2.1 Preliminaries

In this section we will collect a few lemmas which are well known for ordinary ‘symmetric forms’ (that is, pairs $(M, \lambda)$ , where M is an abelian group and $\lambda : M \times M \rightarrow \mathbb{Z}$ is a symmetric bilinear function) (see, for example, [Reference Milnor and HusemollerMH73]) and can be directly applied in the extended setting (see [Nagy21, Lemmas 4.1.10–13]). But first, we recall the following elementary lemma, which we will often use to check that a subgroup of an abelian group is a direct summand.

1. (i) B is a direct summand in A and $\mathop{\mathrm{Tor}}\nolimits A \leqslant B$ ;

2. (ii) $A / B$ is free;

3. (iii) for every $a \in A$ and $k \in \mathbb{Z} \setminus \{ 0 \}$ , if $ka \in B$ then $a \in B$ .

For any subset X, the subgroup $X^{\perp}$ contains $\mathop{\mathrm{Tor}}\nolimits M$ and it is a direct summand in M. Moreover, if X is a subgroup of M, then $\mathop{\mathrm{rk}}\nolimits X + \mathop{\mathrm{rk}}\nolimits X^{\perp} = \mathop{\mathrm{rk}}\nolimits M$ . Therefore, if X is a subset of M such that $\mu \big| _X = 0$ , then X is a T-Lagrangian if and only if $X = X^{\perp}$ .

1. (a) $\mathop{\mathrm{ad}}\nolimits \lambda$ defines an isomorphism $Y^{\perp} \cong X^*$ ;

2. (b) $M = X^{\perp} \oplus Y^{\perp}$ (as an internal direct sum);

3. (c) and if $X \leqslant X^{\perp}$ (hence, $X \oplus Y^{\perp}$ is a subgroup of M by (b)), then $\lambda \big| _{(X \oplus Y^{\perp}) \times (X \oplus Y^{\perp})}$ is non-singular.

Lemma 2.9. Let $\underline{M} = (M, \lambda, \mu)$ be a free metabolic form of rank 2k, and let $L \leqslant M$ be a Lagrangian. Then M has a basis $e_1, e_2, \ldots , e_k, f_1, f_2, \ldots , f_k$ such that $L = \left\langle e_1, e_2, \ldots , e_k \right\rangle$ , and in this basis $\lambda$ is given by the block matrix

\[\begin{bmatrix}0\;\;\;\; & I_k \\I_k\;\;\;\; & D\end{bmatrix} \]

where $I_k$ is the $k \times k$ identity matrix and D is a diagonal matrix such that each of its entries is either 0 or 1.

2.2 Basic constructions

Definition 2.10. Let Q and N be abelian groups, let $\underline{M} = (M, \lambda, \mu)$ be an extended quadratic form over Q, and let $h : N \rightarrow M$ be a homomorphism. The pullback of $\underline{M}$ by h is

\[h^*(\underline{M}) = (N, h^*\lambda, h^*\mu) \]

where $h^*\lambda(x,y) = \lambda(h(x),h(y))$ and $h^*\mu = \mu \circ h$ ; this is also an extended quadratic form over Q.

If $\underline{M} = (M, \lambda_M, \mu_M)$ and $\underline{N} = (N, \lambda_N, \mu_N)$ , then $\underline{N} = h^*(\underline{M})$ if and only if $\lambda_M(h(x),h(y)) = \lambda_N(x,y)$ and $\mu_M(h(x)) = \mu_N(x)$ for every $x,y \in N$ .

Composition of morphisms can be defined in the obvious way, and extended quadratic forms over a fixed group Q together with the above-defined morphisms form a category. In this category a morphism $h : \underline{N} \rightarrow \underline{M}$ is an isomorphism if and only if the underlying homomorphism $h : N \rightarrow M$ is an isomorphism.

Since the third component of $\underline{H}_{2k}$ is zero, it can be regarded as an extended quadratic form over any abelian group Q.

1. (i) $\underline{M}$ is hyperbolic;

2. (ii) $\underline{M}$ is free, metabolic, even and $\mu = 0$ ;

3. (iii) $\underline{M}$ is isomorphic to $\underline{H}_{2k}$ , where $2k = \mathop{\mathrm{rk}}\nolimits \underline{M}$ .

Definition 2.15. Let $\underline{M} = (M, \lambda_M, \mu_M)$ and $\underline{N} = (N, \lambda_N, \mu_N)$ be extended quadratic forms over some abelian group Q. Their direct sum is

\[\underline{M} \oplus \underline{N} = (M \oplus N, \lambda_M \oplus \lambda_N, \mu_M \oplus \mu_N) \]

where $(\lambda_M \oplus \lambda_N)((x_1,y_1),(x_2,y_2)) = \lambda_M(x_1,x_2) + \lambda_N(y_1,y_2)$ and $(\mu_M \oplus \mu_N)(x,y) = \mu_M(x) + \mu_N(y)$ ; this is also an extended quadratic form over Q.

Definition 2.16. Let $\underline{M} = (M, \lambda, \mu)$ be an extended quadratic form over some abelian group Q. We define

\[\begin{aligned}-\underline{M} &= (M, -\lambda, -\mu), \\ \underline{M}{}^* &= (M, \lambda, -\mu),\end{aligned} \]

where $(-\lambda)(x,y) = -\lambda(x,y)$ and $(-\mu)(x) = -\mu(x)$ .

If $\underline{M}$ and $\underline{N}$ are free/non-singular/metabolic/hyperbolic/even/geometric (with respect to some homomorphism), then the same property also holds for $\underline{M} \oplus \underline{N}$ , $-\underline{M}$ and $\underline{M}{}^*$ . If $\underline{M}$ is full, then $\underline{M} \oplus \underline{N}$ , $-\underline{M}$ and $\underline{M}{}^*$ are full too.

Definition 2.17. Let $I : A \rightarrow B$ be an isomorphism of abelian groups. We define the diagonal and anti-diagonal with respect to I:

\[\begin{aligned}\Delta_I &= \{ (x,I(x)) \mid x \in A \} \leqslant A \oplus B, \\ \Delta^*_I &= \{ (x,-I(x)) \mid x \in A \} \leqslant A \oplus B.\end{aligned} \]

1. (a) $\Delta_I$ is a Lagrangian in $\underline{M} \oplus (-\underline{N})$ ;

2. (b) $\Delta^*_I$ is a Lagrangian in $\underline{M} \oplus (-\underline{N}{}^*)$ .

Verifying that $\Delta_I$ and $\Delta^*_I$ have the required properties is straightforward (or see [Nagy21, Lemma 4.1.22]). See also [Nagy21, Lemmas 4.1.23–25] for more detailed proofs of Lemmas 2.19–2.21 below.

2.3 Full geometric metabolic forms

In this section we prove Theorem 2.27, which states that free full geometric metabolic forms are unique up to stable isomorphism. We will need two lemmas.

The composition $M \rightarrow L^*$ of $\mathop{\mathrm{ad}}\nolimits \lambda$ and the dual of the inclusion $L \rightarrow M$ has kernel L, hence induces an isomorphism $M/L \cong L^*$ . Since $\mu \big| _L = 0$ , $\mu$ induces a map $M/L \rightarrow Q$ . The composition of the inverse of the isomorphism $M/L \cong L^*$ and the map $M/L \rightarrow Q$ will be denoted by $\hat{\mu}_L$ .

