Khovanov Homology
$begingroup$
I'm reading this and trying to understand how he computed the Khovanov homology of the Hopf link. The construction of the chain complexes and the maps look fine to me but the only problem is, I do not know how to get the table on page 3 (with homological grading and quantum grading). Can anyone give an explanation?
knot-theory low-dimensional-topology knot-invariants
asked Dec 20 '18 at 20:11
JaxJax
383217
$endgroup$
add a comment |
$begingroup$
I'm reading this and trying to understand how he computed the Khovanov homology of the Hopf link. The construction of the chain complexes and the maps look fine to me but the only problem is, I do not know how to get the table on page 3 (with homological grading and quantum grading). Can anyone give an explanation?
knot-theory low-dimensional-topology knot-invariants
asked Dec 20 '18 at 20:11
JaxJax
383217
$endgroup$
add a comment |
$begingroup$
I'm reading this and trying to understand how he computed the Khovanov homology of the Hopf link. The construction of the chain complexes and the maps look fine to me but the only problem is, I do not know how to get the table on page 3 (with homological grading and quantum grading). Can anyone give an explanation?
knot-theory low-dimensional-topology knot-invariants
asked Dec 20 '18 at 20:11
JaxJax
383217
$endgroup$
I'm reading this and trying to understand how he computed the Khovanov homology of the Hopf link. The construction of the chain complexes and the maps look fine to me but the only problem is, I do not know how to get the table on page 3 (with homological grading and quantum grading). Can anyone give an explanation?
knot-theory low-dimensional-topology knot-invariants
knot-theory low-dimensional-topology knot-invariants
asked Dec 20 '18 at 20:11
JaxJax
383217
asked Dec 20 '18 at 20:11
JaxJax
383217
asked Dec 20 '18 at 20:11
JaxJax
383217
asked Dec 20 '18 at 20:11
JaxJax
383217
asked Dec 20 '18 at 20:11
JaxJax
383217
383217
add a comment |
add a comment |
1 Answer
1
active
oldest
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$begingroup$
Each generator $x$ comes equipped with two gradings: the (co)homological grading $gr(x)$ and the quantum grading $q(x)$. A generator is a choice of marking $v_+$ or $v_-$ for each component of one of the Kauffman states. See the picture below for the situation with the Hopf link.
The homological grading $gr(x)$ of any generator only depends on the Kauffman state it is associated to. It is the number of $1$-resolutions in the Kauffman state minus the number of negative crossings in the link diagram. Generators associated to the leftmost state above have $gr(x)=-2$, generators associated to the middle two states have $gr(x)=-1$, and generators associated to the rightmost state have $gr(x)=0$.
The quantum grading $q(x)$ can be computed as follows. Let $p(x)$ be the difference in the number of $v_+$'s and $v_-$'s. So if $x=v_+otimes v_+$, then $p(x)=2$, and if $x=v_+otimes v_-$, then $p(x)=0$, etc. Let $n_+(D)$ and $n_-(D)$ be the number of positive and negative crossings in the link diagram $D$. Finally let $c_1(x)$ be the number of $1$'s in the Kauffman state associated to $x$. Then
$$q(x) = p(x) +c_1(x) +n_+(D) - 2n_-(D).$$
Let $D$ be the diagram pictured above. Consider the state $x=v_-otimes v_-$ associated to the leftmost Kauffman state. Then $gr(x) = - 2$ and
$$q(x) = -2 + 0 + 0 - 2(2) = -6.$$
answered Dec 20 '18 at 20:54
Adam LowranceAdam Lowrance
2,1921914
$endgroup$
$begingroup$
Thank you very much for the explanation. Considering the leftmost Kauffman state (the same as you considered), since there are two circles, we can have $x=v_{+}otimes v_{+}$. In this case, $q(x)=2+0+0-2(2)=-2$. But if I look at the table, when $r=-2$ and $q=-2$, the homology is trivial. What did I do wrong?
