$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q_{+})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\textup{Hom}(G,SQ)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q_{+})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q^{+})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\textup{Hom}(G,\mathbb{Z}_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

consisting of short exact sequences. In particular the second square in this
diagram is a pullback square.

######  Remark 4.12.

Part b) of Proposition 4.11 was proven by Baues [Bau1, Remark 2.10 (4)] in the
special case $Q=\mathbb{Z}^{P}$ (when
$G\otimes_{\mathbb{Z}}\mathbb{Z}^{P}\cong P^{2}(G)$ and $G\otimes
S\mathbb{Z}^{P}\cong G$).

###### Proof.

a) We get this by applying Proposition 4.10 to the unique morphism of form
parameters $Q\rightarrow Q^{+}$, which is given by ${\rm h}\colon
Q_{e}\rightarrow\mathbb{Z}$. Note that $SQ^{+}=\mathbb{Z}_{2}$ and the map
$I(SQ)\rightarrow I(\mathbb{Z}_{2})$ is $I(v_{Q})$, because $v_{Q}$ is induced
by ${\rm h}$ (see also Lemma 2.10).

b) The exactness of the bottom sequence was proved by Baues. It follows from
part a) that the diagram commutes. This also implies that the map
$S^{2}(G)\rightarrow G\otimes_{\mathbb{Z}}Q$ is injective (because its
composition with $\textup{Id}_{G}\otimes_{\mathbb{Z}}{\rm h}\colon
G\otimes_{\mathbb{Z}}Q\rightarrow\Gamma(G)$ is the injective map
$S^{2}(G)\rightarrow\Gamma(G)$).

The map $S^{2}(G)\rightarrow G\otimes_{\mathbb{Z}}Q$ sends $gh$ to
$[g,h]\otimes 1$, so $G\otimes_{\mathbb{Z}}Q/S^{2}(G)$ is the group generated
by the symbols $g\otimes q$, linear in both $g\in G$ and $q\in Q_{e}$, and
subject to the relation $g\otimes{\rm p}(1)=0$, therefore
$G\otimes_{\mathbb{Z}}Q/S^{2}(G)\cong G\otimes SQ$. Moreover, this isomorphism
is induced by the map $G\otimes_{\mathbb{Z}}Q\rightarrow G\otimes SQ$, because
it sends $[g,h]\otimes 1$ to $0$ and $g\otimes q$ to $g\otimes\pi(q)$.
Therefore the top sequence is exact too.

c) First we show that the bottom sequence is exact. An element
$(G,\lambda,\mu)$ of $\mathrm{Hom}_{\mathbb{Z}}(G,Q^{+})$ is determined by the
symmetric bilinear form $(G,\lambda)$ (because $\mu$ satisfies
$\mu(x)=\lambda(x,x)$). The image of $(G,\lambda)$ is
$f\in\textup{Hom}(G,\mathbb{Z}_{2})$ given by $f(x)=\lambda(x,x)$ mod $2$. For
a free $G$ the map
$\mathrm{Hom}_{\mathbb{Z}}(G,Q^{+})\rightarrow\textup{Hom}(G,\mathbb{Z}_{2})$
is surjective, because for a homomorphism $f$ and a basis $x_{1},\ldots,x_{k}$
of $G$ we can define a $\lambda$ in the preimage of $f$ by
$\lambda(x_{i},x_{i})=0$ if $f(x_{i})=0$, $\lambda(x_{i},x_{i})=1$ if
$f(x_{i})=1$ and $\lambda(x_{i},x_{j})=0$ if $i\neq j$. The kernel of this map
consists of even forms on $G$, which are exactly the elements of
$\mathrm{Hom}_{\mathbb{Z}}(G,Q_{+})$ (because if
$(G,\lambda,\mu^{\prime})\in\mathrm{Hom}_{\mathbb{Z}}(G,Q_{+})$, then
$2\mu^{\prime}(x)=\lambda(x,x)$). Moreover, the map
$\mathrm{Hom}_{\mathbb{Z}}(G,Q_{+})\rightarrow\mathrm{Hom}_{\mathbb{Z}}(G,Q^{+})$
sends $(G,\lambda,\mu^{\prime})$ to $(G,\lambda,\mu)$ (where
$2\mu^{\prime}(x)=\lambda(x,x)=\mu(x)$), therefore the bottom sequence is
exact.

By Lemma 2.29 the second square is a pullback square (note that
$\mathrm{Hom}_{\mathbb{Z}}(X,Q^{+}\oplus
SQ)\cong\mathrm{Hom}_{\mathbb{Z}}(G,Q^{+})\oplus\textup{Hom}(G,SQ)$). It
follows that the map
$\mathrm{Hom}_{\mathbb{Z}}(G,Q)\rightarrow\textup{Hom}(G,SQ)$ is surjective
too, and its kernel is also $\mathrm{Hom}_{\mathbb{Z}}(G,Q_{+})$, therefore
the top sequence is exact too. ∎

######  Corollary 4.13.

For every symmetric form parameter $Q$ and abelian group $G$ there is a short
exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{S^{2}(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes_{\mathbb{Z}}Q\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes
SQ\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

For every symmetric form parameter $Q$ and free abelian group $G$ there is a
short exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q_{+})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\textup{Hom}(G,SQ)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

These sequences are natural in $Q$ and $G$.

###### Proof.

These sequences are obtained from the sequence of Proposition 4.10 by applying
$G\otimes_{\mathbb{Z}}{-}$ and $\mathrm{Hom}_{\mathbb{Z}}(G,{-})$
respectively. So they are natural in $Q$ by Proposition 4.10 and they are
natural in $G$, because $G\otimes_{\mathbb{Z}}{-}$ and
$\mathrm{Hom}_{\mathbb{Z}}(G,{-})$ are bifunctors. Exactness was proven in
Proposition 4.11. ∎

######  Proposition 4.14.

For every anti-symmetric form parameter $Q=(Q_{e},{\rm h},{\rm p})$ there is a
sequence of quadratic modules

$\textstyle{I(Q_{e})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Q\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Q^{-},}$

which is natural in $Q$.

###### Proof.

The first map is given by $\textup{Id}_{Q_{e}}$ and the second one is the
unique morphism of form parameters $Q\rightarrow Q^{-}$, both of these are
natural. ∎

######  Proposition 4.15.

For every anti-symmetric form parameter $Q=(Q_{e},{\rm h},{\rm p})$ and
abelian group $G$ there are short exact sequences

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes
Q_{e}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes_{\mathbb{Z}}Q\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

and

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\textup{Hom}(G,Q_{e})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q^{-})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0.}$

These sequences are natural in $Q$ and $G$.

###### Proof.

These sequences are obtained from the sequence of Proposition 4.14 by applying
$G\otimes_{\mathbb{Z}}{-}$ and $\mathrm{Hom}_{\mathbb{Z}}(G,{-})$
respectively, so naturality follows as in Corollary 4.13. It remains to prove
the exactness of the sequences:

a) The map $G\otimes_{\mathbb{Z}}Q\rightarrow\Lambda(G)$ sends $g\otimes q$ to
$0$ and $[g,h]\otimes a$ to $ag\wedge h$, hence it is surjective. The map
$G\otimes Q_{e}\rightarrow G\otimes_{\mathbb{Z}}Q$ sends $g\otimes q$ to
$g\otimes q$, so its cokernel is generated by the symbols $[g,h]\otimes a$
(linear in all variables) subject to the relation $[g,g]\otimes a=0$, hence it
is isomorphic to $\Lambda(G)$. This shows that the sequence is exact at
$G\otimes_{\mathbb{Z}}Q$. If $G$ is written as the direct sum of cyclic groups
with generators $g_{1},\ldots,g_{k}$, then we can define a map
$G\otimes_{\mathbb{Z}}Q\rightarrow G\otimes Q_{e}$ by $g\otimes q\mapsto
g\otimes q$ (for all $g$), $[g_{i},g_{j}]\otimes a\mapsto 0$ if $i\neq j$ and
$[g_{i},g_{i}]\otimes a\mapsto g_{i}\otimes{\rm p}(a)$. This map is a left
inverse of $G\otimes Q_{e}\rightarrow G\otimes_{\mathbb{Z}}Q$, so the sequence
is also exact at $G\otimes Q_{e}$.

b) The homomorphism
$\mathrm{Hom}_{\mathbb{Z}}(G,Q)\rightarrow\mathrm{Hom}_{\mathbb{Z}}(G,Q^{-})$
is the forgetful map $(G,\lambda,\mu)\mapsto(G,\lambda)$. It is surjective,
because for every bilinear form $\lambda$ on $G$ with $\lambda(x,x)=0$ we can
define a refinement $\mu\colon G\rightarrow Q_{e}$. Namely, if $G$ is written
as the direct sum of cyclic groups with generators $g_{1},\ldots,g_{k}$, then
let $\mu(\sum_{i}a_{i}g_{i})={\rm
p}(\sum_{i<j}a_{i}a_{j}\lambda(g_{i},g_{j}))$. This is well-defined (because
if $g_{i}$ has infinite order, then $a_{i}\in\mathbb{Z}$ is well-defined, and
$\lambda(g_{i},g_{j})=0$ otherwise) and we can check that it satisfies
$\mu(x+y)=\mu(x)+\mu(y)+{\rm p}(\lambda(x,y))$ (using that ${\rm
p}\circ\lambda\colon G\times G\rightarrow Q_{e}$ is symmetric, because $2{\rm
p}=0$).

The kernel of the map
$\mathrm{Hom}_{\mathbb{Z}}(G,Q)\rightarrow\mathrm{Hom}_{\mathbb{Z}}(G,Q^{-})$
consists of forms $(G,\lambda,\mu)$ with $\lambda=0$ and $\mu\colon
G\rightarrow Q_{e}$ a homomorphism. Therefore the sequence is exact. ∎

######  Corollary 4.16.

a) For every anti-symmetric form parameter $Q=(Q_{e},{\rm h},{\rm p})$ there
is a commutative diagram of quadratic modules

$\textstyle{I(\mathbb{Z}_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Q_{-}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Q^{-}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{I(Q_{e})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Q\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Q^{-}}$

b) Given an abelian group $G$, if we apply the functor
$G\otimes_{\mathbb{Z}}{-}$ to the above diagram, then we get a commutative
diagram of abelian groups

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes\mathbb{Z}_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda_{1}(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes
Q_{e}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes_{\mathbb{Z}}Q\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

consisting of short exact sequences. In particular the first square in this
diagram is a pushout square.

c) Given an abelian group $G$, if we apply the functor
$\mathrm{Hom}_{\mathbb{Z}}(G,{-})$ to the diagram in part a), then we get a
commutative diagram of abelian groups

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\textup{Hom}(G,\mathbb{Z}_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q_{-})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q^{-})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\textup{Hom}(G,Q_{e})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Hom}_{\mathbb{Z}}(G,Q^{-})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

consisting of short exact sequences. In particular the first square in this
diagram is a pushout square.

###### Proof.

This follows from Propositions 4.14 and 4.15, applied to the unique morphism
of form parameters $Q_{-}\rightarrow Q$ given by
$v^{\prime}_{Q}\colon\mathbb{Z}_{2}\rightarrow Q_{e}$. ∎

###  4.3. Reduced Witt groups

Now we consider a split form parameter $Q\oplus G$, where $Q$ is a form
parameter and $G$ is an abelian group. This subsection is about the problem of
computing $W_{0}(Q\oplus G)$, assuming that $W_{0}(Q)$ is known. We have seen
that taking Witt groups defines a functor
$W_{0}\colon\mathbf{FP}\to\mathbf{Ab}$. Now there are obvious morphisms
$i\colon Q\to Q\oplus G$ and $r\colon Q\oplus G\to Q$ with $ri={\rm Id}_{Q}$,
and so $W_{0}(Q)$ is naturally identified with a summand of $W_{0}(Q\oplus
G)$.

######  Definition 4.17.

We define the reduced Witt group of $Q\oplus G$ by

$W_{0}^{Q}(G):={\rm Ker}(W_{0}(r))\leq W_{0}(Q\oplus G).$

Evidently $W_{0}(Q\oplus G)\cong W_{0}(Q)\oplus W_{0}^{Q}(G)$ and our problem
reduces to computing $W_{0}^{Q}(G)$.

######  Definition 4.18.

For a form parameter $Q$ and a finitely generated abelian group $G$ we define
a map

$F\colon\mathrm{Hom}_{\mathbb{Z}}^{\rm ns}(Q\oplus G)\rightarrow
G\otimes_{\mathbb{Z}}Q.$

If $(X,\lambda,(\mu_{Q},\mu_{G}))$ is a nonsingular $(Q\oplus G)$-form and
$x_{1},\ldots,x_{k}$ is a basis of $X$, then let $y_{1},\ldots,y_{k}$ be the
basis of $X$ which is dual to $x_{1},\ldots,x_{k}$ with respect to $\lambda$
(i.e. $\lambda(x_{i},y_{i})=1$ and $\lambda(x_{i},y_{j})=0$ if $i\neq j$), and
set

$F(X,\lambda,(\mu_{Q},\mu_{G})):=\sum_{i<j}\left[\mu_{G}(x_{i}),\mu_{G}(x_{j})\right]\otimes\lambda(y_{i},y_{j})+\sum_{i}\mu_{G}(x_{i})\otimes\mu_{Q}(y_{i}).$

######  Proposition 4.19.

$F$ is well-defined; i.e. $F(X,\lambda,(\mu_{Q},\mu_{G}))$ does not depend on
the choice of the basis.

###### Proof.

We need to check that the value of $F$ does not change when we apply
elementary transformations to the basis $x_{1},\ldots,x_{k}$.

If we swap $x_{1}$ and $x_{2}$, then $y_{1}$ and $y_{2}$ also get swapped. In
$F$ the only change is that
$[\mu_{G}(x_{1}),\mu_{G}(x_{2})]\otimes\lambda(y_{1},y_{2})$ is replaced with
$[\mu_{G}(x_{2}),\mu_{G}(x_{1})]\otimes\lambda(y_{2},y_{1})$, and these are
equal, because $[\cdot,\cdot]$ and $\lambda$ are both
$\epsilon_{Q}$-symmetric.

