a map of spaces inducing isomorphism $H_{1}(K;\mathbb{Z}_{2})\cong
H_{1}(B;\mathbb{Z}_{2})$. Then the induced map on the sets of spin structures
$\operatorname{Spin}(\zeta)\to\operatorname{Spin}(i^{*}\zeta)$ is a bijection.

We use the previous theorem by Kirby-Taylor in the proof of the following.

######  Proposition 6.11.

Let $F$, $F_{1}$, and $F_{2}$ be nonorientable surfaces, with connected
nonempty boundary.

  1. (1)

Let $\xi=(\mathbb{R}^{2}\to\mathring{E}\to F)$ be a vector bundle with
orientable total space. Write $\mathring{Y}$ for the total space of the
associated $S^{1}$-bundle $S(\xi)$. Fix a choice of section $s\colon
F\to\mathring{Y}$ and a choice of $S^{1}$-fibre $\mu\subseteq\mathring{Y}$.
There is a bijection of sets

$\begin{array}[]{rcl}\Psi_{s,\mu}\colon\operatorname{Spin}(\mathring{Y})&\xrightarrow{1:1}&\operatorname{Spin}(T\mathring{Y}|_{\mu})\times\left\\{\begin{array}[]{cc}\text{$\mathbb{Z}_{4}$-quadratic
refinements of}\\\
\text{$(H_{1}(s(F);\mathbb{Z}_{2}),Q_{s(F)})$}\end{array}\right\\}\\\
\mathfrak{s}&\mapsto&(\mathfrak{s}|_{\mu},q_{KT}(\mathfrak{s})).\end{array}$

To define $q_{KT}(\mathfrak{s})$, we take
$\zeta:=\nu_{s(F)\subseteq\mathring{Y}}$ and use the spin structure on
$\mathring{Y}$ to induce a spin structure on
$Ts(F)\oplus\nu_{s(F)\subseteq\mathring{Y}}\cong T\mathring{Y}|_{s(F)}$.

  2. (2)

Let $\Phi\colon\xi_{1}\cong\xi_{2}$ be an isomorphism of nonorientable 2-plane
vector bundles with orientable total space, covering a homeomorphism
$\varphi\colon F_{1}\to F_{2}$. Write $\mathring{Y}_{i}$ for the total space
of the associated $S^{1}$-bundles $S(\xi_{i})$. Fix a choice of section
$s_{1}\colon F\to\mathring{Y}_{1}|_{F}$ and a choice of $S^{1}$-fibre
$\mu_{1}\subseteq\mathring{Y}_{1}$. Write $s_{2}=\Phi\circ
s_{1}\circ\varphi^{-1}$ and $\mu_{2}:=\Phi(\mu_{1})$. Then there is a
commutative diagram

${\operatorname{Spin}(\mathring{Y}_{2})}$${\operatorname{Spin}(T\mathring{Y}_{2}|_{\mu_{2}})\times\left\\{\parbox{128.0374pt}{\centering\leavevmode\nobreak\
$\mathbb{Z}_{4}$-quadratic refinements of\\\ \leavevmode\nobreak\
$(H_{1}(s_{2}(F);\mathbb{Z}_{2}),Q_{s_{2}(F)})$\@add@centering}\right\\}}$${\operatorname{Spin}(\mathring{Y}_{1})}$${\operatorname{Spin}(\mathring{Y}_{1}|_{\mu_{1}})\times\left\\{\parbox{128.0374pt}{\centering\leavevmode\nobreak\
$\mathbb{Z}_{4}$-quadratic refinements of\\\ \leavevmode\nobreak\
$(H_{1}(s_{1}(F);\mathbb{Z}_{2}),Q_{s_{1}(F)})$\@add@centering}\right\\},}$$\scriptstyle{\Psi_{s_{2},\mu_{2}}}$$\scriptstyle{\Phi^{*}}$$\scriptstyle{\widehat{\Phi}}$$\scriptstyle{\Psi_{s_{1},\mu_{1}}}$

where the $\Psi_{s_{i},\mu_{i}}$ are the bijections of sets from the first
item, and

${\widehat{\Phi}}(\mathfrak{t},q):=((\Phi|_{\mu_{2}})^{*}\mathfrak{t},q\circ\Phi_{*}).$

###### Proof.

We begin with the first item. Take the standard cell structure for $F$ (see
the conventions at the end of the introduction), with its basis of loops
$\gamma_{1},\dots,\gamma_{h}$. Let $A\subseteq\mathring{Y}$ be a connected
1-complex comprising a single $0$-cell, the $S^{1}$-fibre loop $\mu$, and the
collection of loops $s(\gamma_{i})$. Applying Lemma 6.10 to the inclusion map
of $A$, we obtain a bijection
$\operatorname{Spin}(\mathring{Y})\to\operatorname{Spin}(T\mathring{Y}|_{A})$.
Here we used Proposition 4.2 and the universal coefficient theorem to compute
the $\mathbb{Z}_{2}$-coefficient homology of $\mathring{Y}$. Now, as $A$ is a
bouquet of circles, there is a bijection

$\operatorname{Spin}(T\mathring{Y}|_{A})\xrightarrow{1:1}\operatorname{Spin}(T\mathring{Y}|_{\mu})\times\prod_{i=1}^{h}\operatorname{Spin}(T\mathring{Y}|_{s(\gamma_{i})}).$

Hence it now suffices to prove that the latter factor is in 1:1 correspondence
with the $\mathbb{Z}_{4}$-quadratic refinement factor from the statement of
the proposition.

For this, first note that the collection of loops $s(\gamma_{i})$ form the
$1$-skeleton of $s(F)$ and so, as $s(F)$ is homotopy equivalent to this
$1$-complex, restriction determines a bijection

$\operatorname{Spin}(T\mathring{Y}|_{s(F)})\xrightarrow{1:1}\prod_{i=1}^{h}\operatorname{Spin}(T\mathring{Y}|_{s(\gamma_{i})}).$

It is clear that there is a bijection between
$\operatorname{Spin}(T\mathring{Y}|_{s(F)})$ and
$\operatorname{Spin}(Ts(F)\oplus\nu_{s(F)\subseteq\mathring{Y}})$. We now wish
to apply Theorem 6.8 using the surface $\Sigma=s(F)$ and the line bundle
$\nu_{s(F)\subseteq\mathring{Y}}$. To see that this theorem is indeed
applicable, observe that as $\mathring{Y}$ is orientable, the product formula
for Stiefel-Whitney classes implies that
$w_{1}(Ts(F))=w_{1}(\nu_{s(F)\subseteq\mathring{Y}})$. Thus, by Theorem 6.8,
the Kirby-Taylor quadratic refinement determines a bijection of sets

$\Psi_{s,\mu}\colon\operatorname{Spin}(Ts(F)\oplus\nu_{s(F)\subseteq\mathring{Y}})\xrightarrow{1:1}\left\\{\begin{array}[]{cc}\text{$\mathbb{Z}_{4}$-quadratic
refinements of}\\\
\text{$(H_{1}(s(F);\mathbb{Z}_{2}),Q_{s(F)})$}\end{array}\right\\},$

and this completes the proof of the statement in the first item.

We will now prove the second claim, namely that
$\Psi_{s_{1},\mu_{1}}\circ\Phi^{*}={\widehat{\Phi}}\circ\Psi_{s_{2},\mu_{2}}$.
In other words, for all
$\mathfrak{s}\in\operatorname{Spin}(\mathring{Y}_{2})$, we will show
$((\Phi^{*}\mathfrak{s})|_{\mu_{1}},q_{KT_{1}}(\Phi^{*}\mathfrak{s}))=((\Phi|_{\mu_{2}})^{*}(\mathfrak{s}|_{\mu_{2}}),q_{KT_{2}}(\mathfrak{s})\circ\Phi_{*}).$
As $\mu_{2}=\Phi(\mu_{1})$, by definition, equality in the first argument is
clear. Given $\mathfrak{s}\in\operatorname{Spin}(\mathring{Y}_{2})$, we will
now show the equality

(12)
$q_{KT_{1}}(\Phi^{*}\mathfrak{s})=q_{KT_{2}}(\mathfrak{s})\circ\Phi_{*}\colon
H_{1}(s_{1}(F_{1}))\to\mathbb{Z}_{4},$

which will complete the proof of the second claim. Let $\gamma_{1}\subseteq
F_{1}$ be an embedded circle and let $\gamma_{2}=\varphi(\gamma_{1})$ be the
corresponding loop in $F_{2}$. This way $s_{1}(\gamma_{1})\subseteq
s_{1}(F_{1})\subseteq\mathring{Y}_{1}$ is an embedded circle and, by
definition of $s_{2}=\Phi\circ s_{1}\circ\varphi^{-1}$, we deduce that
$\Phi(s_{1}(\gamma_{1}))=s_{2}(\gamma_{2})\subseteq\mathring{Y}_{2}$ is
similarly an embedded circle. Pick a framing
$f_{2}\colon\nu_{s_{2}(\gamma_{2})\subseteq\mathring{Y}_{2}}\cong
s_{2}(\gamma_{2})\times\mathbb{R}^{2}$ that is odd with respect to the spin
structure $\mathfrak{s}$. Since $\Phi$ is a bundle isomorphism, one checks
that the framing defined by the composition

${f_{1}\colon\nu_{s_{1}(\gamma_{1})\subseteq\mathring{Y}_{1}}}$${\nu_{s_{2}(\gamma_{2})\subseteq\mathring{Y}_{2}}}$${s_{2}(\gamma_{2})\times\mathbb{R}^{2}}$${s_{1}(\gamma_{1})\times\mathbb{R}^{2}}$$\scriptstyle{\Phi}$$\scriptstyle{\cong}$$\scriptstyle{f_{2}}$$\scriptstyle{\cong}$$\scriptstyle{\Phi^{-1}\times\operatorname{Id}_{\mathbb{R}^{2}}}$$\scriptstyle{\cong}$

is odd with respect to the spin structure $\Phi^{*}\mathfrak{s}$. Choose an
orientation on $\gamma_{1}$ and use the induced orientations on
$s_{1}(\gamma_{1})$ and $s_{2}(\gamma_{2})$. Now
$q_{KT_{1}}(\Phi^{*}\mathfrak{s})(s_{1}(\gamma_{1}))$ is calculated by
counting right-handed half-twists of $\nu_{s_{1}(\gamma_{1})\subseteq
s_{1}(F_{1})}\subseteq\nu_{s_{1}(\gamma_{1})\subseteq\mathring{Y}_{1}}$ with
respect to the odd framing $f_{1}$. On the other hand,
$q_{KT_{2}}(\mathfrak{s})\circ\Phi_{*}(s_{1}(\gamma_{1}))=q_{KT_{2}}(\mathfrak{s})(\Phi(s_{1}(\gamma_{1})))$
is calculated by counting right-handed half-twists of
$\nu_{\Phi(s_{1}(\gamma))\subseteq
s_{2}(F_{2})}\subseteq\nu_{\Phi(s_{1}(\gamma))\subseteq\mathring{Y}_{2}}$ with
respect to the odd framing $f_{2}$. These counts agree thanks to our choice of
framings. This establishes (12) and therefore proves the second claim. ∎

######  Remark 6.12.

A similar statement to Proposition 6.11 holds when the surfaces are closed. To
modify the statement to the case of $F$ closed, one must consider the
punctured surface $\mathring{F}$ in order to choose the initial section
$s\colon\mathring{F}\to\mathring{Y}|_{\mathring{F}}$ (similarly for $F_{1}$
and $F_{2}$). The remainder of the statement holds verbatim and the proof
proceeds entirely similarly.

###  6.2. The Guillou-Marin quadratic refinement

Let $F\subseteq D^{4}$ be a properly embedded nonorientable surface with
connected nonempty boundary. We review the definition of the Guillou-Marin
[GuillouMarin] quadratic refinement
$H_{1}(F;\mathbb{Z}_{2})\to\mathbb{Z}_{4}$, refining the
$\mathbb{Z}_{2}$-valued intersection form of $F$.

Let $\gamma$ be an embedded circle in the interior of $F$ and let
$D\looparrowright D^{4}$ be an immersed disc whose boundary meets $F$
transversely in $\partial D=\gamma$, and whose interior meets $F$ transversely
away from $\gamma$, and in double points. Choose an orientation on $\gamma$,
and thus a positive tangent direction. There are two possible choices of
framing $\nu_{D\subseteq D^{4}}\cong D\times\mathbb{R}^{2}$, one for each
orientation of the fibres. Choose the framing with the property that, writing
$\mathbb{R}^{2}=\mathbb{R}\langle e_{1},e_{2}\rangle$, the frame

$\text{(outwards-facing normal to $\gamma$ in $D$)}\times\text{(positive
tangent to\leavevmode\nobreak\ $\gamma$)}\times e_{1}\times
e_{2}\in\operatorname{Fr}(T_{p}D^{4})$

agrees with the ambient orientation of $D^{4}$ at all points $p\in\gamma$.

######  Definition 6.13.

Given an embedded loop $\gamma\subseteq F$, choose an orientation on $\gamma$
and a disc $D$ as above. We then denote by $W(D)$ the number of right-handed
half-twists completed by the sub-bundle $\nu_{\gamma\subseteq
F}\subseteq\nu_{D\subseteq D^{4}}|_{\gamma}$ in one complete rotation around
$\gamma$. Define333Guillou-Marin define $q_{GM}$ for closed surfaces in closed
$4$-manifolds but their definition extends to properly embedded surfaces in
$4$-manifolds with boundary, see e.g. [GilmerLivingston].

$q_{GM}(\gamma):=W(D)+2(D\cdot F)\in\mathbb{Z}_{4}.$

A $\mathbb{Z}_{4}$-quadratic form $(P,Q,q)$ over $\mathbb{Z}_{2}$ has a Brown-
Kervaire invariant $\beta(P,Q,q)\in\mathbb{Z}_{8}$; see e.g. [Brown,
KirbyMelvin]. The following theorem is due to Gilmer-Livingston
[GilmerLivingston, Theorem 7], building on earlier work of Guillou-Marin
[GuillouMarin] that concerned closed nonorientable surfaces.

######  Theorem 6.14 (Guillou-Marin [GuillouMarin], Gilmer-Livingston
[GilmerLivingston]).

The Guillou-Marin function $q_{GM}$ is well defined on homology and gives rise
to a $\mathbb{Z}_{4}$-quadratic refinement of the
$\mathbb{Z}_{2}$-intersection form $(H_{1}(F;\mathbb{Z}_{2});Q_{F})$.
Moreover, if $F$ has normal Euler number $e$ $($which is necessarily even by
Proposition 2.5$)$, then

$\beta(H_{1}(F;\mathbb{Z}_{2}),Q_{F},q_{GM})\equiv-\frac{1}{2}e+4\operatorname{Arf}(\partial
F)\pmod{8}.$

Note that Guillou-Marin and Gilmer-Livingston work in the smooth category, but
their arguments extend to the topological category.

######  Corollary 6.15.

If $F_{1},F_{2}\subseteq D^{4}$ have the same nonorientable genus, connected,
nonempty boundary $K\subseteq S^{3}$, and normal Euler number, then there
exists a homeomorphism $f\colon F_{1}\to F_{2}$, restricting to the identity
map on $K$, and inducing an isometry between the Guillou-Marin forms

$f_{*}\colon(H_{1}(F_{1};\mathbb{Z}_{2}),Q_{F_{1}},q_{GM_{1}})\xrightarrow{\cong}(H_{1}(F_{2};\mathbb{Z}_{2}),Q_{F_{2}},q_{GM_{2}}).$

###### Proof.

Since both $F_{1}$ and $F_{2}$ have the same nonorientable genus, there is an
isometry of intersection forms
$(H_{1}(F_{1};\mathbb{Z}_{2}),Q_{F_{1}})\cong(H_{1}(F_{2};\mathbb{Z}_{2}),Q_{F_{2}})$.
As the surfaces have the same normal Euler number and connected boundary $K$,
Theorem 6.14 implies that their Guillou-Marin forms have the same Brown-
Kervaire invariant. As mentioned in [GuillouMarin, final remark of Section
II], this implies there is an isometry of $\mathbb{Z}_{4}$-quadratic forms
$(H_{1}(F_{1};\mathbb{Z}_{2}),Q_{F_{1}},q_{GM_{1}})\cong(H_{1}(F_{2};\mathbb{Z}_{2}),Q_{F_{2}},q_{GM_{2}})$;
this remark concerns abstract non-degenerate quadratic forms and therefore
applies since $\partial F_{1}=\partial F_{2}$ is connected. Finally, it is
shown in [GadgilPancholi, Theorem 6.1] that every isometry
$(H_{1}(F_{1};\mathbb{Z}_{2}),Q_{F_{1}})\cong(H_{1}(F_{2};\mathbb{Z}_{2}),Q_{F_{2}})$
of intersection forms of compact, nonorientable surfaces with connected
nonempty boundary is induced by some homeomorphism between the surfaces that
fixes the boundary pointwise. ∎

###  6.3. The exterior Guillou-Marin form, in the presence of a nice section

In this section we show it is possible to compute the Guillou-Marin quadratic
refinement of a $\mathbb{Z}_{2}$-surface using immersed discs whose exteriors
lie entirely in the exterior of $F$. Before doing this, we introduce a
mechanism that will allow us to choose immersed discs so that their interiors
miss the tubular neighbourhood.

Throughout this subsection, let $F\subseteq D^{4}$ be a properly embedded
surface with connected nonempty boundary $K\subseteq S^{3}$. Recall the
standing convention that we have fixed a tubular neighbourhood
$\alpha\colon(D(\nu_{F\subseteq D^{4}}),S(\nu_{F\subseteq
D^{4}}))\xrightarrow{\cong}(\alpha(D(\nu_{F\subseteq D^{4}})),\partial
X_{F}\setminus\mathring{X}_{K})$, where $\nu_{F\subseteq D^{4}}$ denotes the
normal bundle of $F$ and $D(\nu_{F\subseteq D^{4}})$ and $S(\nu_{F\subseteq
D^{4}})$ respectively denote its disc and circle bundles.

######  Lemma 6.16.

The bundle $S(\nu_{F\subseteq D^{4}})$ admits a nice section.

###### Proof.

Write $h$ for the nonorientable genus of $F$, write
$\bigvee_{i=1}^{h}\gamma_{i}\subseteq F$ for the $1$-subcomplex that is a
deformation retract of $F$, using the standard cell decomposition of $F$, and
let $b_{i}:=[\gamma_{i}]\in H_{1}(F)$ be the corresponding homology
generators. Note that $w_{1}(F)(b_{i})\neq 0$ for all $i=1,\dots,h$. Choose a
section $s\colon F\to{S}(\nu_{F\subseteq D^{4}})$ and consider the composition

$\mathbb{Z}\langle b_{1},\dots,b_{h}\rangle\cong
H_{1}(F)\xrightarrow{{s}_{*}}H_{1}(S(\nu_{F\subseteq
D^{4}}))\xrightarrow{\alpha_{*}}H_{1}(\partial
X_{F})\xrightarrow{\iota_{*}}H_{1}(X_{F}),$

where $\iota\colon\partial X_{F}\to X_{F}$ is the inclusion. By Proposition
4.2 we may write this map as

(13) ${\mathbb{Z}\langle
b_{1},\dots,b_{h}\rangle}$${\mathbb{Z}_{2}\langle\mu\rangle\oplus\mathbb{Z}\langle
s_{*}(b_{1}),\dots,s_{*}(b_{h})\rangle}$${\mathbb{Z}_{2}.}$$\scriptstyle{s_{*}}$$\scriptstyle{\iota_{*}\circ\alpha_{*}}$

Here, $\mu$ is a generator of the first homology of a fibre of
$S(\nu_{F\subseteq D^{4}})$. We now intend to define a new section
$s^{\prime}\colon F\to S(\nu_{F\subseteq D^{4}})$, which will be the promised
nice section. Because $F$ is homotopy equivalent to a wedge of circles, it is
sufficient to describe our new section individually on the 1-cells
$\gamma_{i}$. Recall that a nonorientable circle bundle $\xi$ over a circle
has exactly two sections up to isotopy. The total space $E(\xi)$ is a Klein
bottle with homology $H_{1}(E(\xi))\cong\mathbb{Z}\oplus\mathbb{Z}_{2}$, and
with respect to some such choice of isomorphism, the two possible sections
represent the classes $(1,0)$ and $(1,1)$. In our case, the choices of
$\gamma_{i}$ and $\mu$ fix such an isomorphism and so the image in
$H_{1}(S(\nu_{F\subseteq D^{4}}))$ of the two possible sections of
$S(\nu_{F\subseteq D^{4}})|_{\gamma_{i}}$ are $s_{*}(b_{i})$ and
$s_{*}(b_{i})+\mu$. For each $i$ such that the composition in (13)
$\iota_{*}\circ\alpha_{*}\circ s_{*}(b_{i})$ is nontrivial, modify
$s|_{\gamma_{i}}$ to $s^{\prime}|_{\gamma_{i}}$ by switching to the other
section of this nontrivial circle bundle. On all other $\gamma_{i}$, i.e.
where $\iota_{*}\circ\alpha_{*}\circ s_{*}(b_{i})$ is trivial, set
$s|_{\gamma_{i}}=s^{\prime}|_{\gamma_{i}}$. This determines the new section
$s^{\prime}\colon F\to S(\nu_{F\subseteq D^{4}})$.

