerkar and Kofman which proposes a nice way to compute the Khovanov homology of a link (diagram) via spaning trees of its Tait graph. The authors actualy define a bigraded chain complexe generated by thoses trees the homology of which is isomorphic to the Khovanov homology up to a change in the graduation. I'm particularly interested in understanding the definition of the differential on that chain complexe which makes use of what the authors call "elementary collapses". Lemma 2 describes elementary collapses in the general case of chain complexes. Lemma 3 uses Lemma 2 to describe the specified elementary collapses the authors will need to define the diffirential. More precisely, its is the proof of Lemma 2 that contains the explicite description of those collapses. The problem is i really dont get that proof because of the notations the authors uses. To be more specific, my question will be the next one:
How did the authors use Lemma 2 to deduce the collapses they got in the proof of Lemma 3 ?
knot-theory knot-invariants
asked Dec 27 '18 at 15:56
Mohammed SabakMohammed Sabak
1118
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add a comment |
$begingroup$
I'm reading this paper by Champanerkar and Kofman which proposes a nice way to compute the Khovanov homology of a link (diagram) via spaning trees of its Tait graph. The authors actualy define a bigraded chain complexe generated by thoses trees the homology of which is isomorphic to the Khovanov homology up to a change in the graduation. I'm particularly interested in understanding the definition of the differential on that chain complexe which makes use of what the authors call "elementary collapses". Lemma 2 describes elementary collapses in the general case of chain complexes. Lemma 3 uses Lemma 2 to describe the specified elementary collapses the authors will need to define the diffirential. More precisely, its is the proof of Lemma 2 that contains the explicite description of those collapses. The problem is i really dont get that proof because of the notations the authors uses. To be more specific, my question will be the next one:
How did the authors use Lemma 2 to deduce the collapses they got in the proof of Lemma 3 ?
knot-theory knot-invariants
asked Dec 27 '18 at 15:56
Mohammed SabakMohammed Sabak
1118
$endgroup$
add a comment |
$begingroup$
I'm reading this paper by Champanerkar and Kofman which proposes a nice way to compute the Khovanov homology of a link (diagram) via spaning trees of its Tait graph. The authors actualy define a bigraded chain complexe generated by thoses trees the homology of which is isomorphic to the Khovanov homology up to a change in the graduation. I'm particularly interested in understanding the definition of the differential on that chain complexe which makes use of what the authors call "elementary collapses". Lemma 2 describes elementary collapses in the general case of chain complexes. Lemma 3 uses Lemma 2 to describe the specified elementary collapses the authors will need to define the diffirential. More precisely, its is the proof of Lemma 2 that contains the explicite description of those collapses. The problem is i really dont get that proof because of the notations the authors uses. To be more specific, my question will be the next one:
How did the authors use Lemma 2 to deduce the collapses they got in the proof of Lemma 3 ?
knot-theory knot-invariants
asked Dec 27 '18 at 15:56
Mohammed SabakMohammed Sabak
1118
$endgroup$
I'm reading this paper by Champanerkar and Kofman which proposes a nice way to compute the Khovanov homology of a link (diagram) via spaning trees of its Tait graph. The authors actualy define a bigraded chain complexe generated by thoses trees the homology of which is isomorphic to the Khovanov homology up to a change in the graduation. I'm particularly interested in understanding the definition of the differential on that chain complexe which makes use of what the authors call "elementary collapses". Lemma 2 describes elementary collapses in the general case of chain complexes. Lemma 3 uses Lemma 2 to describe the specified elementary collapses the authors will need to define the diffirential. More precisely, its is the proof of Lemma 2 that contains the explicite description of those collapses. The problem is i really dont get that proof because of the notations the authors uses. To be more specific, my question will be the next one:
How did the authors use Lemma 2 to deduce the collapses they got in the proof of Lemma 3 ?
knot-theory knot-invariants
knot-theory knot-invariants
asked Dec 27 '18 at 15:56
Mohammed SabakMohammed Sabak
1118
asked Dec 27 '18 at 15:56
Mohammed SabakMohammed Sabak
1118
asked Dec 27 '18 at 15:56
Mohammed SabakMohammed Sabak
1118
asked Dec 27 '18 at 15:56
Mohammed SabakMohammed Sabak
1118
asked Dec 27 '18 at 15:56
Mohammed SabakMohammed Sabak
1118
1118
add a comment |
add a comment |
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Lemma 2 states that if a generator $y$ appears with coefficient $pm 1$ in the differential of generator $x$ in a chain complex $C$, then there is a new chain complex $C'$ that has all the same generators as $C$ except for $x$ and $y$ such that $C$ and $C'$ are chain homotopic.
Lemma 3 essentially says that if $U$ is a twisted unknot, then its Khovanov complex can be taken to be generated by two generators. They prove this by repeated applications of Lemma 2. A twisted unknot can be transformed into the standard diagram of the unknot via Reidemeister 1 twists. The invariance proof of Khovanov homology under Reidemeister 1 moves mimics the conditions in Lemma 2. The complex without the Reidemeister 1 twist is $C'$ in Lemma 2 while the complex in Lemma 2 is $C$.
