An Introduction to Khovanov Homology

Dan Jones

GAndAlF / RADAGAST

20th March 2014

Dan Jones An Introduction to Khovanov Homology 1 Construct a cube of resolutions using

0 1

So if D has n crossings, there are 2n different (complete) resolutions. We present them on the vertices of the cube [0.1]n. 2 Apply a (1 + 1)-dimensional TQFT to this cube. This associates a vector space to every circle, and map between vector spaces to a cobordism between circles.

The Khovanov Complex

In order to construct the Khovanov complex, we have to:

Dan Jones An Introduction to Khovanov Homology So if D has n crossings, there are 2n different (complete) resolutions. We present them on the vertices of the cube [0.1]n. 2 Apply a (1 + 1)-dimensional TQFT to this cube. This associates a vector space to every circle, and map between vector spaces to a cobordism between circles.

The Khovanov Complex

In order to construct the Khovanov complex, we have to: 1 Construct a cube of resolutions using

0 1

Dan Jones An Introduction to Khovanov Homology 2 Apply a (1 + 1)-dimensional TQFT to this cube. This associates a vector space to every circle, and map between vector spaces to a cobordism between circles.

The Khovanov Complex

In order to construct the Khovanov complex, we have to: 1 Construct a cube of resolutions using

0 1

So if D has n crossings, there are 2n different (complete) resolutions. We present them on the vertices of the cube [0.1]n.

Dan Jones An Introduction to Khovanov Homology This associates a vector space to every circle, and map between vector spaces to a cobordism between circles.

The Khovanov Complex

In order to construct the Khovanov complex, we have to: 1 Construct a cube of resolutions using

0 1

So if D has n crossings, there are 2n different (complete) resolutions. We present them on the vertices of the cube [0.1]n. 2 Apply a (1 + 1)-dimensional TQFT to this cube.

Dan Jones An Introduction to Khovanov Homology The Khovanov Complex

In order to construct the Khovanov complex, we have to: 1 Construct a cube of resolutions using

0 1

So if D has n crossings, there are 2n different (complete) resolutions. We present them on the vertices of the cube [0.1]n. 2 Apply a (1 + 1)-dimensional TQFT to this cube. This associates a vector space to every circle, and map between vector spaces to a cobordism between circles.

Dan Jones An Introduction to Khovanov Homology Example: The trefoil

Dan Jones An Introduction to Khovanov Homology So a cobordism on an edge should correspond to a change from a 0-resolution to a 1-resolution:

So every cobordism will contain this saddle-cobordism.

Cobordisms on edges

We think of the resolution cube as increasing from 000 to 111. Then going along each edge of the cube results in exactly one coordinate changing from 0 to 1.

Dan Jones An Introduction to Khovanov Homology Cobordisms on edges

We think of the resolution cube as increasing from 000 to 111. Then going along each edge of the cube results in exactly one coordinate changing from 0 to 1. So a cobordism on an edge should correspond to a change from a 0-resolution to a 1-resolution:

So every cobordism will contain this saddle-cobordism.

Dan Jones An Introduction to Khovanov Homology Then a (1 + 1)-dimensional TQFT A assigns to each circle in the resolution cube, the vector space V . Each additional circle corresponds to a tensor factor of V . A assigns to a cobordism of circles, a map between vector spaces. Since the cobordisms on edges change one crossing from a 0-smoothing to a 1-smoothing, we will either be merging circles, or splitting them. The corresponding maps are

m :V ⊗ V → V ∆ :V → V ⊗ V

v+ ⊗ v+ 7→ v+ v+ 7→ v+ ⊗ v− + v− ⊗ v+

v+ ⊗ v− 7→ v− v− 7→ v− ⊗ v−

v− ⊗ v+ 7→ v−

v− ⊗ v− 7→ 0

A (1 + 1)-dimensional TQFT

Let V be a graded vector space spanned by v+ and v−.

