# What does the tensor product in the definition of Combinatorial Floer knot homology look like?

I am working on a project that involves summarizing the article A combinatorial description of knot Floer homology (http://arxiv.org/abs/math/0607691) and doing some example computations with the theorems. I will attempt to explain the background info for my question but I am not sure if it will make sense without being familiar with the article.

I have calculated the $\mathbb{Z}_2$-homology of the bigraded chain complex $(C(\Gamma),\partial)$ associated with the grid diagram $\Gamma$ representing a knot $K$. Now that I have $H_*(C(\Gamma),\partial)$ I want to recover the Floer homology of $K$, $HFK(K)$.

Both $H_*(C(\Gamma),\partial)$ and $HFK(K)$ are bigraded direct sums of $\mathbb{Z}_2$'s: each generator of each group is assigned an Alexander grading and a Maslov grading, and they all generate a $\mathbb{Z}_2$ in their group. The bigradings are given by

$HFK(K) = \bigoplus_{m,s} HFK_m(K,s)$ and

$H_*(C(\Gamma),\partial) = \bigoplus_{i,j}H_*(C_{i,j}(\Gamma),\partial)$,

where $C_{i,j}(\Gamma)$ is the subgroup of $C(\Gamma)$ generated by generators with maslov grade $i$ and Alexander grade $j$ (we know that $\partial$ preserves Alexander grade and reduces Maslov grade by one, and that $\partial^2 = 0$, so the bigrading gives a series of chain complexes, one for each Alexander grade, where the "dimensions" are the Maslov gradings at that Alexander grade).

Now that I have the homology of $C(\Gamma)$ at each bigrading, I want to recover $HFK(K)$ using theorem 1.1 (page 5) of the article. It states that

$H_*(C(\Gamma),\partial) = HFK(K) \otimes V^{\otimes (n-1)}$

where $V$ is the bigraded group consisting of one $\mathbb{Z}_2$ at bigrading $(-1,-1)$ and one $\mathbb{Z}_2$ at $(0,0)$ (actually a chain complex with trivial boundary operator). I assume the $\otimes$ denotes a tensor product, and the superscript means the tensor product of $V$ $(n-1)$ times.

$n$ is the dimension of the grid used; here $n=5$. I am not very familiar with the structure of a tensor product in general. Am I correct in assuming that in this case the tensor product acts the same as a direct product, so that I just remove $(n-1)$ copies of $\mathbb{Z}_2$ from the bigradings $(-1,-1)$ and $(0,0)$ in $H_*(C(\Gamma),\partial)$ in order to get the structure of $HFK(K)$?

## 1 Answer

Since $V$ is a two-dimensional $\mathbb{Z}_2$-vector space, tensoring $\widehat{HFK}(K)$ with $V$ doubles the dimension of $\widehat{HFK}(K)$. One can view $\widehat{HFK}(K)\otimes V$ as the direct sum of two copies of $\widehat{HFK}(K)$, where one summand has the original gradings of $\widehat{HFK}(K)$ and where the other summand has both the Maslov and Alexander gradings shifted down by one.

Since we are working over the field $\mathbb{Z}_2$, the vector space $\widehat{HFK}(K)$ is determined by its Poincare polynomial $P(u,t)$, defined as follows. Let $\widehat{HFK}_m(K,s)$ be the summand of $\widehat{HFK}(K)$ in Maslov grading $m$ and Alexander grading $s$. Then the Poincare polynomial is defined as $$P(u,t) = \sum_{(m,s)\in\mathbb{Z}\oplus\mathbb{Z}} \dim \widehat{HFK}_m(K,s)\cdot u^m t^s.$$ The Poincare polynomial of $V$ is $1 + u^{-1}t^{-1}$, and tensoring $\widehat{HFK}(K)$ with $V$ corresponds to multiplying the Poincare polynomial of $\widehat{HFK}(K)$ by $(1 + u^{-1}t^{-1})$. Tensoring with $V^{\otimes(n-1)}$ corresponds to multiplying the Poincare polynomial by $(1 + u^{-1}t^{-1})^{n-1}$.

Let's work through an example. Suppose that $\Gamma$ is the grid diagram of the (left-handed) trefoil $K$ that appears on page 2 of the linked paper. The Poincare polynomial for $\widehat{HFK}(K)$ is $$P(u,t) = t^{-1} + u + u^2t.$$ The homology $H_*(C(\Gamma),\partial)$ is isomorphic to $\widehat{HFK}(K)\otimes V^{\otimes 4}$. The Poincare polynomial of $H_*(C(\Gamma),\partial)$ is \begin{align*}&(t^{-1} + u + u^2t)(1+u^{-1}t{-1})^4\\ =& u^{-4}t^{-5}+5u^{-3}t^{-4}+11u^{-2}t^{-3}+14u^{-1}t^{-2}+11t^{-1}+5u + 11u^2t \end{align*}

Of course, you can do this entire process in reverse. Given the Poincare polynomial of $H_*(C(\Gamma),\partial)$, you simply factor out $(1+u^{-1}t^{-1})^{n-1}$ to obtain the Poincare polynomial of $\widehat{HFK}(K)$.