Hence, $\hat{\mu}'_{L'} = \hat{\mu}_L \circ I_0^*$ .

For the other direction, suppose that $\hat{\mu}'_{L'} = \hat{\mu}_L \circ I_0^* : (L')^* \rightarrow Q$ . Since $L \leqslant M$ and $L' \leqslant M'$ are direct summands, they have direct complements $N \leqslant M$ and $N' \leqslant M'$ . By Lemma 2.8, $N \cong L^*$ , and similarly $N' \cong (L')^*$ . Hence, $I_0^*$ can be regarded as an isomorphism $I_0^* : N' \rightarrow N$ , and the condition means that $\mu' \big| _{N'} = \mu \big| _N \circ I_0^* : N' \rightarrow Q$ .

Let $r = \mathop{\mathrm{rk}}\nolimits N (= \mathop{\mathrm{rk}}\nolimits N' = \mathop{\mathrm{rk}}\nolimits L = \mathop{\mathrm{rk}}\nolimits L')$ . Let $f'_1, f'_2, \ldots , f'_r$ be a basis of N’, and let $f_i = I_0^*(f'_i)$ . The basis $f_1, f_2, \ldots , f_r$ of $N \cong L^*$ determines a dual basis $e_1, e_2, \ldots , e_r$ of L (satisfying $\lambda(e_i,f_j)=0$ if $i \neq j$ and $\lambda(e_i,f_i)=1$ ). Similarly, L’ has a basis $e'_1, e'_2, \ldots , e'_r$ which is dual to the basis $f'_1, f'_2, \ldots , f'_r$ of N’.

The isomorphism $N \cong L^*$ is given by the composition $N \rightarrow M \rightarrow M^* \rightarrow L^*$ , hence $f_j \in N$ corresponds to $\mathop{\mathrm{ad}}\nolimits \lambda(f_j) \big| _L \in L^*$ . So we have $\lambda'(f'_j,I_0(e_i)) = (\mathop{\mathrm{ad}}\nolimits \lambda'(f'_j) \big| _{L'})(I_0(e_i)) = (I_0^*(\mathop{\mathrm{ad}}\nolimits \lambda'(f'_j) \big| _{L'}))(e_i) = (\mathop{\mathrm{ad}}\nolimits \lambda(f_j) \big| _L)(e_i) = \lambda(f_j,e_i) = \lambda'(f'_j,e'_i)$ for every i,j (using the definition of $f_j$ and the dual bases). Since $N' \cong (L')^*$ , this implies that $I_0(e_i)=e'_i$ for every i.

Next we define the elements $\bar{f}_1, \bar{f}_2, \ldots , \bar{f}_r \in M$ recursively. Let $\bar{f}_1 = f_1$ , and let $\bar{f}_{i+1} = f_{i+1} - \sum_{j=1}^i \lambda(\bar{f}_j,f_{i+1})e_j - \left\lfloor {\lambda(f_{i+1},f_{i+1})}/{2} \right\rfloor e_{i+1}$ . By induction we see that $e_1, e_2, \ldots , e_r, \bar{f}_1, \bar{f}_2, \ldots , \bar{f}_i$ is a basis of $L \oplus \left\langle f_1, f_2, \ldots , f_i \right\rangle$ , therefore $e_1, e_2, \ldots , e_r, \bar{f}_1, \bar{f}_2, \ldots , \bar{f}_r$ is a basis of M. By construction $\lambda(e_i,\bar{f}_j)=\lambda(e_i,f_j)$ for every i,j, $\lambda(\bar{f}_i,\bar{f}_j)=0$ if $i \neq j$ , and $\lambda(\bar{f}_i,\bar{f}_i)=0$ if $\lambda(f_i,f_i)$ is even and $\lambda(\bar{f}_i,\bar{f}_i)=1$ if $\lambda(f_i,f_i)$ is odd. Furthermore, $\mu(\bar{f}_i)=\mu(f_i)$ . We define the elements $\bar{f}'_1, \bar{f}'_2, \ldots , \bar{f}'_r \in M'$ analogously.

Let $I : M \rightarrow M'$ be the homomorphism given by $I(e_i) = e'_i$ and $I(\bar{f}_i) = \bar{f}'_i$ . This is an isomorphism of abelian groups, and $I \big| _L = I_0$ . We need to check that I is a morphism of extended quadratic forms. For every i we have $\mu(e_i) = 0 = \mu'(e'_i)$ and $\mu(\bar{f}_i) = \mu(f_i) = \mu \circ I_0^*(f'_i) = \mu'(f'_i) = \mu'(\bar{f}'_i)$ , therefore $\mu = \mu' \circ I$ . For every i,j we have $\lambda(e_i,e_j) = 0 = \lambda'(e'_i,e'_j)$ . If $i \neq j$ , then $\lambda(e_i,\bar{f}_j) = 0 = \lambda'(e'_i,\bar{f}'_j)$ and $\lambda(\bar{f}_i,\bar{f}_j) = 0 = \lambda'(\bar{f}'_i,\bar{f}'_j)$ . For every i we have $\lambda(e_i,\bar{f}_i) = 1 = \lambda'(e'_i,\bar{f}'_i)$ . Finally, both $\lambda(\bar{f}_i,\bar{f}_i)$ and $\lambda'(\bar{f}'_i,\bar{f}'_i)$ are either 0 or 1. And since $\underline{\tilde{M}}$ and $\underline{\tilde{M}}{}'$ are geometric, $\varrho_2 \circ \lambda(\bar{f}_i,\bar{f}_i) = v \circ \mu(\bar{f}_i) = v \circ \mu' \circ I(\bar{f}_i) = v \circ \mu'(\bar{f}'_i) = \varrho_2 \circ \lambda'(\bar{f}'_i,\bar{f}'_i)$ , so this means that $\lambda(\bar{f}_i,\bar{f}_i) = \lambda'(\bar{f}'_i,\bar{f}'_i)$ . Therefore, $\lambda(x,y) = \lambda'(I(x),I(y))$ for every $x,y \in M$ , so I is an isomorphism of extended quadratic forms.

We will obtain a uniqueness result for arbitrary (Q,v) by requiring the forms also to be full; see Theorem 2.27. We will get uniqueness only in a weaker sense (up to stable isomorphism, and instead of extending a given isomorphism between Lagrangians, we can only prescribe the image of a Lagrangian), but this will still be sufficient for our applications in § 4.

An immediate corollary is the following result.

We will also need the following lemma, part (a) of which can be extracted from the proof of [Reference KreckKre99, Proposition 8(ii)].

1. (a) There exist free abelian groups F’, G’, and an isomorphism $h : F \oplus F' \rightarrow G \oplus G'$ such that $(f+0) = (g+0) \circ h : F \oplus F' \rightarrow A$ .

2. (b) If, in addition, A is free and $\mathop{\mathrm{rk}}\nolimits F = \mathop{\mathrm{rk}}\nolimits G$ , then there is an isomorphism $h : F \rightarrow G$ such that $f = g \circ h$ .