We now verify that $s^{\prime}$ is nice. Note that for $i$ such that
$\iota_{*}\circ\alpha_{*}\circ s_{*}(b_{i})$ is nontrivial, we have that
$s^{\prime}_{*}(b_{i})=s_{*}(b_{i})+\mu\in H_{1}(S(\nu_{F\subseteq D^{4}}))$.
Thus $\iota_{*}\circ\alpha_{*}\circ s^{\prime}_{*}(b_{i})$ is now trivial for
all $i$, and so $s^{\prime}$ is a nice section, as desired. ∎

Now let $F\subseteq D^{4}$ be a $\mathbb{Z}_{2}$-surface and suppose that we
have chosen a tubular neighbourhood $\alpha$ for $F\subseteq D^{4}$, together
with a nice section $s\colon F\to S(\nu_{F\subseteq D^{4}})$. For any embedded
circle $\gamma\subseteq s(F)$, the facts that $s$ is a nice section and that
$F$ is a $\mathbb{Z}_{2}$-surface, imply $\alpha(\gamma)$ is null-homotopic in
$X_{F}$. So there exists a properly immersed disc $D^{\prime}\looparrowright
X_{F}$, whose boundary meets $F$ transversely in $\partial
D^{\prime}=\alpha(\gamma)$.

######  Definition 6.17.

Let $\alpha$ be a tubular neighbourhood for $F\subseteq D^{4}$, together with
a nice section $s\colon F\to S(\nu_{F\subseteq D^{4}})$. Let
$D^{\prime}\looparrowright X_{F}$ be a properly immersed disc whose boundary
meets $\partial X_{F}$ transversely in $\partial D^{\prime}=\alpha(\gamma)$,
for some embedded loop $\gamma\subseteq s(F)$. Choose an orientation on
$\gamma$ and choose a framing $\nu_{D^{\prime}\subseteq
X_{F}}|_{\alpha(\gamma)}\cong\alpha(\gamma)\times\mathbb{R}\langle
e_{1},e_{2}\rangle$ such that the frame

$\text{(outwards facing normal to $\alpha(\gamma)$ in\leavevmode\nobreak\
$D^{\prime}$)}\times\text{(positive tangent to\leavevmode\nobreak\
$\alpha(\gamma)$)}\times e_{1}\times e_{2}\in\operatorname{Fr}(T_{p}X_{F})$

agrees with the ambient orientation of $X_{F}$ at all points
$p\in\alpha(\gamma)$. Define $W(D^{\prime})$ as the number of right-handed
half-twists of
$\nu_{\alpha(\gamma)\subseteq\alpha(s(F))}\subseteq\nu_{D^{\prime}\subseteq
X_{F}}|_{\alpha(\gamma)}$ in one complete loop around $\gamma$ in the positive
direction. Define

$\widehat{q}_{GM}(\gamma):=W(D^{\prime}).$

As there is a basis of $H_{1}(s(F);\mathbb{Z}_{2})$ by oriented embedded
loops, this extends linearly to a function $\widehat{q}_{GM}\colon
H_{1}(s(F);\mathbb{Z}_{2})\to\mathbb{Z}_{4}$, which we call the _exterior_
Guillou-Marin function. The proof that this is well defined on
$H_{1}(s(F);\mathbb{Z}_{2})$ is similar to the one for the Guillou-Marin form;
see Theorem 6.14.

The next proposition relates $\widehat{q}_{GM}$ to the Guillou-Marin form, and
in particular shows that its isometry type does not depend on the choice of
the tubular neighbourhood $\alpha$ of $F$.

######  Proposition 6.18.

Let $F\subseteq D^{4}$ be a $\mathbb{Z}_{2}$-surface, and fix a nice section
$s\colon F\to S(\nu_{F\subseteq D^{4}})$. Then the exterior Guillou-Marin
quadratic function is a $\mathbb{Z}_{4}$-quadratic refinement of the
$\mathbb{Z}_{2}$-intersection form $(H_{1}(s(F);\mathbb{Z}_{2}),Q_{s(F)})$.

Moreover, the isometry of $\mathbb{Z}_{2}$-intersection forms $s_{*}\colon
H_{1}(F;\mathbb{Z}_{2})\to H_{1}(s(F);\mathbb{Z}_{2})$ induces an isometry
between the Guillou-Marin quadratic form and the exterior Guillou-Marin
quadratic form.

###### Proof.

Let $\alpha$ be a tubular neighbourhood for $F\subseteq D^{4}$, let
$\delta\subseteq F$ be an embedded loop, and let $D^{\prime}\looparrowright
X_{F}$ be a properly immersed disc whose boundary meets $F$ transversely in
$\partial D^{\prime}=\alpha(s(\delta))$. For each $p\in\delta$, write $I_{p}$
for the radial line between $p$ and $s(p)$ in the fibre $\nu_{F\subseteq
D^{4}}|_{p}$. The union $A:=\coprod_{p\in\delta}I_{p}$ is an embedded annulus
in $\nu_{F\subseteq D^{4}}$. The union
$D:=D^{\prime}\cup_{\alpha(s(\delta))}\alpha(A)$ may be used to compute
$q_{GM}(\delta)$. The interior of the disc $D$ does not intersect $F$ and
therefore $q_{GM}(\delta)=W(D)$. As $\alpha(A)$ is an embedded annulus with
trivial normal bundle in $D^{4}$, and the framing conventions for computing
$W(D)$ and $W(D^{\prime})$ are the same, we also have $W(D)=W(D^{\prime})$.
Thus, $q_{GM}(\delta)=\widehat{q}_{GM}(s(\delta))$. This shows both that
$\widehat{q}_{GM}$ is a $\mathbb{Z}_{4}$-quadratic refinement of the
$\mathbb{Z}_{2}$-intersection form on $s(F)$ and that $s_{*}$ is indeed an
isometry of $\mathbb{Z}_{4}$-quadratic forms. ∎

There is an isometry between the (exterior) Guillou-Marin form for $F$ and the
particular Kirby-Taylor forms that correspond to the spin structures on
$\partial X_{F}$ that extend to $X_{F}$:

######  Lemma 6.19.

Let $F\subseteq D^{4}$ be a $\mathbb{Z}_{2}$-surface, and let $s\colon F\to
S(\nu_{F\subseteq D^{4}})$ be a nice section. Write
$\mathring{Y}\subseteq\partial X_{F}$ for the boundary of the tubular
neighbourhood of $F$. Choose a spin structure $\mathfrak{s}$ for
$\mathring{Y}$ that extends to a spin structure on $X_{F}$. Then there is an
equality of $\mathbb{Z}_{4}$-quadratic forms

$((H_{1}(s(F);\mathbb{Z}_{2}),Q_{s(F)},\widehat{q}_{GM})=((H_{1}(s(F);\mathbb{Z}_{2}),Q_{s(F)},q_{KT}(\mathfrak{s})),$

where the latter form $q_{KT}(\mathfrak{s})$ is furnished by Proposition 6.11.

###### Proof.

To ease notation, in this proof we will write $X:=X_{F}$ and $\Sigma:=s(F)$,
and we will henceforth identify $\Sigma$ with $\alpha(\Sigma)$. Let
$\gamma\subseteq\Sigma$ be an oriented, embedded loop. We unravel the
definition of $q_{KT}(\mathfrak{s})(\gamma)$ from Proposition 6.11. In the
notation of that section, set $\zeta=\nu_{\Sigma\subseteq\mathring{Y}}$. The
spin structure $\mathfrak{s}$ on $\mathring{Y}$ gives a framing of

$T\mathring{Y}|_{\Sigma}\cong T\Sigma\oplus\nu_{\Sigma\subseteq\mathring{Y}}.$

This, together with the choice of orientation on $\gamma$, is used to define
the concept of an odd framing for the 2-plane bundle

$\nu_{\gamma\subseteq\Sigma}\oplus\nu_{\Sigma\subseteq\mathring{Y}}\cong\nu_{\gamma\subseteq\mathring{Y}}.$

Then $q_{KT}(\mathfrak{s})(\gamma)$ is the number of right-hand half twists
modulo $4$ made by
$\nu_{\gamma\subseteq\Sigma}\subseteq\nu_{\gamma\subseteq\mathring{Y}}$ as it
rotates around $\gamma$, with respect to a choice of odd framing on this
ambient $2$-plane bundle.

We compare with the process for computing the exterior Guillou-Marin form.
Choose a properly immersed disc $D^{\prime}\looparrowright X_{F}$ for
computing $\widehat{q}_{GM}(\gamma)$. Write

(14) $f\colon\nu_{D^{\prime}\subseteq
X}|_{\gamma}\cong\gamma\times\mathbb{R}^{2}$

for the framing coming from restricting the framing on
$\nu_{D^{\prime}\subseteq X}$ determined by the convention described in
Definition 6.17. Then $\widehat{q}_{GM}(\gamma)$ is the number of right-hand
half-twists modulo $4$ in
$\nu_{\gamma\subseteq\Sigma}\subseteq\nu_{D^{\prime}\subseteq X}|_{\gamma}$
with respect to $f$. Since $D^{\prime}$ is transverse to $\mathring{Y}$, there
is a bundle isomorphism
$\nu_{\gamma\subseteq\mathring{Y}}\cong\nu_{D^{\prime}\subseteq X}|_{\gamma}$.

As both $q_{KT}(\mathfrak{s})(\gamma)$ and $\widehat{q}_{GM}(\gamma)$ are
counts of right-handed half twists of
$\nu_{\gamma\subseteq\Sigma}\subseteq\nu_{D^{\prime}\subseteq X}|_{\gamma}$,
to show they are equal it suffices to prove that $f$ is an odd framing. In
other words, we must show that the following two framings induce the same
orientation, but disagree as framings:

$\mathcal{F}\colon
T\mathring{Y}|_{\gamma}\cong\gamma\times\mathbb{R}^{3},\qquad\text{and}\qquad\mathcal{G}_{f}\colon
T\mathring{Y}|_{\gamma}\cong\gamma\times\mathbb{R}^{3},$

where $\mathcal{F}$ is obtained by using the spin structure $\mathfrak{s}$ on
$\mathring{Y}$ and, recall, $\mathcal{G}_{f}$ is defined as the sequence of
bundle isomorphisms

$T\mathring{Y}|_{\gamma}\cong
T\gamma\oplus\nu_{\gamma\subseteq\mathring{Y}}\cong
T\gamma\oplus\nu_{D^{\prime}\subseteq
X}|_{\gamma}\cong\gamma\times\mathbb{R}^{3}.$

Writing $n$ for the outwards-facing normal to $\gamma$ in $D^{\prime}$, $t$
for the tangent to $\gamma$, and $\langle e_{1},e_{2}\rangle$ for the frame
specified by $f$, the frame $n\times t\times e_{1}\times e_{2}$ agrees with
the ambient orientation of $X$, by definition of $f$. Using the “outwards
normal first” convention, the orientation on $X$ induces an orientation on
$\mathring{Y}$. Our frame includes $n$ as the first vector already, so the
frame $t\times e_{1}\times e_{2}$, (i.e. the frame specified by
$\mathcal{G}_{f}$) induces this ambient orientation on $\mathring{Y}$. Now
consider that $\mathcal{F}$ comes from a spin structure on $\mathring{Y}$, and
this spin structure extends to a spin structure on $X$, by hypothesis. It must
be that the orientations underlying these spin structures agree with the
orientations on $X$ and $\mathring{Y}$. Hence the orientation underlying
$\mathcal{F}$ is the ambient orientation on $\mathring{Y}$, and the
orientations coming from $\mathcal{F}$ and $\mathcal{G}_{f}$ agree, as
claimed.

Next, we show that $\mathcal{F}$ and $\mathcal{G}_{f}$ disagree, as framings.
Using the outwards-facing normal $n$, complete each of $\mathcal{F}$ and
$\mathcal{G}_{f}$ to a framing of

$TX|_{\gamma}\cong\nu_{\gamma\subseteq D^{\prime}}\oplus
T\mathring{Y}|_{\gamma}.$

We refer to the completed framings as $\mathcal{F}^{\prime}$ and
$\mathcal{G}_{f}^{\prime}$, respectively. We now show that these two framings
disagree, as this will imply that $\mathcal{F}$ and $\mathcal{G}_{f}$ also
disagree.

The framing $\mathcal{F}^{\prime}$ extends to a framing of $TX|_{D^{\prime}}$
because the spin structure on $\mathring{Y}$ used to define $\alpha$ extends
to $X$ (this uses the “outwards-facing normal first” convention, again). The
framing $\mathcal{G}_{f}^{\prime}$ splits as a framing of
$TD^{\prime}|_{\gamma}\cong\nu_{\gamma\subseteq D^{\prime}}\oplus T\gamma$ and
of $\nu_{D^{\prime}\subseteq X}|_{\gamma}$. The latter of these framings is
$f$ and extends to $\nu_{D^{\prime}\subseteq X}$, by definition; recall (14).
The former of these framings is given by $n\times t$. An ordered pair of
framings of a $2$-plane bundle over a circle, which correspond to the same
orientation, determines an element of $\pi_{1}(\operatorname{SO}(2))$. Fix the
(unique) framing that extends to a framing of $\nu_{D^{\prime}\subseteq X}$ as
our reference framing; this framing determines the trivial class in
$\pi_{1}(\operatorname{SO}(2))$. On the other hand, the framing $n\times t$
determines a generator of $\pi_{1}(\operatorname{SO}(2))\cong\mathbb{Z}$, and
so determines the Lie framing. Therefore this latter framing does not extend
to a framing of $\nu_{D^{\prime}\subseteq X}$. Thus, overall,
$\mathcal{G}_{f}^{\prime}$ does not extend to $TX|_{D^{\prime}}$. This proves
that $\mathcal{F}^{\prime}$ and $\mathcal{G}^{\prime}_{f}$ are different
framings of $TX|_{\gamma}$.

But the framings $\mathcal{F}^{\prime}$ and $\mathcal{G}^{\prime}_{f}$ were
obtained from $\mathcal{F}$ and $\mathcal{G}_{f}$ by the addition of the same
line bundle (given by the outwards facing normal), from which we deduce that
$\mathcal{F}$ and $\mathcal{G}_{f}$ are different framings. This completes the
proof that $f$ is an odd framing, which completes the proof of Lemma 6.19. ∎

######  Corollary 6.20.

Let $F\subseteq D^{4}$ be a $\mathbb{Z}_{2}$-surface, and write
$\mathring{Y}\subseteq\partial X_{F}$ for the boundary of the tubular
neighbourhood of $F$. Let $\mathfrak{s}\in\operatorname{Spin}(\mathring{Y})$
be a spin structure and let $s$ be a nice section. Then $\mathfrak{s}$ extends
to $X_{F}$ if and only if the quadratic refinement $q_{KT}(\mathfrak{s})$
associated to the spin structure $\mathfrak{s}$ by Proposition 6.11 agrees
with the exterior Guillou-Marin quadratic refinement.

###### Proof.

Write $S$ for the set of spin structures on $\mathring{Y}$ that extend to
$X_{F}$. Write $T$ for the set of
$\mathfrak{t}\in\operatorname{Spin}(\mathring{Y})$ such that
$q_{KT}(\mathfrak{t})(\gamma)=\widehat{q}_{GM}(\gamma)$ for every $\gamma\in
H_{1}(s(F);\mathbb{Z}_{2})$. The corollary is equivalent to the statement
$S=T$. Lemma 6.19 shows that if $\mathfrak{t}$ extends to $X_{F}$ then we have
an identification between $\widehat{q}_{GM}$ and $q_{KT}(\mathfrak{t})$, and
so $S\subseteq T$.

Now note that by Proposition 6.11 (1), exactly two spin structures on
$\mathring{Y}$ correspond to the Guillou-Marin quadratic refinement, so we
have $|T|=2$. The corollary will be proved if we can show $|S|=2$. We know
$|S|\leq|T|=2$ because $S\subseteq T$. As noted earlier there are two spin
structures on $X_{F}$ and these induce at most two spin structures on
$\mathring{Y}$. We write $S=\\{\mathfrak{s}_{1},\mathfrak{s}_{2}\\}$ where
$\mathfrak{s}_{1}$ and $\mathfrak{s}_{2}$ are the restrictions of the two
distinct spin structures on $X_{F}$ to $\mathring{Y}$. These spin structures
on $X_{F}$ differ on their restriction to a meridian $\mu\subseteq X_{F}$,
which is nontrivial in $H_{1}(\mathring{Y};\mathbb{Z}_{2})$. Thus by
Proposition 6.11 (1) we have $\mathfrak{s}_{1}\neq\mathfrak{s}_{2}$ and so
$|S|=2$ as required. ∎

###  6.4. The proof of Proposition 6.3

We begin with a lemma that will allow us to align choices of nice sections. We
then move on to the proof of Proposition 6.3.

Given a surface $\Sigma$ with connected, nonempty boundary and a bundle
$\mathring{\xi}=(\mathbb{R}^{2}\to\mathring{N}\to\Sigma)$, we write
$\mathring{N}|_{\partial\Sigma}$ for the total space of
$\mathring{\xi}|_{\partial\Sigma}$. Since $[\partial\Sigma]=0\in
H_{1}(\Sigma;\mathbb{Z}/2)$, it follows that
$w_{1}(\mathring{\xi})([\partial\Sigma])=0$, and therefore
$\mathring{\xi}|_{\partial\Sigma}$ is a trivial bundle.

######  Lemma 6.21.

For $i=0,1$, let $F_{i}\subseteq D^{4}$ be nonorientable surfaces with the
same nonorientable genus, and the same connected boundary $K\subseteq S^{3}$.
For brevity, write $\mathring{\xi_{i}}=(\mathbb{R}^{2}\to\mathring{N_{i}}\to
F_{i})$ for the respective normal bundles. Suppose $s_{0},s_{1}\colon
F_{i}\to\mathring{N_{i}}$ are nonvanishing sections.

  1. (1)

Given a homeomorphism $\varphi\colon F_{0}\to F_{1}$, restricting to the
identity map on $K$, there exists a vector bundle isomorphism
$\Phi\colon\mathring{\xi_{0}}\to\mathring{\xi_{1}}$ covering $\varphi$ and
sending $s_{0}$ to $s_{1}$.

  2. (2)

Suppose moreover that $s_{0}|_{K}$ and $s_{1}|_{K}$ are isotopic as sections
of $\nu_{K\subseteq S^{3}}$. Then, after an isotopy of $s_{1}$ supported in a
collar of $K\subseteq F_{0}$, the isomorphism $\Phi$ may be assumed to
restrict to the identity on $\nu_{K\subseteq S^{3}}$.

###### Proof.

We prove the first item. For $i=0,1$, the section $s_{i}$ determines a
(trivial) line bundle $\zeta_{i}=(\mathbb{R}\to
E_{i}\xrightarrow{\pi_{i}}F_{i})$. Namely $\zeta_{i}$ is the subbundle of
$\mathring{\xi}_{i}$ with fibre over $x\in F_{i}$ given by $\mathbb{R}\langle
s_{i}(x)\rangle$. Every $v\in E_{i}$ can be written uniquely as
$v=\lambda_{v}^{i}s_{i}(\pi_{i}(v))\in(E_{i})_{\pi_{i}(v)}$ for some
$\lambda_{v}^{i}\in\mathbb{R}$. It follows that the assignment
$v\mapsto\lambda_{v}^{0}s_{1}(\varphi(\pi_{0}(v)))$ defines a bundle
isomorphism $\alpha\colon\zeta_{0}\cong\zeta_{1}$, that covers $\varphi$ and
satisfies $\alpha\circ s_{0}=s_{1}\circ\varphi$.

We now extend $\alpha$ to a bundle isomorphism
$\Phi\colon\mathring{\xi}_{0}\cong\mathring{\xi}_{1}$, using orthogonal
complements. Choose Riemannian metrics on $\mathring{\xi}_{0}$ and
$\mathring{\xi}_{1}$ which agree over $K$, and consider the perpendicular
bundles $\zeta_{0}^{\perp}$ and $\zeta_{1}^{\perp}$. Given an embedded loop
$\gamma\subseteq F_{i}$, the line bundle $\zeta_{i}^{\perp}|_{\gamma}$ is
orientable or nonorientable according to whether $w_{1}(\mathring{\xi}_{i})$
is trivial on $\gamma$ or not. Since $\varphi$ is a homeomorphism, for
$\gamma\subseteq F_{0}$ we have that
$w_{1}(F_{0})(\gamma)=w_{1}(F_{1})(\varphi(\gamma))$. Therefore since the
total space of $\mathring{\xi}_{i}$ is orientable, we have that
$w_{1}(\mathring{\xi}_{0})(\gamma)=w_{1}(\mathring{\xi}_{1})(\varphi(\gamma))$.
Thus in both cases there is a bundle isomorphism
$\zeta_{0}^{\perp}|_{\gamma}\cong\zeta_{1}^{\perp}|_{\varphi(\gamma)}$.
Because line bundles are classified by $w_{1}$, this implies there exists a
bundle isomorphism $\beta\colon\zeta_{0}^{\perp}\to\zeta_{1}^{\perp}$ covering
$\varphi$. Combining $\alpha$ and $\beta$, we obtain a bundle isomorphism

$\Phi\colon\mathring{\xi}_{0}\xrightarrow{\cong}\zeta_{0}\oplus\zeta_{0}^{\perp}\xrightarrow{\alpha\oplus\beta}\zeta_{1}\oplus\zeta_{1}^{\perp}\xrightarrow{\cong}\mathring{\xi}_{1},$

which covers $\varphi$ and satisfies $\Phi\circ s_{0}=s_{1}\circ\varphi$.