Dan Jones An Introduction to Khovanov Homology Each additional circle corresponds to a tensor factor of V . A assigns to a cobordism of circles, a map between vector spaces. Since the cobordisms on edges change one crossing from a 0-smoothing to a 1-smoothing, we will either be merging circles, or splitting them. The corresponding maps are

m :V ⊗ V → V ∆ :V → V ⊗ V

v+ ⊗ v+ 7→ v+ v+ 7→ v+ ⊗ v− + v− ⊗ v+

v+ ⊗ v− 7→ v− v− 7→ v− ⊗ v−

v− ⊗ v+ 7→ v−

v− ⊗ v− 7→ 0

A (1 + 1)-dimensional TQFT

Let V be a graded vector space spanned by v+ and v−. Then a (1 + 1)-dimensional TQFT A assigns to each circle in the resolution cube, the vector space V .

Dan Jones An Introduction to Khovanov Homology A assigns to a cobordism of circles, a map between vector spaces. Since the cobordisms on edges change one crossing from a 0-smoothing to a 1-smoothing, we will either be merging circles, or splitting them. The corresponding maps are

m :V ⊗ V → V ∆ :V → V ⊗ V

v+ ⊗ v+ 7→ v+ v+ 7→ v+ ⊗ v− + v− ⊗ v+

v+ ⊗ v− 7→ v− v− 7→ v− ⊗ v−

v− ⊗ v+ 7→ v−

v− ⊗ v− 7→ 0

A (1 + 1)-dimensional TQFT

Let V be a graded vector space spanned by v+ and v−. Then a (1 + 1)-dimensional TQFT A assigns to each circle in the resolution cube, the vector space V . Each additional circle corresponds to a tensor factor of V .

Dan Jones An Introduction to Khovanov Homology Since the cobordisms on edges change one crossing from a 0-smoothing to a 1-smoothing, we will either be merging circles, or splitting them. The corresponding maps are

m :V ⊗ V → V ∆ :V → V ⊗ V

v+ ⊗ v+ 7→ v+ v+ 7→ v+ ⊗ v− + v− ⊗ v+

v+ ⊗ v− 7→ v− v− 7→ v− ⊗ v−

v− ⊗ v+ 7→ v−

v− ⊗ v− 7→ 0

A (1 + 1)-dimensional TQFT

Let V be a graded vector space spanned by v+ and v−. Then a (1 + 1)-dimensional TQFT A assigns to each circle in the resolution cube, the vector space V . Each additional circle corresponds to a tensor factor of V . A assigns to a cobordism of circles, a map between vector spaces.

Dan Jones An Introduction to Khovanov Homology The corresponding maps are

m :V ⊗ V → V ∆ :V → V ⊗ V

v+ ⊗ v+ 7→ v+ v+ 7→ v+ ⊗ v− + v− ⊗ v+

v+ ⊗ v− 7→ v− v− 7→ v− ⊗ v−

v− ⊗ v+ 7→ v−

v− ⊗ v− 7→ 0

A (1 + 1)-dimensional TQFT

Let V be a graded vector space spanned by v+ and v−. Then a (1 + 1)-dimensional TQFT A assigns to each circle in the resolution cube, the vector space V . Each additional circle corresponds to a tensor factor of V . A assigns to a cobordism of circles, a map between vector spaces. Since the cobordisms on edges change one crossing from a 0-smoothing to a 1-smoothing, we will either be merging circles, or splitting them.

Dan Jones An Introduction to Khovanov Homology A (1 + 1)-dimensional TQFT

Let V be a graded vector space spanned by v+ and v−. Then a (1 + 1)-dimensional TQFT A assigns to each circle in the resolution cube, the vector space V . Each additional circle corresponds to a tensor factor of V . A assigns to a cobordism of circles, a map between vector spaces. Since the cobordisms on edges change one crossing from a 0-smoothing to a 1-smoothing, we will either be merging circles, or splitting them. The corresponding maps are

m :V ⊗ V → V ∆ :V → V ⊗ V

v+ ⊗ v+ 7→ v+ v+ 7→ v+ ⊗ v− + v− ⊗ v+

v+ ⊗ v− 7→ v− v− 7→ v− ⊗ v−

v− ⊗ v+ 7→ v−

v− ⊗ v− 7→ 0

Dan Jones An Introduction to Khovanov Homology Example: The Trefoil

Dan Jones An Introduction to Khovanov Homology So it corresponds to a collection of k circles, or equivalently the k-tensor product V ⊗k . Denote w as the image of v under an edge-cobordism. Then:

Define a homological grading as h(v) = |v| − n−. h(w) = h(v) + 1.

