Tags: #projects/notes/seminars #geomtop\ Refs: Conference and Seminar Talks
Joint GT/UGA Topology Seminar: Monday January 13th
Talk 1: Knot Floer Homology and Cosmetic Surgeries
Dehn surgery : fundamental procedure for building 3manifolds.
Outline
 Background on problem
 Results (known and new)
 Tools
 Proof
Basics
Let \(K\hookrightarrow S^3\) be a knot. Pick a rational number \(p/q\) or \(\infty\). Can perform \(p/q{\hbox{}}\)surgery (i.e. Dehn surgery) to obtain \(S_{p/q}^3(K) \coloneqq(S^3\setminus K) {\textstyle\coprod}_f (D^2 \times S^1)\) where \({{\partial}}D^2 \times{\operatorname{pt}}\mapsto p\mu + q\lambda\) where \(\mu = ?, \lambda\) is the Seifert fiber (?).
Is this a unique operation? I.e. do different knots yield different 3manifolds?
Question: Can different surgeries on the same knot yield different 3manifolds?
Definition: Two surgeries are purely cosmetic iff there is an orientationpreserving diffeomorphism between them. If there is an orientation reversing diffeomorphism, they are said to be chirally cosmetic.
Conjecture: There are no purely cosmetic surgeries.
Remark: The conjecture can be stated for \(K \hookrightarrow Y^3\)
Note: don’t know what \(Y^3\) is.
Remark: There exist chirally cosmetic surgeries.
Example: \(+9\) and \(+\frac 9 2\) surgery on \(T_{2, 3}\), or \(+r, r\) on any ampichiral knot.
Remark: Meant to generalize the “knot complement problem”, i.e. are knots determined by their complements?
Theorem (GordonLuecke 89): If \(S_r^3(K) = S^3 = S_\infty^3(K)\), then \(r=\infty\). (I.e. the only trivial surgery really is the trivial surgery?)
Known Results
Suppose \(K\subset S^3\)is nontrivial, and two Dehn surgeries with different slopes are diffeomorphic.
 Computing \(H^1 = {\mathbf{Z}}_p\) forces \(p=p'\).
 By BoyerLines ’91, the Alexander polynomial satisfies \(\Delta_k''(1) = 0\).

By OsvathSzaboWu (’05, ’09), \(q\) and \(q'\) have opposite signs (not necessarily \(q=q'\)).
 So there at most two ways of getting the same manifold from cosmetic surgeries.
 By Wang ’06, \(g\neq 1\)
 By NiWu ’10, \(\tau(K) = 0, q' = q\), and \(q^2 = 1 \operatorname{mod}p\).
Theorem: Let \(q > 0\), so \(q' < 0\). Then
 \(p = 1, 2\)
 If \(p=2\), then \(q = 1\) and \(g=2\).
 If \(p=1\), then \(q \leq \frac{t+2g}{2g(g1)}\)
where \(g\) is the genus and \(t\) is the HeegardFloer thickness.
Moreover, the knot Floer homology satisfies some further conditions (stronger than e.g. \(\tau(K) = 0\)).
Note that if \(t < 4\), then the last condition forces \(q=1, g=2\). We then only have to consider two slopes.
Corollary: The corollary holds for thin knots (i.e. thickness zero), e.g. all alternating and quasialternating knots.
For knots up to 16 crossings, \(t\leq 2\) (from computations of knotFloer homology on 1.6 million knots?) When \(K\) is thin, this condition can be stated in terms of the Alexander polynomial.
Theorem: If \(K\) is thin and has purely cosmetic surgery, then
 \(g(K) = 2\)
 The slopes are \(\pm 1, \pm 2\)
 The coefficients of the Alexander polynomial occur in ratios: \(\Delta_k(t) = nt^2  4nt + (6n+1)  4nt^{1}+ nt^{2}\) for some \(n\),
This is computationally effective:
 Number of prime knots with at most 16 crossings: 1.7 million
 Number with \(\tau = 0\): 450,000
 Number satisfying the conditions in the theorem: 337
For each of these, need to consider \(\pm 1, \pm 2\). Noting from HF will distinguish these surgeries. Can use SnapPea to compute hyperbolic invariants – most are distinguished by hyperbolic volume, or ChernSimons invariant.
Could potentially take connect sums of the above knots, but eventually they stop satisfying the necessary condition. In fact, the conjecture holds if all prime summands of \(K\) have less than 16 crossings.
Theorem (IchiharaSongMattmanSaito): The conjecture holds for all 2bridge knots.
Theorem (Tao): Conjecture holds for arbitrary connected sums.
So if there is a knot with cosmetic surgery, it is not prime.
Remark: FuterPurcellSchleimer independently proved a similar result using hyperbolic techniques.
Tools
What we’ll want
 Need some 3manifold invariant to distinguish different surgeries
 A knot invariant
 A surgery formula computing (1) from (2) and the slope \(p, q\).
