Khovanov homology over vector spaces, Z-modules or groups?
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When I read various papers on Khovanov homology, sometimes it is defined in terms of graded vector spaces, sometimes as graded $mathbb{Z}$-modules. Is there a difference? E.g. can the vector field definition find torsion in the homology or does it just return the free part?
Why isn't the Khovanov constructed using Abelian groups, wouldn't this be the simplest?
knot-theory knot-invariants
asked Jan 8 at 17:03
Jake B.Jake B.
1786
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add a comment |
$begingroup$
When I read various papers on Khovanov homology, sometimes it is defined in terms of graded vector spaces, sometimes as graded $mathbb{Z}$-modules. Is there a difference? E.g. can the vector field definition find torsion in the homology or does it just return the free part?
Why isn't the Khovanov constructed using Abelian groups, wouldn't this be the simplest?
knot-theory knot-invariants
asked Jan 8 at 17:03
Jake B.Jake B.
1786
$endgroup$
add a comment |
$begingroup$
When I read various papers on Khovanov homology, sometimes it is defined in terms of graded vector spaces, sometimes as graded $mathbb{Z}$-modules. Is there a difference? E.g. can the vector field definition find torsion in the homology or does it just return the free part?
Why isn't the Khovanov constructed using Abelian groups, wouldn't this be the simplest?
knot-theory knot-invariants
asked Jan 8 at 17:03
Jake B.Jake B.
1786
$endgroup$
When I read various papers on Khovanov homology, sometimes it is defined in terms of graded vector spaces, sometimes as graded $mathbb{Z}$-modules. Is there a difference? E.g. can the vector field definition find torsion in the homology or does it just return the free part?
Why isn't the Khovanov constructed using Abelian groups, wouldn't this be the simplest?
knot-theory knot-invariants
knot-theory knot-invariants
asked Jan 8 at 17:03
Jake B.Jake B.
1786
asked Jan 8 at 17:03
Jake B.Jake B.
1786
asked Jan 8 at 17:03
Jake B.Jake B.
1786
asked Jan 8 at 17:03
Jake B.Jake B.
1786
asked Jan 8 at 17:03
Jake B.Jake B.
1786
1786
add a comment |
add a comment |
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Khovanov homology is usually defined over the fields $mathbb{Q}$ or $mathbb{Z}_p$ for some prime $p$, or over $mathbb{Z}$. There is a difference, and Khovanov homology is potentially a stronger invariant when taken with $mathbb{Z}$ coefficients than when taken with field coefficients. This is because there are almost always nontrivial torsion summands in Khovanov homology over $mathbb{Z}$. In fact, Shumakovitch (arXiv link) conjectures that every knot other than the unknot has torsion of order two in its Khovanov homology over $mathbb{Z}$.
If the Khovanov homology of the knot is homologically thin, that is, if it is supported in bigradings $(i,j)$ such that $2i-j = spm 1$ for some integer $s$, then Shumakovitch (arXiv link) proved that the only torsion in the Khovanov homology of the knot over $mathbb{Z}$ is of order two, and that torsion is determined by the free part. However, computations have shown that torsion of order other than two can appear in Khovanov homology. For example, the $(5,6)$-torus knot has torsion of order three and five in its Khovanov homology (see the bottom of this Knot Atlas link). More recently a paper by Mukherjee, Przytycki, Silvero, Wang, and Yang (arXiv link) found plenty of examples with interesting torsion in their Khovanov homology.
answered Jan 8 at 18:31
Adam LowranceAdam Lowrance
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$begingroup$
Khovanov homology is usually defined over the fields $mathbb{Q}$ or $mathbb{Z}_p$ for some prime $p$, or over $mathbb{Z}$. There is a difference, and Khovanov homology is potentially a stronger invariant when taken with $mathbb{Z}$ coefficients than when taken with field coefficients. This is because there are almost always nontrivial torsion summands in Khovanov homology over $mathbb{Z}$. In fact, Shumakovitch (arXiv link) conjectures that every knot other than the unknot has torsion of order two in its Khovanov homology over $mathbb{Z}$.