$endgroup$
– Jax
Dec 20 '18 at 21:29
1
$begingroup$
You've got the grading of that generator correct. However, it is not in the kernel of the differential because $d(v_+ otimes v_+) = (1,1) v_+ neq 0$. Therefore, it cannot generate a homology class.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:33
$begingroup$
I did not understand what you meant. Could you please explain how to get the table? (sorry, this is very new to me)
$endgroup$
– Jax
Dec 20 '18 at 21:38
1
$begingroup$
The homology of a complex $(C,d)$ is the quotient group $operatorname{kernel}(d_i)/operatorname{image}(d_{i-1})$. On page 2, the author computes the boundary matrices, and on page 3, he uses those boundary matrices to compute the kernel and image of each map. The centered equations above the table shows all of the generators of the Khovanov homology of the Hopf link. It's those generators that correspond to the four summands of $mathbb{Z}$ appearing in the table.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:45
1
$begingroup$
That's correct. In general, $q$ of a sum may not make sense, if the individual terms in the sum are evaluated differently by $q$. For $x_1$, we have that $q(v_+otimes v_-) = 0 + 0 + 0 - 4$ and similarly $q(v_-otimes v_+) = 0 + 0 + 0 - 4$. Thus $q(x_1)=-4$.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 22:24
|
show 2 more comments
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$begingroup$
Each generator $x$ comes equipped with two gradings: the (co)homological grading $gr(x)$ and the quantum grading $q(x)$. A generator is a choice of marking $v_+$ or $v_-$ for each component of one of the Kauffman states. See the picture below for the situation with the Hopf link.
The homological grading $gr(x)$ of any generator only depends on the Kauffman state it is associated to. It is the number of $1$-resolutions in the Kauffman state minus the number of negative crossings in the link diagram. Generators associated to the leftmost state above have $gr(x)=-2$, generators associated to the middle two states have $gr(x)=-1$, and generators associated to the rightmost state have $gr(x)=0$.
The quantum grading $q(x)$ can be computed as follows. Let $p(x)$ be the difference in the number of $v_+$'s and $v_-$'s. So if $x=v_+otimes v_+$, then $p(x)=2$, and if $x=v_+otimes v_-$, then $p(x)=0$, etc. Let $n_+(D)$ and $n_-(D)$ be the number of positive and negative crossings in the link diagram $D$. Finally let $c_1(x)$ be the number of $1$'s in the Kauffman state associated to $x$. Then
$$q(x) = p(x) +c_1(x) +n_+(D) - 2n_-(D).$$
Let $D$ be the diagram pictured above. Consider the state $x=v_-otimes v_-$ associated to the leftmost Kauffman state. Then $gr(x) = - 2$ and
$$q(x) = -2 + 0 + 0 - 2(2) = -6.$$
answered Dec 20 '18 at 20:54
Adam LowranceAdam Lowrance
2,1921914
$endgroup$
$begingroup$
Thank you very much for the explanation. Considering the leftmost Kauffman state (the same as you considered), since there are two circles, we can have $x=v_{+}otimes v_{+}$. In this case, $q(x)=2+0+0-2(2)=-2$. But if I look at the table, when $r=-2$ and $q=-2$, the homology is trivial. What did I do wrong?
$endgroup$
– Jax
Dec 20 '18 at 21:29
1
$begingroup$
You've got the grading of that generator correct. However, it is not in the kernel of the differential because $d(v_+ otimes v_+) = (1,1) v_+ neq 0$. Therefore, it cannot generate a homology class.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:33
$begingroup$
I did not understand what you meant. Could you please explain how to get the table? (sorry, this is very new to me)
$endgroup$
– Jax
Dec 20 '18 at 21:38
1
$begingroup$
The homology of a complex $(C,d)$ is the quotient group $operatorname{kernel}(d_i)/operatorname{image}(d_{i-1})$. On page 2, the author computes the boundary matrices, and on page 3, he uses those boundary matrices to compute the kernel and image of each map. The centered equations above the table shows all of the generators of the Khovanov homology of the Hopf link. It's those generators that correspond to the four summands of $mathbb{Z}$ appearing in the table.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:45
1
$begingroup$
That's correct. In general, $q$ of a sum may not make sense, if the individual terms in the sum are evaluated differently by $q$. For $x_1$, we have that $q(v_+otimes v_-) = 0 + 0 + 0 - 4$ and similarly $q(v_-otimes v_+) = 0 + 0 + 0 - 4$. Thus $q(x_1)=-4$.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 22:24
|
show 2 more comments
$begingroup$
Each generator $x$ comes equipped with two gradings: the (co)homological grading $gr(x)$ and the quantum grading $q(x)$. A generator is a choice of marking $v_+$ or $v_-$ for each component of one of the Kauffman states. See the picture below for the situation with the Hopf link.