We prove the second item. The normal bundles $\mathring{\xi}_{0}$ and
$\mathring{\xi}_{1}$ both restrict to $\nu_{K\subseteq S^{3}}$ over $K$, and
recall that our choices of Riemannian metric for $\mathring{\xi}_{0}$ and
$\mathring{\xi}_{1}$ were constructed to agree over $K$. An isotopy of $s_{1}$
over $K$ that makes it agree with $s_{0}$ thus determines an isotopy between
the line bundles $\zeta_{1}|_{K}$ and $\zeta_{0}|_{K}$, as subbundles of
$\nu_{K\subseteq S^{3}}$. Also, because the metrics agree over the boundary,
this isotopy induces an isotopy between $\zeta_{1}^{\perp}|_{K}$ and
$\zeta_{0}^{\perp}|_{K}$ as subbundles of $\nu_{K\subseteq S^{3}}$. This means
we obtain an isotopy from $\Phi|_{\nu_{K\subseteq S^{3}}}$ to a bundle
automorphism of $\nu_{K\subseteq S^{3}}$ that sends $\zeta_{0}|_{K}$ to itself
and $\zeta_{0}^{\perp}|_{K}$ to itself. Since $\nu_{K\subseteq
S^{3}}\cong\zeta_{0}|_{K}\oplus\zeta_{0}^{\perp}|_{K}$, it follows that this
automorphism is in fact the identity. Taper the isotopies of
$\Phi|_{\nu_{K\subseteq S^{3}}}$ and of $s_{1}$ over a collar neighbourhood of
$K$ in $F_{0}$ and $F_{1}$, to obtain the desired isotopy of $\Phi$ to a new
bundle isomorphism
$\Phi^{\prime}\colon\mathring{\xi}_{0}\to\mathring{\xi}_{1}$ that is the
identity over $K$, and to obtain the claimed isotopy of $s_{1}$. ∎

######  Lemma 6.22.

For $i=0,1$, let $F_{i}\subseteq D^{4}$ be nonorientable surfaces with the
same nonorientable genus, the same connected boundary $K\subseteq S^{3}$ and
each with normal Euler number $e$. Let $s_{i}\colon
F_{i}\to\nu_{F_{i}\subseteq D^{4}}$ be nonvanishing sections. Then
$s_{0}|_{K}$ and $s_{1}|_{K}$ are isotopic.

###### Proof.

Recall that, by convention in this section, the tubular neighbourhood of any
surface with boundary a given $K\subseteq S^{3}$ extends a given tubular
neighbourhood for $K$. So there are assumed identifications
$\nu_{F_{i}\subseteq D^{4}}|_{K}=\nu_{K\subseteq S^{3}}$ for $i=0,1$. Isotopy
classes of nonvanishing sections of $\nu_{K\subseteq S^{3}}$ are in one-to-one
correspondence with framings of $\nu_{K\subseteq S^{3}}\cong
S^{1}\times\mathbb{R}^{2}$, and we will identify the two, using the same
notation for a section and the corresponding framing. Write $\tau\colon
K\to\nu_{K\subseteq S^{3}}$ for a nonvanishing section corresponding to the
Seifert framing of $K$. The set of framings is in affine correspondence with
$\mathbb{Z}$, so for $i=0,1$, the difference between framings determines an
integer $d(\tau,s_{i}|_{K})\in\mathbb{Z}$. We claim that for $i=0,1$ there is
an equality $d(\tau,s_{i}|_{K})=e$. To see this, consider that if we add
$-k_{i}$ full turns to $\tau$ to make the difference 0, and thus to make
$\tau_{i}$ isotopic to $s_{i}|_{K}$, this introduces $-k_{i}$ signed
intersections between a generic extension of $\tau$ to $F_{i}$ and the
0-section of $\nu_{F_{i}\subseteq D^{4}}$. As $s_{i}|_{K}$ extends to a
nonvanishing section, i.e. $e(\nu_{F_{i}\subseteq D^{4}},s_{i}|_{K})$=0, this
implies that $d(\tau_{i},s_{i}|_{K})=k_{i}=e(\nu_{F_{i}\subseteq
D^{4}},\tau_{i})=:e$ (using the interpretation of $e(\nu_{F_{i}\subseteq
D^{4}},\tau_{i})$ as a signed intersection count from Remark 2.1).

We thus have that $d(\tau,s_{1}|_{\nu_{K}})=d(\tau,s_{1}|_{\nu_{K}})$, from
which we deduce

$d(s_{0}|_{\nu_{K}},s_{1}|_{\nu_{K}})=d(\tau,s_{0}|_{\nu_{K}})-d(\tau,s_{1}|_{\nu_{K}})=0.$

So $s_{0}|_{\nu_{K}}$ is isotopic to $s_{1}|_{\nu_{K}}$ as claimed. ∎

We can now prove the main result of this section.

###### Proof of Proposition 6.3.

Since the surfaces $F_{0},F_{1}\subseteq D^{4}$ have the same normal Euler
number and nonorientable genus, by Corollary 6.15 we can, and will, choose a
homeomorphism $\varphi\colon F_{0}\to F_{1}$ inducing an isometry of their
Guillou-Marin forms and restricting to the identity map on $K$. The strategy
for the remainder of the proof is to extend $\varphi$ to a homeomorphism
$\nu_{F_{0}\subseteq D^{4}}\cong\nu_{F_{1}\subseteq D^{4}}$, and then to
restrict to $S(\nu_{F_{0}\subseteq D^{4}})\cong S(\nu_{F_{1}\subseteq
D^{4}})$. Finally, extending by the identity on $X_{K}$ leads to the sought-
for homeomorphism $\partial X_{F_{0}}\cong\partial X_{F_{1}}$. Now, the
details.

Recall that for $i=0,1$, we fixed a tubular neighbourhood $\alpha_{i}$ for
$\nu_{F_{i}\subseteq D^{4}}$, so that $\alpha_{i}(S(\nu_{F_{i}\subseteq
D^{4}}))=\partial X_{F_{i}}\setminus\mathring{X}_{K}$. By Lemma 6.16, we may
pick nice sections $s_{i}\colon F_{i}\to S(\nu_{F_{i}\subseteq
D^{4}})|_{F_{i}}$ with respect to this tubular neighbourhood. Thus we may
apply Lemma 6.21 (1) to obtain a bundle isomorphism

$\Phi\colon\nu_{F_{0}\subseteq D^{4}}\cong\nu_{F_{1}\subseteq D^{4}}$

covering $\varphi$. Because the normal Euler numbers of $F_{0}$ and $F_{1}$
agree, Lemma 6.22 shows $s_{0}|_{K}$ and $s_{1}|_{K}$ are isotopic as sections
of $\nu_{K\subseteq S^{3}}$. As $\varphi$ restricts to the identity map on
$K$, this means we may apply Lemma 6.21 (2). This achieves that $\Phi$
restricts to the identity map on $\nu_{K\subseteq S^{3}}$, at the expense of
changing $s_{1}$ by an isotopy. We will continue to call this modified section
$s_{1}$, and we note that the new $s_{1}$ is still a nice section as this
property is preserved by isotopy.

Our chosen tubular neighbourhoods $\alpha_{0}$ and $\alpha_{1}$ satisfy
$\alpha_{i}\colon S(\nu_{F_{i}\subseteq D^{4}})\hookrightarrow\partial
X_{F_{i}}$ for $i=0,1$, and are such that each embedding extends the given
tubular neighbourhood of the knot $S(\nu_{K\subseteq S^{3}})\cong\partial
X_{K}$. We thus have a homeomorphism

$g:=\alpha_{1}\circ\Phi\circ(\alpha_{0})^{-1}\colon\alpha_{0}(S(\nu_{F_{0}\subseteq
D^{4}}))\xrightarrow{\cong}\alpha_{1}(S(\nu_{F_{0}\subseteq D^{4}})).$

As $\Phi$ restricts to the identity bundle map over $K$, this homeomorphism
$g$ restricts to the identity map on the tubular neighbourhood of the knot
$K$. Thus we may extend $g$ by the identity map $\operatorname{Id}\colon
X_{K}\to X_{K}$, to define a homeomorphism

(15) $f\colon\partial X_{F_{0}}\xrightarrow{\cong}\partial X_{F_{1}}.$

We claim $f$ has the properties listed in the statement of the proposition.

Property (2), that $f|_{X_{K}}$ is the identity, is clear from construction.
Properties (3) and (4), that $f$ restricts to a map that is $\nu$-extendable
rel. boundary, and that meridians are preserved, follow because $f$ was
constructed from the restriction of a vector bundle isomorphism
$\nu_{F_{0}\subseteq D^{4}}\cong\nu_{F_{1}\subseteq D^{4}}$. Property (5),
which was about $f$ preserving nice sections, is also automatic from the
construction because we defined $f$ using choices of nice sections.

To finish the proof we prove that $f$ has property (1). Pick a spin structure
$\mathfrak{s}\in\operatorname{Spin}(\partial X_{F_{1}})$ that extends to the
exterior $\operatorname{Spin}(X_{F_{1}})$. Write
$\mathfrak{t}:=\alpha_{1}^{*}\mathfrak{s}$ for the induced spin structure on
$S(\nu_{F_{1}\subseteq D^{4}})$. Choose an $S^{1}$-fibre $\mu_{0}\subseteq
S(\nu_{F_{0}\subseteq D^{4}})$ and write $\mu_{1}:=\Phi(\mu_{0})\subseteq
S(\nu_{F_{1}\subseteq D^{4}})$ for the corresponding $S^{1}$-fibre of
$S(\nu_{F_{1}\subseteq D^{4}})$. Using the fact that $\Phi$ satisfies
$\Phi\circ s_{0}=s_{1}\circ\varphi$, the second item of Proposition 6.11
implies that

(16)
${\widehat{\Phi}}(\mathfrak{t}|_{\mu_{1}},q_{KT}(\mathfrak{t}))=((\Phi^{*}\mathfrak{t})|_{\mu_{0}},q_{KT}(\Phi^{*}\mathfrak{t})).$

On the left-hand side, the quadratic refinement is, by definition
$q_{KT}(\mathfrak{t})\circ\Phi_{*}$, whereas on the right-hand side the
quadratic refinement $q_{KT}(\Phi^{*}\mathfrak{t})$ is assigned by the
construction in Proposition 6.11. Both of these quadratic refinements are
defined on $H_{1}(s_{0}(F_{0});\mathbb{Z}_{2})$.

We claim that we have the following sequence of equalities of quadratic
refinements:

(17)
$\widehat{q}_{GM_{0}}=\widehat{q}_{GM_{1}}\circ\Phi_{*}=q_{KT_{1}}(\mathfrak{t})\circ\Phi_{*}=q_{KT_{0}}(\Phi^{*}\mathfrak{t}).$

The last equality follows from (16), as just explained. To see the central
equality in (17) we use Lemma 6.19, which implies that
$q_{KT_{1}}(\mathfrak{t})=\widehat{q}_{GM_{1}}$, the exterior Guillou-Marin
form of $s_{1}(F_{1})$.

It remains to see that the isometry of $\mathbb{Z}_{2}$-forms $\Phi_{*}$
induces an isometry between the exterior Guillou-Marin forms
$\widehat{q}_{GM_{0}}$ and $\widehat{q}_{GM_{1}}$. As $\Phi$ satisfies
$\Phi\circ s_{0}=s_{1}\circ\varphi$, the isometry $\Phi_{*}$ factors as a
sequence of isomorphisms

$\Phi_{*}\colon
H_{1}(s_{0}(F_{0});\mathbb{Z}_{2})\xrightarrow{(s_{0})_{*}^{-1}}H_{1}(F_{0};\mathbb{Z}_{2})\xrightarrow{\varphi_{*}}H_{1}(F_{1};\mathbb{Z}_{2})\xrightarrow{(s_{1})}H_{1}(s_{1}(F_{1});\mathbb{Z}_{2}).$

By Proposition 6.18, the outer two isomorphisms induce isometries between the
$\widehat{q}_{GM}$ forms on $H_{1}(s_{i}(F_{i});\mathbb{Z}_{2})$ and the
$q_{GM}$ forms on $H_{1}(F_{i};\mathbb{Z}_{2})$. One of the defining
properties of the central isomorphism $\varphi_{*}$ is that it induces an
isometry of $q_{GM}$ forms. Thus $\Phi_{*}$ induces an isometry of exterior
Guillou-Marin forms, concluding the proof of the claim.

We therefore have from (17) that
$\widehat{q}_{GM_{0}}=q_{KT_{0}}(\Phi^{*}\mathfrak{t})$. From Corollary 6.20
it follows that $f^{*}\mathfrak{s}$, corresponding to the spin structure
$\Phi^{*}\mathfrak{t}$ under the tubular neighbourhood map $\alpha_{0}$,
extends to $X_{F_{0}}$. This completes the proof of property (1) and therefore
of Proposition 6.3. ∎

##  7\. The odd-dimensional $\ell$-monoid

We recall the definition and some facts surrounding the monoid
$\ell_{2q+1}(R)$. We will introduce quasi-formations, define $\ell_{2q+1}(R)$,
study elementary quasi-formations and recall how a quadratic form $v$ over $R$
determines a subset $\ell_{2q+1}(v,v)\subseteq\ell_{2q+1}(R)$.

For the remainder of the section we fix an integer $q$ and set
$\varepsilon:=(-1)^{q}$. Since we will ultimately work with
$R=\mathbb{Z},\mathbb{Z}[\mathbb{Z}_{2}]$ which are both groups rings of
groups with vanishing Whitehead torsion, we work in less generality than in
[KreckSurgeryAndDuality, CrowleySixt] in order to avoid technicalities related
to bases and Whitehead torsion. As a consequence, throughout this section we
assume that $R=\mathbb{Z}[G]$ is a group ring of a group $G$ with vanishing
Whitehead group, referring to [CrowleySixt, Section 3] for the discussion in
full generality. The involution $x\mapsto\overline{x}$ on $R$ is $\sum
n_{g}g\mapsto\sum n_{g}g^{-1}$.

###  7.1. Quasi-formations

We introduce quasi-formations and the $\ell$-monoid. The original reference
for these monoids is [KreckSurgeryAndDuality], but we also refer the reader to
[CrowleySixt] as we will work with the equivalent formulation developed there.
It might help to refer back to Section 3.1 on hermitian and quadratic forms.

######  Definition 7.1.

An $\varepsilon$-quadratic _quasi-formation_ is a triple $((P,\psi);F,V)$,
where $(P,\psi)$ is a nonsingular $\varepsilon$-quadratic form, $F\subseteq P$
is a lagrangian, and $V\subseteq P$ is a half rank direct summand; $P,F$ and
$V$ are assumed to be stably free.

This definition and terminology are due to Crowley and Sixt [CrowleySixt,
Section 3] but are inspired by work of Kreck [KreckSurgeryAndDuality].

######  Remark 7.2.

In both [KreckSurgeryAndDuality] and [CrowleySixt], the authors work with
stably based modules throughout, as this is needed in the most general
applications of the theory. However, we have made the assumption that
$R=\mathbb{Z}[G]$ where $G$ has vanishing Whitehead torsion, meaning we can
avoid using based modules, as we now briefly explain. We assume some
familiarity with the definition of the first reduced $K$-group
$\widetilde{K}_{1}(R)$ and refer to e.g. [CrowleySixt, Section 3] for more
details and definitions. An $R$-module $P$ is _stably based_ if $P$ is stably
free and is equipped with an equivalence class of a basis for a free
stabilisation $R^{n}\cong P\oplus R^{k}$. Two bases are defined to be
equivalent if there is a common free stabilisation in which the change of
basis matrix defines an element of $Z:=\\{\pm g\mid g\in
G\\}\subseteq\widetilde{K}_{1}(R)$. Assuming that $\operatorname{Wh}(G)=0$
implies that $Z=\widetilde{K}_{1}(R)$, so that all stable bases are
equivalent, and so do need not be kept track of. For this reason, we do not
work with stably based modules, merely stably free modules.

######  Example 7.3.

Formations (see e.g. [CrowleyLuckMacko, Section 8.2.1] for some background on
formations) are examples of quasi-formations: they are those quasi-formations
where the half rank direct summand is a lagrangian. Here and in what follows,
a formation will refer to what is sometimes called a nonsingular formation,
i.e. a quasi-formation $((P,\psi);F,G)$ with $G\subseteq P$ a lagrangian.

######  Definition 7.4.

  * •

Two quasi-formations $((P,\psi);F,V)$ and
$((P^{\prime},\psi^{\prime}),F^{\prime},V^{\prime})$ are _isomorphic_ if there
is an isometry $f\colon(P,\psi)\to(P^{\prime},\psi^{\prime})$ such that
$f(F)=F^{\prime}$ and $f(V)=V^{\prime}$.

  * •

A quasi-formation is _trivial_ if it is isomorphic to a formation of the type
$(H_{\varepsilon}(F);F,F^{*})$. Here, and afterwards when it causes no
confusion, we use the abbreviation $F,F^{*}\subseteq F\oplus F^{*}$ instead of
$F\oplus 0,0\oplus F^{*}\subseteq F\oplus F^{*}$.

  * •

The _boundary $\varepsilon$-quadratic quasi-formation $\delta(F,\theta)$ of a
sesquilinear form $(F,\kappa)$_ is the quasi-formation
$(H_{\varepsilon}(F),F,\Gamma_{\kappa})$, where $\Gamma_{\kappa}\subseteq
F\oplus F^{*}$ denotes the graph of $\kappa\colon F\to F^{*}$. Note that the
sesquilinear form $\kappa$ need not be hermitian.

  * •

A quasi-formation is a _boundary_ if it is isomorphic to the boundary
$\delta(F,\theta)$ of some sesquilinear form $(F,\kappa)$.

  * •

Two quasi-formations are _stably isomorphic_ if they become isomorphic after
some number of trivial formations is added to each. We use $\cong$ to denote
isomorphism and $\cong_{s}$ to denote stable isomorphism.

  * •

If a sesquilinear form $(F,\kappa)$ is even ($\kappa$ can be written as
$\theta-\varepsilon\theta^{*}$ for some $\theta$) and
$(-\varepsilon)$-hermitian, then the graph $\Gamma_{\kappa}\subseteq F\oplus
F^{*}$ is a lagrangian for $H_{\varepsilon}(F)$. If $(F,[\theta])$ is a
$(-\varepsilon)$-quadratic form, then
$(H_{\varepsilon}(F),F,\Gamma_{\theta-\varepsilon\theta^{*}})$ is called the
_boundary $\varepsilon$-quadratic formation_ of $(F,[\theta])$. Note that
$\theta-\varepsilon\theta^{*}$ is tautologically even and is also
$(-\varepsilon)$-hermitian.

  * •

An $\varepsilon$-quadratic formation is a _boundary_ if it is isomorphic to
the boundary $\delta(F,\theta)$ of some $(-\varepsilon)$-quadratic form
$(F,[\theta])$.

Every quasi-formation $((P,\psi);F,V)$ is isomorphic to one of the form
$(H_{\varepsilon}(F);F,V^{\prime})$ for some half rank direct summand
$V^{\prime}\subseteq F\oplus F^{*}$. Indeed the Fundamental Lemma of
$L$-theory (see e.g. [CrowleyLuckMacko, Lemma 7.92]) implies that the
embedding $(F,0)\hookrightarrow(P,\psi)$ extends to an isometry $f\colon
H_{\varepsilon}(F)\xrightarrow{\cong}(P,\psi)$, which therefore induces an
isomorphism

$(H_{\varepsilon}(F);F,f^{-1}(V))\cong((P,\psi);F,V).$

######  Definition 7.5.

The _$\ell$ -monoid_ $\ell_{2q+1}(R)$ is the unital abelian monoid (under
direct sum) of stable isomorphism classes of $(-1)^{q}$-quadratic quasi-
formations modulo the relation

(18) $((P,\psi);F,G)\oplus((P,\psi),G,V)\sim((P,\psi);F,V),$

where $F,G\subseteq P$ are both lagrangians.

######  Definition 7.6.

The _$L$ -group_ $L_{2q+1}(R)$ is the group (under direct sum) of stable
isomorphism classes of $(-1)^{q}$-quadratic formations modulo the equivalence
relation of cobordism, where $x$ and $x^{\prime}$ are said to be _cobordant_
if there are boundary formations $b$ and $b^{\prime}$ such that $x\oplus b$
and $x^{\prime}\oplus b^{\prime}$ are stably isomorphic.

######  Remark 7.7.

An alternative definition of the $L$-group $L_{2q+1}(R)$ is as the abelian
group (under direct sum) of stable isomorphism classes of $(-1)^{q}$-quadratic
formations modulo the relation (18); see e.g.
[RanickiAnIntroductionToAlgebraic, Remark 9.15]. Thus
$L_{2q+1}(R)\subseteq\ell_{2q+1}(R)$.

###  7.2. Elementary quasi-formations

We recall the definition of elementary quasi-formations and the submonoid of
$\ell_{2q+1}(R)$ that they define. References for the material include
[KreckSurgeryAndDuality] and [CrowleySixt, Section 3].