If we replace $x_{1}$ with $-x_{1}$, then $y_{1}$ is also replaced with
$-y_{1}$. In the first part of $F$ the terms
$[\mu_{G}(x_{1}),\mu_{G}(x_{j})]\otimes\lambda(y_{1},y_{j})$ for $j\geq 2$ are
replaced with

$[\mu_{G}(-x_{1}),\mu_{G}(x_{j})]\otimes\lambda(-y_{1},y_{j})=[\mu_{G}(x_{1}),\mu_{G}(x_{j})]\otimes\lambda(y_{1},y_{j}).$

In the second part $\mu_{G}(x_{1})\otimes\mu_{Q}(y_{1})$ is replaced with
$\mu_{G}(-x_{1})\otimes\mu_{Q}(-y_{1})$, we will show that these are equal. We
have $0=\mu_{Q}(y_{1}-y_{1})=\mu_{Q}(y_{1})+\mu_{Q}(-y_{1})+{\rm
p}(\lambda(y_{1},-y_{1}))$ and also
$\lambda(y_{1},-y_{1})=-\lambda(-y_{1},-y_{1})=-{\rm h}(\mu_{Q}(-y_{1}))$.
Hence

$0=\mu_{G}(x_{1})\otimes\mu_{Q}(y_{1}-y_{1})=\mu_{G}(x_{1})\otimes\mu_{Q}(y_{1})+\mu_{G}(x_{1})\otimes\mu_{Q}(-y_{1})-\mu_{G}(x_{1})\otimes{\rm
p}({\rm h}(\mu_{Q}(-y_{1}))).$

We also have $[\mu_{G}(x_{1}),\mu_{G}(-x_{1})]\otimes{\rm
h}(\mu_{Q}(-y_{1}))=-[\mu_{G}(x_{1}),\mu_{G}(x_{1})]\otimes{\rm
h}(\mu_{Q}(-y_{1}))=-\mu_{G}(x_{1})\otimes{\rm p}({\rm h}(\mu_{Q}(-y_{1})))$,
hence

$0=(\mu_{G}(x_{1})+\mu_{G}(-x_{1}))\otimes\mu_{Q}(-y_{1})=\mu_{G}(x_{1})\otimes\mu_{Q}(-y_{1})+\mu_{G}(-x_{1})\otimes\mu_{Q}(-y_{1})-\mu_{G}(x_{1})\otimes{\rm
p}({\rm h}(\mu_{Q}(-y_{1}))).$

By comparing these two equalities we see that
$\mu_{G}(x_{1})\otimes\mu_{Q}(y_{1})=\mu_{G}(-x_{1})\otimes\mu_{Q}(-y_{1})$.

Finally if we replace $x_{1}$ with $x_{1}+x_{2}$, then $y_{2}$ is replaced
with $y_{2}-y_{1}$. In $F$ the terms

$[\mu_{G}(x_{1}),\mu_{G}(x_{j})]\otimes\lambda(y_{1},y_{j})+[\mu_{G}(x_{2}),\mu_{G}(x_{j})]\otimes\lambda(y_{2},y_{j})$

are replaced by

$[\mu_{G}(x_{1}+x_{2}),\mu_{G}(x_{j})]\otimes\lambda(y_{1},y_{j})+[\mu_{G}(x_{2}),\mu_{G}(x_{j})]\otimes\lambda(y_{2}-y_{1},y_{j})=\\\
[\mu_{G}(x_{1}),\mu_{G}(x_{j})]\otimes\lambda(y_{1},y_{j})+[\mu_{G}(x_{2}),\mu_{G}(x_{j})]\otimes\lambda(y_{2},y_{j})$

for every $j\geq 3$. Further,
$[\mu_{G}(x_{1}),\mu_{G}(x_{2})]\otimes\lambda(y_{1},y_{2})+\mu_{G}(x_{1})\otimes\mu_{Q}(y_{1})+\mu_{G}(x_{2})\otimes\mu_{Q}(y_{2})$
is replaced by
$[\mu_{G}(x_{1}+x_{2}),\mu_{G}(x_{2})]\otimes\lambda(y_{1},y_{2}-y_{1})+\mu_{G}(x_{1}+x_{2})\otimes\mu_{Q}(y_{1})+\mu_{G}(x_{2})\otimes\mu_{Q}(y_{2}-y_{1})$.
The difference is

$[\mu_{G}(x_{2}),\mu_{G}(x_{2})]\otimes\lambda(y_{1},y_{2})+[\mu_{G}(x_{1}),\mu_{G}(x_{2})]\otimes\lambda(y_{1},-y_{1})+[\mu_{G}(x_{2}),\mu_{G}(x_{2})]\otimes\lambda(y_{1},-y_{1})+\\\
+\mu_{G}(x_{2})\otimes\mu_{Q}(y_{1})+[\mu_{G}(x_{1}),\mu_{G}(x_{2})]\otimes{\rm
h}(\mu_{Q}(y_{1}))+\mu_{G}(x_{2})\otimes\mu_{Q}(-y_{1})+\mu_{G}(x_{2})\otimes{\rm
p}(\lambda(y_{2},-y_{1}))=\\\
=[\mu_{G}(x_{2}),\mu_{G}(x_{2})]\otimes\lambda(y_{1},-y_{1})+\mu_{G}(x_{2})\otimes\mu_{Q}(y_{1})+\mu_{G}(x_{2})\otimes\mu_{Q}(-y_{1})=\\\
=\mu_{G}(x_{2})\otimes{\rm
p}(\lambda(y_{1},-y_{1}))+\mu_{G}(x_{2})\otimes\mu_{Q}(y_{1})+\mu_{G}(x_{2})\otimes\mu_{Q}(-y_{1})=\mu_{G}(x_{2})\otimes\mu_{Q}(y_{1}-y_{1})=0,$

where we used that ${\rm h}(\mu_{Q}(y_{1}))=-\lambda(y_{1},-y_{1})$ and ${\rm
p}(\lambda(y_{2},-y_{1}))=-{\rm p}(\lambda(y_{1},y_{2}))$.

Therefore the value of $F$ does not change when we change the basis. ∎

######  Proposition 4.20.

$F$ is a monoid homomorphism.

###### Proof.

Given $(X,\lambda,(\mu_{Q},\mu_{G}))$ and
$(X^{\prime},\lambda^{\prime},(\mu^{\prime}_{Q},\mu^{\prime}_{G}))\in\mathrm{Hom}_{\mathbb{Z}}^{\rm
ns}(Q\oplus G)$ let $x_{1},\ldots,x_{k}$ be a basis of $X$ and
$x^{\prime}_{1},\ldots,x^{\prime}_{l}$ be a basis of $X^{\prime}$ (with dual
bases $y_{1},\ldots,y_{k}$ and $y^{\prime}_{1},\ldots,y^{\prime}_{l}$,
respectively). Then

$F((X,\lambda,(\mu_{Q},\mu_{G}))\oplus(X^{\prime},\lambda^{\prime},(\mu^{\prime}_{Q},\mu^{\prime}_{G})))=\\\
F(X,\lambda,(\mu_{Q},\mu_{G}))+F(X^{\prime},\lambda^{\prime},(\mu^{\prime}_{Q},\mu^{\prime}_{G}))+\sum_{i\leq
k,j\leq
l}[\mu_{G}(x_{i}),\mu^{\prime}_{G}(x^{\prime}_{j})]\otimes(\lambda\oplus\lambda^{\prime})(y_{i},y^{\prime}_{j})=\\\
F(X,\lambda,(\mu_{Q},\mu_{G}))+F(X^{\prime},\lambda^{\prime},(\mu^{\prime}_{Q},\mu^{\prime}_{G})),$

because
$(\lambda\oplus\lambda^{\prime})(y,y^{\prime})=\lambda(y,0)+\lambda^{\prime}(0,y^{\prime})=0$
for every $y\in X$ and $y^{\prime}\in X^{\prime}$. ∎

######  Proposition 4.21.

The map $F$ induces a homomorphism $\tilde{F}\colon W_{0}(Q\oplus
G)\rightarrow G\otimes_{\mathbb{Z}}Q$.

###### Proof.

It is enough to prove that $F$ vanishes on metabolic forms. So suppose that
$(X,\lambda,(\mu_{Q},\mu_{G}))$ is a metabolic form over $Q\oplus G$, let
$L<X$ be a lagrangian and fix an identification $X\cong L\oplus L^{*}$. Let
$x_{1},\ldots,x_{k}$ be a basis of $L$, and let $\alpha_{1},\ldots,\alpha_{k}$
be the dual basis of $L^{*}$. Then
$x_{1},\ldots,x_{k},\alpha_{1},\ldots,\alpha_{k}$ is a basis of $X$, and the
dual basis of $X$ with respect to $\lambda$ is
$y_{1},\ldots,y_{k},\beta_{1},\ldots,\beta_{k}$, where
$y_{i}=\alpha_{i}+z_{i}$ for some $z_{i}\in L$ and
$\beta_{i}=\epsilon_{Q}x_{i}$. Since $L$ is a lagrangian, $\mu_{G}(x_{i})=0$,
$\lambda(\beta_{i},\beta_{j})=0$ and $\mu_{Q}(\beta_{i})=0$ for every $i,j$,
therefore in $F(X,\lambda,(\mu_{Q},\mu_{G}))$ every term vanishes. ∎

######  Definition 4.22.

We define a homomorphism $\gamma\colon G\otimes_{\mathbb{Z}}Q\rightarrow
W_{0}(Q\oplus G)$ by

$\gamma([g_{1},g_{2}]\otimes
1):=\left(\mathbb{Z}^{2},\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}g_{1}\\\
g_{2}\end{array}\right)\right)\quad\text{and}\quad\gamma(g\otimes
q):=\left(\mathbb{Z}^{2},\left[\begin{array}[]{cc}0&1\\\ \epsilon_{Q}&-{\rm
h}(q)\end{array}\right],\left(\begin{array}[]{c}g\\\
R(q)\end{array}\right)\right),$

where $R=\textup{Id}-{\rm p}{\rm h}\colon Q_{e}\rightarrow Q_{e}$.

######  Proposition 4.23.

The map $\gamma$ is a well-defined homomorphism.

###### Proof.

These triples denote $(Q\oplus G)$-forms, because ${\rm h}(g_{1})={\rm
h}(g_{2})={\rm h}(g)=0$ and ${\rm h}(R(q))=-{\rm h}(q)$; see Remark 2.20. The
matrices are nonsingular, so these $(Q\oplus G)$-forms represent elements of
$W_{0}(Q\oplus G)$. It remains to check that the values of $\gamma$ satisfy
the relations coming from the relations in $G\otimes_{\mathbb{Z}}Q$.

The first relation we consider is $[g_{1},g_{2}]\otimes
1+[g_{1}^{\prime},g_{2}]\otimes 1=[g_{1}+g_{1}^{\prime},g_{2}]\otimes 1$.
Since

$\left(\mathbb{Z}^{2}\left<e_{1},f_{1}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}g_{1}\\\
g_{2}\end{array}\right)\right)\oplus\left(\mathbb{Z}^{2}\left<e_{2},f_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}g_{1}^{\prime}\\\
g_{2}\end{array}\right)\right)=\\\
\left(\mathbb{Z}^{2}\left<e_{1}+e_{2},f_{1}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}g_{1}+g_{1}^{\prime}\\\
g_{2}\end{array}\right)\right)\oplus\left(\mathbb{Z}^{2}\left<e_{2},f_{2}-f_{1}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}g_{1}^{\prime}\\\
0\end{array}\right)\right)\sim\\\
\left(\mathbb{Z}^{2}\left<e_{1}+e_{2},f_{1}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}g_{1}+g_{1}^{\prime}\\\
g_{2}\end{array}\right)\right)$

(where $\mathbb{Z}^{2}\left<e_{1},f_{1}\right>$ denotes a copy of the group
$\mathbb{Z}^{2}$ with standard basis $e_{1},f_{1}$), we get that

$\gamma([g_{1},g_{2}]\otimes 1)+\gamma([g_{1}^{\prime},g_{2}]\otimes
1)=\gamma([g_{1}+g_{1}^{\prime},g_{2}]\otimes 1),$

i.e. the corresponding relation is satisfied by the values of $\gamma$.
Similarly, we get that

$\gamma([g_{1},g_{2}]\otimes 1)+\gamma([g_{1},g_{2}^{\prime}]\otimes
1)=\gamma([g_{1},g_{2}+g_{2}^{\prime}]\otimes 1).$

The above calculation also shows that for any $n\in\mathbb{Z}$ we have

$\gamma([g_{1},g_{2}]\otimes
n)=\left(\mathbb{Z}^{2},\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}ng_{1}\\\
g_{2}\end{array}\right)\right).$

We also have $\gamma(g\otimes q)+\gamma(g\otimes
q^{\prime})=\gamma(g\otimes(q+q^{\prime}))$, because

$\left(\mathbb{Z}^{2}\left<e_{1},f_{1}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-{\rm h}(q)\end{array}\right],\left(\begin{array}[]{c}g\\\
R(q)\end{array}\right)\right)\oplus\left(\mathbb{Z}^{2}\left<e_{2},f_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-{\rm
h}(q^{\prime})\end{array}\right],\left(\begin{array}[]{c}g\\\
R(q^{\prime})\end{array}\right)\right)=\\\
\left(\mathbb{Z}^{2}\left<e_{1},f_{1}+f_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-{\rm
h}(q+q^{\prime})\end{array}\right],\left(\begin{array}[]{c}g\\\
R(q+q^{\prime})\end{array}\right)\right)\oplus\\\
\left(\mathbb{Z}^{2}\left<e_{2}-e_{1},f_{2}+{\rm
h}(q^{\prime})e_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&\epsilon_{Q}{\rm
h}(q^{\prime})\end{array}\right],\left(\begin{array}[]{c}0\\\ q^{\prime}+{\rm
h}(q^{\prime})g\end{array}\right)\right)\sim\\\
\left(\mathbb{Z}^{2}\left<e_{1},f_{1}+f_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-{\rm
h}(q+q^{\prime})\end{array}\right],\left(\begin{array}[]{c}g\\\
R(q+q^{\prime})\end{array}\right)\right).$

Next, we have $\gamma([g,g]\otimes 1)=\gamma(g\otimes{\rm p}(1))$, because

$\left(\mathbb{Z}^{2}\left<e,f\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}g\\\
g\end{array}\right)\right)=\left(\mathbb{Z}^{2}\left<e,f-e\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-(1+\epsilon_{Q})\end{array}\right],\left(\begin{array}[]{c}g\\\
-{\rm p}(1)\end{array}\right)\right)$

and $-(1+\epsilon_{Q})=-{\rm h}{\rm p}(1)$ and $-{\rm p}(1)=R({\rm p}(1))$.