In order to completely understand the differential in the spanning tree complex, one needs to follow all of these elementary collapses as well as keep track of the decomposition of the Khovanov complex into iterated mapping cone complexes of the twisted unknots. The rule of thumb is that the fewer number of generators for the complex, the more difficult the differential is to understand (and vice versa).
answered Dec 27 '18 at 22:57
Adam LowranceAdam Lowrance
2,19711014
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$begingroup$
Lemma 2 states that if a generator $y$ appears with coefficient $pm 1$ in the differential of generator $x$ in a chain complex $C$, then there is a new chain complex $C'$ that has all the same generators as $C$ except for $x$ and $y$ such that $C$ and $C'$ are chain homotopic.
Lemma 3 essentially says that if $U$ is a twisted unknot, then its Khovanov complex can be taken to be generated by two generators. They prove this by repeated applications of Lemma 2. A twisted unknot can be transformed into the standard diagram of the unknot via Reidemeister 1 twists. The invariance proof of Khovanov homology under Reidemeister 1 moves mimics the conditions in Lemma 2. The complex without the Reidemeister 1 twist is $C'$ in Lemma 2 while the complex in Lemma 2 is $C$.
In order to completely understand the differential in the spanning tree complex, one needs to follow all of these elementary collapses as well as keep track of the decomposition of the Khovanov complex into iterated mapping cone complexes of the twisted unknots. The rule of thumb is that the fewer number of generators for the complex, the more difficult the differential is to understand (and vice versa).
answered Dec 27 '18 at 22:57
Adam LowranceAdam Lowrance
2,19711014
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add a comment |
$begingroup$
Lemma 2 states that if a generator $y$ appears with coefficient $pm 1$ in the differential of generator $x$ in a chain complex $C$, then there is a new chain complex $C'$ that has all the same generators as $C$ except for $x$ and $y$ such that $C$ and $C'$ are chain homotopic.
Lemma 3 essentially says that if $U$ is a twisted unknot, then its Khovanov complex can be taken to be generated by two generators. They prove this by repeated applications of Lemma 2. A twisted unknot can be transformed into the standard diagram of the unknot via Reidemeister 1 twists. The invariance proof of Khovanov homology under Reidemeister 1 moves mimics the conditions in Lemma 2. The complex without the Reidemeister 1 twist is $C'$ in Lemma 2 while the complex in Lemma 2 is $C$.
In order to completely understand the differential in the spanning tree complex, one needs to follow all of these elementary collapses as well as keep track of the decomposition of the Khovanov complex into iterated mapping cone complexes of the twisted unknots. The rule of thumb is that the fewer number of generators for the complex, the more difficult the differential is to understand (and vice versa).
answered Dec 27 '18 at 22:57
Adam LowranceAdam Lowrance
2,19711014
$endgroup$
add a comment |
$begingroup$
Lemma 2 states that if a generator $y$ appears with coefficient $pm 1$ in the differential of generator $x$ in a chain complex $C$, then there is a new chain complex $C'$ that has all the same generators as $C$ except for $x$ and $y$ such that $C$ and $C'$ are chain homotopic.
Lemma 3 essentially says that if $U$ is a twisted unknot, then its Khovanov complex can be taken to be generated by two generators. They prove this by repeated applications of Lemma 2. A twisted unknot can be transformed into the standard diagram of the unknot via Reidemeister 1 twists. The invariance proof of Khovanov homology under Reidemeister 1 moves mimics the conditions in Lemma 2. The complex without the Reidemeister 1 twist is $C'$ in Lemma 2 while the complex in Lemma 2 is $C$.
In order to completely understand the differential in the spanning tree complex, one needs to follow all of these elementary collapses as well as keep track of the decomposition of the Khovanov complex into iterated mapping cone complexes of the twisted unknots. The rule of thumb is that the fewer number of generators for the complex, the more difficult the differential is to understand (and vice versa).
answered Dec 27 '18 at 22:57
Adam LowranceAdam Lowrance
2,19711014
$endgroup$
Lemma 2 states that if a generator $y$ appears with coefficient $pm 1$ in the differential of generator $x$ in a chain complex $C$, then there is a new chain complex $C'$ that has all the same generators as $C$ except for $x$ and $y$ such that $C$ and $C'$ are chain homotopic.
Lemma 3 essentially says that if $U$ is a twisted unknot, then its Khovanov complex can be taken to be generated by two generators. They prove this by repeated applications of Lemma 2. A twisted unknot can be transformed into the standard diagram of the unknot via Reidemeister 1 twists. The invariance proof of Khovanov homology under Reidemeister 1 moves mimics the conditions in Lemma 2. The complex without the Reidemeister 1 twist is $C'$ in Lemma 2 while the complex in Lemma 2 is $C$.
In order to completely understand the differential in the spanning tree complex, one needs to follow all of these elementary collapses as well as keep track of the decomposition of the Khovanov complex into iterated mapping cone complexes of the twisted unknots. The rule of thumb is that the fewer number of generators for the complex, the more difficult the differential is to understand (and vice versa).
answered Dec 27 '18 at 22:57
Adam LowranceAdam Lowrance
2,19711014
answered Dec 27 '18 at 22:57
Adam LowranceAdam Lowrance
2,19711014
answered Dec 27 '18 at 22:57
Adam LowranceAdam Lowrance
2,19711014
answered Dec 27 '18 at 22:57
Adam LowranceAdam Lowrance
2,19711014
2,19711014
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