We prove that $\bar{f}_1$ is surjective. Let $y \in G$ be an arbitrary element; then there is an $x \in F$ such that $f(x) = g(y)$ (because f is surjective). Let $y_0 = \bar{f}(x) \in \mathop{\mathrm{Im}}\nolimits \bar{f} \subseteq \mathop{\mathrm{Im}}\nolimits \bar{f}_1$ ; then $y-y_0 \in \mathop{\mathrm{Ker}}\nolimits g = \mathop{\mathrm{Im}}\nolimits \bar{f}_0 \subseteq \mathop{\mathrm{Im}}\nolimits \bar{f}_1$ , therefore $y = y_0 + (y-y_0) \in \mathop{\mathrm{Im}}\nolimits \bar{f}_1$ .

Since G is free, $\bar{f}_1$ has a right inverse, that is, there is a homomorphism $c : G \rightarrow F_1$ such that $\bar{f}_1 \circ c = \mathop{\mathrm{id}}\nolimits_G$ . This also implies that $F_1 = \mathop{\mathrm{Ker}}\nolimits \bar{f}_1 \oplus \mathop{\mathrm{Im}}\nolimits c \cong \mathop{\mathrm{Ker}}\nolimits \bar{f}_1 \oplus G$ .

Let $F' = F_0 \oplus G$ and $G' = F_1$ . Then $F \oplus F' = F \oplus (F_0 \oplus G) \cong F_1 \oplus G$ and $G \oplus G' = G \oplus F_1$ , and we can define the homomorphism $h : F_1 \oplus G \rightarrow G \oplus F_1$ by the formula $h(x,y)= (\bar{f}_1(x), x+c(y))$ .

To show that h is an isomorphism, we use the decomposition $F_1 \cong \mathop{\mathrm{Ker}}\nolimits \bar{f}_1 \oplus G$ . For $(x_1,y_2) \in \mathop{\mathrm{Ker}}\nolimits \bar{f}_1 \oplus G \cong F_1$ we have $\bar{f}_1(x_1,y_2)=y_2$ , and for $y \in G$ we have $c(y) = (0,y) \in \mathop{\mathrm{Ker}}\nolimits \bar{f}_1 \oplus G \cong F_1$ . Hence, under the decompositions $F \oplus F' \cong F_1 \oplus G \cong (\mathop{\mathrm{Ker}}\nolimits \bar{f}_1 \oplus G) \oplus G$ and $G \oplus G' = G \oplus F_1 \cong G \oplus (\mathop{\mathrm{Ker}}\nolimits \bar{f}_1 \oplus G)$ the map h is given by the formula $h((x_1,y_2),y) = (y_2, (x_1,y_2+y))$ , therefore it is an isomorphism.

Finally, for any $x \in F$ , $x_0 \in F_0$ and $y \in G$ we have $(g+0) \circ h((x,x_0),y) = g \circ \bar{f}_1(x,x_0) = g \circ (\bar{f} + \bar{f}_0)(x,x_0) = g \circ \bar{f}(x) = f(x) = (f+0)(x,(x_0,y))$ , where we used that $g \circ \bar{f}_0 = 0$ , because $\mathop{\mathrm{Im}}\nolimits \bar{f}_0 = \mathop{\mathrm{Ker}}\nolimits g$ . Therefore, $(f+0) = (g+0) \circ h$ .

(b) Since A is free, there are maps $c : A \rightarrow F$ and $d : A \rightarrow G$ such that $f \circ c = g \circ d = \mathop{\mathrm{id}}\nolimits_A$ . We get the decompositions $F = \mathop{\mathrm{Ker}}\nolimits f \oplus \mathop{\mathrm{Im}}\nolimits c$ and $G = \mathop{\mathrm{Ker}}\nolimits g \oplus \mathop{\mathrm{Im}}\nolimits d$ , where $\mathop{\mathrm{Im}}\nolimits c \cong \mathop{\mathrm{Im}}\nolimits d \cong A$ . The subgroups $\mathop{\mathrm{Ker}}\nolimits f$ and $\mathop{\mathrm{Ker}}\nolimits g$ are free, and $\mathop{\mathrm{rk}}\nolimits \mathop{\mathrm{Ker}}\nolimits f = \mathop{\mathrm{rk}}\nolimits F - \mathop{\mathrm{rk}}\nolimits A = \mathop{\mathrm{rk}}\nolimits G - \mathop{\mathrm{rk}}\nolimits A = \mathop{\mathrm{rk}}\nolimits \mathop{\mathrm{Ker}}\nolimits g$ , therefore there is an isomorphism $h_0 : \mathop{\mathrm{Ker}}\nolimits f \rightarrow \mathop{\mathrm{Ker}}\nolimits g$ . Then we can take $h = h_0 \oplus (d \circ f) \big| _{\mathop{\mathrm{Im}}\nolimits c} : F = \mathop{\mathrm{Ker}}\nolimits f \oplus \mathop{\mathrm{Im}}\nolimits c \rightarrow G = \mathop{\mathrm{Ker}}\nolimits g \oplus \mathop{\mathrm{Im}}\nolimits d$ .

1. (a) For some integers $k,l \geqslant 0$ there is an isomorphism $I : \underline{M} \oplus \underline{H}_{2k} \rightarrow \underline{M}{}' \oplus \underline{H}_{2l}$ such that $I(L \oplus (\{ 0 \} \times \mathbb{Z}^k)) = L' \oplus (\{ 0 \} \times \mathbb{Z}^l)$ .

2. (b) If, in addition, Q is free and $\mathop{\mathrm{rk}}\nolimits \underline{M} = \mathop{\mathrm{rk}}\nolimits \underline{M}{}'$ , then there is an isomorphism $I : \underline{M} \rightarrow \underline{M}{}'$ such that $I(L) = L'$ .

Let $\underline{\tilde{M}} = (\tilde{M}, \tilde{\lambda}, \tilde{\mu})$ denote the direct sum $\underline{M} \oplus \underline{H}_{2k}$ , that is, $\tilde{M} = M \oplus \mathbb{Z}^{2k}$ , $\tilde{\lambda} = \lambda \oplus \bigl[ \begin{smallmatrix} 0 & I_k \\ I_k & 0 \end{smallmatrix} \bigr]$ and $\tilde{\mu} = \mu \oplus 0$ . Similarly, let $\underline{\tilde{M}}{}' = (\tilde{M}', \tilde{\lambda}', \tilde{\mu}') = \underline{M}{}' \oplus \underline{H}_{2l}$ . Let $\tilde{L} = L \oplus (\{ 0 \} \times \mathbb{Z}^k)$ and $\tilde{L}' = L' \oplus (\{ 0 \} \times \mathbb{Z}^l)$ ; these are Lagrangians in $\underline{\tilde{M}}$ and $\underline{\tilde{M}}{}'$ , respectively. Hence, the new extended quadratic forms $\underline{\tilde{M}}$ and $\underline{\tilde{M}}{}'$ are again metabolic, free, (full) and geometric. Moreover, we have $\hat{\tilde{\mu}}_{\tilde{L}} = \hat{\mu}_L + 0 : \tilde{L}^* \cong L^* \oplus (\{ 0 \} \times \mathbb{Z}^k)^* \rightarrow Q$ and $\hat{\tilde{\mu}}'_{\tilde{L}'} = \hat{\mu}'_{L'} + 0 : (\tilde{L}')^* \cong (L')^* \oplus (\{ 0 \} \times \mathbb{Z}^l)^* \rightarrow Q$ .

