# Tuesday July 7th Recall that we had recursive rules for computing the Kausffman bracket, and a normalization factor for the Jones polynomial that made it into an invariant. We'd like a closed formula for these. We do this by ordering the crossings of the unoriented link $1, \cdots, n$, then there is a correspondence \[ \theset{0, 1}^n &\iff \text{Complete resolutions} \\ (\alpha_1, \cdots, \alpha_n) &\iff \alpha_i \text{ resolves the $i$th crossing} .\] :::{.example} ![](figures/image_2020-07-07-11-07-33.png) ::: :::{.claim} \[ \gens{D} = \sum_{\alpha \in \theset{0, 1}^n} (-1)^{\abs \alpha} v^{\abs \alpha} (v+v\inv)^{c_\alpha(D)} ,\] where $\abs \alpha$ is the number of 1-resolutions and $c_\alpha$ is the number of circles in the resolution corresponding to $\alpha$. ::: :::{.proof} Idea: look at resolving the $n$th crossing locally and apply the recursive relation. Then rewrite the sum by appending $\alpha_n = 0$ and $\alpha_n = 1$ respectively. Note that we can rewrite the sum as \[ \sum_{r=0}^n (-1)^r \sum_{\abs \alpha = r} v^r (v+v\inv)^{c_\alpha(D)} .\] This amounts to summing over the "columns" in the previous diagram: ![](figures/image_2020-07-07-11-22-49.png) Here this yields \[ (v+v\inv)^2 + (-1)2v(v+v\inv) + v^2 (v+v\inv)^2 .\] ::: Note that this formula starts to resemble an Euler characteristic! :::{.remark} *Problem*: The coefficient \[ \sum v^r(v+v\inv)^{c_\alpha(D)} \in \ZZ^{\geq 0}[v, v\inv] \] is a Laurent polynomial instead of a natural number, so this can't immediately be interpreted as a dimension of a vector space. *Solution*: Replace finite-dimensional $\CC\dash$vector spaces by $\ZZ\dash$graded vector spaces. The category consists of objects given by $V = \bigoplus_{i\in \ZZ} V_i$ and linear maps $f:V\to W$ such that $f(V_i) \subseteq W_i$ for all $i$. ::: We previously had vector spaces categorifying the natural numbers by taking the dimension, so for graded vector spaces, we take the **graded dimension**: :::{.definition title="Graded Dimension"} \[ \grdim \bigoplus_{i\in\ZZ}V_i = \sum_{i\in \ZZ}\qty{\dim V_i}v^i \in \ZZ^{\geq 0}[v, v\inv] .\] ::: **Goal**: We want to associate to an oriented link diagram $D$ a cochain complex of finite-dimensional graded $\CC\dash$vector spaces $C_i(D) \mapsvia{\bd} C_{i+1}(D) \to \cdots$. Since each chain space decomposes, the differential does as well, and we get a large collection of chain complexes \begin{tikzcd} \cdots \arrow[r] & {C_{i, 1}} \arrow[r, "{\bd_{i, 1}}"] & {C_{i+1, 1}} \arrow[r] & \cdots \\ \cdots \arrow[r] & {C_{i, 0}} \arrow[r, "{\bd_{i, 0}}"] & {C_{i+1, 0}} \arrow[r] & \cdots \\ \cdots \arrow[r] & {C_{i, -1}} \arrow[r, "{\bd_{i, -1}}"] & {C_{i+1, -1}} \arrow[r] & \cdots \end{tikzcd} This yields two gradings: the first is homological, the second is "internal". :::{.remark} We want the following: 1. If $D, D'$ are related by a finite sequence of Reidemeister moves, then \[ H_{i, j}(C_{\wait, \wait}(D)) = H_{i, j}(C_{\wait, \wait}(D')) = \ker \bd_{i, j} / \im \bd_{i-1, j} \text{ for all }i, j. \] 2. Additionally, \[ J(D) = \chi_{\gr}(C_{*, *}(D)) = \sum_{i\in \ZZ} (-1)^i \grdim \qty{ C_{*, *}(D) } \] Note that you can also take the dimension of the homology instead, and at the end of the day this yields $\sum_{i, j\in \ZZ} (-1)^i v^j \dim(H_{i, j})$. ::: :::{.definition title="Homogeneous elements"} For $A = \bigoplus A_i, B = \bigoplus B_i$, $a\in A$ is called *homogeneous* of degree $k$ if $a\in A_k$, i.e. it is a sum of basis elements from only the $k$th graded piece. In this case we say $\abs a = k$. ::: :::{.proposition title="Bases for various combinations of graded spaces"} We can union bases over all graded pieces to get a basis for the entire space. For direct sums $A\oplus B$, a basis is given by $(\alpha_i, 0)$ and $(0, \beta_j)$. We put $\alpha_i$ in degree $\abs{\alpha_i}$, in which case \[ \grdim(A\oplus B) &= \sum_{k\in \ZZ} \dim(\qty{A\oplus B}_k) v^k \\ &= \sum_k \dim A_k v^k + \sum \dim B_k v^k \\ &= \grdim(A) + \grdim(B) ,\] so taking direct sums commutes with taking graded dimensions. Similarly for tensor products $A\tensor_\CC B$, we get a basis $\alpha_i \tensor \beta_j$ placed in degree $\abs{\alpha_i} + \abs{\beta_j}$. ::: :::{.exercise title="?"} Show that \[ \grdim(A\tensor_\CC B) = \grdim(A) \cdot \grdim(B) .\] ::: We also have degree shifts by $i$ for any $i$, denote $A(i)$, where $A_j \mapsto A_{j+i}$ for every $j$. Thus the $k$th graded piece is given by $(A(i))_k = A_{k-i}$, thus \[ \grdim(A(i)) = v^i \grdim(A) \] :::{.example title="Important"} \[ H^*(S^2; \CC) = \begin{cases} \CC & * = 0, 2 \\ 0 & \text{else}. \end{cases} \] Let $A\definedas H^*(S^2; \CC)(-1)$, which now has $\CC$ in degrees $\pm 1$, and $\grdim A = v + v\inv$. Note that - $(v+v\inv)^2$ corresponds to $A^{\tensor 2}$. - $2v(v+v\inv)$ corresponds to $A(1)^{\oplus 2}$. - $v^2(v+v\inv)$ corresponds to $A^{\tensor 2}(2)$. So we can assemble these into a chain complex and take the Euler characteristic in order to recover the Kauffman bracket in the earlier example. ::: :::{.theorem title="?"} There exists a unique isotopy invariant of oriented links in $\RR^3$ called $P(D) \in \CC(a, v)$, a rational function in two variables, the HOMFLY-PT polynomial. It satisfies ![](figures/image_2020-07-07-11-56-40.png) ::: :::{.example title="The Hopf Link"} Yields \[ P(D) = -a(aia\inv) + a^2 \qty{a-a\inv \over v-v\inv}^2 .\] :::