Previously used
 CassemGordon and CassemWalker invariants, and
 Alexander polynomial (\(\implies \Delta''(1) = 0\))
For (1), we’ll use HeegardFloer homology for the 3manifold invariant: Associated to a closed oriented 3manifold \(Y\) a graded vector space \(\widehat{HF}(Y)\). In our case, it will be over \({\mathbf{Z}}/(2)\), and splits over \(\mathrm{Spin}^c\) structures as \(\widehat{HF}(Y) = \bigoplus_{s\in \mathrm{Spin}^c(Y)} \widehat{HF}(Y; s)\).
Note that \(\mathrm{Spin}^c(Y)\) can be put in correspondence with \(H^1(Y)\).
For (2), we’ll use knot Floer homology, namely a reformulation following HRasmussenWatson. To a knot \(K \subset S^3\), associate

An immersed collection of closed curves \(\Gamma = (\gamma_0, \cdots, \gamma_n)\) in the punctured torus \(T\):
 I.e. a graded immersed Lagrangian
 Defined up to homotopy equivalence, where homotopies can’t cross the puncture.

Some grading data (\(\times 2\), Alexander and Maslov) amounting to labeling each component of \(\Gamma\) with an integer. (Important to proof!)

(From world of immersed Lagrangians) A bounding cochain, i.e. a subset of selfintersection points of \(\Gamma\).
Interpret the Alexander grading as specifying a lift \(\mkern 1.5mu\overline{\mkern1.5mu\Gamma\mkern1.5mu}\mkern 1.5mu\) of \(\Gamma\) from \(T\) to \(\mkern 1.5mu\overline{\mkern1.5muT\mkern1.5mu}\mkern 1.5mu\), a \({\mathbf{Z}}{\hbox{}}\)fold covering space of \(T\):
Examples (These are curves wrapped around cylinders):
Somehow, this last example is representative.
A surgery formula : \(\widehat{HF}(S_{p/q}^3(K))\) is floer homology in \(T\) of \(\Gamma\) with \(\ell_{p,q}\) a line of slope \(p/q\), i.e. it is generated by minimal intersection points \(\Gamma \cap\ell_{p, q}\). (This gives a chain complex, count bigons.)
For the \(\mathrm{spin}^c\) decomposition, look at \(\mkern 1.5mu\overline{\mkern1.5muT\mkern1.5mu}\mkern 1.5mu\) with different lifts of \(\ell_{p, q}\).
Talk 2: Branched Covers Bounding \({\mathbf{Q}}H B^4\)
Joint work with Aceto, Meier, A. Miller, M. Miller, Stipsicz.
Definition: Two knots \(K_0, K_1\) are concordant iff they cobound an annulus, i.e. there exists a smooth cylinder \(S^1 \times I\) embedded in \(S^3 \times I\) such that \(S^1 \times\left\{{i}\right\}\) in \(S^3 \times\left\{{i}\right\}\) is \(K_i\).
The concordance group is an abelian group defined by \(C = \left\{{\text{knots in } S^3}\right\}/\sim\) where we identify knots that are concordant.
Theorem (FoxMilnor 66): If \(K\) is slice, then \(\Delta_K(t) = f(t) f(t^{1}) \in {\mathbf{Z}}[t^{\pm 1}]\).
Remark: Define \(A(k) \coloneqq H_1(\tilde{ S^3\setminus K}, {\mathbf{Z}})\) as the integral homology of the infinite cyclic cover as a \({\mathbf{Z}}[t^{\pm 1}]{\hbox{}}\)module. This is equal to \(H_1(S^3\setminus K; {\mathbf{Z}}[t^{\pm 1}])\). Then \(\Delta_k(t) \coloneqq\mathrm{ord}(A(k))\). Given an element of \(M_{n, m}\) we get \({\mathbf{Z}}[t^{\pm 1}]^m \to {\mathbf{Z}}[t^{\pm 1}]^n \to A(k) \to 0\). We can consider the ideal generated by all the minors (the order ideal), and if this ideal is principal we call the generator \(\Delta_k(t)\).
\(V  tV^t\) (the Seifert matrix?) is a square presentation matrix for \(A(k)\), so \(\Delta_k(t) = \operatorname{det}(V  tV^t)\). Note that this is easy to compute. Example: for the figure 8, \(\Delta_{4, 1}(t) = \operatorname{det}([1t, t; 1, 1+t]) = t^2 + 3t  1\).
There is a notion of algebraically slice, and an algebraically slice knot implies FoxMilnor.
Theorem (CassonGordon, 78): If \(K\) is slice and \(p\) prime then the \(p^r{\hbox{}}\)fold branched cover \(\Sigma_{p^r}(K)\) is a rational homology 3sphere \({\mathbf{Q}}H S^3\) and bounds a rational homology 4ball \({\mathbf{Q}}HB^4\).
Remark: \(\Sigma_{p^r}(K_1 {\sharp}K_2) = \Sigma_{p^r} K_1 {\sharp}\Sigma_{p^r} K_2\). The following map measures the obstruction to being slice: \(\beta_{p^r}: \rho \to \Theta^3_{\mathbf{Q}}\), where \([K] \to [\Sigma_{p^r} K]\).
Question: How good is \(\beta_{p^r}\) as a slice obstruction?
\(\beta_2\) is pretty good for 2bridge knots, i.e.
