Discrete and continuous symmetries in monotone Floer theory

Abstract

This paper studies the self-Floer theory of a monotone Lagrangian submanifold L of a symplectic manifold X in the presence of various kinds of symmetry. First we suppose L is K-homogeneous and compute the image of low codimension K-invariant subvarieties of X under the length-zero closed–open string map. Next we consider the group $\mathrm {Symp}(X, L)$ of symplectomorphisms of X preserving L setwise, and extend its action on the Oh spectral sequence to coefficients of arbitrary characteristic, incorporating its action on the classes of holomorphic discs. This imposes constraints on the differentials which force them to vanish in certain situations. These techniques are combined to study a family of homogeneous Lagrangians in products of projective spaces, which exhibit some unusual properties.

Appendix A: Orientation computations

1.1 A.1: Setup

In this appendix we verify the signs in Theorem 3.5.3. So assume throughout that: $L \subset X$ is sharply K-homogeneous and equipped with an arbitrary orientation and the standard spin structure; $M \subset X$ is a K-invariant complex submanifold of complex codimension k (in practice this is the smooth locus of the invariant subvariety Z); $A \in H_2(X, L; \mathbb {Z})$ is a class of index 2k; and $y_\mathrm {min}$ is an arbitrary point in L. We need to show that each axial disc u in

$$\begin{aligned} M \times _X {\mathcal {M}}(A, J_\mathrm {std}) \times _L \{y_\mathrm {min}\} \end{aligned}$$

carries a positive sign, where ${\mathcal {M}} = {\mathcal {M}}(A, J_\mathrm {std})$ maps to the L on the left by ${\text {ev}}_0$ and the L on the right by ${\text {ev}}_1$.

From [4, Section A.1.8] the sign of u is positive if and only if the map

$$\begin{aligned} T_{u(0)}M \oplus T_{u(0)}X \oplus T_u{\mathcal {M}} \oplus T_{y_\mathrm {min}}L \rightarrow T_{u(0)}X \oplus T_{u(0)}X \oplus T_{y_\mathrm {min}}L \oplus T_{y_\mathrm {min}}L \end{aligned}$$

given by

$$\begin{aligned} (v_1, v_2, v_3, v_4) \mapsto (v_1-v_2, v_2-D_u{\text {ev}}_0 (v_3), D_u{\text {ev}}_1(v_3)-v_4, v_4) \end{aligned}$$

is orientation-preserving. Here TX and TM carry their complex orientations, whilst TL carries the orientation we chose on L. Using row and column operations this can be reduced to the orientation sign of

$$\begin{aligned} D_u{\text {ev}}_0 \oplus D_u{\text {ev}}_1 : T_u{\mathcal {M}} \rightarrow T_{u(0)}X/T_{u(0)}M \oplus T_{y_\mathrm {min}}L. \end{aligned}$$

(40)

We will explicitly construct a positively-oriented basis of the left-hand side and show that this map is orientation-preserving.

1.2 A.2: Orienting disc moduli spaces

First we need to understand some properties of the orientations of disc moduli spaces constructed in [17, Chapter 8], so take an arbitrary disc u in ${\mathcal {M}}$. Recall from Proposition 3.3.2 that we have a decomposition of bundle pairs

$$\begin{aligned} (u^*TX, u|_{\partial D}^*TL) \cong \bigoplus _j ({\underline{\mathbb {C}}}, z^{\kappa _j/2}{\underline{\mathbb {R}}}), \end{aligned}$$

where the $\kappa _j \in \mathbb {Z}$ are the partial indices of u, all of which are in fact non-negative. We abbreviate the left-hand side to (E, F). The tangent space $T_u{\mathcal {M}}$ is the kernel of the Cauchy–Riemann operator

$$\begin{aligned} \overline{\partial }_{(E, F)} : \Gamma \big ( (D, \partial D), (E, F) \big ) \rightarrow \Gamma (D, \Omega ^{0, 1}(E)). \end{aligned}$$

A choice of orientation and relative spin structure on L induces a homotopy class of trivialisation of F, and it is this trivialisation which is used to define the orientation on $T_u {\mathcal {M}}$.