Finally $\gamma(g\otimes q)+\gamma(g^{\prime}\otimes
q)+\gamma([g,g^{\prime}]\otimes{\rm h}(q))=\gamma((g+g^{\prime})\otimes q)$,
because

$\left(\mathbb{Z}^{2}\left<e_{1},f_{1}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-{\rm h}(q)\end{array}\right],\left(\begin{array}[]{c}g\\\
R(q)\end{array}\right)\right)\oplus\left(\mathbb{Z}^{2}\left<e_{2},f_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-{\rm
h}(q)\end{array}\right],\left(\begin{array}[]{c}g^{\prime}\\\
R(q)\end{array}\right)\right)\oplus\\\
\left(\mathbb{Z}^{2}\left<e_{3},f_{3}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}{\rm h}(q)g\\\
g^{\prime}\end{array}\right)\right)=\left(\mathbb{Z}^{2}\left<e_{1}+e_{2},f_{1}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-{\rm
h}(q)\end{array}\right],\left(\begin{array}[]{c}g+g^{\prime}\\\
R(q)\end{array}\right)\right)\oplus\\\
\left(\mathbb{Z}^{4}\left<e_{2},f_{2}-f_{1}-{\rm
h}(q)e_{1},e_{3},f_{3}\right>,\left[\begin{array}[]{cccc}0&1&0&0\\\
\epsilon_{Q}&0&0&0\\\ 0&0&0&1\\\
0&0&\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}g^{\prime}\\\
-{\rm h}(q)g\\\ {\rm h}(q)g\\\ g^{\prime}\end{array}\right)\right)\sim\\\
\left(\mathbb{Z}^{2}\left<e_{1}+e_{2},f_{1}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-{\rm
h}(q)\end{array}\right],\left(\begin{array}[]{c}g+g^{\prime}\\\
R(q)\end{array}\right)\right),$

where we used that $(\epsilon_{Q}-1){\rm h}(q)=0$ (because ${\rm h}=0$ if
$\epsilon_{Q}=-1$) and that $\left<e_{2}-f_{3},f_{2}-f_{1}-{\rm
h}(q)e_{1}+e_{3}\right>$ is a lagrangian in
$\mathbb{Z}^{4}\left<e_{2},f_{2}-f_{1}-{\rm h}(q)e_{1},e_{3},f_{3}\right>$.

So the values of $\gamma$ satisfy the appropriate relations and therefore we
have a well-defined homomorphism $G\otimes_{\mathbb{Z}}Q\rightarrow
W_{0}(Q\oplus G)$. ∎

Having shown that $\gamma$ is a homomorphism, we next determine its image.

######  Proposition 4.24.

The homomorphism $\gamma\colon G\otimes_{\mathbb{Z}}Q\rightarrow W_{0}(Q\oplus
G)$ has image $\mathrm{Im}(\gamma)=W_{0}^{Q}(G)$.

###### Proof.

The natural map $\mathrm{Hom}_{\mathbb{Z}}^{\rm ns}(Q\oplus
G)\rightarrow\mathrm{Hom}_{\mathbb{Z}}^{\rm ns}(Q)$ sends both
$\gamma([g_{1},g_{2}]\otimes 1)$ and $\gamma(g\otimes q)$ to metabolic forms,
so the composition of $\gamma$ with the natural map $W_{0}(Q\oplus
G)\rightarrow W_{0}(Q)$ is trivial, therefore $\mathrm{Im}(\gamma)\subseteq
W_{0}^{Q}(G)$.

For the other direction recall that every element of $W_{0}^{Q}(G)$ has a
representative $(X,\lambda,(\mu_{Q},\mu_{G}))$ such that $(X,\lambda,\mu_{Q})$
is metabolic and that every metabolic $Q$-form can be written as a sum of
metabolic rank-$2$ $Q$-forms, so $(X,\lambda,(\mu_{Q},\mu_{G}))$ is the sum of
$(Q+G)$-forms of the form

$\left(\mathbb{Z}^{2},\left[\begin{array}[]{cc}0&1\\\ \epsilon_{Q}&{\rm
h}(q)\end{array}\right],\left(\begin{array}[]{c}g_{1}\\\
q+g_{2}\end{array}\right)\right).$

These elements are contained in $\mathrm{Im}(\gamma)$, because

$\gamma([g_{1},g_{2}]\otimes 1)+\gamma(g_{1}\otimes R(q))=\\\
\left(\mathbb{Z}^{2}\left<e_{1},f_{1}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&0\end{array}\right],\left(\begin{array}[]{c}g_{1}\\\
g_{2}\end{array}\right)\right)\oplus\left(\mathbb{Z}^{2}\left<e_{2},f_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&{\rm h}(q)\end{array}\right],\left(\begin{array}[]{c}g_{1}\\\
q\end{array}\right)\right)=\\\
\left(\mathbb{Z}^{2}\left<e_{1},f_{1}+f_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&{\rm h}(q)\end{array}\right],\left(\begin{array}[]{c}g_{1}\\\
q+g_{2}\end{array}\right)\right)\oplus\\\
\left(\mathbb{Z}^{2}\left<e_{2}-e_{1},f_{2}-{\rm
h}(q)e_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&-\epsilon_{Q}{\rm
h}(q)\end{array}\right],\left(\begin{array}[]{c}0\\\ R(q)-{\rm
h}(q)g_{1}\end{array}\right)\right)\sim\\\
\left(\mathbb{Z}^{2}\left<e_{1},f_{1}+f_{2}\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&{\rm h}(q)\end{array}\right],\left(\begin{array}[]{c}g_{1}\\\
q+g_{2}\end{array}\right)\right).$

Therefore $\mathrm{Im}(\gamma)\supseteq W_{0}^{Q}(G)$. ∎

We now relate the homomorphisms $\tilde{F}$ and $\gamma$.

######  Proposition 4.25.

The homomorphisms $\tilde{F}$ and $\gamma$ fit into the following split short
exact sequence:

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{W_{0}(Q)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{W_{0}(Q\oplus
G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tilde{F}}$$\textstyle{G\otimes_{\mathbb{Z}}Q\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\gamma}$$\textstyle{0}$

###### Proof.

We already noted at the start of this subsection that the map
$W_{0}(Q)\rightarrow W_{0}(Q\oplus G)$ and the splitting map $W_{0}(Q\oplus
G)\rightarrow W_{0}(Q)$ are induced by the morphisms $Q\rightarrow Q\oplus G$
and $Q\oplus G\rightarrow Q$ of form parameters. It follows immediately from
the definition that for any
$(X,\lambda,\mu_{Q})\in\mathrm{Hom}_{\mathbb{Z}}^{\rm ns}(Q)$ we have
$F(X,\lambda,(\mu_{Q},0))=0$, therefore the composition $W_{0}(Q)\rightarrow
W_{0}(Q\oplus G)\rightarrow G\otimes_{\mathbb{Z}}Q$ is trivial. A
straightforward calculation shows that
$\tilde{F}\circ\gamma=\textup{Id}_{G\otimes_{\mathbb{Z}}Q}$, i.e. $\gamma$ is
a splitting map. This implies that $\tilde{F}$ is surjective, and that the
restriction of $\tilde{F}$ to $\mathrm{Im}(\gamma)=W_{0}^{Q}(G)$ is injective,
so $\mathrm{Ker}(\tilde{F})=W_{0}(Q)$, which means that the sequence is exact.
∎

By observing that the map $F$ is natural in $Q$ and $G$, Proposition 4.25
immediately gives the main result of this section:

######  Theorem 4.26.

Let $Q$ be a form parameter and $G$ be an abelian group. Then the restriction
of $\tilde{F}$ and $\gamma$ define isomorphisms

$W_{0}^{Q}(G)\cong G\otimes_{\mathbb{Z}}Q,$

which are natural in $G$ and $Q$. ∎

###  4.4. Computing Witt groups via splittings of form parameters

By combining the results of Sections 4.1 and 4.3 we can calculate the Witt
group $W_{0}(P)$ for any form parameter $P$. The necessary ingredients are
summarised in the following table:

Table 1

$\begin{array}[]{c|cccc}~{}~{}~{}Q{}{}{}&\mathbb{Z}\otimes_{\mathbb{Z}}Q&\mathbb{Z}_{n}\otimes_{\mathbb{Z}}Q&~{}~{}~{}~{}~{}G\otimes_{\mathbb{Z}}Q{}{}{}{}{}&~{}~{}~{}~{}~{}W_{0}(Q){}{}{}{}{}\\\
\hline\cr\hline\cr Q_{+}&\mathbb{Z}&\mathbb{Z}_{n}&S^{2}(G)&\mathbb{Z}\\\
\mathbb{Z}^{P}&\mathbb{Z}\oplus\mathbb{Z}&\mathbb{Z}_{(n,2)n}\oplus\mathbb{Z}_{n/(2,n)}&P^{2}(G)&\mathbb{Z}\oplus\mathbb{Z}\\\
\mathbb{Z}^{P}_{k}&~{}~{}~{}\mathbb{Z}\oplus\mathbb{Z}_{2^{k}}&\mathbb{Z}_{(n,2)n}\oplus\mathbb{Z}_{(n/(2,n),2^{k})}&-&\qquad\\!\mathbb{Z}\oplus\mathbb{Z}_{2^{k-1}}\\\
Q^{+}&\mathbb{Z}&\mathbb{Z}_{(2,n)n}&\Gamma(G)&\mathbb{Z}\\\ \hline\cr
Q_{-}&\mathbb{Z}_{2}&\mathbb{Z}_{(2,n)}&\Lambda_{1}(G)&\mathbb{Z}_{2}\\\
\mathbb{Z}^{\Lambda}_{k}&\mathbb{Z}_{2^{k}}&\mathbb{Z}_{(2^{k},n)}&-&0\\\
Q^{-}&0&0&\Lambda(G)&0\\\ \end{array}$

Table 1 lists the values of the quadratic tensor product
$G\otimes_{\mathbb{Z}}Q$ for every cyclic group $G$ and indecomposable form
parameter $Q$ (where $(a,b)$ denotes the greatest common divisor of $a$ and
$b$), which can be computed directly from Definition 4.6 (or using the results
of Section 4.2). In some cases we also have a simple description of the
functor ${-}\otimes_{\mathbb{Z}}Q$ (see Example 4.7). Finally, we include the
Witt groups $W_{0}(Q)$ which were computed in Theorem 4.3.

Now suppose that $P$ is an arbitrary form parameter. By Theorem 3.3 it has a
maximal splitting $P\cong Q\oplus G$ (and the isomorphism classes of $Q$ and
$G$ are uniquely determined by that of $P$). Then we have the following:

######  Theorem 4.27.

a) A splitting $P\cong Q\oplus G$ of a form parameter $P$ defines an
isomorphism

$W_{0}(P)\cong W_{0}(Q\oplus G)\cong W_{0}(Q)\oplus W_{0}^{Q}(G)\cong
W_{0}(Q)\oplus(G\otimes_{\mathbb{Z}}Q).$

b) For any pair of abelian groups $G_{1}$ and $G_{2}$, and any form parameter
$Q$ we have

$(G_{1}\oplus
G_{2})\otimes_{\mathbb{Z}}Q\cong(G_{1}\otimes_{\mathbb{Z}}Q)\oplus(G_{2}\otimes_{\mathbb{Z}}Q)\oplus(G_{1}\otimes
G_{2}).$

###### Proof.

Part a) follows from Theorem 4.26, and Part b) follows from [Bau1, Proposition
4.2 (3)], since a form parameter is a quadratic module $Q$ with
$Q_{ee}=\mathbb{Z}$. ∎

Next we give some simple examples of computations using Theorem 4.27.

######  Example 4.28.

For a positive integer $l$, define
$\bar{\delta}(l),\delta(l)\in\mathbb{Z}^{+}$ by

$\bar{\delta}(l)=\begin{cases}1&\text{if $l$ is odd,}\\\ 2&\text{if $l$ is
even,}\end{cases}\qquad\text{and}\qquad\delta(l)=\bar{\delta}(l)l.$

  1. a)

There is an isomorphism
$W_{0}(Q_{+}\oplus\mathbb{Z}_{l})\cong\mathbb{Z}(8\sigma^{*})\oplus\mathbb{Z}_{l}$,
where the torsion summand is generated by the Witt class of the form

$\left(\mathbb{Z}^{2},\left[\begin{array}[]{cc}0&1\\\
1&0\end{array}\right],\left(\begin{array}[]{c}(0,1)\\\
(0,1)\end{array}\right)\right).$

  2. b)

There is an isomorphism
$W_{0}(Q^{+}\oplus\mathbb{Z}_{l})\cong\mathbb{Z}(\sigma^{*})\oplus\mathbb{Z}_{\delta(l)}$,
where the torsion summand is generated by the Witt class of the form

$\bigl{(}(\mathbb{Z},[1],(1,0))\oplus(\mathbb{Z},[-1],(-1,1)\bigr{)}\cong\left(\mathbb{Z}^{2},\left[\begin{array}[]{cc}0&1\\\
1&1\end{array}\right],\left(\begin{array}[]{c}(0,1)\\\
(0,1)\end{array}\right)\right).$

  3. c)

There is an isomorphism
$W_{0}(Q_{-}\oplus\mathbb{Z}_{l})\cong\mathbb{Z}_{2}(c^{*})\oplus\mathbb{Z}_{\bar{\delta}(l)}$,
where the second summand is generated by the Witt class of the form

$\left(\mathbb{Z}^{2},\left[\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right],\left(\begin{array}[]{c}(0,1)\\\
(0,1)\end{array}\right)\right).$

  4. d)

We have $W_{0}(Q^{-}\oplus\mathbb{Z}_{l})=0$.

To consider the naturality of the isomorphisms in Theorem 4.27, we recall from
Definition 2.12 that a morphism $\alpha\colon Q_{1}\oplus G_{1}\to Q_{2}\oplus
G_{2}$ of split form parameters is split if
$\alpha=\alpha_{Q}\oplus\alpha_{G}$ for a morphism $\alpha_{Q}\colon Q_{1}\to
Q_{2}$ and a homomorphism $\alpha_{G}\colon G_{1}\to G_{2}$. A morphism
$\alpha\colon P_{1}\to P_{2}$ is splittable if there are splittings of $P_{1}$
and $P_{2}$ with respect to which $\alpha$ is split.

######  Proposition 4.29.

Let $\alpha\colon P_{1}\to P_{2}$ be a splittable morphism of form parameters
and let $\alpha=\alpha_{Q}\oplus\alpha_{G}$ be a splitting of $\alpha$ with
respect to splittings $P_{1}\cong Q_{1}\oplus G_{1}$ and $P_{2}=Q_{2}\oplus
G_{2}$. Then

$W_{0}(\alpha)=W_{0}(\alpha_{Q})\oplus(\alpha_{G}\otimes_{\mathbb{Z}}\alpha_{Q})\colon
W_{0}(P_{1})\cong W_{0}(Q_{1})\oplus(G_{1}\otimes_{\mathbb{Z}}Q_{1})\to
W_{0}(P_{2})\cong W_{0}(Q_{2})\oplus(G_{2}\otimes_{\mathbb{Z}}Q_{2}).$

Remark. If the splittings of $P_{1}$ and $P_{2}$ in Proposition 4.29 are
maximal, then $\alpha_{Q}$ is the composition of some standard morphisms and
isomorphisms by Theorem 3.24, so $W_{0}(\alpha_{Q})$ can be determined using
Theorem 4.3.