If we fix some isomorphisms $A \cong (\{ 0 \} \times \mathbb{Z}^k)^*$ and $A' \cong (\{ 0 \} \times \mathbb{Z}^l)^*$ , then h can be regarded as an isomorphism $h : (\tilde{L}')^* \rightarrow \tilde{L}^*$ such that $\hat{\tilde{\mu}}'_{\tilde{L}'} = \hat{\tilde{\mu}}_{\tilde{L}} \circ h$ . Let $I_0 = h^* : \tilde{L} \rightarrow \tilde{L}'$ . Then by Lemma 2.23 there is an isomorphism $I : \underline{\tilde{M}} \rightarrow \underline{\tilde{M}}{}'$ such that $I \big| _{\tilde{L}} = I_0$ , in particular $I(\tilde{L}) = \tilde{L}'$ .

(b) In this case we can use part (b) of Lemma 2.26 instead of part (a). Hence, we can take $A = A' = 0$ , so the statement of part (a) holds with $k = l = 0$ .

4.1 Definitions

Definition 4.1. A quasi-formation over an abelian group Q is a triple

\[(\underline{M}; L, V) \]

where

• – $\underline{M} = (M, \lambda, \mu)$ is a metabolic extended quadratic form over Q,

• – $L \leqslant M$ is a T-Lagrangian,

• – $V \leqslant M$ is a free half-rank direct summand.

Definition 4.7. For an abelian group Q with a homomorphism $v : Q \rightarrow \mathbb{Z}_2$ , a torsion abelian group R and an even integer $q \geqslant 0$ , let

\[\mathfrak{g}^T_{2q+1}(Q,v,R) \]

denote the class of quasi-formations $(\underline{M}; L, V)$ over Q such that $\underline{M} = (M, \lambda, \mu)$ is geometric (with respect to v), full and $\mathop{\mathrm{Tor}}\nolimits M \cong R$ .

Note that the class $\mathfrak{g}^T_{2q+1}(Q,v,R)$ does not depend on the choice of q.

Definition 4.8. For an abelian group Q with a homomorphism $v : Q \rightarrow \mathbb{Z}_2$ , a torsion abelian group R and an even integer $q \geqslant 0$ , let

\[\mathscr{g}^T_{2q+1}(Q,v,R) = \mathfrak{g}^T_{2q+1}(Q,v,R) / {\cong} \]

denote the set of isomorphism classes of quasi-formations $(\underline{M}; L, V)$ over Q such that $\underline{M} = (M, \lambda, \mu)$ is geometric (with respect to v), full and $\mathop{\mathrm{Tor}}\nolimits M \cong R$ .

Definition 4.9. For an abelian group Q with a homomorphism $v : Q \rightarrow \mathbb{Z}_2$ , a torsion abelian group R and an even integer $q \geqslant 0$ , let

\[\mathscr{s}^T_{2q+1}(Q,v,R) = \mathfrak{g}^T_{2q+1}(Q,v,R) / {\mathrel{\underset{^{\scriptstyle{s}}}{\cong}}} \]

denote the set of stable isomorphism classes of quasi-formations $(\underline{M}; L, V)$ over Q such that $\underline{M} = (M, \lambda, \mu)$ is geometric (with respect to v), full and $\mathop{\mathrm{Tor}}\nolimits M \cong R$ .

Definition 4.10. For an abelian group Q with a homomorphism $v : Q \rightarrow \mathbb{Z}_2$ , a torsion abelian group R and an even integer $q \geqslant 0$ , let

\[\ell^T_{2q+1}(Q,v,R) = \mathfrak{g}^T_{2q+1}(Q,v,R) / {\sim} \]

be the set of quasi-formations $(\underline{M}; L, V)$ over Q such that $\underline{M} = (M, \lambda, \mu)$ is geometric (with respect to v), full and $\mathop{\mathrm{Tor}}\nolimits M \cong R$ up to the equivalence relation $\sim$ , which is generated by the following elementary equivalences:

• – $(\underline{M}; L, V) \sim (\underline{M}{}'; L', V')$ if $(\underline{M}; L, V) \mathrel{\underset{^{\scriptstyle{s}}}{\cong}} (\underline{M}{}'; L', V')$ ;

• – $(\underline{M} \oplus \underline{H}_2; L \oplus (\{ 0 \} \times \mathbb{Z}), V) \sim (\underline{M} \oplus \underline{H}_2; L \oplus (\mathbb{Z} \times \{ 0 \}), V)$ , where L is a T-Lagrangian in $\underline{M}$ and V is a free half-rank direct summand in $\underline{M} \oplus \underline{H}_2$ .

The equivalence class of $(\underline{M}; L, V)$ will be denoted by $[\underline{M}; L, V]$ .

Pairs (Q,v) are the objects in the slice category of abelian groups over $\mathbb{Z}_2$ , where a morphism $(Q,v) \rightarrow (Q',v')$ is a homomorphism $f : Q \rightarrow Q'$ such that $v' \circ f = v$ . Post-composition with such an f turns an extended quadratic form over Q into one over Q’, and determines a map $f_* : \bar{\mathfrak{g}}^T_{2q+1}(Q,v,R) \rightarrow \bar{\mathfrak{g}}^T_{2q+1}(Q',v',R)$ . Then $\sim$ is natural in the sense that if $x \sim y \in \bar{\mathfrak{g}}^T_{2q+1}(Q,v,R)$ , then $f_*(x) \sim f_*(y) \in \bar{\mathfrak{g}}^T_{2q+1}(Q',v',R)$ .

Definition 4.12. For an abelian group Q with a homomorphism $v : Q \rightarrow \mathbb{Z}_2$ and an even integer $q \geqslant 0$ , let

\[\begin{aligned}\mathfrak{g}_{2q+1}(Q,v) &= \mathfrak{g}^T_{2q+1}(Q,v,0) \text{,} \\ \mathscr{g}_{2q+1}(Q,v) &= \mathscr{g}^T_{2q+1}(Q,v,0) \text{,} \\ \mathscr{s}_{2q+1}(Q,v) &= \mathscr{s}^T_{2q+1}(Q,v,0) \text{,} \\ \ell_{2q+1}(Q,v) &= \ell^T_{2q+1}(Q,v,0) .\end{aligned} \]

To prove the equivalence of $\ell^T_{2q+1}(Q,v,R)$ and $\ell_{2q+1}(Q,v)$ we will need the following notation.

Note that $\pi_N : \underline{N} \rightarrow \underline{\bar{N}}$ is a morphism of extended quadratic forms. If $\underline{N}$ is non-singular, metabolic, full or geometric (with respect to some homomorphism), then the same property also holds for $\underline{\bar{N}}$ .