The homological grading $gr(x)$ of any generator only depends on the Kauffman state it is associated to. It is the number of $1$-resolutions in the Kauffman state minus the number of negative crossings in the link diagram. Generators associated to the leftmost state above have $gr(x)=-2$, generators associated to the middle two states have $gr(x)=-1$, and generators associated to the rightmost state have $gr(x)=0$.
The quantum grading $q(x)$ can be computed as follows. Let $p(x)$ be the difference in the number of $v_+$'s and $v_-$'s. So if $x=v_+otimes v_+$, then $p(x)=2$, and if $x=v_+otimes v_-$, then $p(x)=0$, etc. Let $n_+(D)$ and $n_-(D)$ be the number of positive and negative crossings in the link diagram $D$. Finally let $c_1(x)$ be the number of $1$'s in the Kauffman state associated to $x$. Then
$$q(x) = p(x) +c_1(x) +n_+(D) - 2n_-(D).$$
Let $D$ be the diagram pictured above. Consider the state $x=v_-otimes v_-$ associated to the leftmost Kauffman state. Then $gr(x) = - 2$ and
$$q(x) = -2 + 0 + 0 - 2(2) = -6.$$
answered Dec 20 '18 at 20:54
Adam LowranceAdam Lowrance
2,1921914
$endgroup$
$begingroup$
Thank you very much for the explanation. Considering the leftmost Kauffman state (the same as you considered), since there are two circles, we can have $x=v_{+}otimes v_{+}$. In this case, $q(x)=2+0+0-2(2)=-2$. But if I look at the table, when $r=-2$ and $q=-2$, the homology is trivial. What did I do wrong?
$endgroup$
– Jax
Dec 20 '18 at 21:29
1
$begingroup$
You've got the grading of that generator correct. However, it is not in the kernel of the differential because $d(v_+ otimes v_+) = (1,1) v_+ neq 0$. Therefore, it cannot generate a homology class.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:33
$begingroup$
I did not understand what you meant. Could you please explain how to get the table? (sorry, this is very new to me)
$endgroup$
– Jax
Dec 20 '18 at 21:38
1
$begingroup$
The homology of a complex $(C,d)$ is the quotient group $operatorname{kernel}(d_i)/operatorname{image}(d_{i-1})$. On page 2, the author computes the boundary matrices, and on page 3, he uses those boundary matrices to compute the kernel and image of each map. The centered equations above the table shows all of the generators of the Khovanov homology of the Hopf link. It's those generators that correspond to the four summands of $mathbb{Z}$ appearing in the table.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:45
1
$begingroup$
That's correct. In general, $q$ of a sum may not make sense, if the individual terms in the sum are evaluated differently by $q$. For $x_1$, we have that $q(v_+otimes v_-) = 0 + 0 + 0 - 4$ and similarly $q(v_-otimes v_+) = 0 + 0 + 0 - 4$. Thus $q(x_1)=-4$.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 22:24
|
show 2 more comments
$begingroup$
Each generator $x$ comes equipped with two gradings: the (co)homological grading $gr(x)$ and the quantum grading $q(x)$. A generator is a choice of marking $v_+$ or $v_-$ for each component of one of the Kauffman states. See the picture below for the situation with the Hopf link.
The homological grading $gr(x)$ of any generator only depends on the Kauffman state it is associated to. It is the number of $1$-resolutions in the Kauffman state minus the number of negative crossings in the link diagram. Generators associated to the leftmost state above have $gr(x)=-2$, generators associated to the middle two states have $gr(x)=-1$, and generators associated to the rightmost state have $gr(x)=0$.
The quantum grading $q(x)$ can be computed as follows. Let $p(x)$ be the difference in the number of $v_+$'s and $v_-$'s. So if $x=v_+otimes v_+$, then $p(x)=2$, and if $x=v_+otimes v_-$, then $p(x)=0$, etc. Let $n_+(D)$ and $n_-(D)$ be the number of positive and negative crossings in the link diagram $D$. Finally let $c_1(x)$ be the number of $1$'s in the Kauffman state associated to $x$. Then
$$q(x) = p(x) +c_1(x) +n_+(D) - 2n_-(D).$$
Let $D$ be the diagram pictured above. Consider the state $x=v_-otimes v_-$ associated to the leftmost Kauffman state. Then $gr(x) = - 2$ and
$$q(x) = -2 + 0 + 0 - 2(2) = -6.$$
answered Dec 20 '18 at 20:54
Adam LowranceAdam Lowrance
2,1921914
$endgroup$
Each generator $x$ comes equipped with two gradings: the (co)homological grading $gr(x)$ and the quantum grading $q(x)$. A generator is a choice of marking $v_+$ or $v_-$ for each component of one of the Kauffman states. See the picture below for the situation with the Hopf link.
