Quasimorphisms from knot invariants

Last week, Michael Brandenbursky from the Technion gave a talk at Caltech on an interesting connection between knot theory and quasimorphisms. Michael’s paper on this subject may be obtained from the arXiv. Recall that given a group $G$ , a quasimorphism is a function $\phi:G \to \mathbb{R}$ for which there is some least real number $D(\phi) \ge 0$ (called the defect) such that for all pairs of elements $g,h \in G$ there is an inequality $|\phi(gh) - \phi(g) - \phi(h)| \le D(\phi)$ . Bounded functions are quasimorphisms, although in an uninteresting way, so one is usually only interested in quasimorphisms up to the equivalence relation that $\phi \sim \psi$ if the difference $|\phi - \psi|$ is bounded. It turns out that each equivalence class of quasimorphism contains a unique representative which has the extra property that $\phi(g^n) = n\phi(g)$ for all $g\in G$ and $n \in \mathbb{Z}$ . Such quasimorphisms are said to be homogeneous. Any quasimorphism may be homogenized by defining $\overline{\phi}(g) = \lim_{n \to \infty} \phi(g^n)/n$ (see e.g. this post for more about quasimorphisms, and their relation to stable commutator length).

Many groups that do not admit many homomorphisms to $\mathbb{R}$ nevertheless admit rich families of homogeneous quasimorphisms. For example, groups that act weakly properly discontinuously on word-hyperbolic spaces admit infinite dimensional families of homogeneous quasimorphisms; see e.g. Bestvina-Fujiwara. This includes hyperbolic groups, but also mapping class groups and braid groups, which act on the complex of curves.

Michael discussed another source of quasimorphisms on braid groups, those coming from knot theory. Let $I$ be a knot invariant. Then one can extend $I$ to an invariant of pure braids on $n$ strands by $I(\alpha) = I(\widehat{\alpha \Delta})$ where $\Delta = \sigma_1 \cdots \sigma_{n-1}$ , and the “hat” denotes plat closure. It is an interesting question to ask: under what conditions on $I$ is the resulting function on braid groups a quasimorphism?

In the abstract, such a question is probably very hard to answer, so one should narrow the question by concentrating on knot invariants of a certain kind. Since one wants the resulting invariants to have some relation to the algebraic structure of braid groups, it is natural to look for functions which factor through certain algebraic structures on knots; Michael was interested in certain homomorphisms from the knot concordance group to $\mathbb{R}$ . We briefly describe this group, and a natural class of homomorphisms.

Two oriented knots $K_1,K_2$ in the $3$ -sphere are said to be concordant if there is a (locally flat) properly embedded annulus $A$ in $S^3 \times [0,1]$ with $A \cap S^3 \times 0 = K_1$ and $A \cap S^3 \times 1 = K_2$ . Concordance is an equivalence relation, and the equivalence classes form a group, with connect sum as the group operation, and orientation-reversed mirror image as inverse. The only subtle aspect of this is the existence of inverses, which we briefly explain. Let $K$ be an arbitrary knot, and let $K^!$ denote the mirror image of $K$ with the opposite orientation. Arrange $K \cup K^!$ in space so that they are symmetric with respect to reflection in a dividing plane. There is an immersed annulus $A$ in $S^3$ which connects each point on $K$ to its mirror image on $K^!$ , and the self-intersections of this annulus are all disjoint embedded arcs, corresponding to the crossings of $K$ in the projection to the mirror. This annulus is an example of what is called a ribbon surface. Connect summing $K$ to $K^!$ by pushing out a finger of each into an arc in the mirror connects the ribbon annulus to a ribbon disk spanning $K \# K^!$ . A ribbon surface (in particular, a ribbon disk) can be pushed into a (smoothly) embedded surface in a $4$ -ball bounding $S^3$ . Puncturing the $4$ -ball at some point on this smooth surface, one obtains a concordance from $K\#K^!$ to the unknot, as claimed.