Roughly speaking, the construction proceeds by degenerating the disc D into a nodal curve $D \cup \mathbb {C}\mathbb {P}^1$ (joined at the point 0 in each component), itself carrying a bundle pair $(E', F')$ and Cauchy–Riemann operator $\overline{\partial }_{(E', F')}$, such that (E, F) and $(E', F')$ are identified over a collar neighbourhood of $\partial D$. Gluing results give a bijection between holomorphic sections of $E'$ over each component of the nodal curve separately, which agree at the joining point, and holomorphic sections of the original bundle E. From this we obtain an isomorphism

$$\begin{aligned} {\text {ker}}\overline{\partial }_{(E, F)} \cong {\text {ker}}\big ( {\text {ker}}\overline{\partial }_{(E'|_D, F')} \oplus {\text {ker}}\overline{\partial }_{E'|_{\mathbb {C}\mathbb {P}^1}} \rightarrow E'_0 \big ), \end{aligned}$$

(41)

where $E_0'$ is the fibre over 0 and the map on the right-hand side sends a pair of sections $(s_1, s_2)$ to $s_1(0)-s_2(0)$. Since the disc u is regular (Proposition 3.3.2) this map is surjective, so we have a short exact sequence

$$\begin{aligned} 0 \rightarrow {\text {ker}}\overline{\partial }_{(E, F)} \rightarrow {\text {ker}}\overline{\partial }_{(E'|_D, F')} \oplus {\text {ker}}\overline{\partial }_{E'|_{\mathbb {C}\mathbb {P}^1}} \rightarrow E'_0 \rightarrow 0. \end{aligned}$$

(42)

The degeneration is done in such a way that $(E'|_D, F')$ is trivialised by our choice of trivialisation of F, giving an identification of ${\text {ker}}\overline{\partial }_{(E'|_D, F')}$ with $\mathbb {R}^n$ by evaluating solutions at a boundary point (and reversing the orientation on L switches the orientation of this identification, as claimed in Lemma 2.2.4). The other two spaces appearing on the right-hand side of (41) carry complex structures and hence canonical orientations. Putting these three orientations together with the short exact sequence (42), we see that there is an induced orientation on ${\text {ker}}\overline{\partial }_{(E, F)}$, which is the space we are really interested in. The auxiliary choices made do not affect this orientation.

The details of this procedure are technical and unimportant for our purposes, but we note two key properties of the construction:

• (i) If the trivialisation of F extends to a holomorphic trivialisation of E then the bundle pair $(E', F')$ is trivial and ${\text {ker}}\overline{\partial }_{(E, F)}$ is oriented directly by its identification with a fibre of F by evaluation at a boundary point.

• (ii) If (E, F) splits as a direct sum $(E^1, F^1) \oplus (E^2, F^2)$, compatible with the trivialisation of F (i.e. so that fibrewise, over each point z, $F_z^1$ corresponds to the first k components of $F_z \cong \mathbb {R}^n$ and $F_z^2$ to the last $n-k$) then we can make the construction respect this splitting and observe that the identification

  $$\begin{aligned} {\text {ker}}\overline{\partial }_{(E, F)} = {\text {ker}}\overline{\partial }_{(E^1, F^1)} \oplus {\text {ker}}\overline{\partial }_{(E^2, F^2)} \end{aligned}$$

  is orientation-preserving.

All we shall need are these two properties and the following explicit calculation:

Lemma A.2.1

Suppose (E, F) is a rank 1 Riemann–Hilbert pair of index 2—which can be identified with the tangent bundle of the pair $(D, \partial D)$—with F oriented. Evaluation at 0 and 1 defines an isomorphism

$$\begin{aligned} f : {\text {ker}}\overline{\partial }_{(E, F)} \xrightarrow {\sim } E_0 \oplus F_1, \end{aligned}$$

and the codomain is oriented by the complex structure on $E_0$ and the choice of orientation on F. This isomorphism is orientation-preserving.

Proof

Let $(E^{(t)}, F^{(t)}) \rightarrow (D^{(t)}, \partial D^{(t)})$ be the family of bundle pairs, parametrised by $t \in [0, 1]$, realising the above degeneration of the domain from $D^{(0)} = D$ to $D^{(1)} = D \cup \mathbb {C}\mathbb {P}^1$. Note that for all $t < 1$ the domain is biholomorphic to the disc D, and we may choose a continuous family $p^{(t)}$ of interior marked points in the domain which converge to the point $\infty$ in $\mathbb {C}\mathbb {P}^1 \subset D^{(1)}$ as $t \rightarrow 1$, as shown in Fig. 6 (recall that 0 in $\mathbb {C}\mathbb {P}^1$ is the point which is glued to 0 in D to construct $D^{(1)}$). Similarly we can choose a continuous family $q^{(t)}$ of boundary marked points.