Define grading α by setting α(v+) = 1 and α(v−) = −1. This is extended to tensors as α(v1 ⊗ · · · ⊗ vk ) = α(v1) + ··· + α(vk ). α(w) = α(v) − 1. We use α to define a quantum grading q as q(v) = α(v) + h(v) + n+ − n−. q(w) = q(v).

Gradings

Let v be a vertex of the resolution cube.

Dan Jones An Introduction to Khovanov Homology Denote w as the image of v under an edge-cobordism. Then:

Define a homological grading as h(v) = |v| − n−. h(w) = h(v) + 1.

Define grading α by setting α(v+) = 1 and α(v−) = −1. This is extended to tensors as α(v1 ⊗ · · · ⊗ vk ) = α(v1) + ··· + α(vk ). α(w) = α(v) − 1. We use α to define a quantum grading q as q(v) = α(v) + h(v) + n+ − n−. q(w) = q(v).

Gradings

Let v be a vertex of the resolution cube. So it corresponds to a collection of k circles, or equivalently the k-tensor product V ⊗k .

Dan Jones An Introduction to Khovanov Homology Then:

Define a homological grading as h(v) = |v| − n−. h(w) = h(v) + 1.

Define grading α by setting α(v+) = 1 and α(v−) = −1. This is extended to tensors as α(v1 ⊗ · · · ⊗ vk ) = α(v1) + ··· + α(vk ). α(w) = α(v) − 1. We use α to define a quantum grading q as q(v) = α(v) + h(v) + n+ − n−. q(w) = q(v).

Gradings

Let v be a vertex of the resolution cube. So it corresponds to a collection of k circles, or equivalently the k-tensor product V ⊗k . Denote w as the image of v under an edge-cobordism.

Dan Jones An Introduction to Khovanov Homology h(w) = h(v) + 1.

Define grading α by setting α(v+) = 1 and α(v−) = −1. This is extended to tensors as α(v1 ⊗ · · · ⊗ vk ) = α(v1) + ··· + α(vk ). α(w) = α(v) − 1. We use α to define a quantum grading q as q(v) = α(v) + h(v) + n+ − n−. q(w) = q(v).

Gradings

Let v be a vertex of the resolution cube. So it corresponds to a collection of k circles, or equivalently the k-tensor product V ⊗k . Denote w as the image of v under an edge-cobordism. Then:

Define a homological grading as h(v) = |v| − n−.

Dan Jones An Introduction to Khovanov Homology Define grading α by setting α(v+) = 1 and α(v−) = −1. This is extended to tensors as α(v1 ⊗ · · · ⊗ vk ) = α(v1) + ··· + α(vk ). α(w) = α(v) − 1. We use α to define a quantum grading q as q(v) = α(v) + h(v) + n+ − n−. q(w) = q(v).

Gradings

Let v be a vertex of the resolution cube. So it corresponds to a collection of k circles, or equivalently the k-tensor product V ⊗k . Denote w as the image of v under an edge-cobordism. Then:

Define a homological grading as h(v) = |v| − n−. h(w) = h(v) + 1.

Dan Jones An Introduction to Khovanov Homology α(w) = α(v) − 1. We use α to define a quantum grading q as q(v) = α(v) + h(v) + n+ − n−. q(w) = q(v).

Gradings

Let v be a vertex of the resolution cube. So it corresponds to a collection of k circles, or equivalently the k-tensor product V ⊗k . Denote w as the image of v under an edge-cobordism. Then:

Define a homological grading as h(v) = |v| − n−. h(w) = h(v) + 1.

Define grading α by setting α(v+) = 1 and α(v−) = −1. This is extended to tensors as α(v1 ⊗ · · · ⊗ vk ) = α(v1) + ··· + α(vk ).

Dan Jones An Introduction to Khovanov Homology We use α to define a quantum grading q as q(v) = α(v) + h(v) + n+ − n−. q(w) = q(v).

Gradings

Let v be a vertex of the resolution cube. So it corresponds to a collection of k circles, or equivalently the k-tensor product V ⊗k . Denote w as the image of v under an edge-cobordism. Then:

Define a homological grading as h(v) = |v| − n−. h(w) = h(v) + 1.