Next we define a function $G : \ell^T_{2q+1}(Q,v,R) \rightarrow \ell_{2q+1}(Q,v)$ . Suppose that $(\underline{N}; K, W) \in \mathfrak{g}^T_{2q+1}(Q,v,R)$ , where $\underline{N} = (N, \lambda, \mu)$ . Since $\mu \big| _{\mathop{\mathrm{Tor}}\nolimits N} = 0$ (because $\underline{N}$ has a T-Lagrangian), we can define $\underline{\bar{N}}$ . Moreover, $\bar{K} = K / \mathop{\mathrm{Tor}}\nolimits N$ is a Lagrangian in $\underline{\bar{N}}$ , and $\bar{W} = (W \oplus \mathop{\mathrm{Tor}}\nolimits N) / \mathop{\mathrm{Tor}}\nolimits N$ is a half-rank direct summand, hence $(\underline{\bar{N}}; \bar{K}, \bar{W}) \in \mathfrak{g}_{2q+1}(Q,v)$ . Let $G([\underline{N}; K, W])=[\underline{\bar{N}}; \bar{K}, \bar{W}]$ .

It is immediate from the definitions that $G \circ F = \mathop{\mathrm{id}}\nolimits_{\ell_{2q+1}(Q,v)}$ . For the other composition, let $[\underline{N}; K, W] \in \ell^T_{2q+1}(Q,v,R)$ ; then $F \circ G([\underline{N}; K, W]) = [\underline{\bar{N}} \oplus \underline{R}; \bar{K} \oplus R, \bar{W} \oplus 0]$ . Since W is a free direct summand, $\pi_N$ has a right inverse $I_0 : \bar{N} \rightarrow N$ such that $W \leqslant \mathop{\mathrm{Im}}\nolimits I_0$ . By choosing an isomorphism $\mathop{\mathrm{Tor}}\nolimits N \cong R$ we can extend $I_0$ to an isomorphism $I : \bar{N} \oplus R \rightarrow N$ . This is also an isomorphism $\underline{\bar{N}} \oplus \underline{R} \cong \underline{N}$ of extended quadratic forms, and by construction $I(\bar{W} \oplus 0) = I_0(\bar{W})=W$ . Since K is a T-Lagrangian, we have $K = (K \cap \mathop{\mathrm{Im}}\nolimits I_0) \oplus \mathop{\mathrm{Tor}}\nolimits N$ , so $I(\bar{K} \oplus R) = I_0(\bar{K}) \oplus \mathop{\mathrm{Tor}}\nolimits N = (K \cap \mathop{\mathrm{Im}}\nolimits I_0) \oplus \mathop{\mathrm{Tor}}\nolimits N = K$ . This shows that $[\underline{\bar{N}} \oplus \underline{R}; \bar{K} \oplus R, \bar{W} \oplus 0] = [\underline{N}; K, W]$ , hence $F \circ G = \mathop{\mathrm{id}}\nolimits_{\ell^T_{2q+1}(Q,v,R)}$ .

Recall that a quasi-formation $(\underline{M}; L, V)$ is called elementary if $M = L \oplus V$ . This property is invariant under stable isomorphism (because $(\underline{M}; L, V)$ is elementary if and only if $(\underline{M}; L, V) \oplus \mathcal{H}_{2k}$ is elementary, for any k). Therefore, an element of $\mathscr{g}^T_{2q+1}(Q,v,R)$ or $\mathscr{s}^T_{2q+1}(Q,v,R)$ has an elementary representative if and only if all of its representatives are elementary.

In contrast to the above, an elementary quasi-formation may be equivalent to a non-elementary one under the equivalence relation $\sim$ .

4.2 The monoid $\ell_{2q+1}(Q,v)$

From now on we will focus on the free version $\ell_{2q+1}(Q,v)$ . We show that it has a natural monoid structure and, using the results on $\mathop{\mathrm{RU}}\nolimits_{\mathrm{st}}(\underline{M},L)$ from § 3, we prove Theorem 4.26.

If $\Psi(L) = L$ , then $\Psi$ is an isomorphism between $(\underline{M}; L, V)$ and $(\underline{M}; L, \Psi(V))$ .

If there is an isomorphism $I : \underline{M} \rightarrow \underline{H}_2 \oplus \underline{M}{}'$ such that $I(L) = ( \{0 \} \times \mathbb{Z}) \oplus L'$ for some Lagrangian L’ in $\underline{M}{}'$ and $\Psi = I^{-1} \circ (\sigma \oplus \mathop{\mathrm{id}}\nolimits_{M'}) \circ I$ , then $I \circ \Psi(L) = (\sigma \oplus \mathop{\mathrm{id}}\nolimits_{M'}) \circ I(L) = (\sigma \oplus \mathop{\mathrm{id}}\nolimits_{M'})(( \{0 \} \times \mathbb{Z}) \oplus L') = (\mathbb{Z} \times \{0 \} ) \oplus L'$ . So

\[\begin{aligned}(\underline{M}; L, V) &\cong (\underline{H}_2 \oplus \underline{M}{}'; I \circ \Psi(L), I \circ \Psi(V)) \\ &= (\underline{H}_2 \oplus \underline{M}{}'; (\mathbb{Z} \times \{0 \} ) \oplus L', I \circ \Psi(V)) \\ &\sim (\underline{H}_2 \oplus \underline{M}{}'; ( \{0 \} \times \mathbb{Z}) \oplus L', I \circ \Psi(V)) \\ &= (\underline{H}_2 \oplus \underline{M}{}'; I(L), I \circ \Psi(V)) \\ &\cong (\underline{M}; L, \Psi(V))\end{aligned} \]

where we used the isomorphisms $I \circ \Psi$ and I. Therefore, the statement holds for every generator of $\mathop{\mathrm{RU}}\nolimits(\underline{M},L)$ .

First, we show that there is a quasi-formation of the form $(\underline{M}; L, L)$ in the class $\mathfrak{g}_{2q+1}(Q,v)$ (in particular, $\mathfrak{g}_{2q+1}(Q,v)$ and $\ell_{2q+1}(Q,v)$ are non-empty). Let $\mu_0 : A \rightarrow Q$ be a surjective homomorphism from some free abelian group A to Q. Let $k = \mathop{\mathrm{rk}}\nolimits A$ , and let $f_1, f_2, \ldots , f_k$ be a basis of A. Let L be another rank-k free abelian group with basis $e_1, e_2, \ldots , e_k$ . Let $M = L \oplus A$ . Let $\lambda : M \times M \rightarrow \mathbb{Z}$ be the bilinear function given by the block matrix

\[\begin{bmatrix}0 & I_k \\I_k & D\end{bmatrix} \]

(in the basis $e_1, e_2, \ldots , e_k, f_1, f_2, \ldots , f_k$ ), where D is a diagonal matrix such that the diagonal entry corresponding to $f_i$ is 0 if $v \circ \mu_0(f_i)=0$ and 1 if $v \circ \mu_0(f_i)=1$ (hence, $\varrho_2 \circ \lambda(f_i,f_i) = v \circ \mu_0(f_i)$ ). Let $\mu : M \rightarrow Q$ be the homomorphism such that $\mu \big| _L = 0$ and $\mu \big| _A = \mu_0$ . Then $\underline{M} = (M, \lambda, \mu)$ is an extended quadratic form over Q, and it is free and full. It is also geometric, because for any $x = \sum_{i=1}^k r_ie_i + \sum_{i=1}^k s_if_i \in M$ we have

\begin{align*}\varrho_2 \circ \lambda(x,x) &= \varrho_2 \bigg(2\sum_{i=1}^k r_is_i+ \sum_{i=1}^k s_i^2 \lambda(f_i,f_i) \bigg) = \sum_{i=1}^k \varrho_2(s_i^2) \varrho_2 \circ \lambda(f_i,f_i) \\&= \sum_{i=1}^k \varrho_2(s_i) \varrho_2 \circ \lambda(f_i,f_i) = \sum_{i=1}^k \varrho_2(s_i) v \circ \mu_0(f_i) = \sum_{i=1}^k v(s_i\mu_0(f_i)) = v \circ \mu(x).\end{align*}

Furthermore, L is a Lagrangian in M, therefore $(\underline{M}; L, L) \in \mathfrak{g}_{2q+1}(Q,v)$ .