The homological grading $gr(x)$ of any generator only depends on the Kauffman state it is associated to. It is the number of $1$-resolutions in the Kauffman state minus the number of negative crossings in the link diagram. Generators associated to the leftmost state above have $gr(x)=-2$, generators associated to the middle two states have $gr(x)=-1$, and generators associated to the rightmost state have $gr(x)=0$.
The quantum grading $q(x)$ can be computed as follows. Let $p(x)$ be the difference in the number of $v_+$'s and $v_-$'s. So if $x=v_+otimes v_+$, then $p(x)=2$, and if $x=v_+otimes v_-$, then $p(x)=0$, etc. Let $n_+(D)$ and $n_-(D)$ be the number of positive and negative crossings in the link diagram $D$. Finally let $c_1(x)$ be the number of $1$'s in the Kauffman state associated to $x$. Then
$$q(x) = p(x) +c_1(x) +n_+(D) - 2n_-(D).$$
Let $D$ be the diagram pictured above. Consider the state $x=v_-otimes v_-$ associated to the leftmost Kauffman state. Then $gr(x) = - 2$ and
$$q(x) = -2 + 0 + 0 - 2(2) = -6.$$
answered Dec 20 '18 at 20:54
Adam LowranceAdam Lowrance
2,1921914
answered Dec 20 '18 at 20:54
Adam LowranceAdam Lowrance
2,1921914
answered Dec 20 '18 at 20:54
Adam LowranceAdam Lowrance
2,1921914
answered Dec 20 '18 at 20:54
Adam LowranceAdam Lowrance
2,1921914
2,1921914
$begingroup$
Thank you very much for the explanation. Considering the leftmost Kauffman state (the same as you considered), since there are two circles, we can have $x=v_{+}otimes v_{+}$. In this case, $q(x)=2+0+0-2(2)=-2$. But if I look at the table, when $r=-2$ and $q=-2$, the homology is trivial. What did I do wrong?
$endgroup$
– Jax
Dec 20 '18 at 21:29
1
$begingroup$
You've got the grading of that generator correct. However, it is not in the kernel of the differential because $d(v_+ otimes v_+) = (1,1) v_+ neq 0$. Therefore, it cannot generate a homology class.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:33
$begingroup$
I did not understand what you meant. Could you please explain how to get the table? (sorry, this is very new to me)
$endgroup$
– Jax
Dec 20 '18 at 21:38
1
$begingroup$
The homology of a complex $(C,d)$ is the quotient group $operatorname{kernel}(d_i)/operatorname{image}(d_{i-1})$. On page 2, the author computes the boundary matrices, and on page 3, he uses those boundary matrices to compute the kernel and image of each map. The centered equations above the table shows all of the generators of the Khovanov homology of the Hopf link. It's those generators that correspond to the four summands of $mathbb{Z}$ appearing in the table.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:45
1
$begingroup$
That's correct. In general, $q$ of a sum may not make sense, if the individual terms in the sum are evaluated differently by $q$. For $x_1$, we have that $q(v_+otimes v_-) = 0 + 0 + 0 - 4$ and similarly $q(v_-otimes v_+) = 0 + 0 + 0 - 4$. Thus $q(x_1)=-4$.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 22:24
|
show 2 more comments
$begingroup$
Thank you very much for the explanation. Considering the leftmost Kauffman state (the same as you considered), since there are two circles, we can have $x=v_{+}otimes v_{+}$. In this case, $q(x)=2+0+0-2(2)=-2$. But if I look at the table, when $r=-2$ and $q=-2$, the homology is trivial. What did I do wrong?