The resulting group is known as the concordance group $\mathcal{C}$ of knots. Since connect sum is commutative, this group is abelian. Notice as above that a slice knot — i.e. a knot bounding a locally flat disk in the $4$ -ball — is concordant to the unknot. Ribbon knots (those bounding ribbon disks) are smoothly slice, and therefore slice, and therefore concordant to the trivial knot. Concordance makes sense for codimension two knots in any dimension. In higher even dimensions, knots are always slice, and in higher odd dimensions, Levine found an algebraic description of the concordance groups in terms of (Witt) equivalence classes of linking pairings on a Seifert surface; (some of) this information is contained in the signature of a knot.

Let $K$ be a knot (in $S^3$ for simplicity) with Seifert surface $\Sigma$ of genus $g$ . If $\alpha,\beta$ are loops in $\Sigma$ , define $f(\alpha,\beta)$ to be the linking number of $\alpha$ with $\beta^+$ , which is obtained from $\beta$ by pushing it to the positive side of $\Sigma$ . The function $f$ is a bilinear form on $H_1(\Sigma)$ , and after choosing generators, it can be expressed in terms of a matrix $V$ (called the Seifert matrix of $K$ ). The signature of $K$ , denoted $\sigma(K)$ , is the signature (in the usual sense) of the symmetric matrix $V + V^T$ . Changing the orientation of a knot does not affect the signature, whereas taking mirror image multiplies it by $-1$ . Moreover, if $\Sigma_1,\Sigma_2$ are Seifert surfaces for $K_1,K_2$ , one can form a Seifert surface $\Sigma$ for $K_1 \# K_2$ for which there is some sphere $S^2 \in S^3$ that intersects $\Sigma$ in a separating arc, so that the pieces on either side of the sphere are isotopic to the $\Sigma_i$ , and therefore the Seifert matrix of $K_1 \# K_2$ can be chosen to be block diagonal, with one block for each of the Seifert matrices of the $K_i$ ; it follows that $\sigma(K_1 \# K_2) = \sigma(K_1) + \sigma(K_2)$ . In fact it turns out that $\sigma$ is a homomorphism from $\mathcal{C}$ to $\mathbb{Z}$ ; equivalently (by the arguments above), it is zero on knots which are topologically slice. To see this, suppose $K$ bounds a locally flat disk $\Delta$ in the $4$ -ball. The union $\Sigma':=\Sigma \cup \Delta$ is an embedded bicollared surface in the $4$ -ball, which bounds a $3$ -dimensional Seifert “surface” $W$ whose interior may be taken to be disjoint from $S^3$ . Now, it is a well-known fact that for any oriented $3$ -manifold $W$ , the inclusion $\partial W \to W$ induces a map $H_1(\partial W) \to H_1(W)$ whose kernel is Lagrangian (with respect to the usual symplectic pairing on $H_1$ of an oriented surface). Geometrically, this means we can find a basis for the homology of $\Sigma'$ (which is equal to the homology of $\Sigma$ ) for which half of the basis elements bound $2$ -chains in $W$ . Let $W^+$ be obtained by pushing off $W$ in the positive direction. Then chains in $W$ and chains in $W^+$ are disjoint (since $W$ and $W^+$ are disjoint) and therefore the Seifert matrix $V$ of $K$ has a block form for which the lower right $g \times g$ block is identically zero. It follows that $V+V^T$ also has a zero $g\times g$ lower right block, and therefore its signature is zero.

The Seifert matrix (and therefore the signature), like the Alexander polynomial, is sensitive to the structure of the first homology of the universal abelian cover of $S^3 - K$ ; equivalently, to the structure of the maximal metabelian quotient of $\pi_1(S^3 - K)$ . More sophisticated “twisted” and $L^2$ signatures can be obtained by studying further derived subgroups of $\pi_1(S^3 - K)$ as modules over group rings of certain solvable groups with torsion-free abelian factors (the so-called poly-torsion-free-abelian groups). This was accomplished by Cochran-Orr-Teichner, who used these methods to construct infinitely many new concordance invariants.