Second, we show that if $(\underline{M}; L, L), (\underline{M}{}'; L', L') \in \mathfrak{g}_{2q+1}(Q,v)$ , then $(\underline{M}; L, L) \sim (\underline{M}{}'; L', L')$ . By Theorem 2.27(a) there is an isomorphism $I : \underline{M} \oplus \underline{H}_{2k} \rightarrow \underline{M}{}' \oplus \underline{H}_{2l}$ such that $I(L \oplus (\{ 0 \} \times \mathbb{Z}^k)) = L' \oplus (\{ 0 \} \times \mathbb{Z}^l)$ for some $k,l \geqslant 0$ . So we have

\[\begin{aligned}(\underline{M}; L, L) &\sim (\underline{M}; L, L) \oplus (\underline{H}_{2k}; \{ 0 \} \times \mathbb{Z}^k, \{ 0 \} \times \mathbb{Z}^k) \\ &= (\underline{M} \oplus \underline{H}_{2k}; L \oplus (\{ 0 \} \times \mathbb{Z}^k), L \oplus (\{ 0 \} \times \mathbb{Z}^k)) \\ &\cong (\underline{M}{}' \oplus \underline{H}_{2l}; L' \oplus (\{ 0 \} \times \mathbb{Z}^l), L' \oplus (\{ 0 \} \times \mathbb{Z}^l)) \\ &= (\underline{M}{}'; L', L') \oplus (\underline{H}_{2l}; \{ 0 \} \times \mathbb{Z}^l, \{ 0 \} \times \mathbb{Z}^l)\\ &\sim (\underline{M}{}'; L', L'),\end{aligned} \]

using Lemma 4.20 and the isomorphism I. Therefore, all quasi-formations of the form $(\underline{M}; L, L)$ are in the same equivalence class, which we will denote by $\alpha \in \ell_{2q+1}(Q,v)$ .

Third, we show that $\alpha$ is a zero element in $\ell_{2q+1}(Q,v)$ . Suppose that $(\underline{M}; L, V) \in \mathfrak{g}_{2q+1}(Q,v)$ ; then $(\underline{M}; L, L) \in \mathfrak{g}_{2q+1}(Q,v)$ and $[\underline{M}; L, L] = \alpha$ . Let $I : \underline{M} \oplus \underline{M} \rightarrow \underline{M} \oplus \underline{H}_{2k}$ be the isomorphism constructed in the proof of Lemma 2.21, satisfying $I(L \oplus L) = L \oplus (\{ 0 \} \times \mathbb{Z}^k)$ . We also define the basis elements $e_i,f_i,\bar{e}_i,\bar{f}_i,a_i,b_i$ and the integers $d_i$ as in Lemma 2.21. Let $v_1, v_2, \ldots , v_k$ be a basis of V; then $v_i = \sum_{j=1}^k (p_{ij}e_j + q_{ij}f_j)$ for some $p_{ij}, q_{ij} \in \mathbb{Z}$ . We have $I(v_i) = \sum_{j=1}^k (p_{ij}I(e_j) + q_{ij}I(f_j)) = \sum_{j=1}^k (p_{ij}(e_j+b_j) + q_{ij}(f_j+d_jb_j)) = \sum_{j=1}^k (p_{ij}e_j + q_{ij}f_j) + \sum_{j=1}^k (p_{ij} + q_{ij}d_j)b_j = v_i + \sum_{j=1}^k (p_{ij} + q_{ij}d_j)b_j$ and $I(\bar{e}_i)=-b_i$ . Thus,

\begin{align*}I(V \oplus L) &= I(\left< v_1, \ldots , v_k, \bar{e}_1, \ldots , \bar{e}_k \right\rangle) \\&= \bigg\langle v_1 + \sum_{j=1}^k (p_{1j} + q_{1j}d_j)b_j, \ldots ,v_k + \sum_{j=1}^k (p_{kj} + q_{kj}d_j)b_j, -b_1, \ldots , -b_k \bigg\rangle \\&= \left< v_1, \ldots , v_k, b_1, \ldots , b_k \right\rangle = V \oplus (\{ 0 \} \times \mathbb{Z}^k) .\end{align*}

Using the isomorphism I and Lemma 4.20, we get

\begin{align*}(\underline{M}; L, V) \oplus (\underline{M}; L, L) &= (\underline{M} \oplus \underline{M}; L \oplus L, V \oplus L) \\ &\cong (\underline{M} \oplus \underline{H}_{2k}; L \oplus (\{ 0 \} \times \mathbb{Z}^k), V \oplus (\{ 0 \} \times \mathbb{Z}^k)) \\ &= (\underline{M}; L, V) \oplus (\underline{H}_{2k}; \{ 0 \} \times \mathbb{Z}^k, \{ 0 \} \times \mathbb{Z}^k) \\ &\sim (\underline{M}; L, V).\end{align*}

This means that $[\underline{M}; L, V] \oplus \alpha = [\underline{M}; L, V]$ . Since this is true for any $[\underline{M}; L, V] \in \ell_{2q+1}(Q,v)$ , $\alpha$ is a zero element (hence the zero element) in $\ell_{2q+1}(Q,v)$ .

If $f : Q \rightarrow Q'$ is a morphism $(Q,v) \rightarrow (Q',v')$ in the slice category, then, by the naturality of $\sim$ , the map $f_* : \bar{\mathfrak{g}}_{2q+1}(Q,v) \rightarrow \bar{\mathfrak{g}}_{2q+1}(Q',v')$ induces a map $\bar{\ell}_{2q+1}(Q,v) \rightarrow \bar{\ell}_{2q+1}(Q',v')$ between the quotients. By pre-composing this map with the inclusion $\ell_{2q+1}(Q,v) \rightarrow \bar{\ell}_{2q+1}(Q,v)$ and post-composing with $\bar{\ell}_{2q+1}(Q',v') \rightarrow \ell_{2q+1}(Q',v')$ , we get the morphism $f_* : \ell_{2q+1}(Q,v) \rightarrow \ell_{2q+1}(Q',v')$ of $\ell$ -monoids induced by f.