###### Proof of Proposition 4.29.

It follows from the assumption that there is a commutative diagram

$\textstyle{Q_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{Q}}$$\textstyle{P_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha}$$\textstyle{Q_{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\alpha_{Q}}$$\textstyle{Q_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{P_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{Q_{2}}$

of form parameters involving the maps $Q_{i}\rightarrow P_{i}$ and
$P_{i}\rightarrow Q_{i}$ coming from the splittings (in particular the
composition $Q_{i}\rightarrow P_{i}\rightarrow Q_{i}$ is the identity map). If
we apply the functor $W_{0}$ to this, we get an analogous commutative diagram
between the Witt groups. Therefore the map $W_{0}(\alpha)$ is of the form
$W_{0}(\alpha_{Q})\oplus f\colon W_{0}(Q_{1})\oplus
W_{0}^{Q_{1}}(G_{1})\rightarrow W_{0}(Q_{2})\oplus W_{0}^{Q_{2}}(G_{2})$ for
some homomorphism $f\colon W_{0}^{Q_{1}}(G_{1})\rightarrow
W_{0}^{Q_{2}}(G_{2})$ (recall that
$W_{0}^{Q_{i}}(G_{i})\cong\mathrm{Ker}(W_{0}(P_{i})\rightarrow
W_{0}(Q_{i}))$). Since the isomorphism $W_{0}^{Q_{i}}(G_{i})\cong
G_{i}\otimes_{\mathbb{Z}}Q_{i}$ of Theorem 4.26 is natural, we have
$f=\alpha_{G}\otimes_{\mathbb{Z}}\alpha_{Q}\colon
G_{1}\otimes_{\mathbb{Z}}Q_{1}\rightarrow G_{2}\otimes_{\mathbb{Z}}Q_{2}$. ∎

######  Example 4.30.

The morphism of form parameters
$Q_{-}\oplus\mathbb{Z}_{2}\xrightarrow{\textup{i}\oplus{\rm
Id}}\mathbb{Z}^{\Lambda}_{2}\oplus\mathbb{Z}_{2}$ induces the zero map

$W_{0}(Q_{-}\oplus\mathbb{Z}_{2})\cong(\mathbb{Z}_{2}(c^{*})\oplus\mathbb{Z}_{2})\xrightarrow{~{}0~{}}(0\oplus\mathbb{Z}_{2})\cong
W_{0}(\mathbb{Z}^{\Lambda}_{2}\oplus\mathbb{Z}_{2}).$

##  5\. A natural description of Witt groups

By the results of the previous section we can determine the isomorphism type
of $W_{0}(P)$ for any form parameter $P$, by choosing a maximal splitting.
Proposition 4.29 can also be used to determine the homomorphism between the
Witt groups induced by a splittable morphism between form parameters. However,
as Example 2.13 shows, not all morphisms are splittable (with respect to some
non-trivial splittings of the form parameters).

In this section we give a natural description of the group $W_{0}(P)$, for any
form parameter $P$, which, in particular, allows us to determine the
homomorphism induced by any morphism of form parameters.

We begin by outlining the strategy. Recall that for any symmetric form
parameter $P$ there is an extended symmetrisation morphism, $\textup{es}\colon
P\to Q^{+}\oplus SP$. In fact, the morphisms es define a natural
transformation from the identity to the extended symmetrisation functor
$\textup{ES}\colon\mathbf{FP}_{+}\rightarrow\mathbf{FP}_{+}$, where
$\textup{ES}(P):=Q^{+}\oplus SP$ on objects and
$\textup{ES}(\alpha):=\textup{Id}_{Q^{+}}\oplus S\alpha$ on morphisms. By
Theorem 4.27, there is a natural isomorphism $W_{0}(Q^{+}\oplus
SP)\cong\mathbb{Z}\oplus\Gamma(SP)$, and we will show that the induced
homomorphism

$W_{0}(\textup{es})\colon W_{0}(P)\to W_{0}(Q^{+}\oplus
SP)\cong\mathbb{Z}\oplus\Gamma(SP)$

is injective. Then we will determine $\mathrm{Im}(W_{0}(\textup{es}))$ as a
natural subgroup of $\mathbb{Z}\oplus\Gamma(SP)$. Then, for any morphism
$\alpha\colon P\rightarrow P^{\prime}$, we can compute the induced map
$W_{0}(\alpha)\colon W_{0}(P)\to W_{0}(P^{\prime})$ as the restriction of
$\textup{Id}_{\mathbb{Z}}\oplus\Gamma(S\alpha)\colon\mathbb{Z}\oplus\Gamma(SP)\rightarrow\mathbb{Z}\oplus\Gamma(SP)$,
which is a homomorphism
$\mathrm{Im}(W_{0}(\textup{es}_{P}))\rightarrow\mathrm{Im}(W_{0}(\textup{es}_{P^{\prime}}))$.

Dually, for any anti-symmetric form parameter $P$ there is an extended
quadratic lift morphism, $\textup{eql}\colon Q_{-}\oplus P_{e}\to P$, and the
morphisms eql define a natural transformation from the extended quadratic lift
functor EQL to the identity, where
$\textup{EQL}\colon\mathbf{FP}_{-}\rightarrow\mathbf{FP}_{-}$ has
$\textup{EQL}(P):=Q_{-}\oplus P_{e}$ on objects and
$\textup{EQL}(\alpha):=\textup{Id}_{Q_{-}}\oplus\alpha$ on morphisms. By
Theorem 4.27, there is a natural isomorphism $W_{0}(Q_{-}\oplus
P_{e})\cong\mathbb{Z}_{2}\oplus\Lambda_{1}(P_{e})$. We will show that the
induced homomorphism

$W_{0}(\textup{eql})\colon W_{0}(Q_{-}\oplus
P_{e})\cong\mathbb{Z}_{2}\oplus\Lambda_{1}(P_{e})\to W_{0}(P)$

is surjective and determine $\mathrm{Ker}(W_{0}(\textup{eql}))$ as a natural
subgroup of $\mathbb{Z}_{2}\oplus\Lambda_{1}(P_{e})$. Then, for any morphism
$\alpha\colon P\rightarrow P^{\prime}$, the homomorphism $W_{0}(\alpha)\colon
W_{0}(P)\to W_{0}(P^{\prime})$ can be computed by identifying it with the
homomorphism
$\mathbb{Z}_{2}\oplus\Lambda_{1}(P_{e})/\mathrm{Ker}(W_{0}(\textup{eql}_{P}))\rightarrow\mathbb{Z}_{2}\oplus\Lambda_{1}(P^{\prime}_{e})/\mathrm{Ker}(W_{0}(\textup{eql}_{P^{\prime}}))$,
which is induced by the homomorphism
$\textup{Id}_{\mathbb{Z}_{2}}\oplus\Lambda_{1}(\alpha)\colon\mathbb{Z}_{2}\oplus\Lambda_{1}(P_{e})\rightarrow\mathbb{Z}_{2}\oplus\Lambda_{1}(P^{\prime}_{e})$.

###  5.1. Symmetric form parameters

First we consider the restriction of the Witt group functor to the category of
symmetric form parameters, $W_{0}\colon\mathbf{FP}_{+}\to\mathbf{Ab}$.

######  Lemma 5.1.

If $Q$ is an indecomposable symmetric form parameter, then the induced
homomorphism $W_{0}(\textup{es})\colon W_{0}(Q)\to W_{0}(Q^{+}\oplus SQ)$ is
injective.

###### Proof.

Since $Q^{+}$ is the terminal object, the composition $Q\rightarrow
Q^{+}\oplus SQ\rightarrow Q^{+}$ is the unique morphism, and the induced map
$W_{0}(Q)\to W_{0}(Q^{+}\oplus SQ)\cong W_{0}(Q^{+})\oplus
W_{0}^{Q^{+}}(SQ)\rightarrow W_{0}(Q^{+})$ is given by Theorem 4.3. To compute
the other component $W_{0}(Q)\rightarrow W_{0}^{Q^{+}}(SQ)\cong
SQ\otimes_{\mathbb{Z}}Q^{+}$ we will use the invariant $F$ from Definition
4.18, which induces the isomorphism $\tilde{F}$ (Theorem 4.26).

If $Q=Q_{+}$, then $SQ_{+}=0$, and the map $W_{0}(Q_{+})\to W_{0}(Q^{+}\oplus
SQ_{+})\cong W_{0}(Q^{+})$ is equal to the inclusion
$\mathbb{Z}(8\sigma^{*})\rightarrow\mathbb{Z}(\sigma^{*})$.

If $Q=Q^{+}$, then $SQ^{+}=\mathbb{Z}_{2}$. The composition $W_{0}(Q^{+})\to
W_{0}(Q^{+}\oplus SQ^{+})\rightarrow W_{0}(Q^{+})$ is of course the identity,
hence the map $W_{0}(Q^{+})\to W_{0}(Q^{+}\oplus SQ^{+})$ is injective.

If $Q=\mathbb{Z}^{P}$, then $S\mathbb{Z}^{P}=\mathbb{Z}$ and
$W_{0}^{Q^{+}}(S\mathbb{Z}^{P})\cong\mathbb{Z}\otimes_{\mathbb{Z}}Q^{+}\cong\mathbb{Z}$
by Theorem 4.27. The generators of
$W_{0}(\mathbb{Z}^{P})\cong\mathbb{Z}(\sigma^{*})\oplus\mathbb{Z}(\rho_{\infty}^{*})$
are (the equivalence classes of) $\sigma^{*}=(\mathbb{Z},1,(1,0))$ and
$\rho_{\infty}^{*}=\left(\mathbb{Z}^{2},\left[\begin{smallmatrix}0&1\\\
1&0\end{smallmatrix}\right],\left(\begin{smallmatrix}(0,1)\\\
(0,-1)\end{smallmatrix}\right)\right)$ (it is easy to verify that for these
forms we have $\sigma(\sigma^{*})=\rho_{\infty}(\rho_{\infty}^{*})=1$ and
$\rho_{\infty}(\sigma^{*})=\sigma(\rho_{\infty}^{*})=0$). The extended
symmetrisation map is given by the homomorphism
$\left(\begin{smallmatrix}1&0\\\
1&2\end{smallmatrix}\right)\colon\mathbb{Z}^{P}_{e}=\mathbb{Z}\oplus\mathbb{Z}\rightarrow(Q^{+}\oplus
S\mathbb{Z}^{P})_{e}=\mathbb{Z}\oplus\mathbb{Z}$, so the generators are mapped
to $(\mathbb{Z},1,(1,1))$ and $(\mathbb{Z}^{2},\left[\begin{smallmatrix}0&1\\\
1&0\end{smallmatrix}\right],\left(\begin{smallmatrix}(0,2)\\\
(0,-2)\end{smallmatrix}\right))$ respectively. Their images under $F$ are
$1\otimes 1$ and $[2,-2]\otimes 1$, and under the isomorphism
$\mathbb{Z}\otimes_{\mathbb{Z}}Q^{+}\cong\mathbb{Z}$ these correspond to $1$
and $-8$. Therefore the extended symmetrisation map induces the homomorphism
$\left(\begin{smallmatrix}1&0\\\ 1&-8\end{smallmatrix}\right)\colon
W_{0}(\mathbb{Z}^{P})=\mathbb{Z}\oplus\mathbb{Z}\rightarrow W_{0}(Q^{+}\oplus
S\mathbb{Z}^{P})=\mathbb{Z}\oplus\mathbb{Z}$, and this is injective.

If $Q=\mathbb{Z}^{P}_{k}$, then $S\mathbb{Z}^{P}_{k}=\mathbb{Z}_{2^{k+1}}$ and
$W_{0}^{Q^{+}}\\!\\!(S\mathbb{Z}^{P}_{k})\cong\mathbb{Z}_{2^{k+1}}\otimes_{\mathbb{Z}}Q^{+}\cong\mathbb{Z}_{2^{k+2}}$
by Theorem 4.27. The generators of
$W_{0}(\mathbb{Z}^{P}_{k})\cong\mathbb{Z}(\sigma^{*})\oplus\mathbb{Z}_{2^{k-1}}(\rho_{k}^{*})$
are $\sigma^{*}=(\mathbb{Z},1,(1,0))$ and
$\rho_{k}^{*}=(\mathbb{Z}^{2},\left[\begin{smallmatrix}0&1\\\
1&0\end{smallmatrix}\right],\left(\begin{smallmatrix}(0,1)\\\
(0,-1)\end{smallmatrix}\right))$. The extended symmetrisation map is given by
$\left(\begin{smallmatrix}1&0\\\
1&2\end{smallmatrix}\right)\colon(\mathbb{Z}^{P}_{k})_{e}=\mathbb{Z}\oplus\mathbb{Z}_{2^{k}}\rightarrow(Q^{+}\oplus
S\mathbb{Z}^{P}_{k})_{e}=\mathbb{Z}\oplus\mathbb{Z}_{2^{k+1}}$, so the
generators are mapped to $(\mathbb{Z},1,(1,1))$ and
$(\mathbb{Z}^{2},\left[\begin{smallmatrix}0&1\\\
1&0\end{smallmatrix}\right],\left(\begin{smallmatrix}(0,2)\\\
(0,-2)\end{smallmatrix}\right))$ respectively. Their images under $F$ are
$1\otimes 1$ and $[2,-2]\otimes 1$, and under the isomorphism
$\mathbb{Z}_{2^{k+1}}\otimes_{\mathbb{Z}}Q^{+}\cong\mathbb{Z}_{2^{k+2}}$ these
correspond to $1$ and $-8$. Therefore the extended symmetrisation map induces
$\left(\begin{smallmatrix}1&0\\\ 1&-8\end{smallmatrix}\right)\colon
W_{0}(\mathbb{Z}^{P}_{k})=\mathbb{Z}\oplus\mathbb{Z}_{2^{k-1}}\rightarrow
W_{0}(Q^{+}\oplus S\mathbb{Z}^{P}_{k})=\mathbb{Z}\oplus\mathbb{Z}_{2^{k+2}}$,
and this map is also injective. ∎

######  Lemma 5.2.

If $P$ is a symmetric form parameter, then $W_{0}(\textup{es})\colon
W_{0}(P)\to W_{0}(Q^{+}\oplus SP)$ is injective.

###### Proof.

By Theorem 3.3 it is enough to prove the statement for form parameters of the
form $P=Q\oplus G$, where $Q$ is an indecomposable form parameter and $G$ is
an abelian group.