The end result of this discussion is the existence of many, many interesting homomorphisms from the knot concordance group to the reals, and by plat closure, many interesting invariants of braids. The connection with quasimorphisms is the following:

Theorem(Brandenbursky): A homomorphism $I:\mathcal{C} \to \mathbb{R}$ gives rise to a quasimorphism on braid groups if there is a constant $C$ so that $|I([K])| \le C\cdot\|K\|_g$ , where $\|\cdot\|_g$ denotes $4$ -ball genus.

The proof is roughly the following: given pure braids $\alpha,\beta$ one forms the knots $\widehat{\alpha\Delta}$ , $\widehat{\beta\Delta}$ and $\widehat{\alpha\beta\Delta}$ . It is shown that the connect sum $L:= \widehat{\alpha \Delta} \# \widehat{\beta\Delta} \# \widehat{\alpha\beta\Delta}^!$ bounds a Seifert surface whose genus may be universally bounded in terms of the number of strands in the braid group. Pushing this Seifert surface into the $4$ -ball, the hypothesis of the theorem says that $I$ is uniformly bounded on $L$ . Properties of $I$ then give an estimate for the defect; qed.

It would be interesting to connect these observations up to other “natural” chiral, homogeneous invariants on mapping class groups. For example, associated to a braid or mapping class $\phi \in \text{MCG}(S)$ one can (usually) form a hyperbolic $3$ -manifold $M_\phi$ which fibers over the circle, with fiber $S$ and monodromy $\phi$ . The $\eta$ -invariant of $M_\phi$ is the signature defect $\eta(M_\phi) = \int_Y p_1/3 - \text{sign}(Y)$ where $Y$ is a $4$ -manifold with $\partial Y = M_\phi$ with a product metric near the boundary, and $p_1$ is the first Pontriagin form on $Y$ (expressed in terms of the curvature of the metric). Is $\eta$ a quasimorphism on some subgroup of $\text{MCG}(S)$ (eg on a subgroup consisting entirely of pseudo-Anosov elements)?

2 Responses to Quasimorphisms from knot invariants

JSE says:

October 23, 2009 at 8:36 pm

Naive question. The dilatation lambda is a map from the mapping class group to R (or at least on pseudo-Anosovs, so replace the mapping class group with a subgroup consisting entirely of p-As if you want.) It satisfies lambda(g^n) = n lambda(g). Is there any chance this is a quasimorphism?

My guess is no, just on grounds that lambda is a bit like a function that sends matrices to the largest absolute value of an eigenvalue, and this isn’t a quasimorphism at all. (Indeed, for Gamma_{1,1} this really IS what lambda is so I guess that ends the question in at least one case.)

- Danny Calegari says:
  
  October 23, 2009 at 9:21 pm
  
  Dilatation satisfies $\lambda(g^n) = n \lambda(g)$ for positive $n$ , but unfortunately not for negative $n$ ; it is more like a norm (as your comment more or less observes). One can think of dilatation as a bit like translation length (in the Teichmuller metric); by contrast, a quasimorphism is less like the length of a geodesic than the integral of some $1$ -form over that geodesic. This idea leads to the concept of counting quasimorphisms, one of the few known general methods for constructing quasimorphisms on groups (under suitable negative curvature hypotheses). The Bestvina-Fujiwara construction of quasimorphisms on mapping class groups is of this nature.
  
  Stable norms — those satisfying $\phi(g^n) = |n|\phi(g)$ — and quasimorphisms are related; for instance, there is a duality between stable commutator length (which is a (pseudo)-norm on the space of homologically trivial group 1-chains) and homogeneous quasimorphisms modulo homomorphisms (with the ”defect” norm). Another interesting relationship between norms and quasimorphisms is discussed in a paper of Burago-Ivanov-Polterovich (http://arxiv.org/abs/0710.1412).

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