We start by recording some terminology that we will use frequently in the
sequel.

######  Definition 7.8.

A _direct complement_ to a submodule $V\subseteq P$ is a submodule $U\subseteq
P$ such that the inclusions $U,V\subseteq P$ extend to an isomorphism $U\oplus
V\cong P$. We say that $U$ and $V$ are _complementary_. If $(P,\psi)$ is a
quadratic form, then a _lagrangian complement_ to $V\subseteq P$ is a direct
complement $U$ that is moreover a lagrangian.

We introduce a geometrically significant notion of triviality for a quasi-
formation.

######  Definition 7.9.

A quasi-formation $((P,\psi);F,V)$ is _elementary_ if $F\subseteq P$ is a
direct complement to $V\subseteq P$. A class in $\ell_{2q+1}(R)$ is
_elementary_ if it admits an elementary representative.

For an elementary quasi-formation $((P,\psi);F,V)$, since $F$ is by definition
a lagrangian, it follows that $F$ is a lagrangian complement to $V$. In
particular, $V$ admits a lagrangian complement. The next proposition proves a
partial converse by working on the level of the monoid, modulo formations.
This proposition was proved by Kreck [KreckSurgeryAndDuality, Proposition 8
(iii)] but we translate the argument into the language of quasi-formations.

######  Proposition 7.10.

If $\Theta\in\ell_{2q+1}(R)$ is represented by a quasi-formation
$x=(H_{\varepsilon}(F);F,V)$ such that $V$ admits a lagrangian complement,
then there exists a formation $y$ such that $\Theta+[y]$ is elementary.

###### Proof.

Let $L\subseteq F\oplus F^{*}$ be a lagrangian complement for $V\subseteq
F\oplus F^{*}$. Note that $(H_{\varepsilon}(F);L,V)$ is elementary. Consider
the formation $y:=(H_{\varepsilon}(F);L,F)$. Working in $\ell_{2q+1}(R)$, we
have

(19)
$[(H_{\varepsilon}(F);L,V)]=[(H_{\varepsilon}(F);L,F)]+[(H_{\varepsilon}(F);F,V)]=[y]+[x]=[y]+\Theta,$

and thus $\Theta+[y]$ is elementary as required. ∎

Proposition 7.10 motivates considering the action of $L_{2q+1}(R)$ on
$\ell_{2q+1}(R)$ by direct sum.

######  Construction 7.11.

The action of $[y]\in L_{2q+1}(R)$ on $[x]\in\ell_{2q+1}(R)$ is by
$[y]\cdot[x]:=[x\oplus y]$. To verify this is well defined, it suffices to
prove that trivial formations and boundary formations act trivially. For
trivial formations, this is clear: in $\ell_{2q+1}(R)$ stably isomorphic
quasi-formations are equal. To verify that boundary formations act trivially,
use that in $\ell_{2q+1}(R)$ a boundary formation
$y=(H_{\varepsilon}(F);F,\Gamma_{\theta-(-1)^{q}\theta^{*}})$ decomposes as
$y_{1}\oplus
y_{2}:=(H_{\varepsilon}(F);F,F^{*})\oplus(H_{\varepsilon}(F);F^{*},\Gamma_{\theta-(-1)^{q}\theta^{*}})$,
which is the sum of two trivial formations. To see that the latter is trivial
one shows that $F^{*}$ and $\Gamma_{\theta-(-1)^{q}\theta^{*}}$ are
complementary. Here is the proof that $(0\oplus
F^{*})\oplus\Gamma_{\theta-(-1)^{q}\theta^{*}}=F\oplus F^{*}$. The fact that
$(0\oplus F^{*})\cap\Gamma_{\theta-(-1)^{q}\theta^{*}}=0$ is verified by
noting that if $(0,\varphi)=(x,(\theta-(-1)^{q}\theta^{*})(x))$, then $x=0$
and therefore $\varphi=({\theta-(-1)^{q}\theta^{*}})(0)=0$. The inclusion
$(0\oplus F^{*})\oplus\Gamma_{\theta-(-1)^{q}\theta^{*}}\subseteq F\oplus
F^{*}$ is clear and the reverse inclusion follows by writing $(x,\varphi)\in
F\oplus F^{*}$ as
$(0,\varphi-(\theta-(-1)^{q}\theta^{*})(x))+(x,({\theta-(-1)^{q}\theta^{*}})(x))$.
Now apply [CrowleyLuckMacko, Lemma 8.10 i)], which states that a formation
$((P,\psi);F,G)$ is trivial if $F\oplus G=P$. This completes the proof that
$(H_{\varepsilon}(F);F^{*},\Gamma_{\theta-(-1)^{q}\theta^{*}})$ is a trivial
formation. Thus in $\ell_{2q+1}(R)$ we obtain $x\oplus y\sim x\oplus
y_{1}\oplus y_{2}\sim x$, as required.

The following proposition describes an equivalent characterisation of
$\Theta\in\ell_{2q+1}(R)$ being elementary when $L_{2q+1}(R)=0$ cf.
[HambletonKreckTeichnerNonOri, Proposition 2].

######  Proposition 7.12.

Assume that $L_{2q+1}(R)=0$. Given $\Theta\in\ell_{2q+1}(R)$, the following
assertions are equivalent.

  1. (1)

$\Theta$ is elementary, i.e. $\Theta$ admits a representative
$(H_{\varepsilon}(F);F,V)$ such that $F$ is a lagrangian complement to $V$.

  2. (2)

$\Theta$ is represented by a quasi-formation $(H_{\varepsilon}(F);F,V)$ such
that $V$ admits a lagrangian complement.

###### Proof.

The implication $(1)\Rightarrow(2)$ is immediate. Assuming that $(2)$ holds,
Proposition 7.10 implies that $\Theta$ is elementary modulo $L_{2q+1}(R)$.
Since $L_{2q+1}(R)=0$, we deduce that $(2)\Rightarrow(1)$. ∎

If a formation $((P,\psi);F,V)$ is elementary then in particular $V$ admits a
lagrangian complement, but it is unclear in general that for _every_
$((P^{\prime},\psi^{\prime});F^{\prime},V^{\prime})$ in the class
$[((P,\psi);F,V)]\in\ell_{2q+1}(R)$, the module $V^{\prime}\subseteq
P^{\prime}$ admits a lagrangian complement. We will show that this property
holds if $L_{2q+1}(R)=0$ and one is willing to add a formation to
$((P^{\prime},\psi^{\prime});F^{\prime},V^{\prime})$.

For brevity in the next lemma and its proof, given a quasiformation
$x=((P,\psi);F,V)$ we will say _$x$ admits a lagrangian complement_ to meant
that $V\subseteq P$ admits a lagrangian complement.

######  Lemma 7.13.

Let $R$ be a ring with $L_{2q+1}(R)=0$. Let $a$ and $b$ be quasiformations
over $R$ that are either stably isomorphic, or related by the defining
relation (18) of $\ell_{2q+1}(R)$. Then there is a formation $f$ such that
$a\oplus f$ admits a lagrangian complement if and only if there is there is a
formation $f^{\prime}$ such that $b\oplus f^{\prime}$ admits a lagrangian
complement.

###### Proof.

First assume that $a$ and $b$ are stably isomorphic, so that there are trivial
formations $t_{1}$ and $t_{2}$ such that $a\oplus t_{1}\cong b\oplus t_{2}$.
If $a\oplus f$ admits a lagrangian complement for some formation $f$, then
$b\oplus(f\oplus t_{2})\cong a\oplus f\oplus t_{1}$ admits a lagrangian
complement because both $a\oplus f$ and $t_{1}$ do. The proof of the converse
is identical.

Next assume that $a=((P,\psi);F,V)$ and $b=((P,\psi);F,G)\oplus((P,\psi);G,V)$
are as in (18). If $a\oplus f$ admits a lagrangian complement for some
formation $f$, then $((P,\psi);G,V)\oplus f$ admits a lagrangian complement.
Since $L_{2q+1}(R)=0$, every formation is stably isomorphic to a boundary
formation [RanickiAnIntroductionToAlgebraic, Corollary 9.12]. In particular,
after adding a trivial formation $t$, the formation $((P,\psi);F,G)$ admits a
lagrangian complement, because every boundary formation does. It follows that

$b\oplus f\oplus t\cong\left(((P,\psi);F,G)\oplus
t\right)\oplus(((P,\psi);G,V)\oplus f)$

admits a lagrangian complement.

For the converse, assume that $b\oplus f$ admits a lagrangian complement for
some formation $f$. More explicitly,
$((P,\psi);G,V)\oplus(((P,\psi);F,G)\oplus f)$ admits a lagrangian complement.
Since admitting a lagrangian complement is a statement that only involves the
half rank direct summand, it follows that
$((P,\psi);F,V)\oplus((P,\psi);F,G)\oplus f\ a\oplus((P,\psi);F,G)\oplus f$
also admits a lagrangian complement. ∎

######  Proposition 7.14.

Let $R$ be a ring with $L_{2q+1}(R)=0$, let $\Theta\in\ell_{2q+1}(R)$ and let
$u$ be any representative quasiformation for $\Theta$. If $\Theta$ is
elementary, then there is a formation $v$ such that in the formation $u\oplus
v=:((P,\psi);F,V)$, the summand $V\subseteq P$ admits a lagrangian complement.

###### Proof.

The definition of $\ell_{2q+1}(R)$ is as the transitive closure of the set of
quasiformations under the relation of being either stably isomorphic or
related as in (18). The proposition is now immediate from Lemma 7.13. ∎

###  7.3. Induced forms

Following [CrowleySixt, Section 5], we describe some additional facts about
the structure of the monoid $\ell_{2q+1}(R)$.

Given a quasi-formation $x=((P,\psi);F,V)$ (a representative for) the
quadratic form $\psi\colon P\times P\to R$ restricts to a quadratic form
$\theta:=\psi|_{V\times V}$ on $V$ leading to a split isometric inclusion
$(V,\theta)\hookrightarrow(P,\psi)$. The same holds for $V^{\perp}$ and we
formalise this as follows.

######  Definition 7.15.

The _induced forms_ of a quasi-formation $((P,\psi);F,V)$ are the quadratic
forms $(V,\theta)$ and $(V^{\perp},-\theta^{\perp})$, where
$\theta:=\psi|_{V\times V}$ and $\theta^{\perp}:=\psi|_{V^{\perp}\times
V^{\perp}}$.

######  Example 7.16.

We describe the induced forms in some simple cases.

  1. (1)

The induced forms of a formation $((P,\psi);F,G)$ are both equal to the zero
form $(G,0)$. This is immediate since $G$ is a lagrangian of $(P,\psi)$ so
$\psi|_{G\times G}=0$ and $G^{\perp}=G$.

  2. (2)

Adding a formation to a quasi-formation has the effect of adding a zero form
to its induced forms: if $x=(H_{\varepsilon}(K),K,V)$ is a quasi-formation and
$y=(H_{\varepsilon}(F);F,G)$ is a formation, then the induced forms of the
quasi-formation $x\oplus y=(H_{\varepsilon}(K\oplus F),K\oplus F,V\oplus G)$
are the quadratic forms $(V\oplus G,\theta\oplus 0)$ and $(V^{\perp}\oplus
G,-\theta^{\perp}\oplus 0)$.

Example 7.16 (2) motivates the following definition from [CrowleySixt].

######  Definition 7.17.

Two quadratic forms $(P,\psi)$ and $(P^{\prime},\psi^{\prime})$ are _$0$
-stably equivalent_ if there are zero forms $(Q,0)$ and $(Q^{\prime},0)$ and
an isometry

$(P,\psi)\oplus(Q,0)\cong(P^{\prime},\psi^{\prime})\oplus(Q^{\prime},0).$

The $0$-stable equivalence class of a form $(P,\psi)$ is denoted
$[P,\psi]_{0}$. Such an equivalence class is called a _0-stabilised quadratic
form_. We write $\mathcal{F}^{\operatorname{zs}}_{2q}(R)$ for the unital
abelian monoid of $0$-stabilised $(-1)^{q}$-quadratic forms over $R$, where
addition is given by direct sum.

######  Definition 7.18.

The _induced $0$-stabilised forms_ of a quasi-formation $((P,\psi);F,V)$ are
the $0$-stable equivalence classes $[V,\theta]_{0}$ and
$[V^{\perp},-\theta^{\perp}]_{0}$ of its induced forms $(V,\theta)$ and
$(V^{\perp},-\theta^{\perp})$.

The notion of induced $0$-stabilised forms descends to $\ell$-monoids, as
first observed in [CrowleySixt].

######  Proposition 7.19.

Given a ring $R$, taking induced $0$-stabilised forms gives rise to a well
defined monoid homomorphism

$\displaystyle b\colon\ell_{2q+1}(R)$
$\displaystyle\to\mathcal{F}^{\operatorname{zs}}_{2q}(R)\times\mathcal{F}^{\operatorname{zs}}_{2q}(R)$
$\displaystyle[((P,\psi);F,V)]$
$\displaystyle\mapsto([V,\theta]_{0},[V^{\perp},-\theta^{\perp}]_{0}).$

###### Proof.

Adding a trivial formation to a quasi-formation adds a zero form to the
induced forms; recall Example 7.16. It therefore remains to study the effect
of the relation in (18):

$((P,\psi);F,G)\oplus((P,\psi);G,V)\sim((P,\psi);F,V).$

Since $((P,\psi);F,G)$ is a formation, its induced forms are zero forms and
therefore the induced forms on both sides of this equivalence are $0$-stably
equivalent. Note that the induced forms of a quasi-formation are independent
of the data of the first lagrangian. ∎

######  Remark 7.20.

Given quadratic forms $v,v^{\prime}$ such that $[v]_{0}\oplus
H_{\varepsilon}(R^{k})=[v^{\prime}]_{0}\oplus H_{\varepsilon}(R^{k})$,
[CrowleySixt, Corollary 5.3] ensures that
$([v],[v^{\prime}])\in\operatorname{\operatorname{Im}}(b)$. Following
[CrowleySixt, Subsection 5.2] we can therefore define

$\ell_{2q+1}(v,v^{\prime}):=b^{-1}([v]_{0},[v^{\prime}]_{0})\subseteq\ell_{2q+1}(R).$

This is the set of all (equivalence classes of) quasi-formations whose induced
forms are $0$-stably equivalent to $v$ and $v^{\prime}$ respectively. If
$[v]_{0}=[v^{\prime}]_{0}$, then we write $\ell_{2q+1}(v)$ instead of
$\ell_{2q+1}(v,v)$. Note that for any $0$-form $(V,0)$, we have
$\ell_{2q+1}(V,0)=L_{2q+1}(R)$ [CrowleySixt, Lemma 6.2].

Crowley and Sixt also show that for every $0$-stable equivalence class of a
quadratic form $v=(V,\theta)$, there exists a unique elementary class
$e(v)\in\ell_{2q+1}(v)$ such that $b\circ e(v)=([v]_{0},[v]_{0})$
[CrowleySixt, Corollary 5.3]. Explicitly, $e(v)$ is the class of the boundary
quasi-formation $\delta(V,\rho)$, where $\rho$ is a sesquilinear form
representing $\theta\in Q_{\varepsilon}(V)$.

######  Definition 7.21.

Given a quadratic form $v$ we say that $\ell_{2q+1}(v)$ is _trivial_ if it
contains a single (necessarily elementary) element.

###  7.4. Linking forms and boundary automorphism sets

Given a quadratic form $v=(V,\theta)$ over $\mathbb{Z}$, we recall Crowley-
Sixt’s description of $\ell_{5}(v)$ in terms of linking forms [CrowleySixt].
We use this to discuss various examples where $\ell_{5}(v)$ is trivial. In
this section, the group $T$ is a finite abelian group and the group $V$ is
always understood to be a finitely generated abelian group.

######  Definition 7.22.

A (symmetric) _linking form_ (over $\mathbb{Z}$) refers to a pair $(T,b)$
where $T$ is a finite abelian group and $b\colon T\times
T\to\mathbb{Q}/\mathbb{Z}$ is a symmetric bilinear form.

######  Definition 7.23 (Ranicki [RanickiExact, §3.4]).

A _split quadratic linking form_ $(T,b,\nu)$ (over $\mathbb{Z}$) is a
symmetric linking form $(T,b)$ together with a map $\nu\colon
T\to\mathbb{Q}/\mathbb{Z}$ such that the following equalities in
$\mathbb{Q}/\mathbb{Z}$ hold for all $x,x_{1},x_{2}\in T$ and all
$r\in\mathbb{Z}$:

  1. (1)

$\nu(rx)=r^{2}\nu(x),$

  2. (2)

$\nu(x_{1}+x_{2})-\nu(x_{1})-\nu(x_{2})=b(x_{1},x_{2})$.

Note that applying the second condition with $x_{1}=x_{2}=x$ yields
$\nu(2x)-2\nu(x)=b(x,x)$, and then applying the first condition we learn that
$2\nu(x)=b(x,x)$.

######  Definition 7.24.

A nondegenerate symmetric bilinear form $\lambda\colon V\times
V\to\mathbb{Z}$, whose adjoint we denote $\widehat{\lambda}\colon V\to V^{*}$,
determines a linking form
$\partial\lambda\colon\operatorname{coker}(\widehat{\lambda})\times\operatorname{coker}(\widehat{\lambda})\to\mathbb{Q}/\mathbb{Z}$
via the formula $([x],[y])\mapsto\frac{1}{s}y(z)$, where
$sx=\widehat{\lambda}(z)$ for some $z\in V$ and some
$s\in\mathbb{Z}\setminus\\{0\\}$. We call $\partial\lambda$ the _boundary
linking form_ of the symmetric form $\lambda$.

We recall the corresponding definition for quadratic forms. Note that if
$\theta:=[\rho]\in Q_{\varepsilon}(V)$ is a quadratic form, then for $z\in V$,
the integer $\rho(z,z)$ only depends on the the class of $\rho$ in
$Q_{\varepsilon}(V)$. As a consequence, we write $\theta(z,z)$ instead of
$\rho(z,z)$.

######  Definition 7.25.

Let $(V,\theta)\in Q_{\varepsilon}(V)$ be a nondegenerate quadratic form over
$\mathbb{Z}$ with symmetrisation $(V,\lambda)$. The _boundary split quadratic
linking form_ of $(V,\theta)$ is the split quadratic linking form
$\partial(V,\theta):=(\operatorname{coker}(\widehat{\lambda}),\partial\lambda,\partial\theta)$
where $(\operatorname{coker}(\widehat{\lambda}),\partial\lambda)$ is the
boundary linking form of the symmetric form $(V,\lambda)$ and

$\displaystyle\partial\theta\colon\operatorname{coker}(\widehat{\lambda})$
$\displaystyle\to\mathbb{Q}/\mathbb{Z}$ $\displaystyle([x])$
$\displaystyle\mapsto\frac{1}{s^{2}}\theta(z,z),$

where $sx=\widehat{\lambda}(z)$ for $z\in V$ and
$s\in\mathbb{Z}\setminus\\{0\\}$.

One verifies that if a nondegenerate symmetric form $(V,\lambda)$ is
represented by a size $n$ matrix $A$, then $\partial\lambda([x],[y])=xA^{-1}y$
where $A^{-1}$ denotes the inverse of $A$ over $\mathbb{Q}$ and
$[x],[y]\in\mathbb{Z}^{n}/A\mathbb{Z}^{n}.$ We record the corresponding
calculation in the quadratic case.

######  Remark 7.26.

If a nondegenerate quadratic form $(V,\theta)$ is represented by a size $n$
matrix $Q$ so that its symmetrisation is represented by $A=Q+Q^{T}$, then

$\partial\theta([x])=x^{T}(A^{-1})^{T}QA^{-1}x=x^{T}A^{-1}QA^{-1}x$

for every $[x]\in\mathbb{Z}^{n}/A\mathbb{Z}^{n}$, where once again we invert
$A$ over $\mathbb{Q}$.

######  Example 7.27.

We use Remark 7.26 to calculate the boundary split quadratic linking form of a
rank one quadratic form. When $V=\mathbb{Z}$ and
$\theta\in\mathbb{Z}\setminus\\{0\\}$, we have $\lambda=2\theta$ and the
boundary forms are therefore defined on $\mathbb{Z}/2\theta$. Remark 7.26
implies that
$\partial\lambda([x],[y])=\frac{1}{2\theta}xy\in\mathbb{Q}/\mathbb{Z}$ and

$\partial\theta([x])=\mbox{\footnotesize$\displaystyle\frac{\theta}{4\theta^{2}}$}x^{2}=\mbox{\footnotesize$\displaystyle\frac{1}{4\theta}$}x^{2}\in\mathbb{Q}/\mathbb{Z}.$

Still building towards the description of $\ell_{5}(v)$ in terms of linking
forms, we now describe isometries of linking forms.

######  Definition 7.28.

An _isometry_ of split quadratic linking forms $(T_{0},b_{0},\nu_{0})$ and
$(T_{1},b_{1},\nu_{1})$ is an isomorphism $f\colon
T_{0}\xrightarrow{\cong}T_{1}$ such that $\nu_{1}(f(x))=\nu_{0}(x)$ for every
$x\in T_{0}$.

