Degree shift operation when constructing chain complexes of the Khovanov Homology
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I'm reading this paper by Dror Bar-Natan, On Khovanov’s categorification of the Jones polynomial (here). On chapter 3 (categorification), he wrote: With every vertex $alphain {0,1 }^chi$ of the cube ${0,1 }^chi$, we associate the graded vector space $V_alpha(L)=V^{otimes k}{r}$ where $k$ is the number of cycles in the smoothing of $L$ corresponding to $alpha$ and $r$ is the height $|alpha|=sum_{i}alpha_i$ of $alpha$.
My question is, why do we need the degree shift operation $cdot {r }$ on $V^{otimes k}$? In most of the computation, we kind of ignore this degree shift but only care about $V^{otimes k}$.
algebraic-topology knot-theory knot-invariants
asked Dec 22 '18 at 2:45
MinahMinah
275
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add a comment |
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I'm reading this paper by Dror Bar-Natan, On Khovanov’s categorification of the Jones polynomial (here). On chapter 3 (categorification), he wrote: With every vertex $alphain {0,1 }^chi$ of the cube ${0,1 }^chi$, we associate the graded vector space $V_alpha(L)=V^{otimes k}{r}$ where $k$ is the number of cycles in the smoothing of $L$ corresponding to $alpha$ and $r$ is the height $|alpha|=sum_{i}alpha_i$ of $alpha$.
My question is, why do we need the degree shift operation $cdot {r }$ on $V^{otimes k}$? In most of the computation, we kind of ignore this degree shift but only care about $V^{otimes k}$.
algebraic-topology knot-theory knot-invariants
asked Dec 22 '18 at 2:45
MinahMinah
275
$endgroup$
add a comment |
$begingroup$
I'm reading this paper by Dror Bar-Natan, On Khovanov’s categorification of the Jones polynomial (here). On chapter 3 (categorification), he wrote: With every vertex $alphain {0,1 }^chi$ of the cube ${0,1 }^chi$, we associate the graded vector space $V_alpha(L)=V^{otimes k}{r}$ where $k$ is the number of cycles in the smoothing of $L$ corresponding to $alpha$ and $r$ is the height $|alpha|=sum_{i}alpha_i$ of $alpha$.
My question is, why do we need the degree shift operation $cdot {r }$ on $V^{otimes k}$? In most of the computation, we kind of ignore this degree shift but only care about $V^{otimes k}$.
algebraic-topology knot-theory knot-invariants
asked Dec 22 '18 at 2:45
MinahMinah
275
$endgroup$
I'm reading this paper by Dror Bar-Natan, On Khovanov’s categorification of the Jones polynomial (here). On chapter 3 (categorification), he wrote: With every vertex $alphain {0,1 }^chi$ of the cube ${0,1 }^chi$, we associate the graded vector space $V_alpha(L)=V^{otimes k}{r}$ where $k$ is the number of cycles in the smoothing of $L$ corresponding to $alpha$ and $r$ is the height $|alpha|=sum_{i}alpha_i$ of $alpha$.
My question is, why do we need the degree shift operation $cdot {r }$ on $V^{otimes k}$? In most of the computation, we kind of ignore this degree shift but only care about $V^{otimes k}$.
algebraic-topology knot-theory knot-invariants
algebraic-topology knot-theory knot-invariants
asked Dec 22 '18 at 2:45
MinahMinah
275
asked Dec 22 '18 at 2:45
MinahMinah
275
asked Dec 22 '18 at 2:45
MinahMinah
275
asked Dec 22 '18 at 2:45
MinahMinah
275
asked Dec 22 '18 at 2:45
MinahMinah
275
275
add a comment |
add a comment |
1 Answer
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The degree shift operator ensures that the differential preserves polynomial/quantum grading. If we ignore an overall shift depending on the number of positive and negative crossings in the diagram, then the polynomial grading is essentially the number of $v_+$'s minus the number of $v_-$'s plus the height $r$.
Let's look at the maps used to construct the differential. The multiplication $m:Votimes V to V$ is defined by
$$m(v_+otimes v_-)=m(v_-otimes v_+) = v_-,~m(v_+otimes v_+)=v_+,~text{and}~m(v_-otimes v_-)=0.$$
The comulitplication $Delta:Vto Votimes V$ is defined by
$$Delta(v_+) = v_+otimes v_- + v_-otimes v_+~text{and}~Delta(v_-)=v_- otimes v_-.$$
In the multiplication map, there is one fewer $v_+$ in each output and the same number of $v_-$'s. But the height of the output is one more than the height of the input. Thus the polynomial grading is preserved. Similarly, in the comultiplication map, there is one more $v_-$ in the output than the input and the same number of $v_+$'s. Again the height of the output is one greater than the height of the input causing the polynomial grading to be preserved.
answered Dec 22 '18 at 3:09
Adam LowranceAdam Lowrance
2,20711014
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Thank you very much. You made it really clear to me.
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– Minah
Dec 22 '18 at 3:33
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
The degree shift operator ensures that the differential preserves polynomial/quantum grading. If we ignore an overall shift depending on the number of positive and negative crossings in the diagram, then the polynomial grading is essentially the number of $v_+$'s minus the number of $v_-$'s plus the height $r$.
