What are the points of Spec(Vassiliev Invariants)? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T07:56:03Z http://mathoverflow.net/feeds/question/10954 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10954/what-are-the-points-of-specvassiliev-invariants What are the points of Spec(Vassiliev Invariants)? Theo Johnson-Freyd 2010-01-06T20:00:29Z 2010-04-01T07:49:00Z <h2>Background</h2> <p>Recall that a (oriented) <em>knot</em> is a smoothly embedded circle $S^1$ in $\mathbb R^3$, up to some natural equivalence relation (which is not quite trivial to write down). The collection of oriented knots has a binary operation called <em>connected sum</em>: if $K_1,K_2$ are knots, then $K_1 \# K_2$ is formed by spatially separating the knots, then connecting them by a very thin rectangle, which is glued on so that all the orientations are correct. Connect sum is commutative and associative, making the space of knots into a commutative monoid. In fact, <a href="http://en.wikipedia.org/wiki/Prime%5Fknot" rel="nofollow">by a theorem of Schubert</a>, this is the free commutative monoid on countably many generators. A ($\mathbb C$-valued) <em>knot invariant</em> is a $\mathbb C$-valued function on this monoid; under "pointwise" multiplication, the space of knot invariants is a commutative algebra $I$, and $\#$ makes $I$ into a cocommutative bialgebra. I.e. $I$ is a commutative monoid object in $(\text{CAlg})^{\rm{op}}$, where $\text{CAlg}$ is the category of commutative algebras.</p> <blockquote> <p><strong>Warm-up question:</strong> Any knot $K$ defines an algebra morphism $I \to \mathbb C$, i.e. a <em>global point</em> of $I \in (\text{CAlg})^{\rm{op}}$. Are there any other global points?</p> </blockquote> <p><strong>Edit:</strong> In response to Ilya N's comment below, I've made this <a href="http://mathoverflow.net/questions/10957/a-hands-on-description-of-a-completion-of-the-free-commutative-monoid-on-counta" rel="nofollow">into its own question</a>.</p> <h2>Finite type invariants</h2> <p>Recall that a <em>singular knot</em> is a smooth map $S^1 \to \mathbb R^3$ with finitely many transverse self-intersections (and otherwise it is an embedding), again up to a natural equivalence. Any knot invariant extends to an invariant of singular knots, as follows: in a singular knot $K_0$, there are two ways to blow up any singularity, and the orientation determines one as the "right-handed" blow-up $K_+$ and the other as the "left-handed" blow-up $K_0$. Evaluate your knot invariant $i$ on each blow-up, and then define $i(K_0) = i(K_+) - i(K_-)$. Note that although the connect-sum of singular knots is not well-defined as a singular knot, if $i\in I$ is a knot-invariant, then it cannot distinguish different connect-sums of singular knots. Note also that the product of knot invariants (i.e. the product in the algebra $I$) is not the point-wise product on singular knots.</p> <p>A <em>Vassiliev (or <a href="http://en.wikipedia.org/wiki/Finite%5Ftype%5Finvariant" rel="nofollow">finite type</a>) invariant of type $\leq n$</em> is any knot invariant that vanishes on singular knots with $&gt; n$ self-intersections. The space of all Vassiliev invariants is a filtered bialgebra $V$ (filtered by type). The corresponding associated-graded bialgebra $W$ (of "weight systems") has been well-studied (some names: Kontsevich, Bar-Natan, Vaintrob, and I'm sure there are others I haven't read yet) and in fact is more-or-less completely understood (e.g. Hinich and Vaintrob, 2002, "Cyclic operads and algebra of chord diagrams", <a href="http://www.ams.org/mathscinet-getitem?mr=1913297" rel="nofollow">MR1913297</a>, where its graded dual $A$ of "chord diagrams" is described as a sort of universal enveloping algebra). In fact, this algebra $W$ is Hopf. I learned from <a href="http://mathoverflow.net/questions/10827/if-associated-graded-of-a-filtered-bialgebra-is-hopf-does-it-follow-that-the-ori" rel="nofollow">this question</a> that this implies that the bialgebra $V$ of Vassiliev invariants is also Hopf. Thus it is a Hopf sub-bialgebra of the algebra $I$ of knot invariants.</p> <p>I believe that it is an open question whether Vassiliev invariants separate knots (i.e. whether two knots all of whose Vassiliev invariants agree are necessarily the same). But perhaps this has been answered &mdash; I feel reasonably caught-up with the state of knowledge in the mid- to late-90s, but I don't know the literature from the 00s.