Observe that an isometry automatically satisfies $b_{1}(f(x),f(y))=b_{0}(x,y)$
because

$b_{1}(f(x),f(y))=\nu_{1}(f(x)+f(y))-\nu_{1}(f(x))-\nu_{1}(f(y))=\nu_{0}(x+y)-\nu_{0}(x)-\nu_{0}(y)=b_{0}(x,y).$

Given an isometry $h\colon(V_{0},\theta_{0})\to(V_{1},\theta_{1})$ of
quadratic forms, one verifies that $(h^{*})^{-1}\colon V_{0}^{*}\to V_{1}^{*}$
induces an isomorphism
$\operatorname{coker}(\widehat{\lambda}_{0})\to\operatorname{coker}(\widehat{\lambda}_{1})$
on the cokernels. This isomorphism, which we denote by $\partial
h:=(h^{*})^{-1}$, is an isometry of the boundary split quadratic linking
forms.

######  Definition 7.29.

The _boundary_ of an isometry $h\colon(V_{0},\theta_{0})\to(V_{1},\theta_{1})$
is the isometry

$\partial
h:=(h^{*})^{-1}\colon\partial(V_{0},\theta_{0})\to\partial(V_{1},\theta_{1}).$

This determines a homomorphism

$\partial\colon\operatorname{Aut}(V,\theta)\to\operatorname{Aut}(\partial(V,\theta)).$

The group $\operatorname{Aut}(V,\theta)$ of isometries of $(V,\theta)$ acts on
the group $\operatorname{Aut}(\partial(V,\theta))$ on the left both via
$h\cdot f:=\partial h\circ f$ and $g\cdot f:=f\circ\partial g^{-1}$ where
$f\in\operatorname{Aut}(\partial(V,\theta))$ and
$g,h\in\operatorname{Aut}(V,\theta)$. Combining these actions, which clearly
commute, leads to a left action of
$\operatorname{Aut}(V,\theta)\times\operatorname{Aut}(V,\theta)$ on
$\operatorname{Aut}(\partial(V,\theta))$.

######  Definition 7.30.

The _boundary automorphism set_ of a nondegenerate quadratic form $(V,\theta)$
is the orbit set

$\operatorname{bAut}(V,\theta)=\operatorname{Aut}(\partial(V,\theta))/\operatorname{Aut}(V,\theta)\times\operatorname{Aut}(V,\theta).$

Observe that an isometry $f\in\operatorname{Aut}(\partial(V,\theta))$ is
trivial in the boundary automorphism set $\operatorname{bAut}(V,\theta)$ if
and only if $f$ extends to an isometry of $(V,\theta)$.

######  Example 7.31.

Using Example 7.27, one verifies that for a rank one form
$(\mathbb{Z},\theta)$, the automorphisms of the module $\mathbb{Z}/2\theta$
are the units of the ring $\mathbb{Z}/2\theta$, while the automorphisms of
$\partial(\mathbb{Z},\theta)$ consist of the units $x$ of $\mathbb{Z}/2\theta$
such that $x^{2}=1$ mod $4\theta$. The corresponding boundary automorphism set
$\operatorname{bAut}(\mathbb{Z},\theta)$ consists of the same
$x\in\mathbb{Z}/2\theta$, but with $x$ and $-x$ identified.

  * •

For $\theta=4$, the units of $\mathbb{Z}/8$ are $\pm 1$ and $\pm 3$, but only
$1$ and $-1$ satisfy $x^{2}=1$ mod $16$. It follows that
$\operatorname{bAut}(\mathbb{Z},4)$ is trivial.

  * •

For $\theta=6$, the units of $\mathbb{Z}/12$ are $\pm 1,\pm 5$ all of which
satisfy $x^{2}=1$ mod $4\theta=24$. It follows that
$\operatorname{bAut}(\mathbb{Z},6)=\\{1,5\\}$.

######  Theorem 7.32 (Crowley-Sixt [CrowleySixt, Section 6.3]).

Given a nondegenerate quadratic form $v=(V,\theta)$ over $\mathbb{Z}$, there
is a bijection between $\ell_{5}(v)$ and $\operatorname{bAut}(v)$.

######  Example 7.33.

We illustrate Theorem 7.32 with some examples that will be relevant to the
study of $\mathbb{Z}_{2}$-surfaces of nonorientable genus $h=1,2$.

  * •

In Example 7.31 we saw $\operatorname{bAut}(\mathbb{Z},4)$ is trivial, and
thus so is $\ell_{5}(\mathbb{Z},4)$, by Theorem 7.32.

  * •

We argue that $\operatorname{bAut}(4H_{+}(\mathbb{Z}))$ is trivial, thus
proving $\ell_{5}(4H_{+}(\mathbb{Z}))$ is trivial. First observe that the
isometries of the quadratic form $4H_{+}(\mathbb{Z})$ are

$\operatorname{Aut}(4H_{+}(\mathbb{Z}))=\operatorname{Aut}(H_{+}(\mathbb{Z}))=\left\\{\left(\begin{smallmatrix}\varepsilon_{1}&0\\\
0&\varepsilon_{2}\end{smallmatrix}\right),\left(\begin{smallmatrix}0&\varepsilon_{1}\\\
\varepsilon_{2}&0\end{smallmatrix}\right)\mid\varepsilon_{1}\varepsilon_{2}=1\right\\}.$

We verify that all the isometries of $\partial(4H_{+}(\mathbb{Z}))$ are
induced by $\operatorname{Aut}(4H_{+}(\mathbb{Z}))$. This was already observed
by Kreck [KreckOnTheHomeomorphism, p. 70]. First, note that the symmetric
linking form is
$b\colon(\mathbb{Z}_{4}\oplus\mathbb{Z}_{4})\times(\mathbb{Z}_{4}\oplus\mathbb{Z}_{4})\to\mathbb{Q}/\mathbb{Z}$,
represented by $B:=\left(\begin{smallmatrix}0&\frac{1}{4}\\\
\frac{1}{4}&0\end{smallmatrix}\right)$, and the quadratic enhancement is
$\nu\colon\mathbb{Z}_{4}\oplus\mathbb{Z}_{4}\to\mathbb{Q}/\mathbb{Z},ke_{1}+\ell
e_{2}\mapsto k\ell/4$. An automorphism $\alpha$ of this is represented by a
matrix $A:=\left(\begin{smallmatrix}a&b\\\ c&d\end{smallmatrix}\right)$ over
$\mathbb{Z}_{4}$ such that $A^{T}BA=B$ over $\mathbb{Q}/\mathbb{Z}$ and such
that
$\nu\circ\alpha=\nu\colon\mathbb{Z}_{4}\oplus\mathbb{Z}_{4}\to\mathbb{Q}/\mathbb{Z}$.
The former condition implies that $ad+bc=1\in\mathbb{Z}_{4}$, while the
conditions that $\nu(e_{1})=\nu(ae_{1}+ce_{2})$ and
$\nu(e_{2})=\nu(be_{1}+de_{2})$ give $ac\equiv 0\equiv bd$. A case analysis
yields exactly the same four matrices for $A$ as in the description of
$\operatorname{Aut}(4H_{+}(\mathbb{Z}))$. It follows that
$\operatorname{Aut}(4H_{+}(\mathbb{Z}))\to\operatorname{Aut}(\partial(4H_{+}(\mathbb{Z})))$
is surjective, as claimed.

###  7.5. Nikulin’s criterion for trivial bAut

Determining whether $\ell_{5}(v)$ is trivial is geometrically significant, as
we will recall in the next section. Theorem 7.32 converts this problem into a
question about linking forms. In order to state a criterion due to Nikulin
[Nikulin] that describes conditions for $\operatorname{bAut}(v)$ to be
trivial, we introduce some notation. Namely, given an abelian group $G$ and a
prime $p$, we write $n_{p}(G)$ for the minimal number of generators for the
$p$-primary part of $G$, or equivalently
$n_{p}(G):=\dim_{\mathbb{F}_{p}}(G\otimes\mathbb{F}_{p})$.

######  Theorem 7.34 (Nikulin [Nikulin, Theorem 1.14.2]).

Let $v=(V,\theta)$ be an indefinite, nondegenerate quadratic form over
$\mathbb{Z}$ with symmetrisation $(V,\lambda)=(V,\theta+\theta^{*})$. Suppose
that

  1. (1)

the inequality $\operatorname{rk}(V)\geq
n_{p}(\operatorname{coker}(\widehat{\lambda}))+2$ holds for all odd primes
$p$, and

  2. (2)

if $\operatorname{rk}(V)=n_{2}(\operatorname{coker}(\widehat{\lambda}))$, then
the split quadratic linking form
$\partial(\mathbb{Z}^{2},[\left(\begin{smallmatrix}0&2\\\
0&0\end{smallmatrix}\right)])$ is a summand of $\partial(V,\theta)$.

Then
$\partial\colon\operatorname{Aut}(V,\theta)\to\operatorname{Aut}(\partial(V,\theta))$
is surjective, and so $\operatorname{bAut}(v)$ consists of one element.

The statement in Nikulin’s paper is not written this way. The fact that
Nikulin’s work could be translated into this language and used in this context
was observed in Crowley-Sixt [CrowleySixt, Proposition 6.18 i)], where they
reported the outcome of that translation. We provide the details of the
translation momentarily, but first record the key consequence for our
purposes.

######  Corollary 7.35.

Under the conditions in the hypotheses of Theorem 7.34, $\ell_{5}(v)$ is
trivial.

###### Proof.

Combine Theorem 7.34 with Theorem 7.32. ∎

We now translate the result stated in [Nikulin, Theorem 1.14.2] to that stated
in Theorem 7.34. First, Nikulin works with the nonsplit version of quadratic
linking forms, whose definition we recall next, in which the quadratic
enhancements $\mu$ are valued in $\mathbb{Q}/2\mathbb{Z}$.

######  Definition 7.36 (Ranicki [RanickiExact, §3.4]).

A _nonsplit quadratic linking form_ $(T,b,\mu)$ (over $\mathbb{Z}$) is a
symmetric linking form $(T,b)$ together with a map $\mu\colon
T\to\mathbb{Q}/2\mathbb{Z}$ such that the following equalities in
$\mathbb{Q}/2\mathbb{Z}$ hold for all $x,x_{1},x_{2}\in T$ and all
$r\in\mathbb{Z}$:

  1. (1)

$\mu(rx)=r^{2}\mu(x),$

  2. (2)

$\mu(x_{1}+x_{2})-\mu(x_{1})-\mu(x_{2})=2b(x_{1},x_{2})$.

In the more general context where Ranicki defined these notions, there is in
general a difference between split and nonsplit quadratic linking forms. For
us, because $1/2\in\mathbb{Q}/\mathbb{Z}$, there is effectively no difference.

######  Proposition 7.37 ([RanickiExact, Proposition 3.4.2]).

A split quadratic linking form $(T,b,\nu)$ determines a nonsplit quadratic
linking form $(T,b,\mu)$ by sending $\nu\colon T\to\mathbb{Q}/\mathbb{Z}$ to
$\mu:=2\nu\colon T\to\mathbb{Q}/2\mathbb{Z}$. This gives rise to an
equivalence of categories

$\left\\{\begin{array}[]{c}\text{split quadratic linking forms over
}\mathbb{Z}\\\ \text{and morphisms
thereof}\end{array}\right\\}\xrightarrow{\cong}\left\\{\begin{array}[]{c}\text{nonsplit
quadratic linking forms over }\mathbb{Z}\\\ \text{and morphisms
thereof}\end{array}\right\\}$

that is the identity on the morphisms.

######  Definition 7.38.

The _boundary_ of a nondegenerate, even symmetric bilinear form $\lambda\colon
V\times V\to\mathbb{Z}$ is the nonsplit quadratic linking form
$\partial(V,\lambda):=(\operatorname{coker}(\widehat{\lambda}),\partial\lambda,\mu)$,
where $(\operatorname{coker}(\widehat{\lambda}),\partial\lambda)$ is the
boundary linking form of the symmetric form $(V,\lambda)$, and

$\mu\colon\operatorname{coker}(\widehat{\lambda})\to\mathbb{Q}/2\mathbb{Z}$

is defined by $\mu([y])=\mbox{\large$\frac{1}{s^{2}}$}\lambda(z,z)$, where
$s\in\mathbb{Z},z\in V$ and $y\in V^{*}$ satisfy $sy=\widehat{\lambda}(z)$.

######  Remark 7.39.

We caution the reader that $\partial(V,\lambda)$ does not refer to a symmetric
linking form but to a nonsplit _quadratic_ linking form. As illustrated in
Example 7.31, quadratic linking forms typically have fewer isometries than
their symmetrisations.

######  Construction 7.40.

Write $\mathbf{Quad},\mathbf{EvenSym},\mathbf{SplitQuadLink}$, and
$\mathbf{NonSplitQuadLink}$ for the categories of nondegenerate quadratic
forms over $\mathbb{Z}$, even symmetric forms over $\mathbb{Z}$, split
quadratic linking forms over $\mathbb{Z}$ and nonsplit quadratic linking forms
over $\mathbb{Z}$, together with morphisms thereof. Consider the following
functors:

(20) ${\mathbf{Quad}}$${\mathbf{SplitQuadLink}}$${\mathbf{EvenSym}}$${\
\mathbf{NonSplitQuadLink}.}$$\scriptstyle{\cong}$$\scriptstyle{\partial}$$\scriptstyle{\cong}$$\scriptstyle{\partial}$

Here, the left vertical functor maps a quadratic form to its symmetrisation
and is the identity on morphisms. It is an equivalence of categories because
every even symmetric form over $\mathbb{Z}$ admits a unique quadratic
refinement. The right hand side vertical functor is the equivalence described
in Proposition 7.37, and the horizontal functors take the boundary.

######  Proposition 7.41.

The square (20) commutes.

###### Proof.

To see that the diagram of categories and functors commutes on the level of
objects, note that if $\theta\in Q_{+}(V)$ is a quadratic form with
symmetrisation $\lambda$, then

$\partial\theta([y])=\mbox{\large$\frac{1}{s^{2}}$}\theta(z,z)=\mbox{\large$\frac{1}{2s^{2}}$}\lambda(z,z)=\mbox{\large$\frac{1}{2}$}\mu([y]).$

The commutativity on the level of morphisms holds because a homomorphism
$f\colon V\to V$ is sent via both routes to $\partial
f\colon\operatorname{coker}(\widehat{\lambda})\to\mathbb{Q}/2\mathbb{Z}$. ∎

######  Corollary 7.42.

The diagram in (20) induces the following commutative diagram of automorphism
groups:

(21)
${\operatorname{Aut}(V,\theta)}$${\operatorname{Aut}(\partial(V,\theta))}$${\operatorname{Aut}(V,\lambda)}$${\operatorname{Aut}(\partial(V,\lambda)).}$$\scriptstyle{\partial}$$\scriptstyle{=}$$\scriptstyle{=}$$\scriptstyle{\partial}$

The vertical equalities refer to equalities as subsets of the automorphisms
$\operatorname{Aut}(V)$ of the group $V$.

###### Proof.

Restrict (20) to the isometries. ∎

Nikulin’s theorem is stated in his paper in terms of the objects on the bottom
row of (21) and Crowley-Sixt’s Theorem 7.32 is stated in terms the objects on
the top row. For our statement in Theorem 7.34, we translated Nikulin’s
statement into the language of quadratic forms and split quadratic linking
forms, in order to be able to apply it in the context of Theorem 7.32.

Summarising, Nikulin’s map
$\partial\colon\operatorname{Aut}(V,\lambda)\to\operatorname{Aut}(\partial(V,\lambda))$
is identified with the map
$\partial\colon\operatorname{Aut}(V,\theta)\to\operatorname{Aut}(\partial(V,\theta))$
from Definition 7.29. Thus for $(V,\lambda)=(V,\theta+\theta^{*})$, Nikulin’s
requirement that the nonsplit quadratic form $\partial(V,\lambda)$ split off a
$\partial(\mathbb{Z}^{2},\left(\begin{smallmatrix}0&2\\\
2&0\end{smallmatrix}\right))$ summand (Nikulin writes $U^{(2)}(2)$ for the
symmetric form $\lambda$ and $u_{+}^{(2)}(2)$ for its nonsplit quadratic
boundary $\partial(V,\lambda)$ [Nikulin, Proposition 1.8.1]) is equivalent to
requiring that the split quadratic linking form $\partial(V,\theta)$ contains
a $\partial(\mathbb{Z}^{2},[\left(\begin{smallmatrix}0&2\\\
0&0\end{smallmatrix}\right)])$ summand.

We conclude this section by illustrating Corollary 7.35 with some examples
that will be useful when studying $\mathbb{Z}_{2}$-surfaces. Given an integer
$n\in\mathbb{Z}$, consider the quadratic form

$X_{n}=\operatorname{sgn}(n)\left[\begin{pmatrix}2&2&2&\ldots&2\\\
0&1&1&\cdots&1\\\ 0&0&\ddots&\ddots&\vdots\\\ \vdots&\ddots&\ddots&\ddots&1\\\
0&\ldots&0&0&1\end{pmatrix}\right]\in Q_{+}(\mathbb{Z}^{n})$

from Proposition 5.12. For nonnegative integers $a,b\in\mathbb{Z}_{\geq 0}$,
set $h:=a+b$ and $\sigma=a-b$. Observe that $h-|\sigma|$ is nonnegative and
even, and note that if $\sigma=0$ then $h$ is even. Consider the quadratic
form

$\theta_{a,b}:=\begin{cases}2H_{+}(\mathbb{Z})\oplus
H_{+}(\mathbb{Z})^{\oplus(\frac{h}{2}-1)}&\quad\text{ if }\sigma=0,\\\
X_{\sigma}\oplus H_{+}(\mathbb{Z})^{\oplus\frac{1}{2}(h-|\sigma|)}&\quad\text{
if }\sigma\neq 0.\end{cases}$

This is a rank $h$, signature $\sigma$, nondegenerate quadratic form. If
$h\neq|\sigma|$ then $h-|\sigma|>0$ so $\theta_{a,b}$ is an indefinite form on
$V:=\mathbb{Z}^{h}$.

######  Proposition 7.43.

Given nonnegative integers $a,b\in\mathbb{Z}_{\geq 0}$, set $h:=a+b$ and
$\sigma=a-b$. If $h\leq 3$ or if $|\sigma|\neq h$, then the set
$\ell_{5}(\mathbb{Z}^{h},2\theta_{a,b})$ is trivial.

###### Proof.

For $h=1$ we have $2\theta_{a,b}=4$ and we already proved in Example 7.33 that
$\ell_{5}(\mathbb{Z},4)$ is trivial. For $h=2$ and $\sigma=0$ we have
$2\theta_{a,b}=4H_{+}(\mathbb{Z})$ and again we already proved in Example 7.33
that $\ell_{5}(4H_{+}(\mathbb{Z}))$ is trivial. The cases where
$|\sigma|=h=2,3$ involve detailed explicit computations, and are treated in
the appendix.

We now assume that $h\geq 3$ and $|\sigma|\neq h$. Write
$\theta:=\theta_{a,b}$. If $h\geq 3$ then $\theta$ has at least one hyperbolic
summand, and therefore $2\theta$ has a $2H_{+}(\mathbb{Z})$-summand. We deduce
that $\partial(\mathbb{Z}^{h},2\theta)$ splits off a
$\partial(\mathbb{Z}^{2},[\left(\begin{smallmatrix}0&2\\\
0&0\end{smallmatrix}\right)])$ summand.

Thanks to Corollary 7.35, the proof therefore reduces to verifying that
$\operatorname{coker}(\widehat{\lambda})$ is $2$-primary and satisfies
$n_{2}(\operatorname{coker}(\widehat{\lambda}))=\operatorname{rk}(V)=h$. Thus
in our case $n_{p}(\operatorname{coker}(\widehat{\lambda}))=0$ for every odd
prime, and so $h\geq 3$ implies that $h=\operatorname{rk}(V)\geq
n_{p}(\operatorname{coker}(\widehat{\lambda}))+2=2$. We have that

$2\theta\cong\begin{cases}4H_{+}(\mathbb{Z})\oplus
2H_{+}(\mathbb{Z})^{\oplus\frac{h}{2}-1}&\quad\text{ if }\sigma=0,\\\
2X_{\sigma}\oplus
2H_{+}(\mathbb{Z})^{\oplus\frac{1}{2}(h-|\sigma|)}&\quad\text{ if }\sigma\neq
0.\end{cases}$

When $\sigma=0$, we have that $\operatorname{coker}(4H^{+}(\mathbb{Z})\oplus
2H^{+}(\mathbb{Z})^{\oplus\frac{h}{2}-1})\cong\mathbb{Z}_{4}\oplus\mathbb{Z}_{4}\oplus\mathbb{Z}_{2}^{\oplus(h-2)}$.
This is 2-primary and generated by no fewer than $h$ elements. Thus
$n_{2}(\operatorname{coker}(\widehat{\lambda}))=h$ in this case. When
$\sigma\neq 0$, we have that
$\operatorname{coker}(2H_{+}(\mathbb{Z})^{\oplus\frac{1}{2}(h-|\sigma|)})\cong(\mathbb{Z}_{2})^{\oplus
h-|\sigma|}$, which is 2-primary and generated by no fewer than $h-|\sigma|$
elements. To complete the $\sigma\neq 0$ case, it therefore suffices to show
that $\operatorname{coker}(2X_{\sigma}+2X_{\sigma}^{T})$ is $2$-primary and
generated by $|\sigma|$ and no fewer elements.