We have $SP\cong SQ\oplus G$, and the map $\textup{es}\colon P=Q\oplus
G\rightarrow Q^{+}\oplus SP\cong(Q^{+}\oplus SQ)\oplus G$ splits as the sum of
$\textup{es}\colon Q\rightarrow Q^{+}\oplus SQ$ and $\textup{Id}_{G}$.
Therefore the induced map

$W_{0}(P)=W_{0}(Q)\oplus W_{0}^{Q}(G)\rightarrow W_{0}(Q^{+}\oplus SP)\cong
W_{0}(Q^{+}\oplus SQ)\oplus W_{0}^{Q^{+}\oplus SQ}(G)$

splits as the sum of $W_{0}(Q)\rightarrow W_{0}(Q^{+}\oplus SQ)$ and the map

$W_{0}^{Q}(G)\cong G\otimes_{\mathbb{Z}}Q\rightarrow W_{0}^{Q^{+}\oplus
SQ}(G)\cong G\otimes_{\mathbb{Z}}(Q^{+}\oplus
SQ)\cong(G\otimes_{\mathbb{Z}}Q^{+})\oplus(G\otimes
SQ)\cong\Gamma(G)\oplus(G\otimes SQ).$

The former is injective by Lemma 5.1. By Proposition 4.11 b) there is a
pullback diagram

$\textstyle{G\otimes_{\mathbb{Z}}Q\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes
SQ\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Gamma(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes\mathbb{Z}_{2}.}$

The map $W_{0}^{Q}(G)\cong G\otimes_{\mathbb{Z}}Q\rightarrow
W_{0}^{Q^{+}\oplus SQ}(G)\cong\Gamma(G)\oplus(G\otimes SQ)$ is the direct sum
of the maps $G\otimes_{\mathbb{Z}}Q\rightarrow\Gamma(G)$ and
$G\otimes_{\mathbb{Z}}Q\rightarrow G\otimes SQ$ in this diagram, hence it is
injective too. ∎

Next we determine the image of $W_{0}(\textup{es})\colon W_{0}(P)\to
W_{0}(Q^{+}\oplus SP)\cong\mathbb{Z}\oplus\Gamma(SP)$. Given an abelian group
$A$, we will use the identification $\Gamma(A)\cong
A\otimes_{\mathbb{Z}}Q^{+}$ to describe elements of $\Gamma(A)$ (in
particular, $\Gamma(A)$ is generated by the symbols $a\otimes 1$ and
$[a,b]\otimes 1$ for $a,b\in A$). Note that there is a short exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{S^{2}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Gamma(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes\mathbb{Z}_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

where the first map sends $ab$ to $[a,b]\otimes 1$, and the second map sends
$[a,b]\otimes 1$ to $0$ and $a\otimes 1$ to $a\otimes 1$ (see Proposition 4.11
b) or [Bau1, Remark 2.10]).

######  Definition 5.3.

Let $v\colon A\to\mathbb{Z}_{2}$ be a homomorphism. We define the subgroup

$\Sigma(v):=\left<(1,x\otimes 1),(0,[k_{1},k_{2}]\otimes 1),(8,0)\bigm{|}x\in
v^{-1}(1),k_{1},k_{2}\in\mathrm{Ker}(v)\right>\leq\mathbb{Z}\oplus\Gamma(A).$

We note that if $\alpha\colon v_{1}\rightarrow v_{2}$ is a morphism in
$\mathbf{Ab}/\mathbb{Z}_{2}$, then the induced map
$\textup{Id}_{\mathbb{Z}}\oplus\Gamma(\alpha)\colon\mathbb{Z}\oplus\Gamma(A_{1})\rightarrow\mathbb{Z}\oplus\Gamma(A_{2})$
sends the generators of $\Sigma(v_{1})$ to generators of $\Sigma(v_{2})$. This
shows that $\Sigma$ defines a functor
$\mathbf{Ab}/\mathbb{Z}_{2}\to\mathbf{Ab}$.

######  Lemma 5.4.

Let $v\colon A\to\mathbb{Z}_{2}$ be a homomorphism.

a) If $v=0$, then $\Sigma(v)=\left<(0,[k_{1},k_{2}]\otimes
1),(8,0)\bigm{|}k_{1},k_{2}\in\mathrm{Ker}(v)\right>$.

b) If $v\neq 0$, then $\Sigma(v)=\left<(1,x\otimes 1)\bigm{|}x\in
v^{-1}(1)\right>$.

###### Proof.

Part a) holds, because $v^{-1}(1)=\emptyset$.

b) Let $\Sigma^{\prime}(v)$ denote the right hand side, we need to show that
it contains the remaining generators of $\Sigma(v)$. First note that if $x\in
v^{-1}(1)$ and $k\in\mathrm{Ker}(v)$, then $[k,x]\otimes 1+k\otimes
1=(x+k)\otimes 1-x\otimes 1$, hence $(0,[k,x]\otimes 1+k\otimes
1)\in\Sigma^{\prime}(v)$. We have $[k_{1},k_{2}]\otimes
1=([k_{1},x+k_{2}]\otimes 1+k_{1}\otimes 1)-([k_{1},x]\otimes 1+k_{1}\otimes
1)$ for any $k_{1},k_{2}\in\mathrm{Ker}(v)$ and $x\in v^{-1}(1)$, therefore
$(0,[k_{1},k_{2}]\otimes 1)\in\Sigma^{\prime}(v)$. Finally, for any $y\in A$,
we have $[2y,2y]\otimes 1=2y\otimes 2=y\otimes 2+y\otimes 2+[y,y]\otimes
2=y\otimes 8$. So, with any $x\in v^{-1}(1)$, we have $(8,0)=8(1,x\otimes
1)-(0,x\otimes 8)\in\Sigma^{\prime}(v)$. ∎

######  Theorem 5.5.

Let $P$ be a symmetric form parameter. Then under the identification
$W_{0}(Q^{+}\oplus SP)\cong\mathbb{Z}\oplus\Gamma(SP)$ we have
$\mathrm{Im}(W_{0}(\textup{es}))=\Sigma(v_{P})$.

###### Proof.

First assume that $v_{P}=0$, i.e. $P\cong Q_{+}\oplus SP$. It follows from
Lemmas 5.1 and 5.2 that the map $W_{0}(\textup{es})\colon W_{0}(P)\cong
W_{0}(Q_{+}\oplus SP)\cong W_{0}(Q_{+})\oplus W_{0}^{Q_{+}}(SP)\rightarrow
W_{0}(Q^{+}\oplus SP)\cong W_{0}(Q^{+})\oplus W_{0}^{Q^{+}}(SP)$ splits as the
sum of the inclusions $W_{0}(Q_{+})\cong\mathbb{Z}(8\sigma^{*})\to
W_{0}(Q^{+})\cong\mathbb{Z}(\sigma^{*})$ and $W_{0}^{Q_{+}}(SP)\cong
S^{2}(SP)\rightarrow W_{0}^{Q^{+}}(SP)\cong\Gamma(SP)$. Therefore the image of
$W_{0}(\textup{es})$ is generated by $(8,0)$ and the elements
$(0,[k_{1},k_{2}]\otimes 1)$ for $k_{1},k_{2}\in SP=\mathrm{Ker}(v_{P})$, so
it is equal to $\Sigma(v_{P})$ by Lemma 5.4 a).

Now assume that $v_{P}\neq 0$. By adding the metabolic $P$-form

$\left(\mathbb{Z}^{2},\left[\begin{array}[]{cc}0&1\\\ 1&{\rm
h}(q)\end{array}\right],\left(\begin{array}[]{c}0\\\
q\end{array}\right)\right)$

(where $q\in v_{P}^{-1}(1)$) we get that every element of $W_{0}(P)$ has a
representative $(X,\lambda,\mu)$ such that $\lambda$ is indefinite and odd. By
the classification of nonsingular indefinite forms (see [M-H, Ch. II. Theorem
(5.3)]), $(X,\lambda,\mu)$ is isomorphic to the sum of $P$-forms of the form
$(\mathbb{Z},[\pm 1],q)$ (where ${\rm h}(q)=\pm 1$). Therefore elements of
this form generate $W_{0}(P)$. The image of such a form under extended
symmetrisation is $(\mathbb{Z},[\pm 1],(\pm 1,\pi_{P}(q)))$, which corresponds
to $(\pm 1,\pi_{P}(q)\otimes(\pm 1))=\pm(1,\pi_{P}(q)\otimes 1)$ under the
isomorphism $W_{0}(Q_{+}\oplus SP)\cong\mathbb{Z}\oplus\Gamma(SP)$ (see
Definition 4.18). Since $\\{\pi_{P}(q)\mid{\rm h}(q)=\pm 1\\}=v_{P}^{-1}(1)$,
this implies that $\mathrm{Im}(W_{0}(\textup{es}))=\left<(1,x\otimes 1)\mid
x\in v_{P}^{-1}(1)\right>$. This is equal to $\Sigma(v_{P})$ by Lemma 5.4 b).
∎

Using Lemma 5.2, Theorem 5.5 and the fact that extended symmetrisation is a
natural transformation, we get

######  Corollary 5.6.

The homomorphisms induced by extended symmetrisation define a natural
isomorphism $W_{0}\cong\Sigma\circ v_{(-)}:\mathbf{FP}_{+}\to\mathbf{Ab}$. ∎

To give a more explicit description of the group $\Sigma(v)$, we include
Theorem 5.8 below, which requires the following

######  Definition 5.7.

Let $v\colon A\to\mathbb{Z}_{2}$ be a homomorphism. We define the subgroups

$\displaystyle\Phi(v)$ $\displaystyle:=\left<x\otimes 1,[k,x]\otimes
1+k\otimes 1,[k_{1},k_{2}]\otimes 1\bigm{|}x\in
v^{-1}(1),k,k_{1},k_{2}\in\mathrm{Ker}(v)\right>\leq\Gamma(A),$
$\displaystyle\Psi(v)$ $\displaystyle:=\left<[k,x]\otimes 1+k\otimes
1,[k_{1},k_{2}]\otimes 1\bigm{|}x\in
v^{-1}(1),k,k_{1},k_{2}\in\mathrm{Ker}(v)\right>\leq\Phi(v),$

and the groups

$\displaystyle C(v)$ $\displaystyle:=\begin{cases}\mathbb{Z}_{8}&\text{if
}v=0,\\\ \Phi(v)/\Psi(v)&\text{if }v\neq 0,\end{cases}$
$\displaystyle\Upsilon(v)$
$\displaystyle:=\begin{cases}\mathbb{Z}_{8}\oplus\Gamma(A)/\Psi(v)&\text{if
}v=0,\\\ \Gamma(A)/\Psi(v)&\text{if }v\neq 0.\end{cases}$

######  Theorem 5.8.

Let $v\colon A\to\mathbb{Z}_{2}$ be a homomorphism.

  1. a)

We have

$C(v)\cong\begin{cases}\mathbb{Z}_{4}&\text{if $v\cong 1_{1}\oplus G$ for some
$G$,}\\\ \mathbb{Z}_{8}&\text{otherwise}.\end{cases}$

Moreover, if $v\neq 0$, then $C(v)$ is generated by $[x\otimes 1]$ for any
$x\in v^{-1}(1)$.

  2. b)

There is a commutative diagram with short exact rows and columns

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Sigma(v)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{Z}\oplus\Phi(v)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota-q}$$\textstyle{C(v)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Sigma(v)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{Z}\oplus\Gamma(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{0+u_{v}}$$\scriptstyle{\iota-\bar{q}}$$\textstyle{\Upsilon(v)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\tilde{u}_{v}}$$\textstyle{0}$$\textstyle{\mathrm{Ker}(v_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathrm{Ker}(v_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0,}$

where the following hold:

     * $\bullet$

$v_{2}=v\otimes\textup{Id}_{\mathbb{Z}_{2}}\colon
A\otimes\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}=\mathbb{Z}_{2}$,
hence $\mathrm{Ker}(v_{2})\leq A\otimes\mathbb{Z}_{2}$;

     * $\bullet$

The homomorphism $\iota\colon\mathbb{Z}\rightarrow C(v)\leq\Upsilon(v)$ is
defined by

$\iota(1)=\begin{cases}1&\text{if }v=0,\\\ [x\otimes 1]\text{~{}for some $x\in
v^{-1}(1)$}&\text{if }v\neq 0;\end{cases}$

     * $\bullet$

The homomorphism $q\colon\Phi(v)\rightarrow C(v)$ is the quotient map
$\Phi(v)\rightarrow\Phi(v)/\Psi(v)$ if $v\neq 0$, and trivial if $v=0$;

     * $\bullet$

The homomorphism $\bar{q}\colon\Gamma(A)\rightarrow\Upsilon(v)$ is the
quotient map $\Gamma(A)\rightarrow\Gamma(A)/\Psi(v)$;

     * $\bullet$

The homomorphism $u_{v}\colon\Gamma(A)\rightarrow A\otimes\mathbb{Z}_{2}$ is
given by $u_{v}([a_{1},a_{2}]\otimes 1)=a_{1}\otimes v(a_{2})+a_{2}\otimes
v(a_{1})$ and $u_{v}(a\otimes 1)=a\otimes 1+a\otimes v(a)$;

     * $\bullet$

The homomorphism $\tilde{u}_{v}\colon\Upsilon(v)\rightarrow
A\otimes\mathbb{Z}_{2}$ is the map $\Gamma(A)/\Psi(v)\rightarrow
A\otimes\mathbb{Z}_{2}$ induced by $u_{v}$ (precomposed with the projection
$\mathbb{Z}_{8}\oplus\Gamma(A)/\Psi(v)\rightarrow\Gamma(A)/\Psi(v)$ if $v=0$);

     * $\bullet$

The unlabelled maps are the obvious inclusions.

######  Remark 5.9.

Theorem 5.8 shows that $\Sigma(v)$ can be described as the kernel of a map
$\mathbb{Z}\oplus\Gamma(A)\rightarrow\Upsilon(v)$, where $\Upsilon(v)$ is an
extension of the $\mathbb{Z}_{2}$-vector space $\mathrm{Ker}(v_{2})$ by the
cyclic group $C(v)$. Alternatively, it is the subgroup of elements
$(n,\alpha)\in\mathbb{Z}\oplus\Phi(v)$ such that $n\equiv q(\alpha)$ mod $4$
or $8$ (depending on $C(v)$), where $\Phi(v)$ is the kernel of the surjective
map $u_{v}\colon\Gamma(A)\rightarrow\mathrm{Ker}(v_{2})$. In particular, we
see that $\Sigma(v)\cap(0\oplus\Gamma(A))=\mathrm{Ker}(\bar{q})=\Psi(v)$, and
that the image of the composition
$\Sigma(v)\rightarrow\mathbb{Z}\oplus\Gamma(A)\rightarrow\Gamma(A)$ is
$\Phi(v)$.

###### Proof of Theorem 5.8.

a) If $v=0$, then the statement holds by definition. In the rest of this proof
we restrict to the $v\neq 0$ case. Then the elements $[x\otimes 1]$ with
$v(x)=1$ generate $\Phi(v)/\Psi(v)$. If $v(x)=v(y)=1$, then $x\otimes
1-y\otimes 1=(x-y)\otimes 1+[y,x-y]\otimes 1\in\Psi(v)$, so all of these
generators coincide. Hence $\Phi(v)/\Psi(v)$ is cyclic.