If the Khovanov homology of the knot is homologically thin, that is, if it is supported in bigradings $(i,j)$ such that $2i-j = spm 1$ for some integer $s$, then Shumakovitch (arXiv link) proved that the only torsion in the Khovanov homology of the knot over $mathbb{Z}$ is of order two, and that torsion is determined by the free part. However, computations have shown that torsion of order other than two can appear in Khovanov homology. For example, the $(5,6)$-torus knot has torsion of order three and five in its Khovanov homology (see the bottom of this Knot Atlas link). More recently a paper by Mukherjee, Przytycki, Silvero, Wang, and Yang (arXiv link) found plenty of examples with interesting torsion in their Khovanov homology.
answered Jan 8 at 18:31
Adam LowranceAdam Lowrance
2,20711014
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add a comment |
$begingroup$
Khovanov homology is usually defined over the fields $mathbb{Q}$ or $mathbb{Z}_p$ for some prime $p$, or over $mathbb{Z}$. There is a difference, and Khovanov homology is potentially a stronger invariant when taken with $mathbb{Z}$ coefficients than when taken with field coefficients. This is because there are almost always nontrivial torsion summands in Khovanov homology over $mathbb{Z}$. In fact, Shumakovitch (arXiv link) conjectures that every knot other than the unknot has torsion of order two in its Khovanov homology over $mathbb{Z}$.
If the Khovanov homology of the knot is homologically thin, that is, if it is supported in bigradings $(i,j)$ such that $2i-j = spm 1$ for some integer $s$, then Shumakovitch (arXiv link) proved that the only torsion in the Khovanov homology of the knot over $mathbb{Z}$ is of order two, and that torsion is determined by the free part. However, computations have shown that torsion of order other than two can appear in Khovanov homology. For example, the $(5,6)$-torus knot has torsion of order three and five in its Khovanov homology (see the bottom of this Knot Atlas link). More recently a paper by Mukherjee, Przytycki, Silvero, Wang, and Yang (arXiv link) found plenty of examples with interesting torsion in their Khovanov homology.
answered Jan 8 at 18:31
Adam LowranceAdam Lowrance
2,20711014
$endgroup$
add a comment |
$begingroup$
Khovanov homology is usually defined over the fields $mathbb{Q}$ or $mathbb{Z}_p$ for some prime $p$, or over $mathbb{Z}$. There is a difference, and Khovanov homology is potentially a stronger invariant when taken with $mathbb{Z}$ coefficients than when taken with field coefficients. This is because there are almost always nontrivial torsion summands in Khovanov homology over $mathbb{Z}$. In fact, Shumakovitch (arXiv link) conjectures that every knot other than the unknot has torsion of order two in its Khovanov homology over $mathbb{Z}$.
If the Khovanov homology of the knot is homologically thin, that is, if it is supported in bigradings $(i,j)$ such that $2i-j = spm 1$ for some integer $s$, then Shumakovitch (arXiv link) proved that the only torsion in the Khovanov homology of the knot over $mathbb{Z}$ is of order two, and that torsion is determined by the free part. However, computations have shown that torsion of order other than two can appear in Khovanov homology. For example, the $(5,6)$-torus knot has torsion of order three and five in its Khovanov homology (see the bottom of this Knot Atlas link). More recently a paper by Mukherjee, Przytycki, Silvero, Wang, and Yang (arXiv link) found plenty of examples with interesting torsion in their Khovanov homology.
answered Jan 8 at 18:31
Adam LowranceAdam Lowrance
2,20711014
$endgroup$
Khovanov homology is usually defined over the fields $mathbb{Q}$ or $mathbb{Z}_p$ for some prime $p$, or over $mathbb{Z}$. There is a difference, and Khovanov homology is potentially a stronger invariant when taken with $mathbb{Z}$ coefficients than when taken with field coefficients. This is because there are almost always nontrivial torsion summands in Khovanov homology over $mathbb{Z}$. In fact, Shumakovitch (arXiv link) conjectures that every knot other than the unknot has torsion of order two in its Khovanov homology over $mathbb{Z}$.
If the Khovanov homology of the knot is homologically thin, that is, if it is supported in bigradings $(i,j)$ such that $2i-j = spm 1$ for some integer $s$, then Shumakovitch (arXiv link) proved that the only torsion in the Khovanov homology of the knot over $mathbb{Z}$ is of order two, and that torsion is determined by the free part. However, computations have shown that torsion of order other than two can appear in Khovanov homology. For example, the $(5,6)$-torus knot has torsion of order three and five in its Khovanov homology (see the bottom of this Knot Atlas link). More recently a paper by Mukherjee, Przytycki, Silvero, Wang, and Yang (arXiv link) found plenty of examples with interesting torsion in their Khovanov homology.
answered Jan 8 at 18:31
Adam LowranceAdam Lowrance
2,20711014
edited Jan 9 at 17:01
answered Jan 8 at 18:31
Adam LowranceAdam Lowrance
2,20711014
answered Jan 8 at 18:31
Adam LowranceAdam Lowrance
2,20711014
answered Jan 8 at 18:31
Adam LowranceAdam Lowrance
2,20711014
2,20711014
add a comment |
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