We consider

$2X_{\sigma}+2X_{\sigma}^{T}=\operatorname{sgn}(\sigma)\begin{pmatrix}8&4&4&\ldots&4\\\
4&4&2&\cdots&2\\\ 4&2&\ddots&\ddots&\vdots\\\ \vdots&\vdots&\ddots&\ddots&2\\\
4&2&\ldots&2&4\end{pmatrix}.$

By performing row and column operations, we compute that

$\operatorname{coker}(2X_{\sigma}+2X_{\sigma}^{T})=\begin{cases}\mathbb{Z}_{2}^{|\sigma|-1}\oplus\mathbb{Z}_{8}&\quad\text{
if }|\sigma|\text{ is odd,}\\\
\mathbb{Z}_{2}^{|\sigma|-2}\oplus\mathbb{Z}_{4}\oplus\mathbb{Z}_{4}&\quad\text{
if }|\sigma|\text{ is even.}\end{cases}$

Thus $\operatorname{coker}(2X_{\sigma}+2X_{\sigma}^{T})$ is indeed $2$-primary
and is generated by no fewer than $|\sigma|$ elements. This concludes the
verification that the assumptions of Corollary 7.35 hold, from which it
follows that $\ell_{5}(\mathbb{Z}^{h},2\theta)$ is trivial. ∎

######  Remark 7.44.

We currently know of no pair $(a,b)$ where
$\ell_{5}(\mathbb{Z}^{h},2\theta_{a,b})$ is nontrivial; here recall that
$h:=a+b$ and $\sigma=a-b$. In particular, it could be that
$\ell_{5}(\mathbb{Z}^{h},2\theta_{a,b})$ is trivial for all $a,b$. It would be
very interesting to know whether or not this holds. Proposition 7.43 implies
that $\ell_{5}(\mathbb{Z}^{h},2\theta_{a,b})$ could only potentially be
nontrivial when $h=|\sigma|\geq 4$. In the appendix, we verify by direct
calculations of isometries that $\ell_{5}(\mathbb{Z}^{h},2\theta_{a,b})$ is
trivial in the cases corresponding to $h=|\sigma|\in\\{2,3\\}$. The
computations grow rapidly in complexity with $h$. For $h=|\sigma|=2$, the
group $\operatorname{Aut}(\mathbb{Z}^{h},\theta_{a,b}))$ has $8$ elements,
whereas for $h=|\sigma|=3$, it has $48$ elements. This computational
complexity explains why, for extremal values of $\sigma$, we only proved
Proposition 7.43 for $h\leq 3$.

##  8\. Modified surgery theory

In this section we recall some elements of Kreck’s modified surgery theory
[KreckSurgeryAndDuality]. In Subsection 8.1, we review the concept of a
_normal $k$-smoothing_ of a manifold $M$. Restricting to dimension $4$,
Subsection 8.2 recalls the definition of the modified surgery obstruction that
takes values in $\ell_{5}(\mathbb{Z}[\pi_{1}(M)])$. In Subsection 8.3 we
narrow down where in the monoid $\ell_{5}(\mathbb{Z}[\pi_{1}(M)])$ the
modified surgery obstruction lives.

###  8.1. Normal smoothings

We recall the notion of a normal smoothing and of the normal $k$-type from
[KreckSurgeryAndDuality]. We then consider the concrete case where the
manifolds are $4$-dimensional and orientable with fundamental group
$\mathbb{Z}_{2}$.

Let $B$ be a topological space with the homotopy type of a CW complex and let
$\xi\colon B\to\operatorname{BSTOP}$ be a fibration. An _$n$ -dimensional
$(B,\xi)$-manifold_ consists of a pair $(M,\overline{\nu})$, where $M$ is an
oriented $n$-manifold and $\overline{\nu}\colon M\to B$ is a lift of the
oriented stable (topological) normal bundle $\nu\colon
M\to\operatorname{BSTOP}$ of $M$, meaning that $\nu=\xi\circ\overline{\nu}$.

A  _$(B,\xi)$ -null-bordism_ for a closed $n$-dimensional $(B,\xi)$-manifold
$(M_{0},\overline{\nu}_{0})$ is an $(n{+}1)$-dimensional $(B,\xi)$-manifold
$(W,\overline{\nu})$ for which
$\partial(W,\overline{\nu})=(M_{0},\overline{\nu}_{0})$. A  _$(B,\xi)$
-cobordism_ between two $n$-dimensional $(B,\xi)$-manifolds
$(M_{0},\overline{\nu}_{0})$ and $(M_{1},\overline{\nu}_{1})$ consists of:

  1. (i)

a homeomorphism $f\colon\partial M_{0}\to\partial M_{1}$;

  2. (ii)

for $i=0,1$, homotopies $\overline{\nu}_{i}\sim\overline{\nu}^{\dagger}_{i}$,
to some maps $\overline{\nu}^{\dagger}_{i}\colon M_{i}\to B$, such that

${\overline{\nu}^{\dagger}_{0}}|_{\partial
M_{0}}={\overline{\nu}^{\dagger}_{1}}|_{\partial M_{1}}\circ f\colon\partial
M_{0}\to B;$

  3. (iii)

an $(n{+}1)$-dimensional $(B,\xi)$-null-bordism of
$(M_{0}\cup_{f}-M_{1},\overline{\nu}^{\prime})$ where
$\overline{\nu}^{\prime}|_{M_{0}}=\overline{\nu}^{\dagger}_{0}$ and
$\overline{\nu}^{\prime}|_{M_{1}}=-\overline{\nu}^{\dagger}_{1}$.

We note that the homotopies in condition (ii) are there to ensure one can
produce a well-defined $B$-structure $\overline{\nu}^{\prime}$ on
$M_{0}\cup_{f}-M_{1}$ in the scenario that the $B$-structures
$\overline{\nu}_{1}\circ f$ and $\overline{\nu}_{0}$ are isomorphic, but not
necessarily equal (equality is required for glueing).

We briefly discuss orientation reversal, to give meaning to the notation
$-\overline{\nu}^{\dagger}_{1}$ above. Given a $(B,\xi)$-manifold
$(M,\overline{\nu})$, the map $\overline{\nu}\circ\operatorname{pr}_{1}\colon
M\times[0,1]\to B$ defines a $(B,\xi)$-manifold
$(M\times[0,1],\operatorname{pr}_{1}\circ\overline{\nu})$. We identify
$(M,\overline{\nu})$ with the restriction to $M\times\\{0\\}$, and define
$-\overline{\nu}\colon M\to B$ as the restriction to the end $M\times\\{1\\}$,
which then produces a $(B,\xi)$-manifold $(-M,-\overline{\nu})$, i.e.
$-\overline{\nu}$ lifts the oppositely oriented stable normal bundle
$\nu_{-M}\colon-M\to\operatorname{BSTOP}$.

Recall that a map of spaces $f\colon X\to Y$ is _$m$ -connected_ if
$f_{*}\colon\pi_{i}(X)\to\pi_{i}(Y)$ is an isomorphism for $i<m$ and is
surjective for $i=m$. A map of spaces $f\colon X\to Y$ is _$m$ -coconnected_
if $f_{*}\colon\pi_{k}(X)\to\pi_{k}(Y)$ is an isomorphism for $i>m$ and is
injective for $i=m$.

######  Definition 8.1.

Let $B$ be a space with the homotopy type of a CW complex with finite
$(k+1)$-skeleton, let $\xi\colon B\to\operatorname{BSTOP}$ be a fibration, and
let $(M,\overline{\nu})$ be a $(B,\xi)$-manifold.

  1. (1)

If $\overline{\nu}$ is $(k{+}1)$-connected, then $(M,\overline{\nu})$ is
called a _normal $k$-smoothing_ into $(B,\xi)$.

  2. (2)

The pair $(B,\xi)$ is a _normal $k$-type_ for $M$ if $\xi$ is
$(k+1)$-coconnected and there exists a normal $k$-smoothing
$(M,\overline{\nu})$ into $(B,\xi)$.

The existence of the Moore-Postnikov factorisation [Baues-obstruction-theory]
of the stable normal bundle $\nu\colon M\to\operatorname{BSTOP}$ ensures that
for all $k\geq 0$, there exists a normal $k$-type for $M$. Moreover, this
theory implies any two normal $k$-types for $M$ are fibre homotopy equivalent.

We now assume that $M$ is a $4$-manifold with fundamental group
$\mathbb{Z}_{2}$, and equipped with a spin structure. This implies the
existence of maps $c\colon M\to B\mathbb{Z}_{2}$ and $\mathfrak{s}\colon
M\to\operatorname{BTOPSpin}$, where $c$ classifies the universal cover of $M$,
and $\mathfrak{s}$ is a lift of the stable normal bundle along the standard
principal fibration
$\gamma\colon\operatorname{BTOPSpin}\to\operatorname{BSTOP}$. These maps are
unique up to homotopy.

######  Lemma 8.2.

Let $M$ be a spin $4$-manifold with fundamental group $\mathbb{Z}_{2}$,
equipped with a spin structure. Write

$(B,\xi):=(\operatorname{BTOPSpin}\times
B\mathbb{Z}_{2},\gamma\circ\operatorname{pr}_{1}).$

Then the map

$\overline{\nu}:=\mathfrak{s}\times c\colon M\to\operatorname{BTOPSpin}\times
B\mathbb{Z}_{2}$

determines a normal $1$-smoothing into $(B,\xi)$, and $(B,\xi)$ is a normal
$1$-type for $M$.

###### Proof.

The space $B$ has finite $2$-skeleton because both
$B\mathbb{Z}_{2}\simeq\mathbb{R}P^{\infty}$ and
$\operatorname{\operatorname{BTOPSpin}}$ do (for the latter, use
$\pi_{i}(\operatorname{BTOPSpin})=0$ for $i=1,2$ together with CW
approximation). The map $\operatorname{BTOPSpin}\to\operatorname{BSTOP}$ is
2-coconnected because $\operatorname{TOPSpin}\to\operatorname{STOP}$ is the
universal covering map. It follows that $\xi$ is 2-coconnected, because
$\pi_{i}(B\mathbb{Z}_{2})=0$ for $i\geq 2$. Next, since $M$ is spin, the
stable normal bundle $\nu\colon M\to\operatorname{BSTOP}$ lifts to
$\mathfrak{s}\colon M\to\operatorname{BTOPSpin}$ and therefore combining with
$c$ we obtain a smoothing $M\to\operatorname{BTOPSpin}\times B\mathbb{Z}_{2}$.
This is 2-connected because $\pi_{i}(\operatorname{BTOPSpin})=0$ for $i=1,2$.
∎

######  Proposition 8.3.

If $F_{0}$ and $F_{1}$ are $\mathbb{Z}_{2}$-surfaces with the same
nonorientable genus and normal Euler number, then there exist normal
$1$-smoothings $(X_{F_{i}},\overline{\nu}_{i})$ for $i=0,1$, that are bordant
over their normal $1$-type, relative to a homeomorphism $f\colon\partial
X_{F_{0}}\to\partial X_{F_{1}}$ that restricts to a homeomorphism $S(\nu
F_{0})\to S(\nu F_{1})$ that is $\nu$-extendable rel. boundary.

###### Proof.

Write $(B,\xi)$ for the normal $1$-type of the $X_{F_{i}}$ that was described
in Lemma 8.2. By Proposition 6.3, there is a homeomorphism $f\colon\partial
X_{F_{0}}\to\partial X_{F_{1}}$ such that $M:=X_{F_{0}}\cup_{f}-X_{F_{1}}$ is
spin and has fundamental group $\mathbb{Z}_{2}$. Fix a spin structure on $M$.
Via Lemma 8.2, we obtain a normal $1$-smoothing $\overline{\nu}\colon M\to B$.
Denote by $\overline{\nu_{0}}\colon X_{F_{0}}\to B$ and
$-\overline{\nu_{1}}\colon-X_{F_{1}}\to B$ the restriction of $\overline{\nu}$
to the respective exteriors. By Lemma 8.2, $(X_{F_{i}},\overline{\nu}_{i})$ is
a normal $1$-smoothing for $i=0,1$. The proposition now reduces to proving
that $[(M,\overline{\nu})]$ vanishes in the bordism group
$\Omega_{4}(B,\xi)\cong\Omega_{4}^{{\operatorname{TOPSpin}}}(\mathbb{Z}_{2})$.
A spectral sequence calculation shows that
$\Omega_{4}^{\operatorname{TOPSpin}}(\mathbb{Z}_{2})\cong\Omega_{4}^{\operatorname{TOPSpin}}\cong\mathbb{Z}$,
detected by the signature divided by 8. See e.g. [OrsonPowellSpines, Proof of
Proposition 5.1] for the details.

For $i=0,1$, the Mayer-Vietoris sequence corresponding to the decomposition
$D^{4}=X_{F_{i}}\cup-\overline{\nu}F_{i}$ implies that the inclusion-induced
map $H_{2}(\partial X_{F_{i}};\mathbb{Z})\to H_{2}(X_{F_{i}};\mathbb{Z})$ is
surjective. As a result, the ordinary intersection pairing of $X_{F_{i}}$
vanishes identically, and so in particular $\sigma(X_{F_{i}})=0$. Then Novikov
additivity of the signature implies that the signature of $M$ also vanishes.
It follows that $(M,\overline{\nu})$ is $(B,\xi)$-null-bordant, as desired. ∎

###  8.2. The surgery obstruction

We recall the definition of the modified surgery obstruction in odd-
dimensions. The main reference is [KreckSurgeryAndDuality]. For simplicity, we
restrict our discussion to $4$-manifolds whose fundamental group has vanishing
Whitehead group. Since $\operatorname{Wh}(\mathbb{Z}_{2})=0$
[WhiteheadSimple], this will suffice for our purposes.

We say that
$(W,\overline{\nu},M_{0},\overline{\nu}_{0},M_{1},\overline{\nu}_{1})$ is a
_modified surgery problem_ if $M_{0}$ and $M_{1}$ are $4$-manifolds with
normal $1$-type $(B,\xi)$, the $\overline{\nu}_{i}\colon M_{i}\to B$ are
normal $1$-smoothings, and $(W,\overline{\nu})$ is a $(B,\xi)$-cobordism
between $(M_{0},\overline{\nu}_{0})$ and $(M_{1},\overline{\nu}_{1})$.

Let $(W,\overline{\nu},M_{0},\overline{\nu}_{0},M_{1},\overline{\nu}_{1})$ be
a modified surgery problem. By surgery below the middle dimension
[KreckSurgeryAndDuality, Proposition 4], we may assume that
$\overline{\nu}\colon W\to B$ is $2$-connected i.e. that $(W,\overline{\nu})$
is a normal $1$-smoothing. It can then be proved that
$K\pi_{2}(W):=\ker(\overline{\nu}_{*})$ is finitely generated; see e.g. [CCPS,
Lemma 3.4]. Choose a finite set $\omega$ of generators for $K\pi_{2}(W)$.
Choose disjoint embeddings $\varphi_{i}\colon S^{2}\times D^{3}\hookrightarrow
W$ with basing paths representing these generators, that are compatible with
the $(B,\xi)$-structure. Set

$U:=\bigsqcup_{i}\varphi_{i}(S^{2}\times D^{3}).$

Since $\partial U\cong\bigsqcup_{i}S^{2}\times S^{2}$, its quadratic
intersection form, denoted $(H_{2}(\partial U;\mathbb{Z}[\pi_{1}(B)]),\psi)$,
is hyperbolic. For concision, we refer to the pair $(\omega,\varphi)$ as a
_set of embedded generators for $K\pi_{2}(W)$_, we set
$\Lambda:=\mathbb{Z}[\pi_{1}(B)]$, and consider the $\Lambda$-modules

(22) $F:=H_{3}(U,\partial U;\Lambda)\quad\text{ and }\quad
V:=H_{3}(W\setminus\mathring{U},M_{0}\sqcup\partial U;\Lambda).$

Via the boundary maps in the long exact sequences of the appropriate pairs,
$F$ and $V$ can be identified with submodules of $P:=H_{2}(\partial
U;\Lambda)$. Under these identifications, $F\subseteq P$ is a lagrangian and
$V\subseteq P$ is a stably free half-rank direct summand
[KreckSurgeryAndDuality, p. 734]. The _kernel quasi-formation_ associated with
these data is defined as

$\Sigma(W,\overline{\nu},\omega,\varphi):=((P,\psi);F,V).$

The class of this quasi-formation in $\ell_{5}(\Lambda)$ only depends on the
$(B,\xi)$-bordism class rel. boundary of $(W,\overline{\nu})$
[KreckSurgeryAndDuality, Theorem 4]. This leads to the following definition
[KreckSurgeryAndDuality, p. 734]; see also [CrowleySixt, p. 494].

######  Definition 8.4.

Let $(W,\overline{\nu},M_{0},\overline{\nu}_{0},M_{1},\overline{\nu}_{1})$ be
a modified surgery problem. The _modified surgery obstruction_ of
$(W,\overline{\nu})$ is

$\Theta(W,\overline{\nu}):=[(\Sigma(W^{\prime},\overline{\nu}^{\prime},\omega^{\prime},\varphi^{\prime}))]\in\ell_{5}(\Lambda),$

where $(W^{\prime},\overline{\nu}^{\prime})$ is a normal $1$-smoothing over
$(B,\xi)$, which is $(B,\xi)$-bordant rel. boundary to $(W,\overline{\nu})$
and $(\omega^{\prime},\varphi^{\prime})$ is a system of embedded generators
for $K\pi_{2}(W^{\prime})$.

######  Remark 8.5.

The class $\Theta(W,\overline{\nu})\in\ell_{5}(\Lambda)$ only depends on the
$(B,\xi)$-bordism class rel. boundary of the $(B,\xi)$-null-bordism
$(W,\overline{\nu})$: it is independent of the choice of the $1$-smoothing
$(W^{\prime},\overline{\nu}^{\prime})$ and of the subsequent choice of a
system of embedded generators for $K\pi_{2}(W^{\prime})$
[KreckSurgeryAndDuality, Theorem 4].

######  Theorem 8.6 (Kreck [KreckSurgeryAndDuality, Theorem 4]).

Let $(W,\overline{\nu},M_{0},\overline{\nu}_{0},M_{1},\overline{\nu}_{1})$ be
a modified surgery problem. The modified surgery obstruction
$\Theta(W,\overline{\nu})\in\ell_{5}(\Lambda)$ is elementary if and only if
$(W,\overline{\nu})$ is $(B,\xi)$-bordant rel. boundary to an $s$-cobordism.

###  8.3. Wall forms

Let $(W^{5},\overline{\nu},M_{0},\overline{\nu}_{0},M_{1},\overline{\nu}_{1})$
be a modified surgery problem as in the previous section. We wish to narrow
down where in the monoid $\ell_{5}(\Lambda)$ the modified surgery obstruction
$\Theta(W,\overline{\nu})$ lives. For this we will show how to choose
nondegenerate quadratic forms $v$ and $v^{\prime}$, on free $\Lambda$-modules,
so that the obstruction lies in the subset
$\ell_{5}(v,v^{\prime})\subseteq\ell_{5}(\Lambda)$ described in Remark 7.20.

Write $(M^{4}_{0},\overline{\nu}_{0})$ for a normal $1$-smoothing into the
normal $1$-type $(B,\xi)$ of $M_{0}$ and recall that

$K\pi_{2}(M_{0}):=\ker((\overline{\nu}_{0})_{*}\colon\pi_{2}(M_{0})\to\pi_{2}(B)).$

Intersections and self-intersections in $M_{0}$ define a quadratic form
$(K\pi_{2}(M_{0}),\psi_{M_{0}})$ called the _Wall form_ of
$(M_{0},\overline{\nu}_{0})$ [KreckSurgeryAndDuality, Section 5]. Write
$\lambda_{M_{0}}$ for the symmetrisation of $\psi_{M_{0}}$ (this agrees with
the restriction of the equivariant intersection form on
$\pi_{2}(M_{0})=H_{2}(M_{0};\Lambda)$) and recall that the _radical_
$\operatorname{rad}(\lambda_{M_{0}})\subseteq K\pi_{2}(M_{0})$ of this pairing
consists by definition of those $x\in K\pi_{2}(M_{0})$ such that
$\lambda_{M_{0}}(x,-)\equiv 0$. We shorten
$K\pi_{2}(M_{0})/\operatorname{rad}(\lambda_{M_{0}})$ to
$K\pi_{2}(M_{0})/\operatorname{rad}$ from now on. The Hermitian form
$\lambda_{M_{0}}$ descends to a nondegenerate Hermitian form
$\lambda_{M_{0}}^{\operatorname{nd}}$ on this quotient and therefore
determines a quadratic form $\psi_{M_{0}}^{\operatorname{nd}}$ on this
quotient. The form
$(K\pi_{2}(M_{0})/\operatorname{rad},\psi_{M_{0}}^{\operatorname{nd}})$ is
called the _nondegenerate Wall form_.

We would like to use the nondegenerate Wall forms for $M_{0}$ and $M_{1}$ as
the forms $v$ and $v^{\prime}$ mentioned at the start of this subsection.
However, the modules $K\pi_{2}(M_{i})$ need not be free, or even stably free.
This is solved using the following idea from [KreckSurgeryAndDuality,
CrowleySixt].