By Theorem 3.14 $v$ is isomorphic to $v_{0}\oplus G$ for some abelian group
$G$, where $v_{0}\colon A_{0}\rightarrow\mathbb{Z}_{2}$ is either
$1_{k}\colon\mathbb{Z}_{2^{k}}\rightarrow\mathbb{Z}_{2}$ for some positive
integer $k$ or $1_{\infty}\colon\mathbb{Z}\rightarrow\mathbb{Z}_{2}$. Since
$\Phi(v)$ and $\Psi(v)$ depend functorially on $v$, the morphism
$v_{0}\rightarrow v_{0}\oplus G\cong v$ induces a homomorphism
$C(v_{0})\rightarrow C(v)$. This map is injective, because the composition
$v_{0}\rightarrow v_{0}\oplus G\rightarrow v_{0}$ is the identity, and
surjective, because it maps a generator of the cyclic group $C(v_{0})$ (of the
form $[x\otimes 1]$) into a generator of $C(v)$. Therefore $C(v)\cong
C(v_{0})$.

We have $\Gamma(A_{0})\cong\mathbb{Z}_{2^{k+1}}$ or
$\Gamma(A_{0})\cong\mathbb{Z}$, generated by $1\otimes 1$. So from the
definitions we see that in all cases $\Phi(v_{0})=\Gamma(A_{0})$ and
$\Psi(v_{0})=8\Gamma(A_{0})$. Moreover,
$\Gamma(A_{0})/8\Gamma(A_{0})\cong\mathbb{Z}_{4}$ if $k=1$, and
$\Gamma(A_{0})/8\Gamma(A_{0})\cong\mathbb{Z}_{8}$ otherwise.

b) First consider the top horizontal sequence. It follows from the definitions
that $\Sigma(v)\leq\mathbb{Z}\oplus\Phi(v)$. The map $\iota$ is well-defined,
because if $v(x)=v(y)=1$, then $x\otimes 1-y\otimes 1=(x-y)\otimes
1+[x-y,y]\otimes 1\in\Psi(v)$. Moreover, $\iota$ is surjective, hence
$\iota-q$ is surjective too. We need to prove that the sequence is exact at
$\mathbb{Z}\oplus\Phi(v)$. If $v=0$, then
$\Sigma(v)=8\mathbb{Z}\oplus\Phi(v)=\mathrm{Ker}(\iota-q)$ by Lemma 5.4, so
the sequence is exact. If $v\neq 0$, then $\Sigma(v)$ is generated by the
elements $(1,x\otimes 1)$ with $x\in v^{-1}(1)$, so
$\Sigma(v)\leq\mathrm{Ker}(\iota-q)$. Every element of
$\mathbb{Z}\oplus\Phi(v)$ can be written as $n(1,x\otimes 1)+(0,\alpha)$ for
some $n\in\mathbb{Z}$ and $\alpha\in\Phi(v)$, and such an element is in
$\mathrm{Ker}(\iota-q)$ if and only if $\alpha\in\mathrm{Ker}(q)=\Psi(v)$. In
the proof of Lemma 5.4 b) we saw that $0\oplus\Psi(v)\leq\Sigma(v)$, and this
implies that $\Sigma(v)\geq\mathrm{Ker}(\iota-q)$. Therefore the sequence is
exact.

Next consider the middle horizontal sequence. The map $\iota-\bar{q}$ is
surjective, because $\bar{q}$ is surjective onto $\Gamma(A)/\Psi(v)$, and if
$v=0$, then $\iota$ is surjective onto $\mathbb{Z}_{8}$. If $v=0$, then
$\Sigma(v)=8\mathbb{Z}\oplus\Phi(v)=8\mathbb{Z}\oplus\Psi(v)=\mathrm{Ker}(\iota-\bar{q})$.
If $v\neq 0$, then, using the observations from the top sequence, $\Sigma(v)$
and $\mathrm{Ker}(\iota-\bar{q})$ are both equal to the subgroup generated by
the elements $(1,x\otimes 1)$ (where $x\in v^{-1}(1)$) and $0\oplus\Psi(v)$.
Therefore the middle sequence is exact too.

Next we consider the vertical sequence in the middle. The homomorphism $u_{v}$
is well-defined, because $u_{v}([a,a]\otimes 1)=0=u_{v}(a\otimes 2)$ and
$u_{v}((a_{1}+a_{2})\otimes 1)=(a_{1}+a_{2})\otimes 1+(a_{1}+a_{2})\otimes
v(a_{1}+a_{2})=u_{v}(a_{1}\otimes 1)+u_{v}(a_{2}\otimes
1)+u_{v}([a_{1},a_{2}]\otimes 1)$ for every $a,a_{1},a_{2}\in A$. Moreover,
$\mathrm{Im}(u_{v})\leq\mathrm{Ker}(v_{2})$, because $v_{2}(a_{1}\otimes
v(a_{2})+a_{2}\otimes v(a_{1}))=v(a_{1})\otimes v(a_{2})+v(a_{2})\otimes
v(a_{1})=0\in\mathbb{Z}_{2}$ and $v_{2}(a\otimes 1+a\otimes
v(a))=v(a)\otimes(1+v(a))=0$.

For the exactness of the sequence we need to show that
$\Phi(v)\leq\mathrm{Ker}(u_{v})$ and that the induced homomorphism
$\Gamma(A)/\Phi(v)\rightarrow\mathrm{Ker}(v_{2})$ is an isomorphism. The
former is verified by evaluating $u_{v}$ on the generators of $\Phi(v)$. For
the latter, we show that the induced homomorphism has an inverse
$\mathrm{Ker}(v_{2})\rightarrow\Gamma(A)/\Phi(v)$. Namely, we have
$\mathrm{Ker}(v_{2})=\\{k\otimes 1\mid k\in\mathrm{Ker}(v)\\}$ and the inverse
map is given by $k\otimes 1\mapsto[k\otimes 1]$. This map is well-defined,
because $k_{1}\otimes 1=k_{2}\otimes 1\in A\otimes\mathbb{Z}_{2}$ if and only
if $k_{2}=k_{1}+2a$ for some $a\in A$, and $(k_{1}+2a)\otimes 1-k_{1}\otimes
1=2a\otimes 1+[k_{1},2a]\otimes 1\in\Phi(v)$. Here we used that
$2a\in\mathrm{Ker}(v)$, so $[k_{1},2a]\otimes 1\in\Phi(v)$, and $2a\otimes
1=4(a\otimes 1)=2([a,a]\otimes 1)\in\Phi(v)$ for both $a\in v^{-1}(1)$ and
$a\in\mathrm{Ker}(v)$. This map is a homomorphism, because
$[(k_{1}+k_{2})\otimes 1]-[k_{1}\otimes 1]-[k_{2}\otimes
1]=[[k_{1},k_{2}]\otimes 1]=0\in\Gamma(A)/\Phi(v)$. It is the inverse of the
homomorphism induced by $u_{v}$, because $u_{v}(k\otimes 1)=k\otimes 1$ for
$k\in\mathrm{Ker}(v)$, and the elements $[k\otimes 1]$ generate
$\Gamma(A)/\Phi(v)$. Therefore $\Gamma(A)/\Phi(v)\cong\mathrm{Ker}(v_{2})$ and
the sequence is exact.

Finally consider the vertical sequence on the right. Since $u_{v}$ induces an
isomorphism $\Gamma(A)/\Phi(v)\cong\mathrm{Ker}(v_{2})$ and
$\Psi(v)\leq\Phi(v)$, the map $\tilde{u}_{v}$ is well-defined. If $v=0$, then
$\Psi(v)=\Phi(v)$, hence the restriction of $\tilde{u}_{v}$ to
$\Gamma(A)/\Psi(v)\leq\Upsilon(v)$ is an isomorphism, so the sequence is
exact. It is also exact if $v\neq 0$, because then it is isomorphic to the
sequence
$\Phi(v)/\Psi(v)\rightarrow\Gamma(A)/\Psi(v)\rightarrow\Gamma(A)/\Phi(v)$. ∎

######  Remark 5.10.

We can give an alternative description of the map $u_{v}$ in terms of
$(Q^{+}\oplus A)$-forms, using that $\Gamma(A)\cong W_{0}^{Q^{+}}(A)$ by
Theorem 4.26. Let $(X,\lambda,(\mu_{0},\mu_{A}))$ be a nonsingular
$(Q^{+}\oplus A)$-form and $\chi,\bar{\chi}\in A$ be characteristic elements
for $\lambda$ and $v\circ\mu_{A}$ respectively; i.e. for all $x\in X$,
$\lambda(x,x)\equiv\lambda(\chi,x)$ mod $2$ and
$v(\mu_{A}(x))=\varrho_{2}(\lambda(\bar{\chi},x))$. Since $\lambda$ is
nonsingular, $\chi$ and $\bar{\chi}$ are well-defined modulo $2X$. We have

$v(\mu_{A}(\chi-\bar{\chi}))=\varrho_{2}(\lambda(\bar{\chi},\chi-\bar{\chi}))=\varrho_{2}(\lambda(\bar{\chi},\chi)-\lambda(\bar{\chi},\bar{\chi}))=0$

so $\mu_{A}(\chi-\bar{\chi})\otimes 1\in\mathrm{Ker}(v_{2})$. By comparing the
definitions we can see that the image of $[X,\lambda,(\mu_{0},\mu_{A})]\in
W_{0}^{Q^{+}}(A)\cong\Gamma(A)$ is
$u_{v}([X,\lambda,(\mu_{0},\mu_{A})])=\mu_{A}(\chi-\bar{\chi})\otimes 1$. In
particular, if $v\circ\mu_{A}$ is characteristic for $\lambda$ (see Definition
2.27), then $\chi\equiv\bar{\chi}$ mod $2X$, showing that, if $v=v_{P}\colon
A=SP\rightarrow\mathbb{Z}_{2}$ for some form parameter $P$, then the image of
the composition $W_{0}(P)\xrightarrow{W_{0}(\textup{es})}W_{0}(Q^{+}\oplus
A)\rightarrow W_{0}^{Q^{+}}(A)$ is in $\mathrm{Ker}(u_{v})$ (see Lemma 2.29).

###  5.2. Anti-symmetric form parameters

Next we consider the restriction of the Witt group functor to the category of
anti-symmetric form parameters, $W_{0}\colon\mathbf{FP}_{-}\to\mathbf{Ab}$.

######  Lemma 5.11.

If $Q$ is an indecomposable anti-symmetric form parameter, then the induced
homomorphism $W_{0}(\textup{eql})\colon W_{0}(Q_{-}\oplus Q_{e})\to W_{0}(Q)$
is surjective.

###### Proof.

If $Q=Q_{-}$, then the composition $Q_{-}\rightarrow
Q_{-}\oplus\mathbb{Z}_{2}\rightarrow Q_{-}$ is the identity, and if $Q\neq
Q_{-}$, then $W_{0}(Q)\cong 0$ by Theorem 4.3. ∎

######  Lemma 5.12.

If $P$ is an anti-symmetric form parameter, then $W_{0}(\textup{eql})\colon
W_{0}(Q_{-}\oplus P_{e})\to W_{0}(P)$ is surjective.

###### Proof.

Again we prove this using a maximal splitting $P=Q\oplus G$. We have
$P_{e}=Q_{e}\oplus G$, and the induced homomorphism

$W_{0}(Q_{-}\oplus Q_{e}\oplus G)\cong W_{0}(Q_{-}\oplus Q_{e})\oplus
W_{0}^{Q_{-}\oplus Q_{e}}(G)\rightarrow W_{0}(P)=W_{0}(Q)\oplus W_{0}^{Q}(G)$

splits as the sum of $W_{0}(Q_{-}\oplus Q_{e})\rightarrow W_{0}(Q)$ and

$W_{0}^{Q_{-}\oplus Q_{e}}(G)\cong G\otimes_{\mathbb{Z}}(Q_{-}\oplus
Q_{e})\cong(G\otimes_{\mathbb{Z}}Q_{-})\oplus(G\otimes
Q_{e})\cong\Lambda_{1}(G)\oplus(G\otimes Q_{e})\rightarrow W_{0}^{Q}(G)\cong
G\otimes_{\mathbb{Z}}Q.$

The former is surjective by Lemma 5.11. By Corollary 4.16 b) there is a
pushout diagram

$\textstyle{G\otimes\mathbb{Z}_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda_{1}(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes
Q_{e}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{G\otimes_{\mathbb{Z}}Q.}$

The map $W_{0}^{Q_{-}\oplus Q_{e}}(G)\cong\Lambda_{1}(G)\oplus(G\otimes
Q_{e})\rightarrow W_{0}^{Q}(G)\cong G\otimes_{\mathbb{Z}}Q$ is the sum of the
maps $\Lambda_{1}(G)\rightarrow G\otimes_{\mathbb{Z}}Q$ and $G\otimes
Q_{e}\rightarrow G\otimes_{\mathbb{Z}}Q$ in this diagram, hence it is
surjective too. ∎

######  Remark 5.13.

Lemma 5.12 has the following alternative proof: Since every nonsingular anti-
symmetric bilinear form is isomorphic to a sum of hyperbolic forms, it is easy
to see that the induced homomorphism $\mathrm{Hom}_{\mathbb{Z}}^{\rm
ns}(X,\textup{eql})\colon\mathrm{Hom}_{\mathbb{Z}}^{\rm ns}(X,Q_{-}\oplus
P_{e})\to\mathrm{Hom}_{\mathbb{Z}}^{\rm ns}(X,P)$ is surjective for any
torsion free abelian group $X$, and hence $W_{0}(\textup{eql})$ induces a
surjection on Witt groups.

Let $A$ be an abelian group. Recall that $\Lambda_{1}(A)=(A\otimes
A)/\bigl{\langle}{x\otimes y+y\otimes x\mid x,y\in A}\bigr{\rangle}$. We will
denote the image of $x\otimes y\in A\otimes A$ in the quotient
$\Lambda_{1}(A)$ by
$x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits y$. Recall
from Example 4.7 that $A\otimes_{\mathbb{Z}}Q_{-}\cong\Lambda_{1}(A)$, under
this identification $a\otimes 1$ and $[a,b]\otimes 1$ correspond to
$a\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits a$ and
$a\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits b$
respectively. By Proposition 4.15 there is a short exact sequence

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{A\otimes\mathbb{Z}_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda_{1}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$

where the first map sends $a\otimes 1$ to
$a\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits a$, and
the second map sends
$a\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits b$ to
$a\wedge b$. By decomposing $A$ as a direct sum of cyclic groups, we can see
that this short exact sequence splits, but the splitting is not natural.