Fig. 6

The degeneration $D^{(t)}$

We obtain a continuous family of kernels of the corresponding Cauchy–Riemann operators, which we denote by ${\text {ker}}\overline{\partial }_t$. At $t=1$ we impose the condition that the sections over D and $\mathbb {C}\mathbb {P}^1$ agree at 0, and continuity is in the sense of the above gluing result. This comes equipped with a family of evaluation maps, giving a continuous family of isomorphisms

$$\begin{aligned} f^{(t)} : {\text {ker}}\overline{\partial }_t \xrightarrow {\sim } E^{(t)}_{p^{(t)}} \oplus F^{(t)}_{q^{(t)}}. \end{aligned}$$

The codomains are naturally oriented, and the definition of the orientation on ${\text {ker}}\overline{\partial }_0$ is such that the orientation sign of $f^{(t)}$ is constant in t. Taking $p^{(0)} = 0$ and $q^{(0)} = 1$, the statement of the lemma amounts to $f^{(0)}$ being orientation-preserving, so it suffices to show that $f^{(1)}$ is orientation-preserving.

The restriction of the bundle $E^{(1)}$ to the sphere component of $D^{(1)}$ has first Chern class given by half the index of our original Riemann–Hilbert pair, so is isomorphic to ${\mathcal {O}}(1)$. We may therefore think of holomorphic sections as affine linear functions on $\mathbb {C}$, say $z \mapsto a+bz$ for complex numbers a and b. Solutions to the Riemann–Hilbert problem on the disc component, meanwhile, are all of the form cs, where c is a real number and s is an arbitrary fixed solution with s(1) pointing in the positive direction in $F^{(1)}_1$. The matching condition at 0 forces $a=\lambda c$ for some fixed $\lambda \in \mathbb {C}^*$, so if $b=b_1+ib_2$ then from (42) we see that $b_1$, $b_2$, c form a positively oriented set of coordinates on ${\text {ker}}\overline{\partial }'$. Choosing an appropriate basis vector for $E'_\infty$, the map

$$\begin{aligned} f' : {\text {ker}}\overline{\partial }' \rightarrow E'_\infty \oplus F'_1 \end{aligned}$$

can be viewed as sending the point $(b_1, b_2, c)$ to $(b_1+ ib_2, c)$. It is therefore orientation-preserving, which is exactly what we needed to show, completing the proof. $\square$

1.3 A.3: Splittings for axial discs

Now suppose u is an axial disc with u(0) in M, as in Appendix A.1. Our next goal is to give an explicit splitting of (E, F) into rank 1 pairs of index 0 or 2 and rank 2 pairs with partial indices (1, 1). Note that in order for the boundary real subbundle to have a trivialisation it must be orientable, which is equivalent to having even total index, so there is no hope of decomposing the problem further into rank 1 pairs of index 1. This splitting allows us to decompose the orientation problem using property (i), and then apply property (ii) and Lemma A.2.1 to orient each summand, after degenerating the (1, 1)-summands into (2, 0)-summands.

In order to construct the splitting, consider the infinitesimal action of ${\mathfrak {g}} = {\mathfrak {k}} \otimes \mathbb {C}$ at u(0) in M, and let its kernel be V. Since M is ${\mathfrak {g}}$-invariant and of complex codimension k we have $\dim _\mathbb {C}V \ge k$. Pick an $\mathbb {R}$-basis $\xi _1, \dots , \xi _a$ for $V \cap {\mathfrak {k}}$, and extend to a $\mathbb {C}$-basis $\xi _1, \dots , \xi _a, \eta _1, \dots , \eta _b$ for V. Each $\eta _j$ can be written as $\alpha _j + i \beta _j$ for unique $\alpha _j, \beta _j \in {\mathfrak {k}}$.

Lemma A.3.1

The set $\xi _1, \dots , \xi _a, \alpha _1, \dots , \alpha _b, \beta _1, \dots , \beta _b$ is $\mathbb {R}$-linearly independent in ${\mathfrak {k}}$.

Proof

Let $\overline{}$ denote the conjugate-linear involution of ${\mathfrak {g}} = {\mathfrak {k}} \otimes \mathbb {C}$ given by complex conjugation on the $\mathbb {C}$ factor. The vectors $\xi _1, \dots , \xi _a, \eta _1, \dots , \eta _b, \overline{\eta }_1, \dots , \overline{\eta }_b$ span $V + \overline{V}$ over $\mathbb {C}$, and we claim that they form a basis. Assuming this, $\xi _1, \dots , \xi _a, \alpha _1, \dots , \alpha _b, \beta _1, \dots , \beta _b$ also form a basis (we can transform between $\eta _j, \overline{\eta }_j$ and $\alpha _j, \beta _j$), so they are linearly independent over $\mathbb {C}$ and hence also over $\mathbb {R}$.