######  Definition 8.7.

Let $(M_{0}^{4},\overline{\nu}_{0})$ be a normal $1$-smoothing into the normal
$1$-type $(B,\xi)$ of $M_{0}$. Given a free $\Lambda$-module $S$ and a
surjection $\varpi\colon S\twoheadrightarrow
K\pi_{2}(M_{0})/\operatorname{rad}$, the pull-back of the nondegenerate Wall
form $(K\pi_{2}(M_{0})/\operatorname{rad},\psi_{M_{0}}^{\operatorname{nd}})$
by $\varpi$ is called a free Wall form of $(M_{0},\overline{\nu}_{0})$ and is
denoted by

$(S,\vartheta):=(S,\varpi^{*}\psi_{M_{0}}^{\operatorname{nd}}).$

A consequence of [KreckSurgeryAndDuality, Proposition 8] is that all free Wall
forms of $(M_{0},\overline{\nu}_{0})$ are $0$-stably equivalent (Definition
7.17). The following proposition is also a consequence of
[KreckSurgeryAndDuality, Proposition 8 ii)]. Recall the map of monoids
$b\colon\ell_{5}(\Lambda)\to\mathcal{F}^{\operatorname{zs}}_{4}(\Lambda)\times\mathcal{F}^{\operatorname{zs}}_{4}(\Lambda)$
from Proposition 7.19, which takes a quasi-formation to its pair of induced
forms, considered up to 0-stable equivalence.

######  Proposition 8.8.

Let $(W,\overline{\nu},M_{0},\overline{\nu}_{0},M_{1},\overline{\nu}_{1})$ be
a modified surgery problem with modified surgery obstruction
$\Theta(W,\overline{\nu})\in\ell_{5}(\Lambda)$. For any choice of free Wall
forms $v(\overline{\nu}_{0})$ and $v(\overline{\nu}_{1})$ for
$(M_{0},\overline{\nu}_{0})$ and $(M_{1},\overline{\nu}_{1})$ respectively, we
have a $0$-stable equivalence:

$b(\Theta(W,\overline{\nu}))=(v(\overline{\nu}_{0})_{0},v(\overline{\nu}_{1})_{0})\in\mathcal{F}^{\operatorname{zs}}_{4}(\Lambda)\times\mathcal{F}^{\operatorname{zs}}_{4}(\Lambda)$

and hence

$\Theta(W,\overline{\nu})\in\ell_{5}(v(\overline{\nu}_{0}),v(\overline{\nu}_{1})).$

##  9\. Forms and quasi-formations over $\mathbb{Z}[\mathbb{Z}_{2}]$

As before we write $\mathbb{Z}_{2}=\langle T\mid T^{2}=1\rangle$. In this
section we specialise to the ring
$\Lambda:=\mathbb{Z}[\mathbb{Z}_{2}]=\mathbb{Z}[T]/\langle T^{2}-1\rangle$ and
develop some results concerning quasi-formations over this ring. The eventual
goal is to describe a tractable criterion guaranteeing that a class
$\Theta\in\ell_{5}(\mathbb{Z}[\mathbb{Z}_{2}])$ is elementary.

In Subsection 9.1 we use $\pm 1$ eigenspaces to associate two forms over
$\mathbb{Z}$ to a form over $\mathbb{Z}[\mathbb{Z}_{2}]$, called the _plus and
minus forms._ In Subsection 9.2, we use plus and minus forms to associate two
quasi-formations over $\mathbb{Z}$ to a quasi-formation over
$\mathbb{Z}[\mathbb{Z}_{2}]$. In Subsection 9.3, we use this construction to
prove the main technical ingredient. In Subsection 9.4, we deduce a criterion
ensuring that a class $\Theta\in\ell_{5}(\mathbb{Z}[\mathbb{Z}_{2}])$ is
elementary; see Proposition 9.14.

In this section we assume that $\mathbb{Z}[\mathbb{Z}_{2}]$-modules are
finitely generated and free; cf. Lemma 3.9.

###  9.1. Forms over $\mathbb{Z}[\mathbb{Z}_{2}]$ and eigenspaces

A $\mathbb{Z}[\mathbb{Z}_{2}]$-module can also be thought of as an abelian
group. Given a $\mathbb{Z}[\mathbb{Z}_{2}]$-module $P$, recall that we write
$\mathcal{E}_{\pm}(P)\subseteq P$ for the abelian subgroup given by the $(\pm
1)$-eigenspaces of $T$.

In this subsection we recall how a hermitian (resp. quadratic) form over
$\mathbb{Z}[\mathbb{Z}_{2}]$ induces a hermitian (resp. quadratic) form over
$\mathbb{Z}$ on $\mathcal{E}_{\pm}(P)$.

######  Remark 9.1.

Recall that $P$ is assumed to be free. In this case, we argue that
$\mathcal{E}_{\pm}(P)$ can equivalently be described as $(1\pm T)P$.
Certainly, we always have $(1\pm T)P\subseteq\mathcal{E}_{\pm}(P)$. For the
reverse inclusion, since $P$ is free we can work coordinate by coordinate,
thereby reducing to the case $P=\mathbb{Z}[\mathbb{Z}_{2}]$. Now if
$a+bT\in\mathbb{Z}[\mathbb{Z}_{2}]$ satisfies $T(a+bT)=\pm(a+bT)$, then
$T(a+bT)=b+Ta$, so $a=\pm b$.

######  Remark 9.2.

In the notation of Subsection 3.1, which concerned $\varepsilon$-quadratic
forms over a ring $R$, we have $R=\mathbb{Z}[\mathbb{Z}_{2}]$ and
$\varepsilon=1$. Note that $\overline{a+bT}=a+bT$ in
$\mathbb{Z}[\mathbb{Z}_{2}]$. It follows that
$Q_{+}(\mathbb{Z}[\mathbb{Z}_{2}])=\mathbb{Z}[\mathbb{Z}_{2}]$, and so
quadratic refinements of hermitian forms take values in
$\mathbb{Z}[\mathbb{Z}_{2}]$.

Given a quadratic form $(P,\lambda,\mu)$ over $\mathbb{Z}[\mathbb{Z}_{2}]$,
since $(1\pm T)\overline{(1\pm T)}=(1\pm T)^{2}=2(1\pm T)$, we have that
$\mu((1\pm T)x)=(1\pm T)\mu(x)\overline{(1\pm T)}=2(1\pm T)\mu(x)$ and
$\lambda((1\pm T)x,(1\pm T)y)=(1\pm T)\lambda(x,y)\overline{(1\pm T)}=2(1\pm
T)\lambda(x,y)$ for every $x,y\in P$. It follows that the restrictions of
$\lambda$ and $\mu$ to $\mathcal{E}_{\pm}(P)$ take values in
$2\mathcal{E}_{\pm}(\mathbb{Z}[\mathbb{Z}_{2}])\subseteq\mathcal{E}_{\pm}(\mathbb{Z}[\mathbb{Z}_{2}])$.
We identify $\mathbb{Z}\cong\mathcal{E}_{\pm}(\mathbb{Z}[\mathbb{Z}_{2}])$
using the isomorphism $n\mapsto(1\pm T)n$. We refer to the inverse as dividing
by $(1\pm T)$. Under this identification
$2\mathcal{E}_{\pm}(\mathbb{Z}[\mathbb{Z}_{2}])$ is identified with
$2\mathbb{Z}$. Dividing by $2$ gives rise to an isomorphism
$2\mathbb{Z}\to\mathbb{Z}$. We will use the composition

$\frac{1}{2(1\pm T)}\colon
2\mathcal{E}_{\pm}(\mathbb{Z}[\mathbb{Z}_{2}])\xrightarrow{/(1\pm
T)}2\mathbb{Z}\xrightarrow{/2}\mathbb{Z}$

in the following construction.

######  Construction 9.3.

Given a quadratic form $(P,\lambda,\mu)$ over $\mathbb{Z}[\mathbb{Z}_{2}]$, we
set

$\displaystyle\mathcal{E}_{\pm}(\lambda)$
$\displaystyle:=\mbox{\footnotesize$\displaystyle\frac{1}{2(1\pm
T)}$}\circ\lambda|\colon\mathcal{E}_{\pm}(P)\times\mathcal{E}_{\pm}(P)\to\mathbb{Z},$
$\displaystyle\mathcal{E}_{\pm}(\mu)$
$\displaystyle:=\mbox{\footnotesize$\displaystyle\frac{1}{2(1\pm
T)}$}\circ\mu|\colon\mathcal{E}_{\pm}(P)\to\mathbb{Z}.$

We write
$\mathcal{E}_{\pm}(P,\lambda,\mu):=(\mathcal{E}_{\pm}(P),\mathcal{E}_{\pm}(\lambda),\mathcal{E}_{\pm}(\mu))$.
Similarly if $\theta=[\rho]\in Q_{\varepsilon}(P)$ is a quadratic form, then
we write $\mathcal{E}_{\pm}(\theta)\in Q_{\varepsilon}(\mathcal{E}_{\pm}(P))$
for the quadratic forms determined by this construction, namely the
equivalence class represented by the pairing

$\mbox{\footnotesize$\displaystyle\frac{1}{2(1\pm
T)}$}\circ\rho|_{\mathcal{E}_{\pm}(P)\times\mathcal{E}_{\pm}(P)}\colon\mathcal{E}_{\pm}(P)\times\mathcal{E}_{\pm}(P)\to\mathbb{Z}.$

######  Remark 9.4.

Note that $\mathcal{E}_{\pm}(P)$ is naturally a
$\mathbb{Z}[\mathbb{Z}_{2}]$-module, where $T$ acts by multiplication by $\pm
1$. But $\mathcal{E}_{\pm}(P,\lambda,\mu)$ is not a quadratic form over
$\mathbb{Z}[\mathbb{Z}_{2}]$ because $\mathcal{E}_{\pm}(P)$ is not stably free
over $\mathbb{Z}[\mathbb{Z}_{2}]$.

######  Remark 9.5.

If $x,y\in\mathcal{E}_{\pm}(P)$, then $\lambda(x,y)=2(1\pm
T)\mathcal{E}_{\pm}(\lambda)(x,y)$.

The following proposition can be straightforwardly verified from the
definitions and the fact that $(P,\lambda,\mu)$ is a quadratic form.

######  Proposition 9.6.

The data $\mathcal{E}_{\pm}(P,\lambda,\mu)$ determines a quadratic form over
$\mathbb{Z}$. If $(P,\lambda,\mu)$ is nonsingular, then so is
$\mathcal{E}_{\pm}(P,\lambda,\mu)$.

######  Definition 9.7.

The _plus form_ and the _minus form_ of a quadratic form $(P,\lambda,\mu)$
over $\mathbb{Z}[\mathbb{Z}_{2}]$ are respectively the quadratic forms
$\mathcal{E}_{+}(P,\lambda,\mu)$ and $\mathcal{E}_{-}(P,\lambda,\mu)$ over
$\mathbb{Z}$.

######  Example 9.8.

Let $\psi\in Q_{+}(\mathbb{Z}[\mathbb{Z}_{2}]^{n})$ be a quadratic form. A
matrix representing the quadratic form $\mathcal{E}_{\pm}(\psi)\in
Q_{+}(\mathbb{Z}^{n})$ can be obtained by plugging in $T=\pm 1$ into a matrix
representing $\psi$.

In particular, given a quadratic form $\theta\in Q_{+}(\mathbb{Z}^{n})$, the
plus form of $(1-T)\theta\in Q_{+}(\mathbb{Z}[\mathbb{Z}_{2}]^{n})$ is zero,
while its minus form is $2\theta\in Q_{+}(\mathbb{Z}^{n})$. To see this, let
$\\{e_{i}\\}_{i=1}^{n}$ be the standard basis for
$\mathbb{Z}[\mathbb{Z}_{2}]^{n}$ and pick a sesquilinear form $\rho$
representing $\psi=[\rho]\in Q_{+}(\mathbb{Z}[\mathbb{Z}_{2}]^{n})$. Then
$\\{(1\pm T)e_{i}\\}_{i=1}^{n}$ is a basis for
$\mathcal{E}_{\pm}(\mathbb{Z}[\mathbb{Z}_{2}]^{n})$. For
$p\in\mathbb{Z}[\mathbb{Z}_{2}]$ we write $p(\pm 1)$ for the evaluation of $p$
at $\pm 1$. Similarly to above we have

$\rho((1\pm T)e_{i},(1\pm T)e_{j})=(1\pm T)\rho(e_{i},e_{j})\overline{(1\pm
T)}=2(1\pm T)\rho(e_{i},e_{j})=2(1\pm T)\big{(}\rho(e_{i},e_{j})(\pm
1)\big{)}.$

Therefore the $(i,j)$ entry of a matrix representing $\mathcal{E}_{\pm}(\rho)$
is

$\mathcal{E}_{\pm}(\theta)((1\pm T)e_{i},(1\pm
T)e_{j})=\mbox{\footnotesize$\displaystyle\frac{1}{2(1\pm T)}$}\Big{(}2(1\pm
T)\big{(}\rho(e_{i},e_{j})(\pm 1)\big{)}\Big{)}=\rho(e_{i},e_{j})(\pm 1),$

which arises by evaluating as asserted.

###  9.2. Plus and minus quasi-formations

Given a quasi-formation over $\mathbb{Z}[\mathbb{Z}_{2}]$, we describe how
plus and minus forms can be used to produce two quasi-formations over
$\mathbb{Z}$.

First of all, observe that if $i\colon V\hookrightarrow P$ is a split
injection then the induced map
$i_{\pm}\colon\mathcal{E}_{\pm}(V)\to\mathcal{E}_{\pm}(P)$ is also a split
injection.

######  Proposition 9.9.

If $x:=((P,\psi);F,V)$ is a quasi-formation over $\mathbb{Z}[\mathbb{Z}_{2}]$,
then the triple

$\mathcal{E}_{\pm}(x)=(\mathcal{E}_{\pm}(P,\psi);\mathcal{E}_{\pm}(F),\mathcal{E}_{\pm}(V))$

is a quasi-formation over $\mathbb{Z}$.

###### Proof.

First observe (e.g. using Example 9.8) that $\mathcal{E}_{\pm}(P,\psi)$ is
hyperbolic over $\mathbb{Z}$. Next we observe that since $F,V\subseteq P$ are
summands, we have inclusions of summands
$\mathcal{E}_{\pm}(F),\mathcal{E}_{\pm}(V)\subseteq\mathcal{E}_{\pm}(P)$.
Finally, one verifies that
$\mathcal{E}_{\pm}(F)\subseteq\mathcal{E}_{\pm}(P,\psi)$ is a lagrangian and
$\mathcal{E}_{\pm}(V)$ is a half rank direct summand. ∎

Using $(V,\theta)$ and $(V^{\perp},-\theta^{\perp})$ to denote the induced
forms of a quasi-formation $((P,\psi);F,V)$ (Definition 7.15), note that the
induced form of $\mathcal{E}_{\pm}(P,\psi)$ on $\mathcal{E}_{\pm}(V)$ and
$\mathcal{E}_{\pm}(V^{\perp})$ coincide with $\mathcal{E}_{\pm}(\theta)$ and
$\mathcal{E}_{\pm}(-\theta^{\perp})$ respectively.

######  Proposition 9.10.

The following assignments define monoid maps

$f_{\pm}\colon\ell_{5}(\mathbb{Z}[\mathbb{Z}_{2}])\to\ell_{5}(\mathbb{Z}),[x]\mapsto[\mathcal{E}_{\pm}(x)].$

###### Proof.

Since $f_{\pm}$ maps trivial formations to trivial formations, we deduce that
it preserves stable isomorphisms. The reader can then verify that $f_{\pm}$
maps quasi-formations of the form $((P,\psi);F,V)$ and
$((P,\psi);F,G)\oplus((P,\psi);G,V)$ (which are equivalent in
$\ell_{5}(\mathbb{Z}[\mathbb{Z}_{2}]$) to equivalent quasi-formations in
$\ell_{5}(\mathbb{Z})$. Additivity follows as well. ∎

In what follows, we will often write
$\mathcal{E}_{\pm}(\Theta)=[\mathcal{E}_{\pm}(x)]$ instead of
$f_{\pm}(\Theta)$ and refer to $\mathcal{E}_{+}(x)$ and $\mathcal{E}_{-}(x)$
as the _plus and minus quasi-formations_ of $x$.

######  Remark 9.11.

Given quadratic forms $v,v^{\prime}$ over $\mathbb{Z}[\mathbb{Z}_{2}]$, the
monoid map $f_{\pm}$ from Proposition 9.9 restricts to a map
$f_{\pm}\colon\ell_{5}(v,v^{\prime})\to\ell_{5}(\mathcal{E}_{\pm}(v),\mathcal{E}_{\pm}(v^{\prime}))$.
This follows from the observation that taking induced forms commutes with
taking $\pm$-forms/formations i.e. the following diagram commutes:

${\ell_{5}(\mathbb{Z}[\mathbb{Z}_{2}])}$${\ell_{5}(\mathbb{Z})}$${\mathcal{F}^{\operatorname{zs}}_{4}(\mathbb{Z}[\mathbb{Z}_{2}])\times\mathcal{F}^{\operatorname{zs}}_{4}(\mathbb{Z}[\mathbb{Z}_{2}]))}$${\mathcal{F}^{\operatorname{zs}}_{4}(\mathbb{Z})\times\mathcal{F}^{\operatorname{zs}}_{4}(\mathbb{Z})).}$$\scriptstyle{f_{\pm}}$$\scriptstyle{b}$$\scriptstyle{b}$$\scriptstyle{f_{\pm}}$

###  9.3. Finding a lagrangian complement

The following lemma is a technical ingredient towards establishing our
criterion for showing that a quasi-formation over
$\Lambda=\mathbb{Z}[\mathbb{Z}_{2}]$ is elementary. The lemma will justify
that, for this purpose, it suffices to understand the plus and minus quasi-
formations. A sketch was given on [KreckOnTheHomeomorphism, p. 70]. We follow
the same overall strategy, but the proof given below is considerably more
detailed, and includes arguments not found in Kreck’s sketch.

######  Lemma 9.12.

Suppose $(P,\lambda)$ is a nonsingular, even, hermitian form on a free
$\Lambda$-module $P$, with a free half-rank direct summand $V\subseteq P$.
Suppose that $\mathcal{E}_{+}(V)$ is a lagrangian for
$(\mathcal{E}_{+}(P),\mathcal{E}_{+}(\lambda))$ and that
$\mathcal{E}_{-}(V)\subseteq\mathcal{E}_{-}(P)$ admits a lagrangian
complement. Then $V\subseteq P$ admits a lagrangian complement.

###### Proof.

We will frequently use the following fact. Let $U$ be a complementary free
direct summand to $V$ in $P$. Let $\\{u_{i}\\}$ be a basis for $U$ and let
$\\{v_{i}\\}$ be a basis for $V$. Add linear combinations of basis elements of
$V$ to the basis elements for $U$, to obtain a set
$\\{u_{i}+\sum_{j}a_{ij}v_{j}\\}$, for some $a_{ij}\in\Lambda$. An elementary
linear algebra argument shows that this set is a basis for a new complementary
direct summand $U^{\prime}$ to $V$ in $P$, for any $a_{ij}\in\Lambda$.

Now we begin the proof. Choose a basis $x_{1},\dots,x_{n}$ for $V$ and
complete it to a basis of $P$ via $v_{1},\ldots,v_{n}$. Since $P$ is free we
may consider the dual basis elements $x_{i}^{*}\in P^{*}$ for $i=1,\ldots,n$
that satisfy $x_{i}^{*}(x_{j})=\delta_{ij}$ and $x^{*}_{i}(v_{j})=0$ for every
$i,j$. Since the form $\lambda$ is nonsingular, we can choose
$u_{1},\ldots,u_{n}\in P$ such that $\widehat{\lambda}(u_{i})=x_{i}^{*}$. In
particular, we have $\lambda(x_{i},u_{j})=\delta_{ij}$ for every $i,j$.

We assert that the $x_{j}$ and $u_{j}$ generate $P$. We will prove this by
showing that each $v_{i}$ can be expressed as a linear combination of the
$u_{j}$ and $x_{j}$. Write $u_{i}=\sum_{k}a_{ik}x_{k}+\sum_{l}b_{il}v_{l}$ for
some $a_{ik},b_{il}\in\Lambda$. Arrange the $u_{i},x_{k},v_{l}$ as column
vectors $\mathbf{u},\mathbf{x},\mathbf{v}$ and the $a_{ij},b_{il}$ as matrices
$A,B$ so that, formally, we can write $\mathbf{u}=A\mathbf{x}+B\mathbf{v}$.
Since we have $\lambda(u_{i},x_{j})=\delta_{ij}$ and since the form
$\mathcal{E}_{+}(\lambda)$ vanishes on
$\mathcal{E}_{+}(V)\times\mathcal{E}_{+}(V)$, we obtain

$\delta_{ij}=\lambda(u_{i},x_{j})=\mathcal{E}_{+}(\lambda)((1+T)u_{i},(1+T)x_{j})=\sum_{l=1}^{n}b_{il}\cdot\mathcal{E}_{+}(\lambda)((1+T)v_{l},(1+T)x_{j}).$

The second equality here uses that $\lambda(u_{i},x_{j})\in\mathbb{Z}$. Write
$L$ for the matrix with coefficients
$L_{lj}=\mathcal{E}_{+}(\lambda)((1+T)v_{l},(1+T)x_{j})$ and $I_{n}$ for the
size $n$ identity matrix so that $I_{n}=BL$. A square matrix with coefficients
in a commutative ring admits an inverse if and only if its determinant is a
unit. That inverse is both a left and a right inverse. The equalities
$1=\det(I_{n})=\det(BL)=\det(B)\det(L)$ imply that $\det(B)$ is a unit. We
deduce that $B$ is invertible with inverse $L$. Therefore
$\mathbf{u}=A\mathbf{x}+B\mathbf{v}$ can be rewritten as
$\mathbf{v}=L(\mathbf{u}-A\mathbf{x})$. Thus the $v_{i}$ can be expressed as
linear combinations of the $u_{j}$ and $x_{j}$. This concludes the proof of
the assertion the $x_{j}$ and $u_{j}$ generate $P$.