######  Definition 5.14.

Let $v^{\prime}\colon\mathbb{Z}_{2}\to A$ be a homomorphism. We define the
subgroup

$K(v^{\prime})=\bigl{\langle}{(1,v^{\prime}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1)),(0,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
x+x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1))\mid x\in
A}\bigr{\rangle}\leq\mathbb{Z}_{2}\oplus\Lambda_{1}(A)$

and the quotient group

$\Lambda(v^{\prime})=(\mathbb{Z}_{2}\oplus\Lambda_{1}(A))/K(v^{\prime}).$

If $\alpha\colon v^{\prime}_{1}\rightarrow v^{\prime}_{2}$ is a morphism in
$\mathbb{Z}_{2}/\mathbf{Ab}$, then the induced map
$\textup{Id}_{\mathbb{Z}_{2}}\oplus\Lambda_{1}(\alpha)\colon\mathbb{Z}_{2}\oplus\Lambda_{1}(A_{1})\rightarrow\mathbb{Z}_{2}\oplus\Lambda_{1}(A_{2})$
sends the generators of $K(v^{\prime}_{1})$ to generators of
$K(v^{\prime}_{2})$. This shows that $K$, and hence $\Lambda$, are functors
$\mathbb{Z}_{2}/\mathbf{Ab}\to\mathbf{Ab}$. Also note that if
$v^{\prime}=0\colon\mathbb{Z}_{2}\rightarrow A$ is the trivial homomorphism,
then $\Lambda(v^{\prime})\cong\Lambda(A)$.

######  Lemma 5.15.

Let $v^{\prime}\colon\mathbb{Z}_{2}\to A$ be a homomorphism and suppose that
$v^{\prime}\cong v^{\prime}_{0}\oplus C\colon\mathbb{Z}_{2}\rightarrow B\oplus
C$ is a maximal splitting of $v^{\prime}$. Then

$K(v^{\prime})\cong\begin{cases}\mathbb{Z}_{2}\oplus(C\otimes\mathbb{Z}_{2})&\text{if
$v^{\prime}_{0}\cong\iota_{1}$, }\\\
\mathbb{Z}_{2}\oplus(A\otimes\mathbb{Z}_{2})&\text{otherwise. }\end{cases}$

###### Proof.

If $v^{\prime}=0$, then $K(v^{\prime})$ is generated by $(1,0)$ and the
elements
$(0,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits x)$ for
every $x\in A$, therefore it is the direct sum of the subgroups
$\mathbb{Z}_{2}\oplus 0$ and $0\oplus A\otimes\mathbb{Z}_{2}$.

If $v^{\prime}_{0}\cong\iota_{1}$, then
$\Lambda_{1}(A)\cong\Lambda_{1}(\mathbb{Z}_{2}\oplus
C)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\otimes C\oplus\Lambda_{1}(C)$, and
$v^{\prime}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1)$ is the generator of the first component
$\Lambda_{1}(\mathbb{Z}_{2})\cong\mathbb{Z}_{2}$. So the subgroup
$K(v^{\prime})\leq\mathbb{Z}_{2}\oplus(\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\otimes
C\oplus\Lambda_{1}(C))$ is generated by $(1,(1,0,0))$ and the elements
$(0,(0,1\otimes
x,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits x))$ for
every $x\in C$. Since the map $\mathbb{Z}_{2}\otimes
C\rightarrow\mathbb{Z}_{2}\oplus(\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\otimes
C\oplus\Lambda_{1}(C))$, $1\otimes x\mapsto(0,(0,1\otimes
x,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits x))$ is
an isomorphism onto its image, we have
$K(v^{\prime})\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\otimes C$.

If $v^{\prime}_{0}\cong\iota_{k}$, $k\geq 2$, then
$\Lambda_{1}(A)\cong\Lambda_{1}(\mathbb{Z}_{2^{k}}\oplus
C)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2^{k}}\otimes C\oplus\Lambda_{1}(C)$.
We have $v^{\prime}(1)=2^{k-1}\in\mathbb{Z}_{2^{k}}$, so
$v^{\prime}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1)=0\in\Lambda_{1}(\mathbb{Z}_{2^{k}})\cong\mathbb{Z}_{2}$. For the
generator $1\in\mathbb{Z}_{2^{k}}$ we have that
$1\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
1+1\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1)=1\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
1$ is the generator of $\Lambda_{1}(\mathbb{Z}_{2^{k}})\cong\mathbb{Z}_{2}$.
Therefore $K(v^{\prime})$ is generated by $(1,(0,0,0))$, $(0,(1,0,0))$ and the
elements $(0,(0,2^{k-1}\otimes
x,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits x))$ for
every $x\in C$. Moreover, the map $\mathbb{Z}_{2}\otimes
A\rightarrow\mathbb{Z}_{2}\oplus\Lambda_{1}(A)$, $1\otimes
x\mapsto(0,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
x+x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1))$ is an isomorphism onto the subgroup generated by $(0,(1,0,0))$
and the elements $(0,(0,2^{k-1}\otimes
x,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits x))$ for
every $x\in C$, showing that
$K(v^{\prime})\cong\mathbb{Z}_{2}\oplus(\mathbb{Z}_{2}\otimes A)$. ∎

######  Lemma 5.16.

Let $P$ be an anti-symmetric form parameter and suppose that $P\cong Q\oplus
G$ is a maximal splitting of $P$. Then

$\mathrm{Ker}\bigl{(}W_{0}(\textup{eql})\colon W_{0}(Q_{-}\oplus P_{e})\to
W_{0}(P)\bigr{)}\cong\begin{cases}\mathbb{Z}_{2}\oplus(G\otimes\mathbb{Z}_{2})&\text{if
$Q\cong Q_{-}$, }\\\
\mathbb{Z}_{2}\oplus(P_{e}\otimes\mathbb{Z}_{2})&\text{otherwise.
}\end{cases}$

###### Proof.

Recall from Lemma 5.12 that $W_{0}(\textup{eql})\colon W_{0}(Q_{-}\oplus
P_{e})\to W_{0}(P)$ is the sum of $W_{0}(Q_{-}\oplus Q_{e})\rightarrow
W_{0}(Q)$ and $W_{0}^{Q_{-}\oplus Q_{e}}(G)\rightarrow W_{0}^{Q}(G)$, where
the second map is the sum of the two maps in the pushout square of Corollary
4.16 b). Since this is also a pullback square (see Remark 4.8), we get that
the kernel of the second map is isomorphic to $G\otimes\mathbb{Z}_{2}$. For
the first map we have $W_{0}(Q_{-}\oplus Q_{e})\cong W_{0}(Q_{-})\oplus
W_{0}^{Q_{-}}(Q_{e})\cong\mathbb{Z}_{2}\oplus\Lambda_{1}(Q_{e})$ and
$W_{0}(Q)\cong\mathbb{Z}_{2}$ if $Q\cong Q_{-}$ and $W_{0}(Q)\cong 0$
otherwise.

If $Q=Q^{-}$, then the first map is $\mathbb{Z}_{2}\rightarrow 0$, so
$\mathrm{Ker}(W_{0}(\textup{eql}))\cong\mathbb{Z}_{2}\oplus
G\otimes\mathbb{Z}_{2}=\mathbb{Z}_{2}\oplus P_{e}\otimes\mathbb{Z}_{2}$.

If $Q=Q_{-}$, then the first map is
$\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{2}$, which is the
identity on the first component, which implies that
$\mathrm{Ker}(W_{0}(\textup{eql}))\cong\mathbb{Z}_{2}\oplus
G\otimes\mathbb{Z}_{2}$. We note that the second component is generated by
$1\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
1\in\Lambda_{1}(Q_{e})$, and by Definition 4.22 the corresponding element of
$W_{0}^{Q_{-}}(Q_{e})$ is represented by the $(Q_{-}\oplus Q_{e})$-form

$\left(\mathbb{Z}^{2},\left[\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right],\left(\begin{array}[]{c}(0,1)\\\
(0,1)\end{array}\right)\right)$

Its image under $W_{0}(\textup{eql})$ has Arf-invariant $1$, so it is the
generator of $W_{0}(Q)$. Therefore the map
$\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{2}$ restricts to the
identity on both components.

Finally, if $Q=\mathbb{Z}^{\Lambda}_{k}$, $k\geq 2$, then the first map is
$\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\rightarrow 0$, so
$\mathrm{Ker}(W_{0}(\textup{eql}))\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\oplus(G\otimes\mathbb{Z}_{2})\cong\mathbb{Z}_{2}\oplus(P_{e}\otimes\mathbb{Z}_{2})$.
∎

######  Theorem 5.17.

Let $P$ be an anti-symmetric form parameter. Then under the identification
$W_{0}(Q_{-}\oplus P_{e})\cong\mathbb{Z}_{2}\oplus\Lambda_{1}(P_{e})$ we have
$\mathrm{Ker}(W_{0}(\textup{eql}))=K(v^{\prime}_{P})$.

###### Proof.

First we consider
$(1,v^{\prime}_{P}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}_{P}(1))\in\mathbb{Z}_{2}\oplus\Lambda_{1}(P_{e})$. If we take a
maximal splitting $P\cong Q\oplus G$ of $P$ as in Lemma 5.16, then this
element is in the first component of $W_{0}(Q_{-}\oplus P_{e})\cong
W_{0}(Q_{-}\oplus Q_{e})\oplus W_{0}^{Q_{-}\oplus Q_{e}}(G)$, so it is mapped
to $W_{0}(Q)$ by $W_{0}(\textup{eql})$. If $Q\not\cong Q_{-}$, then
$W_{0}(Q)\cong 0$. If $Q\cong Q_{-}$, then
$(1,v^{\prime}_{P}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}_{P}(1))$ corresponds to
$(1,1)\in\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\cong W_{0}(Q_{-})\oplus
W_{0}^{Q_{-}}(Q_{e})$, which generates the kernel of the map to
$\mathbb{Z}_{2}\cong W_{0}(Q)$ by the observations in the proof of Lemma 5.16.
Therefore
$(1,v^{\prime}_{P}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}_{P}(1))\in\mathrm{Ker}(W_{0}(\textup{eql}))$ in both cases.

Next consider
$x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
x+x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}_{P}(1)=x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits(x+v^{\prime}_{P}(1))\in\Lambda_{1}(P_{e})$
for some $x\in P_{e}$. By Definition 4.22 the corresponding element of
$W_{0}^{Q_{-}}(P_{e})$ is represented by the $(Q_{-}\oplus P_{e})$-form

$\left(\mathbb{Z}^{2}\left<e,f\right>,\left[\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right],\left(\begin{array}[]{c}(0,x)\\\
(0,x+v^{\prime}_{P}(1))\end{array}\right)\right).$

This is mapped by $W_{0}(\textup{eql})$ to the element of $W_{0}(P)$
represented by the $P$-form

$\left(\mathbb{Z}^{2}\left<e,f\right>,\left[\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right],\left(\begin{array}[]{c}x\\\
x+v^{\prime}_{P}(1)\end{array}\right)\right)$

which is metabolic, because $\left<e-f\right>$ is a lagrangian. This shows
that $W_{0}(\textup{eql})$ sends every generator of $K(v^{\prime}_{P})$ to
$0$, hence $K(v^{\prime}_{P})\leq\mathrm{Ker}(W_{0}(\textup{eql}))$. Lemmas
5.15 and 5.16 show that $K(v^{\prime}_{P})$ and
$\mathrm{Ker}(W_{0}(\textup{eql}_{P}))$ are isomorphic finite groups,
therefore $K(v^{\prime}_{P})=\mathrm{Ker}(W_{0}(\textup{eql}))$. ∎

Using Lemma 5.12, Theorem 5.17 and the fact that eql is a natural
transformation, we get the following:

######  Corollary 5.18.

The homomorphisms $W_{0}(\textup{eql})$ induce a natural isomorphism
$\Lambda\circ v^{\prime}_{(-)}\cong W_{0}:\mathbf{FP}_{-}\to\mathbf{Ab}$.

We end this section with Theorem 5.20 below, which is the dual of Theorem 5.8,
and gives an explicit description of $\Lambda(v^{\prime})$.

######  Definition 5.19.

Let $v^{\prime}\colon\mathbb{Z}_{2}\to A$ be a homomorphism. We define the
subgroup

$L(v^{\prime})=\bigl{\langle}{x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
x+x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1)\mid x\in A}\bigr{\rangle}\leq\Lambda_{1}(A)$

and the quotient group

$\Xi(v^{\prime})=\Lambda_{1}(A)/L(v^{\prime}).$

Note that since $0\oplus L(v^{\prime})\leq K(v^{\prime})$, the quotient map
$\mathbb{Z}_{2}\oplus\Lambda_{1}(A)\rightarrow\Lambda(v^{\prime})$ factors as
a composition
$\mathbb{Z}_{2}\oplus\Lambda_{1}(A)\rightarrow\mathbb{Z}_{2}\oplus\Xi(v^{\prime})\rightarrow\Lambda(v^{\prime})$.

######  Theorem 5.20.

Let $v^{\prime}\colon\mathbb{Z}_{2}\to A$ be a homomorphism. There is a
commutative diagram with short exact rows and columns

$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\textup{Coker}(v^{\prime}_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u^{\prime}_{v^{\prime}}}$$\textstyle{\textup{Coker}(v^{\prime}_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{u^{\prime}_{v^{\prime}}}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{K(v^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{r}$$\textstyle{\mathbb{Z}_{2}\oplus\Lambda_{1}(A)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda(v^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\mathbb{Z}_{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\iota}$$\textstyle{\mathbb{Z}_{2}\oplus\Xi(v^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\Lambda(v^{\prime})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{0}$$\textstyle{0}$$\textstyle{0}$

where

  * $\bullet$

$v^{\prime}_{2}=v^{\prime}\otimes\textup{Id}_{\mathbb{Z}_{2}}\colon\mathbb{Z}_{2}\otimes\mathbb{Z}_{2}=\mathbb{Z}_{2}\rightarrow
A\otimes\mathbb{Z}_{2}$, hence $v^{\prime}_{2}(1)=v^{\prime}(1)\otimes 1$;

  * $\bullet$

The homomorphism
$\iota\colon\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{2}\oplus\Xi(v^{\prime})$ is
given by
$\iota(1)=(1,[v^{\prime}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1)])$;

  * $\bullet$

The homomorphism $r\colon K(v^{\prime})\rightarrow\mathbb{Z}_{2}$ is the
composition
$K(v^{\prime})\rightarrow\mathbb{Z}_{2}\oplus\Lambda_{1}(A)\rightarrow\mathbb{Z}_{2}$
of the natural inclusion and projection maps;

  * $\bullet$

The homomorphism
$u^{\prime}_{v^{\prime}}\colon\textup{Coker}(v^{\prime}_{2})\rightarrow
K(v^{\prime})\leq\mathbb{Z}_{2}\oplus\Lambda_{1}(A)$ is induced by the map
$A\otimes\mathbb{Z}_{2}\rightarrow K(v^{\prime})$, $x\otimes
1\mapsto(0,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
x+x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1))$;

  * $\bullet$

The unlabelled maps are the obvious inclusions/quotient maps.