To prove the claim it suffices to show that $\dim _\mathbb {C}(V + \overline{V}) \ge a+2b$, and we know that

$$\begin{aligned} \dim _\mathbb {C}(V + \overline{V}) = \dim _\mathbb {C}V + \dim _\mathbb {C}\overline{V} - \dim _\mathbb {C}(V\cap \overline{V}) = 2a+2b - \dim _\mathbb {C}(V\cap \overline{V}), \end{aligned}$$

so it’s enough to show that $\dim _\mathbb {C}(V \cap \overline{V}) \le a$. In other words, we’re done if we can find a set of size a which spans the intersection $V \cap \overline{V}$. We’ll show that $\xi _1, \dots , \xi _a$ works. Suppose then that v is an element of this intersection, so both v and $\overline{v}$ lie in V. We can write v uniquely in the form $v_\mathbb {R}+ i v_{\mathbb {I}}$ with $v_\mathbb {R}$ and $v_\mathbb {I}$ in ${\mathfrak {k}}$, and we see that both $v_\mathbb {R}+ i v_{\mathbb {I}}$ and $v_\mathbb {R}- i v_{\mathbb {I}}$ lie in V. Thus

$$\begin{aligned} v_\mathbb {R}= \frac{(v_\mathbb {R}+ i v_{\mathbb {I}})+(v_\mathbb {R}- i v_{\mathbb {I}})}{2} \quad \text {and} \quad v_\mathbb {I} = \frac{(v_\mathbb {R}+ i v_{\mathbb {I}})-(v_\mathbb {R}- i v_{\mathbb {I}})}{2i} \end{aligned}$$

lie in V as well as in ${\mathfrak {k}}$. They therefore lie in the real span of the $\xi _j$, so v itself lies in the complex span, completing the proof. $\square$

We can now pick $\theta _1, \dots , \theta _c$ which extend $\xi _1, \dots , \xi _a, \alpha _1, \dots , \alpha _b, \beta _1, \dots , \beta _b$ to a basis of ${\mathfrak {k}}$. We obtain a trivialising frame for F

$$\begin{aligned} \xi _1 \cdot u , \dots , \xi _a \cdot u, \alpha _1 \cdot u, \beta _1 \cdot u, \dots , \alpha _b \cdot u, \beta _b \cdot u, \theta _1 \cdot u, \dots , \theta _c \cdot u, \end{aligned}$$

where, for example, $\xi _1 \cdot u$ denotes the section $z \mapsto \xi _1 \cdot u(z) \in T_{u(z)}L$. Since the actual choice of orientation on L is irrelevant we may assume this frame is positively oriented. The frame is compatible with the global trivialisation of TL defining the standard spin structure, so its homotopy class is precisely that induced by this spin structure.

We can then split E as

$$\begin{aligned}&\left\langle \frac{\xi _1}{z}\cdot u \right\rangle _\mathbb {C}\oplus \dots \oplus \left\langle \frac{\xi _a}{z}\cdot u \right\rangle _\mathbb {C}\oplus \left\langle \frac{(1+z)\alpha _1+i(1-z)\beta _1}{z}\cdot u, \frac{(1+z)\beta _1-i(1-z)\alpha _1}{z}\cdot u \right\rangle _\mathbb {C}\nonumber \\&\quad \oplus \dots \oplus \left\langle \frac{(1+z)\alpha _b+i(1-z)\beta _b}{z}\cdot u, \frac{(1+z)\beta _b-i(1-z)\alpha _b}{z}\cdot u \right\rangle _\mathbb {C}\oplus \left\langle \theta _1 \cdot u \right\rangle _\mathbb {C}\nonumber \\&\quad \oplus \dots \oplus \left\langle \theta _c \cdot u \right\rangle _\mathbb {C}, \end{aligned}$$