Theorem (Lisca 07): If \(K\) is a connected sum of 2bridge knots, then \(\beta_2(K) = 0 \implies K\) is slice.
Note: there are nonslice knots with \(\beta_2 =0\).
Theorem (CassonHaner 81): For each \(s>0\), \(\Sigma(2, 2s1, 2s+1) \cong \Sigma_2(T_{2s1, 2s+1})\) bounds a contractible manifold.
Theorem (Litherland 78): Torus knots are \(L_\delta I_0\) in \(\rho\). (??)
These together imply that \(\ker \beta_2 \geq {\mathbf{Z}}^\infty\).
Theorem (AcetoLarson 18): \(\ker \beta_2 \cong {\mathbf{Z}}^\infty \oplus G\).
Main Theorem: \(\cap_{p \text{ prime }, r\in {\mathbb{N}}} \ker \beta_{p^r} \geq ({\mathbf{Z}}/(2))^4\).
Knots in here have arf invariant zero, and are torsion in the concordance group.
Step 1: Construction
Define \(K_n \coloneqq(\sigma_1 \cdot \sigma_2^{1})^n\) for \(\sigma_i\) in the braid group \(B_3\).
Example:
Definition: \(K\) is strongly negative ampichiral if there exists an orientation reversing involution \(\tau:S^3 \to S^3\) such that \(\tau(K) = K\).
Lemma (Kawauchi 09): If \(K\) is strongly negative ampichiral, then \(K\) bounds a disc in a \(X = {\mathbf{Q}}H B^4\) with only 2torsion in \(H_1(X; {\mathbf{Z}})\).
Proof (sketch): Let \(M_k = S_0^3(K)\) be zero surgery on the knot.
Then \(\tau: M_k \to M_k\) is fixedpoint free.
Can then consider the map \(\pi: M_k \to M_k/\tau\) and the associated twisted \(I{\hbox{}}\)bundle \(I \to Z \to M_k \to M_k/\tau\). Then:
Theorem: If \(K\) is slice, the \(\Sigma_{p^r}(K)\) bounds a \({\mathbf{Q}}H B^4\) if \(p\) is prime.
Proof (Milnor 68).
There is an exact sequence
\begin{align*} \tilde H_i(\tilde X; {\mathbf{Z}}_p) \xrightarrow{t^{p^r}  \operatorname{id}} \tilde H_i(\tilde X; {\mathbf{Z}}_p) \to \tilde H(\Sigma_{p^r}(D); {\mathbf{Z}}_p) (= 0) \to 0 \\ \tilde H_i(\tilde X; {\mathbf{Z}}_p) \xrightarrow{t  \operatorname{id}} \tilde H_i(\tilde X; {\mathbf{Z}}_p) \to \tilde H(B^4; {\mathbf{Z}}_p) (= 0) \to 0 ,\end{align*}
where if \(t\operatorname{id}\) is an isomorphism, \((t^{p^r}  \operatorname{id}) = (t\operatorname{id})^{p^r}\) is an isomorphism as well (note we’re in \({\mathbf{Z}}_p\).)
Corollary: If \(p\) is an odd prime, then \(\Sigma_{p^r}(K)\) bounds a \({\mathbf{Q}}H B^4\).
Thus \(\Sigma_{2^r}(K_n) \cong \Sigma_n(K_{2^r})\) by this symmetry.
In conclusion, if \(n\) is an odd prime power, then \(K_n \in \cap\ker \beta_{p^r}\).
Step 2: Obstruction
Theorem (Brandenbursky 16): \(K_n\) is algebraically slice iff \(n\) odd.
Uses a twisted Alexander polynomial: Take \(M_k = S_0^3(K)\) a zero surgery, \(G = \pi_1(M_k)\), and \(A(k) = G^{(1)}/G^{(2)}\) (?).
The input is a map \(X: H_1(\Sigma_{p^r}(K); {\mathbf{Z}}) \to {\mathbf{Z}}_q\). This lets you define a character
\begin{align*} \alpha(X): G \to G/G^{(1)} \cong {\mathbf{Z}}\rtimes A(k) \to {\mathbf{Z}}\rtimes A(t) / t^{p^r}  \operatorname{id} \cong {\mathbf{Z}}\rtimes H_1(\Sigma_{p^r}(K)) \to \operatorname{GL}(K, {\mathbf{Q}}(\zeta_q)[t^{\pm 1}] ) .\end{align*}
Then \(\tilde M_k\) is the universal cover of \(M_k\). Consider \(C_*(\tilde M_k) \otimes{{\mathbf{Z}}[G]} Y\) for \(Y = ({\mathbf{Q}}(\zeta_q)[t^{\pm 1}] )^k\), then define the twisted Alexander polynomial \(\Delta_k^{\tilde X}(t) = \mathrm{ord}(H_1(M_k, Y))\).
Theorem: If \(K\) is slice, then there exists some \(X\) such that \(\delta_K^{\tilde X}(t) = f(t) f(t^{1})\) in \({\mathbf{Q}}(\zeta_q)[t^{\pm 1}]\).
Open question: Are there infinite order elements in this group?