2
$\begingroup$

I'm reading Paul Seidel's book "Fukaya Categories and Picard-Lefschetz Theory", chapter 12, and I'm currently trying to understand the differential on Floer cohomology in terms of orientation lines $o_y$ for $y \in \mathcal{C}(L_0, L_1)$, in Seidel's notation. $L_0$ and $L_1$ are exact Lagrangians in $(M, \omega = d\theta)$ equipped with brane structures, and $\mathcal{C}(L_0, L_1)$ are time-1 Hamiltonian chords which start at $L_0$ and end at $L_1$. $o_y$ is a certain one-dimensional real vector space associated to $y$, which can either be seen as a tensor product of certain one-dimensional spaces obtained as intersections of Lagrangian subspaces at $y(1)$ and their duals, or as the determinant line of a certain Fredholm operator on the upper half-plane (the precise definitions are in equations $(11.20)$ and $(11.25)$ in the book, as well as lemma $11.11$).

For a field $\mathbb{K}$, $|o_y|$ is defined by taking the direct sum of the rank-1 vector spaces generated by the two orientations of $o_y$, and then setting their sum to be $0$. The Floer cochain space is defined as $$CF^* (L_0, L_1) = \bigoplus_{y} |o_y|_{\mathbb{K}}.$$ The grading on each Lagrangian induces a grading on $CF^*(L_0, L_1)$.

Now suppose I have $y_0$, $y_1 \in \mathcal{C}(L_0, L_1)$ with $y_0$ of degree $1$ greater than $y_1$. From what I understood, Proposition $11.13$ implies that there is a canonical isomorphism $$\lambda^{top}(T_u\mathcal{M}(y_0, y_1)) \otimes o_{y_1} \cong o_{y_0},$$

where by $\mathcal{M}(y_0, y_1)$ I mean the "unquotiented" moduli space (although it can easily be replaced by the quotiented one). If $u$ is a regular point, then the first term on the left vanishes and we obtain an isomorphism $$c_u: o_{y_1} \to o_{y_0}.$$ This induces an isomorphism of $\mathbb{K}$-normalizations $$|c_u|_{\mathbb{K}}: |o_{y_1}|_{\mathbb{K}} \to |o_{y_0}|_{\mathbb{K}}.$$ Now when defining the differential, $u$ contributes a factor of $\pm 1$ depending on the "sign" of $c_u$.

My question: what is the sign of $c_u$? If there were a canonical orientation of the orientation lines $o_y$ for each $y$ I would understand this, but I was not able to find something like that.

$\endgroup$

0

You must log in to answer this question.