Define grading α by setting α(v+) = 1 and α(v−) = −1. This is extended to tensors as α(v1 ⊗ · · · ⊗ vk ) = α(v1) + ··· + α(vk ). α(w) = α(v) − 1.

Dan Jones An Introduction to Khovanov Homology q(w) = q(v).

Gradings

Let v be a vertex of the resolution cube. So it corresponds to a collection of k circles, or equivalently the k-tensor product V ⊗k . Denote w as the image of v under an edge-cobordism. Then:

Define a homological grading as h(v) = |v| − n−. h(w) = h(v) + 1.

Define grading α by setting α(v+) = 1 and α(v−) = −1. This is extended to tensors as α(v1 ⊗ · · · ⊗ vk ) = α(v1) + ··· + α(vk ). α(w) = α(v) − 1. We use α to define a quantum grading q as q(v) = α(v) + h(v) + n+ − n−.

Dan Jones An Introduction to Khovanov Homology Gradings

Let v be a vertex of the resolution cube. So it corresponds to a collection of k circles, or equivalently the k-tensor product V ⊗k . Denote w as the image of v under an edge-cobordism. Then:

Define a homological grading as h(v) = |v| − n−. h(w) = h(v) + 1.

Define grading α by setting α(v+) = 1 and α(v−) = −1. This is extended to tensors as α(v1 ⊗ · · · ⊗ vk ) = α(v1) + ··· + α(vk ). α(w) = α(v) − 1. We use α to define a quantum grading q as q(v) = α(v) + h(v) + n+ − n−. q(w) = q(v).

Dan Jones An Introduction to Khovanov Homology Since the quantum grading does not change passing over edges, this complex splits as a direct sum of complexes; one for each quantum grading. This gives a bi-graded chain complex.

Example: The Trefoil

For the trefoil, n− = 3, so homological grading is h(v) = |v| − 3.

Dan Jones An Introduction to Khovanov Homology Example: The Trefoil

For the trefoil, n− = 3, so homological grading is h(v) = |v| − 3.

Since the quantum grading does not change passing over edges, this complex splits as a direct sum of complexes; one for each quantum grading. This gives a bi-graded chain complex.

Dan Jones An Introduction to Khovanov Homology To ensure that we do get a chain complex (d2 = 0), we just sprinkle the cube with (−)-signs so that every face anti-commutes:

Dan Jones An Introduction to Khovanov Homology To ensure that we do get a chain complex (d2 = 0), we just sprinkle the cube with (−)-signs so that every face anti-commutes:

Dan Jones An Introduction to Khovanov Homology i,j Then Kh ( ; Z) is given by:

j \i -3 -2 -1 0 -1 Z -3 Z -5 Z -7 Z/2Z -9 Z

To prove invariance of Khovanov homology, you take two diagrams D1 and D2 of the same knot (so differ by a sequence of Reidemeister moves), and produce a quasi-isomorphism on ∗,∗ ∗,∗ C (D1) → C (D2). It is sufficient to provide a quasi-isomorphism for each of the three Reidemeister moves.

So for the trefoil, the Khovanov chain complex CKhi,j is given as before, with the additional (−)-signs to ensure we can take homology.

Dan Jones An Introduction to Khovanov Homology To prove invariance of Khovanov homology, you take two diagrams D1 and D2 of the same knot (so differ by a sequence of Reidemeister moves), and produce a quasi-isomorphism on ∗,∗ ∗,∗ C (D1) → C (D2). It is sufficient to provide a quasi-isomorphism for each of the three Reidemeister moves.

So for the trefoil, the Khovanov chain complex CKhi,j is given as before, with the additional (−)-signs to ensure we can take i,j homology. Then Kh ( ; Z) is given by:

j \i -3 -2 -1 0 -1 Z -3 Z -5 Z -7 Z/2Z -9 Z

Dan Jones An Introduction to Khovanov Homology It is sufficient to provide a quasi-isomorphism for each of the three Reidemeister moves.

So for the trefoil, the Khovanov chain complex CKhi,j is given as before, with the additional (−)-signs to ensure we can take i,j homology. Then Kh ( ; Z) is given by:

j \i -3 -2 -1 0 -1 Z -3 Z -5 Z -7 Z/2Z -9 Z

To prove invariance of Khovanov homology, you take two diagrams D1 and D2 of the same knot (so differ by a sequence of Reidemeister moves), and produce a quasi-isomorphism on ∗,∗ ∗,∗ C (D1) → C (D2).