(43)

where $\left\langle \cdot \right\rangle _\mathbb {C}$ denotes $\mathbb {C}$-linear span and, for example, $(\xi _1/z)\cdot u$ denotes the holomorphic section of E given by $z \mapsto (\xi _1/z) \cdot u(z)$. Note that the $a+2b+c = n$ sections of E listed in (43) are holomorphic (since the $\xi _j \cdot u$ and $(\alpha _j + i \beta _j) \cdot u$ vanish at 0), and are fibrewise $\mathbb {C}$-linearly independent. To see the latter, consider the wedge product (over $\mathbb {C}$) of the sections: it is

$$\begin{aligned}&\frac{1}{z^{a+b}} (\xi _1 \cdot u) \wedge \dots \wedge (\xi _a \cdot u) \wedge (\alpha _1 \cdot u) \wedge (\beta _1 \cdot u) \wedge \dots \wedge (\alpha _b \cdot u) \\&\quad \wedge (\beta _b \cdot u) \wedge (\theta _1 \cdot u) \wedge \dots \wedge (\theta _c \cdot u), \end{aligned}$$

which is clearly non-zero on $D {\setminus } \{0\}$. Since u has index 2k, this expression vanishes to order $k-(a+b)$ at 0, but we know that $a+b = \dim _\mathbb {C}V \ge k$. We thus have equality $a+b=k$ and the sections remain independent at 0. We also deduce that $T_{u(0)}M = {\mathfrak {g}} \cdot u(0)$.

For $j=1, \dots , a+b+c$ let $E^j$ denote the jth summand of (43). For each j, the intersection $F^j {:=}F \cap E^j|_{\partial D}$ is a subbundle of F whose rank is half the real rank of $E^j$. Moreover one can write down explicit frames for each $F^j$ in terms of the given trivialisations of the $E^j$:

$$\begin{aligned} F^j= & {} z \left\langle \frac{\xi _j}{z} \cdot u \right\rangle _\mathbb {R}\text { for } 1 \le j \le a\\ F^{j+a}= & {} z^{1/2} \left\langle \frac{(1+z)\alpha _j+i(1-z)\beta _j}{z}\cdot u, \frac{(1+z)\beta _j-i(1-z)\alpha _j}{z}\cdot u \right\rangle _\mathbb {R}\text { for } 1 \le j \le b\\ F^{j+a+b}= & {} \left\langle \theta _j \cdot u \right\rangle _\mathbb {R}\text { for } 1 \le j \le c, \end{aligned}$$

where $\left\langle \cdot \right\rangle _\mathbb {R}$ denotes $\mathbb {R}$-linear span. In particular, the splitting of E is in fact a splitting of the pair (E, F) compatible with our trivialisation of F, and the first a summands are rank 1, index 2; the next b are rank 2, partial indices (1, 1); the final c are rank 1, index 0. We thus have the desired decomposition of (E, F).

1.4 A.4: Computing the signs

Now return to the map (40) which we hope to prove is orientation-preserving. Using the above splitting and property (i), $T_u{\mathcal {M}}$ decomposes as $\oplus _j {\text {ker}}\overline{\partial }_{(E^j, F^j)}$. We always order such sums from left to right. Similarly $T_{y_\mathrm {min}}L$ decomposes as $\oplus _j F^j_1$ (the sum of the fibres over $1 \in \partial D$). These decompositions are all as oriented vector spaces.

Lemma A.4.1

$T_{u(0)}M$ decomposes as $\oplus _j (E_0^j \cap T_{u(0)}M)$.

Proof

We saw in Appendix A.3 that $T_{u(0)}M = {\mathfrak {g}} \cdot u(0)$. By definition, the space $V \subset {\mathfrak {g}}$ of complex codimension k acts trivially, so giving a basis for $T_{u(0)}M$ is equivalent to giving a basis for ${\mathfrak {g}}/V$. Consider the basis $\alpha _1-i\beta _1, \dots , \alpha _b-i\beta _b, \theta _1, \dots , \theta _c$ for the latter. We claim that the induced splitting of $T_{u(0)}M$ is compatible with the splitting of $E_0$ into the $E_0^j$.

