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I'm trying to determine the validity of a property of the morphisms on Khovanov homology that are induced by oriented link cobordisms. These maps are defined with fixed diagrams for the boundary links, but what changes when we choose different diagrams? Is there a relation between the induced morphisms? More explicitly:

Suppose we have an oriented link cobordism $\Sigma \subset \mathbb{R}^3 \times [0,1]$ from $L_0$ to $L_1$ and that we have a pair of diagrams for each link: $D_0, D_0'$ for $L_0$ and $D_1, D_1'$ for $L_1$. There are induced morphisms (defined on the chain level via surface diagrams) on the Khovanov homology:

$$F : \text{Kh}(D_0) \to \text{Kh}(D_1)$$ $$F' : \text{Kh}(D_0') \to \text{Kh}(D_1')$$

How are these maps related? I'm tempted to say that because there are Reidemeister-induced isomorphisms $\varphi_0 : \text{Kh}(D_0) \to \text{Kh}(D_0')$ and $\varphi_1 : \text{Kh}(D_1) \to \text{Kh}(D_1')$, then the following diagram should commute (up to sign). $\require{AMScd}$

\begin{CD} \text{Kh}(D_0) @>{\varphi_0}>> \text{Kh}(D_0') \\ @V{F}VV @VV{F'}V \\ \text{Kh}(D_1) @>{\varphi_1}>> \text{Kh}(D_1') \end{CD}

I haven't found any resources discussing this, and I'm having a hard time (dis)proving it. All of the maps I've come up with have commuted, and I haven't had much of a breakthrough in proving it.

Here's what I've considered:

Jacobsson proved that the induced map $F$ is invariant up to an overall multiplication by -1 under ambient isotopy of $\Sigma$ leaving $\partial \Sigma$ setwise fixed. The same can be said for $F'$, or any induced map defined in this manner (i.e. by choosing a generic surface diagram beginning/ending with the given diagrams for the boundary links of $\Sigma$ and composing the Reidemeister/Morse induced chain maps that the surface diagram records).

In the spirit of Jacobsson's theorem, one could try to argue that the maps $F'\varphi_0$ and $\varphi_1F$ are induced by equivalent cobordisms (i.e. cobordisms that are related by a boundary-preserving isotopy). This would imply that these morphisms are the same (up to sign). But constructing these cobordisms is tricky, as they need to have fairly specific surface diagrams. Moreover, surface diagrams (as I understand them) are defined with respect to a projection $\mathbb{R}^3 \to \mathbb{R}$, which is already not consistent between the surface diagrams that induce $F$ and $F'$.

octocat
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1 Answers1

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Your diagram should commute and this is a sketch of an argument for that using Jacobsson's theorem.

The diagrams $D_0$ and $D_0'$ are related by a sequence of Reidemeister moves. Each Reidemeister move can be thought of as a cobordism so combining them gives a cobordism $A$ from $D_0$ to $D_0'$. Similarly, a sequence of Reidemeister moves gives a cobordism $B$ from $D'_1$ to $D_1$. These cobordisms $A$ and $B$ induce the maps $\varphi_0$ and $\varphi_1^{-1}$.

Now you have a new cobordism from $D_0$ to $D_1$ given by the composition $B \circ \Sigma_1 \circ A$ which is inducing the map $\varphi_1^{-1} \circ F' \circ \varphi_0 $. The cobordisms $\Sigma_0$ and $B \circ \Sigma_1 \circ A$ are isotopic relative the boundaries because $A$ and $B$ are cobordisms corresponding to the isotopies relating the different diagrams and $\Sigma_0$, $\Sigma_1$ are representations of the same cobordism for these different diagrams. Then Jacobsson's theorem says that the maps $F$ and $\varphi_1^{-1} \circ F' \circ \varphi_0 $ agree up to a sign. This implies that the diagram you've drawn commutes.

GageMartin
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  • This is the direction I was heading with the proof, but it's the "should be isotopic relative the boundaries because $A$ and $B$ are cobordisms corresponding to isotopies" that I'm not convinced about. I don't think it's true in general that if you add the trace of an isotopy to a cobordism that you'll stay in the same boundary preserving isotopy class. For example, take a product cobordism on the two component unlink and add the trace of an isotopy that swaps the two components. I don't think we're considering any isotopy though, just those obtained from changing the projection of the link. – octocat Jul 29 '20 at 17:12
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    Yes sorry, I looked at the wrong parts of your post to get some notation. I will fix it now – GageMartin Jul 29 '20 at 17:19
  • So I think the issue about being isotopic is then that how are the surfaces $\Sigma_0$ from $D_0$ to $D_1$ and $\Sigma_1$ from $D_0'$ to $D_1'$ defined. From the abstract link cobordism $\Sigma$ from $L_0$ to $L_1$. The differences between them should exactly cancel out with the cobordisms $B$ and $A$. – GageMartin Jul 29 '20 at 17:24
  • In fact one way to define $\Sigma_1$ is as $B^{-1} \circ \Sigma_0 \circ A^{-1} $ – GageMartin Jul 29 '20 at 17:25
  • I've been thinking of $\Sigma_0$ and $\Sigma_1$ as surface diagrams, so projections of $\Sigma$ into $\mathbb{R}^2 \times [0,1]$. These projections are (I'm assuming) consistent, meaning $\Sigma_i$ is the image of $\Sigma \subset \mathbb{R}^3 \times [0,1]$ under the map $p_i \times I$, where $p_i: \mathbb{R}^3 \to \mathbb{R}^2$ is a chosen projection. In this case, $A$ and $B$ could be traces of rotations of $\mathbb{R}^3$ taking $p_0$ to $p_1$, maybe? Perhaps you're thinking of it differently, so I'm not seeing how the surface diagram $\Sigma_1$ is defined from $B \circ \Sigma_0 \circ A$. – octocat Jul 29 '20 at 17:52
  • So I guess in this language $A$ and $B$ should be thought of as exactly the isotopies needed to chance the projections from $p_0$ to $p_1$ and these isotopies are realized by a sequence of Reidemeister moves. So one way to then think of how $\Sigma_1$ looks in projection $p_1$ by using $\Sigma_0$ is the following. First do $A^{-1}$ so you are in projection $p_0$ then do $\Sigma_0$ in $p_0$ then do $B^{-1}$ to get back to projection $p_1$ – GageMartin Jul 29 '20 at 21:21