The next lemma expands on Lemma 4.20 and Proposition 4.21 by showing that every quasi-formation of the form $(\underline{N}; K, K)$ has a trivial effect on $\ell_{2q+1}(Q,v)$ , even if $\underline{N}$ is not full (so $(\underline{N}; K, K)$ does not represent an element of $\ell_{2q+1}(Q,v)$ ).

We can now prove the stable version of Lemma 4.19.

Lemma 4.25. Suppose that $(\underline{M}; L, V) \in \mathfrak{g}_{2q+1}(Q,v)$ . If $\Psi \in \mathop{\mathrm{RU}}\nolimits_{\mathrm{st}}(\underline{M},L)$ , then

\[(\underline{M}; L, V) \sim (\underline{M}; L, \Psi(V)) . \]

Theorem 4.26. Suppose that $(\underline{M}; L, V) \in \mathfrak{g}_{2q+1}(Q,v)$ , and let K be a Lagrangian in $\underline{M}$ (so $(\underline{M}; K, L), (\underline{M}; K, V) \in \mathfrak{g}_{2q+1}(Q,v)$ ). Then

\[[\underline{M}; K, L] \oplus [\underline{M}; L, V] = [\underline{M}; K, V] \in \ell_{2q+1}(Q,v) . \]

By Theorem 3.7 $\Phi \oplus \Phi^{-1} \in \mathop{\mathrm{RU}}\nolimits_{\mathrm{st}}(\underline{\tilde{M}} \oplus \underline{\tilde{M}}, \tilde{K} \oplus \tilde{K})$ . Therefore, we have

\[\begin{aligned}(\underline{\tilde{M}}; \tilde{K}, \tilde{L}) \oplus (\underline{\tilde{M}}; \tilde{L}, \tilde{V}) &\cong (\underline{\tilde{M}}; \tilde{K}, \tilde{L}) \oplus (\underline{\tilde{M}}; \tilde{K}, \Phi(\tilde{V})) \\ &= (\underline{\tilde{M}} \oplus \underline{\tilde{M}}; \tilde{K} \oplus \tilde{K}, \tilde{L} \oplus \Phi(\tilde{V})) \\ &\sim (\underline{\tilde{M}} \oplus \underline{\tilde{M}}; \tilde{K} \oplus \tilde{K}, \tilde{K} \oplus \tilde{V}) \\ &= (\underline{\tilde{M}}; \tilde{K}, \tilde{K}) \oplus (\underline{\tilde{M}}; \tilde{K}, \tilde{V}) \\ &\sim (\underline{\tilde{M}}; \tilde{K}, \tilde{V});\end{aligned} \]

using the isomorphism $\Phi$ and Lemmas 4.25 and 4.24.

These results generalise the relations that were used to define the classical L-groups and Kreck’s monoid $l_{2q+1}(e)$ . In [Reference KreckKre99, § 6] the equivalence relation on pairs $(\underline{H}_{2k}, V)$ is given by the group $\mathop{\mathrm{RU}}\nolimits^+(\mathbb{Z})$ , which is analogous to the relation in Lemma 4.25 (see Proposition 4.29 below for the correspondence between pairs and quasi-formations). The analogue of the identity of Theorem 4.26 has been used to reformulate the definitions of the classical L-groups and the l-monoids in terms of (quasi-)formations (see, for example, [Reference RanickiRan80, Reference Crowley and SixtCS11]).

4.3 The group $L_{2q+1}(Q,v)$

In this section we consider the extended version of the L-groups.

Definition 4.27. For an abelian group Q with a homomorphism $v : Q \rightarrow \mathbb{Z}_2$ , a torsion abelian group R and an even integer $q \geqslant 0$ , let

\[L^T_{2q+1}(Q,v,R) = \{ [\underline{M}; L, K] \in \ell^T_{2q+1}(Q,v,R) \mid {\textit{K}} \text{is a free Lagrangian in $\underline{M}$} \} \]

and

\[L_{2q+1}(Q,v) = L^T_{2q+1}(Q,v,0). \]

The first important application of Theorem 4.26 shows that the elements of $L_{2q+1}(Q,v)$ are invertible, hence that it is a group.

Next, we show that every invertible element is contained in $L_{2q+1}(Q,v)$ . For an $(\underline{M}; L, V) \in \mathfrak{g}_{2q+1}(Q,v)$ let $\alpha(\underline{M}; L, V)$ denote the subgroup $\mu(V) \leqslant Q$ (where $\underline{M} = (M, \lambda, \mu)$ ). Then $\alpha(\underline{M}; L, V)$ is invariant under the elementary equivalences that define $\sim$ , so it induces a well-defined function on $\ell_{2q+1}(Q,v)$ . Moreover, we have $\alpha([\underline{M}; L, V] \oplus [\underline{M}{}'; L', V']) = \alpha([\underline{M}; L, V]) + \alpha([\underline{M}{}'; L', V'])$ and $\alpha([\underline{M}; L, L]) = 0 \leqslant Q$ . This implies that if $[\underline{M}; L, V]$ is invertible, then $\alpha([\underline{M}; L, V]) = 0$ , that is, $\mu \big| _V = 0$ . Similarly, let $\beta(\underline{M}; L, V) \leqslant \mathbb{Z}$ denote the image of the homomorphism $V \otimes V \rightarrow \mathbb{Z}$ corresponding to the bilinear function $\lambda \big| _{V \times V} : V \times V \rightarrow \mathbb{Z}$ . Then $\beta$ also induces a well-defined function on $\ell_{2q+1}(Q,v)$ , and we have $\beta([\underline{M}; L, V] \oplus [\underline{M}{}'; L', V']) = \beta([\underline{M}; L, V]) + \beta([\underline{M}{}'; L', V'])$ and $\beta([\underline{M}; L, L]) = 0 \leqslant \mathbb{Z}$ . So if $[\underline{M}; L, V]$ is invertible, then $\beta([\underline{M}; L, V]) = 0$ , that is, $\lambda \big| _{V \times V} = 0$ . This means that if $[\underline{M}; L, V]$ is invertible, then V is a Lagrangian, therefore $[\underline{M}; L, V] \in L_{2q+1}(Q,v)$ .

We will now prove that if Q is free, then $L_{2q+1}(Q,v) \cong 0$ (Theorem 4.33). The first step towards that is the following proposition.

We define $F : l_{2q+1}(e) \rightarrow \ell_{2q+1}(0,0)$ by $F([\underline{H}_{2k}, V]) = [\underline{H}_{2k}; \{0 \} \times \mathbb{Z}^k, V]$ ; this map is well-defined by Lemma 4.19.