Let's look at the maps used to construct the differential. The multiplication $m:Votimes V to V$ is defined by
$$m(v_+otimes v_-)=m(v_-otimes v_+) = v_-,~m(v_+otimes v_+)=v_+,~text{and}~m(v_-otimes v_-)=0.$$
The comulitplication $Delta:Vto Votimes V$ is defined by
$$Delta(v_+) = v_+otimes v_- + v_-otimes v_+~text{and}~Delta(v_-)=v_- otimes v_-.$$
In the multiplication map, there is one fewer $v_+$ in each output and the same number of $v_-$'s. But the height of the output is one more than the height of the input. Thus the polynomial grading is preserved. Similarly, in the comultiplication map, there is one more $v_-$ in the output than the input and the same number of $v_+$'s. Again the height of the output is one greater than the height of the input causing the polynomial grading to be preserved.
answered Dec 22 '18 at 3:09
Adam LowranceAdam Lowrance
2,20711014
$endgroup$
$begingroup$
Thank you very much. You made it really clear to me.
$endgroup$
– Minah
Dec 22 '18 at 3:33
add a comment |
$begingroup$
The degree shift operator ensures that the differential preserves polynomial/quantum grading. If we ignore an overall shift depending on the number of positive and negative crossings in the diagram, then the polynomial grading is essentially the number of $v_+$'s minus the number of $v_-$'s plus the height $r$.
Let's look at the maps used to construct the differential. The multiplication $m:Votimes V to V$ is defined by
$$m(v_+otimes v_-)=m(v_-otimes v_+) = v_-,~m(v_+otimes v_+)=v_+,~text{and}~m(v_-otimes v_-)=0.$$
The comulitplication $Delta:Vto Votimes V$ is defined by
$$Delta(v_+) = v_+otimes v_- + v_-otimes v_+~text{and}~Delta(v_-)=v_- otimes v_-.$$
In the multiplication map, there is one fewer $v_+$ in each output and the same number of $v_-$'s. But the height of the output is one more than the height of the input. Thus the polynomial grading is preserved. Similarly, in the comultiplication map, there is one more $v_-$ in the output than the input and the same number of $v_+$'s. Again the height of the output is one greater than the height of the input causing the polynomial grading to be preserved.
answered Dec 22 '18 at 3:09
Adam LowranceAdam Lowrance
2,20711014
$endgroup$
$begingroup$
Thank you very much. You made it really clear to me.
$endgroup$
– Minah
Dec 22 '18 at 3:33
add a comment |
$begingroup$
The degree shift operator ensures that the differential preserves polynomial/quantum grading. If we ignore an overall shift depending on the number of positive and negative crossings in the diagram, then the polynomial grading is essentially the number of $v_+$'s minus the number of $v_-$'s plus the height $r$.
Let's look at the maps used to construct the differential. The multiplication $m:Votimes V to V$ is defined by
$$m(v_+otimes v_-)=m(v_-otimes v_+) = v_-,~m(v_+otimes v_+)=v_+,~text{and}~m(v_-otimes v_-)=0.$$
The comulitplication $Delta:Vto Votimes V$ is defined by
$$Delta(v_+) = v_+otimes v_- + v_-otimes v_+~text{and}~Delta(v_-)=v_- otimes v_-.$$
In the multiplication map, there is one fewer $v_+$ in each output and the same number of $v_-$'s. But the height of the output is one more than the height of the input. Thus the polynomial grading is preserved. Similarly, in the comultiplication map, there is one more $v_-$ in the output than the input and the same number of $v_+$'s. Again the height of the output is one greater than the height of the input causing the polynomial grading to be preserved.
answered Dec 22 '18 at 3:09
Adam LowranceAdam Lowrance
2,20711014
$endgroup$
The degree shift operator ensures that the differential preserves polynomial/quantum grading. If we ignore an overall shift depending on the number of positive and negative crossings in the diagram, then the polynomial grading is essentially the number of $v_+$'s minus the number of $v_-$'s plus the height $r$.
Let's look at the maps used to construct the differential. The multiplication $m:Votimes V to V$ is defined by
$$m(v_+otimes v_-)=m(v_-otimes v_+) = v_-,~m(v_+otimes v_+)=v_+,~text{and}~m(v_-otimes v_-)=0.$$
The comulitplication $Delta:Vto Votimes V$ is defined by
$$Delta(v_+) = v_+otimes v_- + v_-otimes v_+~text{and}~Delta(v_-)=v_- otimes v_-.$$
In the multiplication map, there is one fewer $v_+$ in each output and the same number of $v_-$'s. But the height of the output is one more than the height of the input. Thus the polynomial grading is preserved. Similarly, in the comultiplication map, there is one more $v_-$ in the output than the input and the same number of $v_+$'s. Again the height of the output is one greater than the height of the input causing the polynomial grading to be preserved.
answered Dec 22 '18 at 3:09
Adam LowranceAdam Lowrance
2,20711014
answered Dec 22 '18 at 3:09
Adam LowranceAdam Lowrance
2,20711014
answered Dec 22 '18 at 3:09
Adam LowranceAdam Lowrance
2,20711014
answered Dec 22 '18 at 3:09
Adam LowranceAdam Lowrance
2,20711014
2,20711014
$begingroup$
Thank you very much. You made it really clear to me.
$endgroup$
– Minah
Dec 22 '18 at 3:33
add a comment |
$begingroup$
Thank you very much. You made it really clear to me.
$endgroup$
– Minah
Dec 22 '18 at 3:33
$begingroup$
Thank you very much. You made it really clear to me.
$endgroup$
– Minah
Dec 22 '18 at 3:33
$begingroup$
Thank you very much. You made it really clear to me.
$endgroup$
– Minah
Dec 22 '18 at 3:33
add a comment |
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