Dan Jones An Introduction to Khovanov Homology So for the trefoil, the Khovanov chain complex CKhi,j is given as before, with the additional (−)-signs to ensure we can take i,j homology. Then Kh ( ; Z) is given by:

j \i -3 -2 -1 0 -1 Z -3 Z -5 Z -7 Z/2Z -9 Z

To prove invariance of Khovanov homology, you take two diagrams D1 and D2 of the same knot (so differ by a sequence of Reidemeister moves), and produce a quasi-isomorphism on ∗,∗ ∗,∗ C (D1) → C (D2). It is sufficient to provide a quasi-isomorphism for each of the three Reidemeister moves.

Dan Jones An Introduction to Khovanov Homology They have the same (unnormalised) Jones polynomials; −3 −5 −7 −15 Jˆ(51) = Jˆ(10132) = q + q + q − q . Khovanov homology, however, can distinguish the two:

Significance of Kh∗,∗

Khovanov homology is a stronger invariant than the Jones polynomial. To see this, consider the knots 51 and 10132:

Dan Jones An Introduction to Khovanov Homology Khovanov homology, however, can distinguish the two:

Significance of Kh∗,∗

Khovanov homology is a stronger invariant than the Jones polynomial. To see this, consider the knots 51 and 10132:

They have the same (unnormalised) Jones polynomials; −3 −5 −7 −15 Jˆ(51) = Jˆ(10132) = q + q + q − q .

Dan Jones An Introduction to Khovanov Homology Significance of Kh∗,∗

Khovanov homology is a stronger invariant than the Jones polynomial. To see this, consider the knots 51 and 10132:

They have the same (unnormalised) Jones polynomials; −3 −5 −7 −15 Jˆ(51) = Jˆ(10132) = q + q + q − q . Khovanov homology, however, can distinguish the two:

Dan Jones An Introduction to Khovanov Homology What I have been looking at is a ‘homotopy type’ of Khovanov Homology. So given a knot K, there is a sequence W j χKh(K) = χKh(K) of CW-complexes assigned to K, with the j property that

i j ∼ i,j He (χKh(K)) = Kh (K) j and each χKh(K) is an invariant of the knot, up to stable homotopy. This is interesting because there are pairs of knots K1, K2 which have the same Khovanov homology, but have distinct Khovanov homotopy types.

What now?

Continuing the study of Khovanov Homology can lead into many directions.

Dan Jones An Introduction to Khovanov Homology So given a knot K, there is a sequence W j χKh(K) = χKh(K) of CW-complexes assigned to K, with the j property that

i j ∼ i,j He (χKh(K)) = Kh (K) j and each χKh(K) is an invariant of the knot, up to stable homotopy. This is interesting because there are pairs of knots K1, K2 which have the same Khovanov homology, but have distinct Khovanov homotopy types.

What now?

Continuing the study of Khovanov Homology can lead into many directions. What I have been looking at is a ‘homotopy type’ of Khovanov Homology.

Dan Jones An Introduction to Khovanov Homology This is interesting because there are pairs of knots K1, K2 which have the same Khovanov homology, but have distinct Khovanov homotopy types.

What now?

Continuing the study of Khovanov Homology can lead into many directions. What I have been looking at is a ‘homotopy type’ of Khovanov Homology. So given a knot K, there is a sequence W j χKh(K) = χKh(K) of CW-complexes assigned to K, with the j property that

i j ∼ i,j He (χKh(K)) = Kh (K) j and each χKh(K) is an invariant of the knot, up to stable homotopy.

Dan Jones An Introduction to Khovanov Homology What now?

Continuing the study of Khovanov Homology can lead into many directions. What I have been looking at is a ‘homotopy type’ of Khovanov Homology. So given a knot K, there is a sequence W j χKh(K) = χKh(K) of CW-complexes assigned to K, with the j property that

i j ∼ i,j He (χKh(K)) = Kh (K) j and each χKh(K) is an invariant of the knot, up to stable homotopy. This is interesting because there are pairs of knots K1, K2 which have the same Khovanov homology, but have distinct Khovanov homotopy types.

Dan Jones An Introduction to Khovanov Homology