To define $G : \ell_{2q+1}(0,0) \rightarrow l_{2q+1}(e)$ , note that if $[\underline{H}; L, V] \in \ell_{2q+1}(0,0)$ , then $\underline{H}$ is hyperbolic (see Lemma 2.13), so by Corollary 2.25 there exist a $k \geqslant 0$ and an isomorphism $I : \underline{H} \rightarrow \underline{H}_{2k}$ such that $I(L) = \{0 \} \times \mathbb{Z}^k$ . Let $G([\underline{H}; L, V]) = [\underline{H}_{2k}, I(V)]$ ; we will check that this is well-defined. If $I : \underline{H} \rightarrow \underline{H}_{2k}$ and $I' : \underline{H} \rightarrow \underline{H}_{2k}$ are two isomorphisms as above, then $I' \circ I^{-1}(\{0 \} \times \mathbb{Z}^k) = \{0 \} \times \mathbb{Z}^k$ , so $I' \circ I^{-1} \in RU(\underline{H}_{2k}, \{ 0 \} \times \mathbb{Z}^k)$ and $[\underline{H}_{2k}, I(V)] = [\underline{H}_{2k}, I'(V)]$ . If $(\underline{H}; L, V) \cong (\underline{H}{}'; L', V')$ , then $I' : \underline{H}{}' \rightarrow \underline{H}_{2k}$ can be chosen to be the isomorphism $\underline{H}{}' \rightarrow \underline{H}$ composed with $I : \underline{H} \rightarrow \underline{H}_{2k}$ , so that $(\underline{H}_{2k}, I(V)) = (\underline{H}_{2k}, I'(V'))$ . If $(\underline{H}{}'; L', V') = (\underline{H}; L, V) \oplus \mathcal{H}_2$ , then we can choose $I' = I \oplus \mathop{\mathrm{id}}\nolimits_{\mathbb{Z}^2}$ , so that $(\underline{H}_{2k+2}, I'(V')) = (\underline{H}_{2k} \oplus \underline{H}_2, I(V) \oplus (\mathbb{Z} \times \{0 \} )) \sim (\underline{H}_{2k}, I(V))$ . Finally, consider the equivalence $(\underline{H}{}' \oplus \underline{H}_2; L \oplus (\{ 0 \} \times \mathbb{Z}), V) \sim (\underline{H}{}' \oplus \underline{H}_2; L \oplus (\mathbb{Z} \times \{ 0 \}), V)$ . If $I : \underline{H}{}' \oplus \underline{H}_2 \rightarrow \underline{H}_{2k}$ is an isomorphism such that $I(L \oplus (\{ 0 \} \times \mathbb{Z})) = \{0 \} \times \mathbb{Z}^k$ , then $I \circ (\mathop{\mathrm{id}}\nolimits_{H'} \oplus \sigma)(L \oplus (\mathbb{Z} \times \{ 0 \})) = \{0 \} \times \mathbb{Z}^k$ and $I \circ (\mathop{\mathrm{id}}\nolimits_{H'} \oplus \sigma) \circ I^{-1} \in RU(\underline{H}_{2k}, \{ 0 \} \times \mathbb{Z}^k)$ , therefore $[\underline{H}_{2k}, I(V)] = [\underline{H}_{2k}, I \circ (\mathop{\mathrm{id}}\nolimits_{H'} \oplus \sigma)(V)]$ .

F and G are obviously inverses of each other, and they preserve direct sums, therefore they are isomorphisms.

The pair $(\underline{H}_{2k}, V)$ represents an element in $L_{2q+1}$ if and only if V is a Lagrangian in $\underline{H}_{2k}$ , which holds if and only if $[\underline{H}_{2k}; \{0 \} \times \mathbb{Z}^k, V] \in L_{2q+1}(0,0)$ , showing that F restricts to an isomorphism $L_{2q+1} \rightarrow L_{2q+1}(0,0)$ . (Alternatively, note that $L_{2q+1}(0,0)$ and $L_{2q+1}$ are both equal to the group of invertible elements in the corresponding monoid.)

Let $J = K \cap L$ . Since K and L are direct summands in M, the same holds for J (see Lemma 2.4). Let X be a direct complement to J in M, $K' = K \cap X$ and $L' = L \cap X$ . Then $K = J \oplus K'$ and $L = J \oplus L'$ .

Let $M' = J \oplus X^{\perp}$ . It follows from Lemma 2.7 that this is a subgroup of M and $\lambda \big| _{M' \times M'}$ is non-singular. Let $H = (M')^{\perp}$ . By Lemma 2.6, $M = M' \oplus H$ . Let $\underline{M}{}' = (M', \lambda \big| _{M' \times M'}, \mu \big| _{M'})$ and $\underline{H} = (H, \lambda \big| _{H \times H}, \mu \big| _H)$ . Then $\underline{M} = \underline{M}{}' \oplus \underline{H}$ , and since $\underline{M}$ is non-singular, $\underline{H}$ is non-singular too.

Next we show that J is a Lagrangian in $\underline{M}{}'$ . It is a direct summand in M’ by the latter’s definition. We have $\lambda \big| _{J \times J} = 0$ and $\mu \big| _J = 0$ , because $J \leqslant L$ . Finally, $\mathop{\mathrm{rk}}\nolimits M' = \mathop{\mathrm{rk}}\nolimits J + \mathop{\mathrm{rk}}\nolimits X^{\perp} = \mathop{\mathrm{rk}}\nolimits J + (\mathop{\mathrm{rk}}\nolimits M - \mathop{\mathrm{rk}}\nolimits X) = \mathop{\mathrm{rk}}\nolimits J + \mathop{\mathrm{rk}}\nolimits J = 2 \mathop{\mathrm{rk}}\nolimits J$ . So J is a Lagrangian in $\underline{M}{}'$ , and this means that $\underline{M}{}'$ is metabolic.

Next we show that L’ is a Lagrangian in $\underline{H}$ . We have $L' = L \cap X = L^{\perp} \cap X \leqslant J^{\perp} \cap X = (J \oplus X^{\perp})^{\perp} = H$ . By construction L’ is a direct summand in L, hence in M, hence in H. We have $\lambda \big| _{L' \times L'} = 0$ and $\mu \big| _{L'} = 0$ , because $L' \leqslant L$ . Finally, $\mathop{\mathrm{rk}}\nolimits L' = \mathop{\mathrm{rk}}\nolimits L - \mathop{\mathrm{rk}}\nolimits J = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits M - \frac{1}{2} \mathop{\mathrm{rk}}\nolimits M' = \frac{1}{2}(\mathop{\mathrm{rk}}\nolimits M - \mathop{\mathrm{rk}}\nolimits M') = \frac{1}{2} \mathop{\mathrm{rk}}\nolimits H$ . Therefore, L’ is a Lagrangian in $\underline{H}$ , and since $\underline{H}$ is non-singular, it is metabolic. We can show similarly that K’ is also a Lagrangian in $\underline{H}$ .

Consider the subgroup $K' \oplus L' \leqslant H$ (note that $K' \cap L' = \{ 0 \}$ ). Since K’ and L’ are Lagrangians in H, we have $\mathop{\mathrm{rk}}\nolimits (K' \oplus L') = \mathop{\mathrm{rk}}\nolimits H$ , therefore $H / (K' \oplus L')$ is a torsion group. The homomorphism $\mu \big| _H : H \rightarrow Q$ induces a homomorphism $H / (K' \oplus L') \rightarrow Q$ , because $\mu \big| _{K' \oplus L'} = 0$ . Since Q is assumed to be free, this induced homomorphism is trivial, therefore $\mu \big| _H = 0$ . This implies that $\underline{M}{}'$ is full (because $\underline{M}$ is full). Moreover, since $\underline{M}$ is geometric (with respect to v), the same holds for $\underline{H}$ , so $\underline{H}$ is even. By applying Lemma 2.13, we get that $\underline{H}$ is hyperbolic.