Well, the $\left\langle \theta _j \cdot u(0) \right\rangle _\mathbb {C}$ summand of $T_{u(0)}M$ clearly lies in the $\left\langle \theta _j \cdot u \right\rangle _\mathbb {C}$ summand of E. Meanwhile, the

$$\begin{aligned} \left\langle \frac{(1+z)\alpha _j+i(1-z)\beta _j}{z}\cdot u, \frac{(1+z)\beta _j-i(1-z)\alpha _j}{z}\cdot u \right\rangle _\mathbb {C}\end{aligned}$$

summand of E contains the section

$$\begin{aligned} \frac{1}{2}\left( \frac{(1+z)\alpha _j+i(1-z)\beta _j}{z}\cdot u\right) - \frac{i}{2}\left( \frac{(1+z)\beta _j-i(1-z)\alpha _j}{z}\cdot u\right) = (\alpha _j - i \beta _j) \cdot u, \end{aligned}$$

and hence contains the $\left\langle (\alpha _j-i\beta _j)\cdot u(0) \right\rangle _\mathbb {C}$ summand of $T_{u(0)}M$. $\square$

We therefore need to understand the orientation sign of the map

$$\begin{aligned} \bigoplus _j {\text {ker}}\overline{\partial }_{(E^j, F^j)} \rightarrow \bigg ( \bigoplus _j E^j_0\Big /\left( E^j_0 \cap T_{u(0)}Z\right) \bigg ) \oplus \bigg ( \bigoplus _j F^j_1 \bigg ), \end{aligned}$$

which is the product of the orientation signs of the maps

$$\begin{aligned} {\text {ker}}\overline{\partial }_{(E^j, F^j)} \rightarrow \left( E^j_0 \Big / \left( E^j_0 \cap T_{u(0)}Z \right) \right) \oplus F^j_1. \end{aligned}$$

Letting $\varepsilon _j$ denote the jth sign, the verification of the signs in Theorem 3.5.3 is completed by

Lemma A.4.2

For all j we have $\varepsilon _j = +1$

Proof

Let $\mu _j$ denote the partial indices of $(E^j, F^j)$, and $\overline{\partial }_j$ the Cauchy–Riemann operator. If $\mu _j = 2$ then the intersection $E^j_0 \cap T_{u(0)}Z$ is zero and the result follows from Lemma A.2.1. On the other hand, if $\mu _j = 0$ then $(E^j, F^j) = (\mathbb {C}\theta \cdot u, \mathbb {R}\theta \cdot u)$, where $\theta = \theta _{j-(a+b)} \in {\mathfrak {k}}$, and the space $E^j_0 / (E^j_0 \cap T_{u(0)}Z)$ is zero. A positively oriented basis for ${\text {ker}}\overline{\partial }_j$ is given by $\theta \cdot u$, whilst a positively oriented basis for $F^j_1$ is given by $\theta \cdot u(1)$, so the required map is indeed orientation-preserving.

Finally suppose $\mu _j = (1, 1)$, so $E^j$ and $F^j$ are spanned by appropriate expressions built from $\alpha = \alpha _{j-a}$ and $\beta = \beta _{j-a}$, with $(\alpha + i \beta ) \cdot u(0) = 0$. To find a positively oriented basis for ${\text {ker}}\overline{\partial }_j$ we degenerate our (1, 1) pair to a (2, 0) as follows. Suppose we replace the condition $(\alpha + i \beta ) \cdot u(0) = 0$ by $(\alpha + ti \beta ) \cdot u(0) = 0$, and deform the real parameter t from 1 down to 0. We obtain a family of Riemann–Hilbert pairs over [0, 1], and bases for the kernels of the corresponding Cauchy–Riemann operators are given by

$$\begin{aligned} \frac{(1+z^2)\alpha + ti(1-z^2)\beta )}{z} \cdot u, \frac{i(1-z^2)\alpha - t(1+z^2)\beta )}{z} \cdot u, \alpha \cdot u, \beta \cdot u. \end{aligned}$$

(44)

By Lemma A.2.1, and the fact that $\alpha \cdot u(1)$, $\beta \cdot u(1)$ is positively oriented as a basis of $F^j_1$, this basis is positively oriented in the limit $t \rightarrow 0$. Hence by continuity of the orientation construction it is also positively oriented at $t=1$.

Now (restricting again to the $t=1$ situation, which is what we actually care about) consider the evaluation map to

$$\begin{aligned}&\left( E^j_0 \Big / \left( E^j_0 \cap T_{u(0)}Z \right) \right) \oplus F^j_1 = \left( \left\langle \lim _{z \rightarrow 0} \frac{\alpha + i\beta }{z}\cdot u(z), \beta \cdot u(0) \right\rangle \Big / \left\langle \beta \cdot u(0) \right\rangle \right) \\&\quad \oplus \left\langle \alpha \cdot u(1), \beta \cdot u(1) \right\rangle _\mathbb {R}. \end{aligned}$$

The basis (44) is sent to a positively oriented basis for this space, and so $\varepsilon _j = +1$ as claimed. $\square$