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I'm attempting the calculate the Khovanov homology of the unknot using the figure eight diagram of the unknot with exactly one crossing going from top left to bottom right as shown below.enter image description here

I also include the convention I'll be using for the resolutions of the crossings.

I calculate that the resulting chain complex given by this diagram is $$\cdots\to 0\to V\otimes V\stackrel{m}{\to} V\to0\to\cdots$$ where $V$ is the graded abelian group $\mathbb{Z}_{(1)}\oplus\mathbb{Z}_{(-1)}$ (I'll drop the gradings from here as I don't think they play any part in my eventual misunderstanding - I could be wrong) generated by $v_+$ and $v_-$ and $m$ is the map which acts on generators by $$\begin{array}{rcl}m(v_-\otimes v_-)&=&0\\ m(v_+\otimes v_-)&=&v_-\\ m(v_-\otimes v_+)&=&v_-\\ m(v_+\otimes v_+)&=&v_+.\end{array}$$

Here we see that the kernal of $m$ is freely generated by $v_-\otimes v_-$ and $v_-\otimes v_+ - v_+\otimes v_-$ and so we get that $H^{Kh}_0\cong\mathbb{Z}\oplus\mathbb{Z}$, and $m$ is surjective so $H^{Kh}_1$ is trivial.

Now consider the mirror image of the above diagram which is still isotopic to the unknot, and with the same choice of crossing resolutions. Because of our choice of resolutions, we now get the chain complex $$\cdots\to 0\to V\stackrel{\Delta}{\to} V\otimes V\to0\to\cdots$$ where $\Delta$ is the map which acts on generators by $$\begin{array}{rcl}\Delta(v_-)&=&v_-\otimes v_-\\ \Delta(v_+)&=&v_+\otimes v_- + v_-\otimes v_+.\end{array}$$

Here we see that the kernal of $\Delta$ is trivial so $H^{Kh}_0$ is trivial, but $\mbox{Im}\Delta\cong \mathbb{Z}\oplus\mathbb{Z}$ and so $H^{Kh}_1\cong\mathbb{Z}\oplus\mathbb{Z}$.

It seems that I'm missing something here as either I've inadvertently shifted homological degrees somewhere where I should not have (or haven't where I should have), or I need to maybe dualise the chain complex for some reason when taking the mirror image of our knot. Which of my calculations above is incorrect (or both?) and am I not understanding some crucial step in the definition of the Khovanov homology of a knot which has lead to this mistake?

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    $\begingroup$ You're probably forgetting to renormalize the grading based on the number of + and - crossings in the diagram. $\endgroup$ Mar 3, 2014 at 17:34
  • $\begingroup$ @GrumpyParsnip Oh wow I need to check my notes more often I guess. Thanks Jim. By introducing a - crossing in place of a + crossing, the homological degree shifts one to the left and everything works out as expected (for some reason I thought crossing parities only affected the grading of the individual graded homology groups, not the homological degrees). $\endgroup$
    – Dan Rust
    Mar 3, 2014 at 21:14
  • $\begingroup$ If you want to write that up as an answer I can accept it and move this off of the unanswered queue. $\endgroup$
    – Dan Rust
    Mar 3, 2014 at 21:17
  • $\begingroup$ Glad to be of help. I'll do that. $\endgroup$ Mar 3, 2014 at 21:19
  • $\begingroup$ hi guys, im reading bar nathans paper about this subject and was wondering how to construct the homology's. Kees $\endgroup$
    – Kees Til
    Apr 29, 2016 at 15:51

1 Answer 1

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You're probably forgetting to renormalize the grading based on the number of $+$ and $-$ crossings in the diagram.

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