</p> <p>Geometrically, then, $V \in (\text{CAlg})^{\rm{op}}$ is a commutative group object, and is a quotient (or something) of the monoid-object $I \in (\text{CAlg})^{\rm{op}}$ of knot invariants. The global points of $V$ (i.e. the algebra maps $V \to \mathbb C$ in $\text{CAlg}$) are a group.</p> <h2>Main Questions</h2> <p>Supposing that Vassiliev invariants separate knots, there must be global points of $V$ that do not correspond to knots, as by <a href="http://en.wikipedia.org/wiki/Eilenberg%E2%80%93Mazur%5Fswindle" rel="nofollow">Mazur's swindle</a> there are no "negative knots" among the monoid $I$. Thus my question.</p> <blockquote> <p><strong>Main question.</strong> What do the global points of $V$ look like? </p> </blockquote> <p>If Vassiliev invariants do separate knots, are there still more global points of $V$ than just the free abelian group on countably many generators (i.e. the group generated by the free monoid of knots)? <strike>Yes: the singular knots.</strike> (<strong>Edit:</strong> The rule for being a global point is that you can evaluate any knot invariant at it, and that the value of the invariant given by pointwise multiplication <em>on knots</em> is the multiplication of the values at the global point. Let $K_0$ be a singular knot with one crossing and with non-singular blow-ups $K_+$ and $K_-$, and let $f,g$ be two knot invariants. Then \begin{aligned} (f\cdot g)(K_0) &amp; = (f\cdot g)(K_+) - (f\cdot g)(K_-) = f(K_+)\cdot g(K_+) - f(K_-)\cdot g(K_-) \neq \\ f(K_0) \cdot g(K_0) &amp; = f(K_+)\cdot g(K_+) - f(K_+)\cdot g(K_-) - f(K_-)\cdot g(K_+) + f(K_-)\cdot g(K_-)\end{aligned}.) What else is there? </p> <p>What can be said without knowing whether Vassiliev invariants separate knots?</p> http://mathoverflow.net/questions/10954/what-are-the-points-of-specvassiliev-invariants/10963#10963 Answer by Ilya Nikokoshev for What are the points of Spec(Vassiliev Invariants)? Ilya Nikokoshev 2010-01-06T20:47:06Z 2010-01-06T22:58:39Z <p>I don't have much knowledge in either knots or Hopf bialgebras, but I think one can answer part of your question (the one formerly asked as a title: "What do negative knots look like?") anyway. </p> <p>About finite type invariants: Wikipedia says it is <strong>still not known</strong> whether finite type invariants separate knots (and I think a discovery of an answer to this question would make pretty big news anyway).</p> <p>It matters whether they do or not if you want to learn about <em>all points of $V$</em>. However, either way you can be certain that the points of $V$ (which I'll denote by $\mathop{\text{Spec}} V$) can be <strong>formally added or subtracted</strong>. In other words, there are points in $\mathop{\text{Spec}} V$ corresponding to $[\text{unknot}]+2\cdot[\text{trefoil}]$ or $-[\text{trefoil}]$. This is implied when you say that those form <a href="http://en.wikipedia.org/wiki/Hopf%5Falgebra" rel="nofollow">Hopf bialgebra</a>.</p> <p>So, for any given knot $K$ there exists an element of $\mathop{\text{Spec}} V$ that has finite invariants negative of finite invariants of knot $K$, but this observation is rather trivial. </p> <p>Thus I think one shouldn't ask what is asked in the title, but rather more general question (emphasized); <strong>update:</strong> Theo Johnson-Freyd did so, rendering the answer rather irrelevant.</p> http://mathoverflow.net/questions/10954/what-are-the-points-of-specvassiliev-invariants/10969#10969 Answer by Daniel Moskovich for What are the points of Spec(Vassiliev Invariants)? Daniel Moskovich 2010-01-06T21:41:15Z 2010-01-07T00:55:38Z <ol> <li>Regarding whether or not Vassiliev invariants separate knots, the sticking point is whether they separate a knot from its reverse, where the reverse of a knot is the same knot with the opposite orientation. If this were false, Vassiliev invariants would not separate knots by a <a href="http://front.math.ucdavis.edu/9712.5188" rel="nofollow">result of Kuperberg</a>. This boils down to the question of whether the space of Jacobi diagrams with an odd number of legs vanishes. As a piece of shameless self-promotion I will refer to <a href="http://front.math.ucdavis.edu/math.GT/0511602" rel="nofollow">my own paper</a> (joint with Tomotoda Ohtsuki), where we prove this for 3-loop Jacobi diagrams. You could ask the same question about homotopy string links, where we know that Vassiliev invariants do separate homotopy string links (see <a href="http://www.math.toronto.edu/~drorbn/papers/homotopy/homotopy.pdf" rel="nofollow">this paper</a> by Bar-Natan).