$endgroup$
– Jax
Dec 20 '18 at 21:29
1
$begingroup$
You've got the grading of that generator correct. However, it is not in the kernel of the differential because $d(v_+ otimes v_+) = (1,1) v_+ neq 0$. Therefore, it cannot generate a homology class.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:33
$begingroup$
I did not understand what you meant. Could you please explain how to get the table? (sorry, this is very new to me)
$endgroup$
– Jax
Dec 20 '18 at 21:38
1
$begingroup$
The homology of a complex $(C,d)$ is the quotient group $operatorname{kernel}(d_i)/operatorname{image}(d_{i-1})$. On page 2, the author computes the boundary matrices, and on page 3, he uses those boundary matrices to compute the kernel and image of each map. The centered equations above the table shows all of the generators of the Khovanov homology of the Hopf link. It's those generators that correspond to the four summands of $mathbb{Z}$ appearing in the table.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:45
1
$begingroup$
That's correct. In general, $q$ of a sum may not make sense, if the individual terms in the sum are evaluated differently by $q$. For $x_1$, we have that $q(v_+otimes v_-) = 0 + 0 + 0 - 4$ and similarly $q(v_-otimes v_+) = 0 + 0 + 0 - 4$. Thus $q(x_1)=-4$.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 22:24
$begingroup$
Thank you very much for the explanation. Considering the leftmost Kauffman state (the same as you considered), since there are two circles, we can have $x=v_{+}otimes v_{+}$. In this case, $q(x)=2+0+0-2(2)=-2$. But if I look at the table, when $r=-2$ and $q=-2$, the homology is trivial. What did I do wrong?
$endgroup$
– Jax
Dec 20 '18 at 21:29
$begingroup$
Thank you very much for the explanation. Considering the leftmost Kauffman state (the same as you considered), since there are two circles, we can have $x=v_{+}otimes v_{+}$. In this case, $q(x)=2+0+0-2(2)=-2$. But if I look at the table, when $r=-2$ and $q=-2$, the homology is trivial. What did I do wrong?
$endgroup$
– Jax
Dec 20 '18 at 21:29
1
1
$begingroup$
You've got the grading of that generator correct. However, it is not in the kernel of the differential because $d(v_+ otimes v_+) = (1,1) v_+ neq 0$. Therefore, it cannot generate a homology class.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:33
$begingroup$
You've got the grading of that generator correct. However, it is not in the kernel of the differential because $d(v_+ otimes v_+) = (1,1) v_+ neq 0$. Therefore, it cannot generate a homology class.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:33
$begingroup$
I did not understand what you meant. Could you please explain how to get the table? (sorry, this is very new to me)
$endgroup$
– Jax
Dec 20 '18 at 21:38
$begingroup$
I did not understand what you meant. Could you please explain how to get the table? (sorry, this is very new to me)
$endgroup$
– Jax
Dec 20 '18 at 21:38
1
1
$begingroup$
The homology of a complex $(C,d)$ is the quotient group $operatorname{kernel}(d_i)/operatorname{image}(d_{i-1})$. On page 2, the author computes the boundary matrices, and on page 3, he uses those boundary matrices to compute the kernel and image of each map. The centered equations above the table shows all of the generators of the Khovanov homology of the Hopf link. It's those generators that correspond to the four summands of $mathbb{Z}$ appearing in the table.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:45
$begingroup$
The homology of a complex $(C,d)$ is the quotient group $operatorname{kernel}(d_i)/operatorname{image}(d_{i-1})$. On page 2, the author computes the boundary matrices, and on page 3, he uses those boundary matrices to compute the kernel and image of each map. The centered equations above the table shows all of the generators of the Khovanov homology of the Hopf link. It's those generators that correspond to the four summands of $mathbb{Z}$ appearing in the table.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 21:45
1
1
$begingroup$
That's correct. In general, $q$ of a sum may not make sense, if the individual terms in the sum are evaluated differently by $q$. For $x_1$, we have that $q(v_+otimes v_-) = 0 + 0 + 0 - 4$ and similarly $q(v_-otimes v_+) = 0 + 0 + 0 - 4$. Thus $q(x_1)=-4$.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 22:24
$begingroup$
That's correct. In general, $q$ of a sum may not make sense, if the individual terms in the sum are evaluated differently by $q$. For $x_1$, we have that $q(v_+otimes v_-) = 0 + 0 + 0 - 4$ and similarly $q(v_-otimes v_+) = 0 + 0 + 0 - 4$. Thus $q(x_1)=-4$.
$endgroup$
– Adam Lowrance
Dec 20 '18 at 22:24
|
show 2 more comments
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