Since $\\{x_{j}\\}\cup\\{u_{j}\\}$ form a rank $n$ generating set of
$P\cong\Lambda^{n}$, they are also linearly independent. This statement
follows from the general fact, which we now prove, that a surjective
$\Lambda$-linear map $f\colon\Lambda^{n}\to\Lambda^{n}$ is necessarily
injective. Let $A$ be a matrix for $f$ with respect to the canonical basis
$e_{1},\ldots,e_{n}$ of $\Lambda^{n}$. Surjectivity implies that there are
column vectors $b_{1},\ldots,b_{n}$ such that $f(b_{i})=e_{i}$. Arrange the
column vectors $b_{i}$ into a matrix $B$ such that $AB=I_{n}$. Taking
determinants, as above we deduce that $f$ is also injective, as claimed.

Since $\\{x_{j}\\}\cup\\{u_{j}\\}$ forms a basis of $P$, it follows that the
$u_{i}$ span a half rank direct summand $U_{0}\subseteq P$ with $P=V\oplus
U_{0}$ and $\lambda(x_{i},u_{j})=\delta_{ij}$.

Let $U_{-}\subseteq\mathcal{E}_{-}(P)$ be the hypothesised complementary
lagrangian to $\mathcal{E}_{-}(V)\subseteq\mathcal{E}_{-}(P)$. First, we will
modify $U_{0}$ to obtain a direct summand $U_{1}\subseteq P$ that is
complementary to $V$ and such that $\mathcal{E}_{-}(U_{1})=U_{-}$. Then we
will improve $U_{1}$ further, to obtain a lagrangian complement $U_{2}$ for
$V$.

Write $b_{i}:=(1-T)x_{i}$ for the basis of $\mathcal{E}_{-}(V)$ obtained from
$\\{x_{i}\\}$. Recall that we have a basis $u_{1},\dots,u_{n}$ for $U_{0}$ and
that $\\{x_{1},\dots,x_{n},u_{1},\dots,u_{n}\\}$ is a basis for $P$. Write
$(1-T)u_{i}=w_{i}+z_{i}\in\mathcal{E}_{-}(U_{0})$ with respect to the
decomposition $\mathcal{E}_{-}(P)\cong\mathcal{E}_{-}(V)\oplus U_{-}$, i.e.
$w_{i}\in\mathcal{E}_{-}(V)$ and $z_{i}\in U_{-}$ for each $i$. Write
$w_{i}=\sum_{j}n_{ij}b_{j}$ for some $n_{ij}\in\mathbb{Z}$. Define
$U_{1}\subseteq P$, a new complementary direct summand to $V$, as the
submodule with basis given by the elements

$y_{i}:=u_{i}-\sum_{j}n_{ij}x_{j}.$

Here we used the fact from the first paragraph of the proof to see that
$U_{1}$ is a direct complement to V.

To confirm that $\mathcal{E}_{-}(U_{1})=U_{-}$, first note that
$\\{(1-T)y_{i}\\}$ forms a basis for $\mathcal{E}_{-}(U_{1})$, because
$\\{y_{i}\\}$ forms a basis for $U_{1}$. Next, note that
$\\{(1-T)x_{1},\dots,(1-T)x_{n},(1-T)u_{1},\dots,(1-T)u_{n}\\}=\\{b_{1},\dots,b_{n},w_{1}+z_{1},\dots,w_{n}+z_{n}\\}$
is a basis for
$\mathcal{E}_{-}(P)=\mathcal{E}_{-}(V)\oplus\mathcal{E}_{-}(U_{0})$, and
therefore so is $\\{b_{1},\dots,b_{n},z_{1},\dots,z_{n}\\}$, because
$w_{i}\in\mathcal{E}_{-}(V)$ for all $i$. Therefore
$\\{b_{1},\dots,b_{n},z_{1},\dots,z_{n}\\}$ is a basis for
$\mathcal{E}_{-}(V)\oplus U_{-}\cong\mathcal{E}_{-}(P)$, and we deduce that
$\\{z_{i}\\}$ forms a basis for $U_{-}$. Now we compute that

$(1-T)y_{i}=(1-T)u_{i}-\textstyle{\sum_{j}}n_{ij}(1-T)x_{j}=w_{i}+z_{i}-\textstyle{\sum_{j}}n_{ij}b_{j}=z_{i}.$

It follows that $\mathcal{E}_{-}(U_{1})=U_{-}$ as asserted. This completes the
$U_{1}$ step. Note that $U_{1}$ need not be a lagrangian, so the proof is not
yet complete.

At this point, we consider the following subspaces with their respective bases
shown underneath each subspace:

$\begin{array}[]{ccccccc}\mathcal{E}_{+}(V)&&\mathcal{E}_{-}(V)&&U_{+}:=\mathcal{E}_{+}(U_{1})&&U_{-}=\mathcal{E}_{-}(U_{1})\\\
\\{a_{i}:=(1+T)x_{i}\\}&&\\{b_{i}:=(1-T)x_{i}\\}&&\\{t_{i}^{+}:=(1+T)y_{i}\\}&&\\{t_{i}^{-}:=(1-T)y_{i}\\}.\end{array}$

Next we modify the bases $t^{\pm}_{i}$ to something more favourable. Consider
the $\mathbb{Z}$-valued forms $\mathcal{E}_{\pm}(\lambda)$ on
$\mathcal{E}_{\pm}(P)=\mathcal{E}_{\pm}(V)\oplus{U}_{\pm}$; this decomposition
holds because $P=V\oplus U_{1}$. With respect to this decomposition, these
forms are represented by block diagonal matrices over $\mathbb{Z}$, say
$\mathcal{E}_{\pm}(\lambda)=\left(\begin{smallmatrix}A_{\pm}&Y_{\pm}\\\
Y_{\pm}^{T}&B_{\pm}\end{smallmatrix}\right)$. Observe for later use that
$\mathcal{E}_{+}(\lambda)|_{U_{+}\times U_{+}}$ is even. To see this note that
$\lambda(y_{i},y_{i})\in 2\Lambda$ because $\lambda$ is even and
$p+\overline{p}\in 2\Lambda$ for every $p\in\Lambda$. Therefore

$\mathcal{E}_{+}(\lambda)(t_{i}^{+},t_{i}^{+})=\frac{1}{2(1+T)}\lambda((1+T)y_{i},(1+T)y_{i})=\varepsilon(\lambda(y_{i},y_{i}))\in
2\mathbb{Z},$

where the last equality uses $\lambda(y_{i},y_{i})\in 2\Lambda$ and
$\varepsilon\colon\Lambda\to\mathbb{Z}$ denotes the augmentation map, defined
by $\varepsilon(a+bT)=a+b$.

In the “$+$” case, $A_{+}=0$ because $\mathcal{E}_{+}(V)$ is a lagrangian.
Thus, since $\mathcal{E}_{+}(\lambda)$ is nonsingular, we deduce that $Y_{+}$
is nonsingular. Using column operations, $Y_{+}$ can be made into the identity
matrix. This corresponds to performing elementary basis change operations on
the basis $\\{t_{i}^{+}\\}$ of ${U}_{+}$ (using only linear combinations of
the $t_{i}^{+}$). We denote the resulting basis of $U_{+}$ by
$c_{1},\dots,c_{n}$. Note that
$\mathcal{E}_{+}(\lambda)(a_{i},c_{j})=\delta_{ij}$. Since $c_{i}\in
U_{+}\subseteq P$ and $a_{i}\in\mathcal{E}_{+}(V)\subseteq P$, we have that
$\lambda(a_{i},c_{j})=2(1+T)\delta_{ij}$ (recall Remark 9.5). In fact, in the
$+$-case, since $\mathcal{E}_{+}(\lambda)|_{U_{+}\times U_{+}}$ is even, by
adding linear combinations of the $a_{i}$, we can assume that
$\mathcal{E}_{+}(\lambda)(c_{i},c_{j})=0$ for all $i,j$. Note that $U_{+}$ may
not satisfy $\mathcal{E}_{+}(U_{1})=U_{+}$ anymore, since we changed the basis
for $U_{+}\subseteq\mathcal{E}_{+}(U_{1})$ but did not make the corresponding
basis change for $U_{1}\subseteq P$, but this does not affect the remainder of
the argument.

In the “$-$” case, $B_{-}=0$ because $U_{-}$ is a lagrangian. Thus, since
$\mathcal{E}_{-}(\lambda)$ is nonsingular, we deduce that $Y_{-}$ is
nonsingular. Arguing as in the “$+$” case, we perform elementary basis change
operations on $t_{i}^{-}$ to obtain a basis $d_{1},\dots,d_{n}$ of ${U}_{-}$
such that $\lambda(b_{i},d_{j})=2(1-T)\delta_{ij}$, again because $d_{j}\in
U_{-}\subseteq P$ and $b_{i}\in\mathcal{E}_{-}(V)\subseteq P$ (recall Remark
9.5).

We summarise the outcome. With respect to the basis
$\\{a_{1},\ldots,a_{n},c_{1},\ldots,c_{n}\\}$ of $\mathcal{E}_{+}(P)$ and
$\\{b_{1},\ldots,b_{n},d_{1},\ldots,d_{n}\\}$ of $\mathcal{E}_{-}(P)$,
matrices for $\mathcal{E}_{+}(\lambda)$ and $\mathcal{E}_{-}(\lambda)$) are
given by:

$\mathcal{E}_{+}(\lambda)=\begin{pmatrix}0&I\\\ I&0\end{pmatrix}\ \
\quad\text{and}\quad\ \ \mathcal{E}_{-}(\lambda)=\begin{pmatrix}*&I\\\
I&0\end{pmatrix}.$

We claim that, for all $i,j$, we have:

(23)
$\lambda(a_{i},a_{j})=0,\,\,\lambda(a_{i},c_{j})=2(1+T)\delta_{ij},\,\,\lambda(b_{i},d_{j})=2(1-T)\delta_{ij},\text{
and }\lambda(c_{i},c_{j})=0=\lambda(d_{i},d_{j}).$

An argument is needed to see that the first and the last two equalities hold.
We will just argue for $\lambda(c_{i},c_{j})=0$; the others proceed similarly.
Since $\mathcal{E}_{+}(\lambda)(c_{i},c_{j})=0$, then $\lambda(c_{i},c_{j})\in
2\mathcal{E}_{+}(\Lambda)$ maps to zero under the identifications
$2\mathcal{E}_{+}(\Lambda)\cong 2\mathbb{Z}\cong\mathbb{Z}$, and therefore
$\lambda(c_{i},c_{j})=0$.

Now, using the same elementary basis change operations we just used to change
$t_{i}^{-}$ into $d_{i}$, transform the basis $\\{y_{i}\\}$ of $U_{1}$ into a
new basis $\\{\rho^{\prime}_{i}\\}$ of $U_{{1}}$. For this we take the basis
changes over $\mathbb{Z}$ and consider them as basis changes over $\Lambda$
via the ring homomorphism $\mathbb{Z}\to\Lambda$. By construction we also have
$(1-T)\rho^{\prime}_{i}=d_{i}$. Write
$(1+T)\rho^{\prime}_{i}=\alpha_{i}+\gamma_{i}\in\mathcal{E}_{+}(V)\oplus
U_{+}$ for some $\alpha_{i}\in\mathcal{E}_{+}(V)$ and $\gamma_{i}\in U_{+}$,
and note that

$\rho^{\prime}_{i}=\frac{\alpha_{i}+\gamma_{i}+d_{i}}{2}.$

Next we claim we can make a further helpful basis change from the
$\rho_{i}^{\prime}$ to a new basis for $U_{1}$, given by

$\rho_{i}=\frac{\alpha_{i}+c_{i}+d_{i}}{2}.$

We must prove $\alpha_{i}+c_{i}+d_{i}$ is indeed divisible by $2$ and that the
resulting $\rho_{i}$ form a basis, then after that we will modify $U_{1}$ to
obtain the desired lagrangian complement $U_{2}$ to $V$.

To begin proving the claim that $\alpha_{i}+c_{i}+d_{i}$ is divisible by $2$
and that $\\{\rho_{i}\\}$ forms a basis for $U_{1}$, first observe that since
the $\\{c_{i}\\}$ form a $\mathbb{Z}$-basis of $U_{+}$, for each $j$ we can
write $\gamma_{j}=\sum_{k}\gamma_{jk}c_{k}$ with $\gamma_{jk}\in\mathbb{Z}$.
It is also the case that the $c_{k}\in\mathcal{E}_{+}(P)$ can be written as
$c_{k}=(1+T)z_{k}$, for some $z_{k}\in P$. We note for later use that

(24)
$2\lambda(x_{i},c_{k})=\lambda(x_{i},(1+T)(1+T)z_{k})=\lambda((1+T)x_{i},(1+T)z_{k})=\lambda(a_{i},c_{k}).$

Similarly $2\lambda(x_{i},\alpha_{j})=\lambda(a_{i},\alpha_{j})$ and
$2\lambda(x_{i},(1-T)\rho^{\prime}_{j})=\lambda(b_{i},(1-T)\rho^{\prime}_{j})$.
Using (24), we observe that

$\displaystyle 4\lambda(x_{i},\rho^{\prime}_{j})$
$\displaystyle=2\lambda(x_{i},\alpha_{j})+2\lambda(x_{i},\gamma_{j})+2\lambda(x_{i},d_{j})$
$\displaystyle=2\lambda(x_{i},\alpha_{j})+2\sum_{k}\lambda(x_{i},\gamma_{jk}c_{k})+2\lambda(x_{i},(1-T)\rho^{\prime}_{j})$
$\displaystyle=\lambda(a_{i},\alpha_{j})+\sum_{k}\lambda(a_{i},\gamma_{jk}c_{k})+\lambda(b_{i},(1-T)\rho^{\prime}_{j})$
$\displaystyle=0+2(1+T)\gamma_{ji}+\lambda(b_{i},d_{j})$
$\displaystyle=2(1+T)\gamma_{ji}+2(1-T)\delta_{ij}\in\Lambda.$

For the last two equalities we applied (23). Since $\Lambda$ is torsion-free
as an abelian group, we deduce that

(25) $2\lambda(x_{i},\rho^{\prime}_{j})=(1+T)\gamma_{ji}+(1-T)\delta_{ij}.$

Write $\lambda(x_{i},\rho^{\prime}_{j})=r_{ij}+s_{ij}(1-T)$ with
$r_{ij},s_{ij}\in\mathbb{Z}$; this is possible because $a+bT=a+b-(1-T)b$ for
every $a+bT\in\mathbb{Z}[\mathbb{Z}_{2}].$ Plugging in $T=1$ into (25) gives
$r_{ij}=\gamma_{ji}$. It follows that setting $T=-1$ into (25), we get
$2\delta_{ij}=2\gamma_{ji}+4s_{ij}$, which we write as
$2s_{ij}=\delta_{ij}-\gamma_{ji}$. Now define

$\rho_{i}:=\rho^{\prime}_{i}+\sum_{k}s_{ki}c_{k}=\frac{\alpha_{i}+d_{i}}{2}+\frac{1}{2}\gamma_{i}+\frac{1}{2}\sum_{k}(\delta_{ki}-\gamma_{ik})c_{k}=\frac{\alpha_{i}+d_{i}}{2}+\frac{1}{2}\sum_{k}\delta_{ki}c_{k}=\frac{\alpha_{i}+c_{i}+d_{i}}{2}.$

Since $c_{i}\in U_{+}=\mathcal{E}_{+}(U_{1})\subseteq U_{1}$ for
$i=1,\ldots,n$ and $\\{\rho^{\prime}_{1},\ldots,\rho^{\prime}_{n}\\}$ forms a
basis of $U_{1}$, it follows that $\\{\rho_{1},\ldots,\rho_{n}\\}$ again forms
a basis for $U_{1}$. This concludes the proof of the claim.

Next we observe some symmetries in the coefficients of
$\alpha_{i}\in\mathcal{E}_{+}(V)$ with respect to the $\mathbb{Z}$-basis
$a_{i}$ of $\mathcal{E}_{+}(V)$. Write

$\alpha_{i}=\sum_{j}\alpha_{ij}a_{j}$

for some $\alpha_{ij}\in\mathbb{Z}$. Recall that
$\lambda(\alpha_{i},\alpha_{j})=0$, $\lambda(c_{i},c_{j})=0$,
$\lambda(d_{i},d_{j})=0$, and $\lambda(a_{i},c_{j})=2(1+T)\delta_{ij}$, by
(23). We use these and the fact that
$\lambda|_{\mathcal{E}_{+}(P)\times\mathcal{E}_{-}(P)}=0$, which follows from
the equality $(1+T)(1-T)=0$, to compute that

$\displaystyle\lambda(\rho_{i},\rho_{j})$
$\displaystyle=\textstyle{\frac{1}{4}}\lambda(\alpha_{i}+c_{i}+d_{i},\alpha_{j}+c_{j}+d_{j})$
$\displaystyle=\textstyle{\frac{1}{4}}\left(\lambda(\alpha_{i},c_{j})+\lambda(c_{i},\alpha_{j})\right)$
$\displaystyle=\textstyle{\frac{1}{4}}\left(\lambda\left(\sum_{k}\alpha_{ik}a_{k},c_{j}\right)+\lambda\left(c_{i},\sum_{s}\alpha_{js}a_{s}\right)\right)$
$\displaystyle=\textstyle{\frac{1}{4}}\cdot
2(1+T)\cdot\left(\alpha_{ij}+\alpha_{ji}\right)$
$\displaystyle=\textstyle{\frac{1}{2}}(1+T)\cdot\left(\alpha_{ij}+\alpha_{ji}\right).$

Because $\lambda(\rho_{i},\rho_{j})\in\Lambda$ and $\alpha_{ij}\in\mathbb{Z}$,
we deduce that $\alpha_{ij}+\alpha_{ji}\in 2\mathbb{Z}.$

We now modify the summand $U_{1}\subseteq P$ into a lagrangian complement to
$V$. As $V_{+}\subseteq V$, so $a_{j}\in V$ for all $j$. Therefore if we
modify $\rho_{i}$ by adding multiples of $a_{j}$ we will obtain a new direct
summand of $P$ that is still complementary to $V$. Of course, the objective is
to do this so that the resulting direct summand is self-annihilating. Consider
that for $p\in\Lambda$ we have

$\rho_{i}+pa_{j}=\frac{\alpha_{i}+2pa_{j}+c_{i}+d_{i}}{2}$

so the proposed operation modifies the coefficient $\alpha_{ij}$ of $a_{j}$ in
$\alpha_{i}$ by the addition of $2p$. Using this operation, and the symmetries
in the $\alpha_{ij}$ (namely $\alpha_{ij}+\alpha_{ji}\in 2\mathbb{Z}$), we may
now modify the coefficients $\alpha_{ij}$ to $\alpha_{ij}^{\prime}$, which
satisfy that $\alpha^{\prime}_{ij}+\alpha^{\prime}_{ji}=0$ for all $i,j$. We
obtain a new direct summand $U_{2}$ of $P$, which is still complementary to
$V$ by the first paragraph of this proof. We write $\rho_{i}^{\prime}$ for the
modified basis element coming from $\rho_{i}$. The same computation as we made
for $\lambda(\rho_{i},\rho_{j})$ above now shows that
$\lambda(\rho^{\prime}_{i},\rho^{\prime}_{j})=0$ for all $i,j$ (here it is
important that $\lambda(a_{i},a_{j})=0$ for all $i,j$), and thus $U_{2}$ is a
complementary sublagrangian to $V$, i.e. $U_{2}\subseteq U_{2}^{\perp}$. Since
$U_{2}\subseteq P$ is a half rank sublagrangian and $\lambda$ is nonsingular,
it must in fact be a lagrangian; see e.g. [RanickiAlgebraicAndGeometric,
Proposition 11.53]. We have therefore found a complementary lagrangian to
$V\subseteq P$. ∎

###  9.4. A criterion for being elementary in
$\ell_{5}(\mathbb{Z}[\mathbb{Z}_{2}])$

We now prove the promised criterion, which ensures that a class
$\Theta\in\ell_{5}(\mathbb{Z}[\mathbb{Z}_{2}])$ is elementary.

######  Lemma 9.13.

If $\Theta\in\ell_{5}(\mathbb{Z}[\mathbb{Z}_{2}])$ is such that
$\mathcal{E}_{-}(\Theta)$ is elementary, then $\Theta$ is represented by a
quasiformation $x^{\prime}=((P^{\prime},\psi^{\prime});F^{\prime},V^{\prime})$
such that $\mathcal{E}_{-}(V^{\prime})\subseteq\mathcal{E}_{-}(P^{\prime})$
admits a lagrangian complement.

###### Proof.

Let $x=((P,\psi);F,V)$ be a representative of $\Theta$. Since