</li> <li>I might be way off-base here, but can't you just take the formal product of an element of $\mathbb{C}$ with the knot? So a negative knot looks like -1 times K. So no, singular knots are not global points of V. Resolutions (via the Vassiliev skein relation) of singular knots as formal sums of knots over $\mathbb{C}$ are global points of V.</li> <li>You sort-of say this, but the Fundamental Theorem of Vassiliev Invariants (proved over $\mathbb{C}$ by Kontsevich and others) tells us that weight systems integrate, which is the statement that any weight system (with some natural conditions) gives rise to a Vassiliev invariant (which is nothing short of miraculous when you think about it). A reference is <a href="http://uk.arxiv.org/PS_cache/q-alg/pdf/9702/9702009v1.pdf" rel="nofollow">this survey paper by Bar-Natan and Stoimenow</a>.</li> </ol> http://mathoverflow.net/questions/10954/what-are-the-points-of-specvassiliev-invariants/10990#10990 Answer by Ryan Budney for What are the points of Spec(Vassiliev Invariants)? Ryan Budney 2010-01-07T00:26:50Z 2010-01-07T06:15:12Z <p>Just some small comments here. Yes, the question of whether or not Vassiliev invariants separate knots is still open. </p> <p>One broader context for the question is to consider how Vassiliev invariants were first observed -- via the Vassiliev spectral sequence for the space of "long knots". These are knots of the form $\mathbb R \to \mathbb R^3$ which restrict to a standard (linear) inclusion outside of an interval, say $[-1,1]$. </p> <p>There's a few standard ways in which the space of long knots can be turned into a topological monoid -- a "Moore loop space" construction which is pretty standard, or a "fat knots" construction from my <a href="http://front.math.ucdavis.edu/0309.5427" rel="nofollow">"little cubes and long knots"</a> paper. So this monoid operation is just the connect-sum operation, suitably jazzed-up to be strictly associative. </p> <p>An observation for which I'm not sure if anyone has written up a complete proof yet is that the Goodwillie embedding calculus (an alternative approach to the Vassiliev spectral sequence, among other things) factors through the group-completion of the space of long knots. By this I mean, if K is the space of long knots, call the "group completion" the map $K \to \Omega BK$. $\Omega BK$ is where the "formal inverse" to a knot lives. The homotopy-type of $\Omega BK$ is computed in <a href="http://front.math.ucdavis.edu/0309.5427" rel="nofollow">"little cubes and long knots"</a>. In particular, you can think of the Vassiliev spectral sequence as being an invariant of $\Omega BK$, not $K$. From the point of view of Vassiliev invariants this isn't such a major insight as $\pi_0 \Omega BK$ is the group-completion of the monoid $\pi_0 K$, so it's just a free abelian group on countably-many generators. </p> <p>In particular, there are classes in $H^1 K$ for which there are no obvious finite-type approximations. $H_1 K$ has some torsion classes for which it's pretty unclear how to detect using the Vassiliev spectral sequence. <a href="http://front.math.ucdavis.edu/0504.5206" rel="nofollow">Here is a mass of torsion computations</a> for the homology of the long knot space. </p> <p>edit: although "little cubes and long knots" describes the homotopy-type of the group completion of the space of long knots, it would be nice to have a more "concrete" embedding-space type description of the group completion. Mostovoy has an attempt <a href="http://front.math.ucdavis.edu/9810.5040" rel="nofollow">here</a>. I believe an idea like this should work and likely Mostovoy's idea works as well, but (again) this is something that needs to be written up for which there hasn't been enough time to start the project.</p> http://mathoverflow.net/questions/10954/what-are-the-points-of-specvassiliev-invariants/20049#20049 Answer by Dylan Thurston for What are the points of Spec(Vassiliev Invariants)? Dylan Thurston 2010-04-01T07:49:00Z 2010-04-01T07:49:00Z <p>The points of Spec $V$ live n a somewhat complicated completion of the space of knots (not formal linear combinations), but I think it's illuminating to think about the case of braids and their "Vassiliev" invariants. In that context the global points of Vassiliev invariants are elements of the completion of the braid group with respect to the lower central series. The braid group embeds in this completion, and so Vassiliev invariants do separate braids. (This is not hard.)</p> <p>By the way, the knot invariants $I$ also form a bi-algebra, where the coproduct is dual to connect sum of knots, although that coproduct plays no role for this question.</p>