###### Proof.

First consider the vertical sequence on the left. The map
$A\otimes\mathbb{Z}_{2}\rightarrow K(v^{\prime})$ sends $v^{\prime}(1)\otimes
1$ to
$(0,v^{\prime}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1)+v^{\prime}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1))=0$, hence $u^{\prime}_{v^{\prime}}$ is well-defined. Suppose
that $v^{\prime}\cong v^{\prime}_{0}\oplus C\colon\mathbb{Z}_{2}\rightarrow
B\oplus C$ is a maximal splitting of $v^{\prime}$. If
$v^{\prime}_{0}\cong\iota_{1}$, then
$A\otimes\mathbb{Z}_{2}\cong\mathbb{Z}_{2}\oplus C\otimes\mathbb{Z}_{2}$ and
$\textup{Coker}(v^{\prime}_{2})\cong C\otimes\mathbb{Z}_{2}$. If
$v^{\prime}=0$ or $v^{\prime}_{0}\cong\iota_{k}$, then
$v^{\prime}_{2}(1)=v^{\prime}(1)\otimes 1=0$, so
$\textup{Coker}(v^{\prime}_{2})\cong A\otimes\mathbb{Z}_{2}$. Moreover, the
proof of Lemma 5.15 shows that $u^{\prime}_{v^{\prime}}$ maps
$\textup{Coker}(v^{\prime}_{2})$ isomorphically onto $0\oplus
L(v^{\prime})\leq K(v^{\prime})$. The quotient $K(v^{\prime})/(0\oplus
L(v^{\prime}))$ is generated by
$[1,v^{\prime}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1)]$, and $r$ induces an isomorphism onto $\mathbb{Z}_{2}$, showing
that the sequence is exact. The map $\mathbb{Z}_{2}\rightarrow K(v^{\prime})$,
$1\mapsto(1,v^{\prime}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1))$ splits $r$.

The vertical sequence in the middle is exact, because the kernel of
$\mathbb{Z}_{2}\oplus\Lambda_{1}(A)\rightarrow\mathbb{Z}_{2}\oplus\Xi(v^{\prime})$
is $0\oplus L(v^{\prime})$. The middle horizontal sequence is exact by the
definition of $K(v^{\prime})$ and $\Lambda(v^{\prime})$. Finally, the bottom
horizontal sequence is exact, because $K(v^{\prime})=(0\oplus
L(v^{\prime}))\oplus\bigl{\langle}{(1,v^{\prime}(1)\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits
v^{\prime}(1))}\bigr{\rangle}$. ∎

The projection $\mathbb{Z}_{2}\oplus\Xi(v^{\prime})\rightarrow\mathbb{Z}_{2}$
splits the bottom exact sequence, hence $\Lambda(v^{\prime})$ is naturally
isomorphic to $\Xi(v^{\prime})$.

If $v^{\prime}=0$, then $L(v^{\prime})$ is the image of the natural embedding
$A\otimes\mathbb{Z}_{2}\rightarrow\Lambda_{1}(A)$, so
$\Xi(v^{\prime})=\Lambda(A)$.

If $v^{\prime}\cong\iota_{1}\oplus C$, then, by the proof of Lemma 5.15,
$L(v^{\prime})$ is the image of the map $\mathbb{Z}_{2}\otimes
C\rightarrow\Lambda_{1}(A)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\otimes
C\oplus\Lambda_{1}(C)$, $1\otimes x\mapsto(0,1\otimes
x,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits x)$. This
map can be extended to an automorphism of $\Lambda_{1}(A)$ by taking the
identity on $\mathbb{Z}_{2}$ and $\Lambda_{1}(C)$. This automorphism shows
that
$\Xi(v^{\prime})=\Lambda_{1}(A)/L(v^{\prime})\cong\Lambda_{1}(A)/\mathbb{Z}_{2}\otimes
C\cong\mathbb{Z}_{2}\oplus\Lambda_{1}(C)$. Note that $\Lambda_{1}(C)\cong
0\oplus
C\otimes\mathbb{Z}_{2}\oplus\Lambda(C)\cong\Lambda(\mathbb{Z}_{2}\oplus
C)\cong\Lambda(A)$, hence
$\Xi(v^{\prime})\cong\mathbb{Z}_{2}\oplus\Lambda(A)$, but this isomorphism is
not natural.

If $v^{\prime}\cong\iota_{k}\oplus C$, $k\geq 2$, then $L(v^{\prime})$ is the
direct sum of the image of the map $\mathbb{Z}_{2}\otimes
C\rightarrow\Lambda_{1}(A)\cong\mathbb{Z}_{2}\oplus\mathbb{Z}_{2^{k}}\otimes
C\oplus\Lambda_{1}(C)$, $1\otimes x\mapsto(0,2^{k-1}\otimes
x,x\mathop{\stackrel{{\scriptstyle{}_{\wedge}}}{{\otimes}}}\nolimits x)$ and
$\mathbb{Z}_{2}\oplus 0\oplus 0$. This time we regard $\mathbb{Z}_{2}\otimes
C$ as a subgroup of $\Lambda_{1}(C)$ via a splitting
$\Lambda_{1}(C)\cong\mathbb{Z}_{2}\otimes C\oplus\Lambda(C)$, and extend the
above map to an automorphism of $\Lambda_{1}(A)$ by taking the identity on
$\mathbb{Z}_{2}$, $\mathbb{Z}_{2^{k}}\otimes C$ and $\Lambda(C)$. We get that
$\Xi(v^{\prime})=\Lambda_{1}(A)/L(v^{\prime})\cong\Lambda_{1}(A)/(\mathbb{Z}_{2}\oplus\mathbb{Z}_{2}\otimes
C)\cong\mathbb{Z}_{2^{k}}\otimes
C\oplus\Lambda(C)\cong\Lambda(\mathbb{Z}_{2^{k}}\oplus C)\cong\Lambda(A)$,
but, again, this isomorphism is not natural.

##  6\. Computing Grothendieck-Witt groups and characterising absorbing forms

In this section we recall the definition of the Grothendieck-Witt group of a
form parameter $Q$ and describe its relationship to the Witt group of $Q$. We
also define and characterise absorbing $Q$-forms.

###  6.1. The groups $GW_{0}(Q)$ and $W_{0}(Q)$

Let $Q$ be a form parameter, $\underline{\mu}$ a nonsingular $Q$-form and
$\\{\underline{\mu}\\}$ the isometry class of $\underline{\mu}$. Recall from
Section 1.2 that $\mathrm{ICHom}^{\rm
ns}_{\mathbb{Z}}(Q)=\big{\\{}\\{\underline{\mu}\\}\big{\\}}$ is the set of
isomorphism classes of nonsingular $Q$-forms, which is a monoid under
orthogonal sum, and that the Grothendieck-Witt group of $Q$ is the
Grothendieck group of this monoid:

$GW_{0}(Q):={\mathcal{G}r}\big{(}\mathrm{ICHom}^{\rm
ns}_{\mathbb{Z}}(Q),\oplus\big{)}.$

There is a natural surjection

$\pi\colon GW_{0}(Q)\to
W_{0}(Q),\quad\bigl{(}\\{\underline{\mu}_{0}\\},\\{\underline{\mu}_{1}\\}\bigr{)}\mapsto[\underline{\mu}_{0}]-[\underline{\mu}_{1}],$

and by definition, there is a short exact sequence of abelian groups

$0\to\mathrm{Ker}(\pi)\xrightarrow{}GW_{0}(Q)\xrightarrow{\pi}W_{0}(Q)\to 0.$
(7)

This exact sequence allows us to compute $GW_{0}(Q)$ in terms of $W_{0}(Q)$,
using the following definition and lemma.

######  Definition 6.1.

For every form parameter $Q$, there is a well-defined rank homomorphism

$\textup{rk}\colon
GW_{0}(Q)\rightarrow\mathbb{Z}\text{\,,}\quad\big{(}\\{\underline{\mu_{0}}\\},\\{\underline{\mu_{1}}\\}\big{)}\mapsto\textup{rk}(\underline{\mu_{0}})-\textup{rk}(\underline{\mu_{1}}).$

If $Q$ is symmetric, then there is also a signature homomorphism

$\sigma\colon
GW_{0}(Q)\rightarrow\mathbb{Z}\text{\,,}\quad\big{(}\\{\underline{\mu_{0}}\\},\\{\underline{\mu_{1}}\\}\big{)}\mapsto\sigma(\underline{\mu_{0}})-\sigma(\underline{\mu_{1}}).$

It is well known that $\mathrm{Ker}(\pi)\cong\mathbb{Z}$ is generated by
$\big{(}\\{H_{\epsilon}(\mathbb{Z})\\},\\{\underline{0}\\}\big{)}$, where we
recall that $\underline{0}$ is the $Q$-form on the trivial group; see e.g. [S,
Lemma 2.31]. Indeed Schlichting [S, Definition 2.14] defines $W_{0}(Q)$ to be
the quotient of $GW_{0}(Q)$ by the subgroup of formal differences of
hyperbolic forms. Hence we have

######  Lemma 6.2.

For every form parameter $Q$ the restriction of the rank homomorphism
$\textup{rk}\colon GW_{0}(Q)\rightarrow\mathbb{Z}$ yields an isomorphism

$\textup{rk}\big{|}_{\mathrm{Ker}(\pi)}\colon\mathrm{Ker}(\pi)\xrightarrow{\cong}2\mathbb{Z}.\qed$

Remark. We present an a priori alternative proof of Lemma 6.2 below.

######  Theorem 6.3.

a) For every form parameter $Q$ the map $\textup{rk}\oplus\pi\colon
GW_{0}(Q)\to\mathbb{Z}\oplus W_{0}(Q)$ is injective.

b) If $Q$ is symmetric, then
$(n,[\underline{\mu}])\in\mathrm{Im}(\textup{rk}\oplus\pi)$ if and only if
$\sigma([\underline{\mu}])\equiv n~{}\mathrm{mod}~{}2$. If $Q$ is anti-
symmetric, then $\mathrm{Im}(\textup{rk}\oplus\pi)=2\mathbb{Z}\oplus
W_{0}(Q)$.

c) The short exact sequence (7) splits.

###### Proof.

a) The restriction of $\textup{rk}\colon GW_{0}(Q)\to\mathbb{Z}$ to
$\mathrm{Ker}(\pi)$ is injective by Lemma 6.2.

b) If $Q$ is symmetric, then we have
$\sigma(\underline{\mu})\equiv\textup{rk}(\underline{\mu})~{}\mathrm{mod}~{}2$
for every $Q$-form $\underline{\mu}$, therefore
$\mathrm{Im}(\textup{rk}\oplus\pi)\subseteq\left\\{(n,[\underline{\mu}])\mid\sigma([\underline{\mu}])\equiv
n~{}\mathrm{mod}~{}2\right\\}$. If $Q$ is anti-symmetric, then every
nonsingular form over $Q$ has even rank, so
$\mathrm{Im}(\textup{rk}\oplus\pi)\subseteq 2\mathbb{Z}\oplus W_{0}(Q)$. In
the other direction, for any form parameter $Q$, every nonsingular $Q$-form
$\underline{\mu}$ and $k\geq 0$, we have
$(\textup{rk}\oplus\pi)([\underline{\mu}\oplus
H_{\epsilon_{Q}}(\mathbb{Z}^{k}),\underline{0}])=(\textup{rk}(\underline{\mu})+2k,[\underline{\mu}])$,
and so equality holds for both symmetries.

c) If $Q$ is symmetric, then the signature map $\sigma\colon
GW_{0}(Q)\to\mathrm{Im}(\sigma)$ has a splitting
$s\colon\mathrm{Im}(\sigma)\rightarrow GW_{0}(Q)$ (because
$\mathrm{Im}(\sigma)\subseteq\mathbb{Z}$), and
$GW_{0}(Q)=\mathrm{Ker}(\sigma)\oplus\mathrm{Im}(s)$. Since
$\mathrm{Ker}(\pi)\subseteq\mathrm{Ker}(\sigma)$ and every form with zero
signature has even rank, we can define a splitting map $f\colon
GW_{0}(Q)\rightarrow\mathrm{Ker}(\pi)\cong 2\mathbb{Z}$ by
$f\big{|}_{\mathrm{Ker}(\sigma)}=\textup{rk}\big{|}_{\mathrm{Ker}(\sigma)}$
and $f\big{|}_{\mathrm{Im}(s)}=0$.

If $Q$ is anti-symmetric and $\mathrm{Ker}(\pi)$ is identified with
$2\mathbb{Z}$ using Lemma 6.2, then the rank homomorphism $\textup{rk}\colon
GW_{0}(Q)\to 2\mathbb{Z}$ is a canonical splitting of the inclusion
$\mathrm{Ker}(\pi)\to GW_{0}(Q)$. ∎

We conclude this subsection with an alternative proof of Lemma 6.2. Our main
motivation for doing this is that we believe Lemma 6.4 is of independent
interest in the study of $Q$-forms; see Remark 1.11.

######  Lemma 6.4.

Let $\underline{\mu}=(X,\lambda,\mu)$ and
$\underline{\mu}^{\prime}=(X^{\prime},\lambda^{\prime},\mu^{\prime})$ be full
metabolic $Q$-forms on isomorphic free abelian groups $X\cong X^{\prime}$.
Then for some integer $m\geq 0$, $\underline{\mu}\oplus
H_{\epsilon_{Q}}(\mathbb{Z}^{m})$ and $\underline{\mu}^{\prime}\oplus
H_{\epsilon_{Q}}(\mathbb{Z}^{m})$ are isometric $Q$-forms.

###### Proof.

If $Q$ is symmetric, then by Lemma 2.29 we can regard a $Q$-form
$(X,\lambda,\mu)$ as a $(Q^{+}\oplus SQ)$-form $(X,\lambda,(\mu_{0},\mu_{1}))$
such that $v(\mu_{1}(x))\equiv\lambda(x,x)$ mod $2$. The analogous statement
for such $(Q^{+}\oplus SQ)$-forms is proved in [N2, Theorem 2.22 a)].

Here we consider the anti-symmetric case. As $\underline{\mu}$ and
$\underline{\mu}^{\prime}$ are metabolic, they have lagrangians $L<X$ and
$L^{\prime}<X^{\prime}$. Let $e_{1},\ldots,e_{k},f_{1},\ldots,f_{k}$ and
$e^{\prime}_{1},\ldots,e^{\prime}_{k},f^{\prime}_{1},\ldots,f^{\prime}_{k}$ be
symplectic bases for the bilinear forms $(X,\lambda)$ and
$(X^{\prime},\lambda^{\prime})$ respectively such that
$L=\left<e_{1},\ldots,e_{k}\right>$ and
$L^{\prime}=\left<e_{1}^{\prime},\ldots,e_{k}^{\prime}\right>$, and let
$N=\left<f_{1},\ldots,f_{k}\right>$ and
$N^{\prime}=\left<f^{\prime}_{1},\ldots,f^{\prime}_{k}\right>$. Since $L$ and
$L^{\prime}$ are lagrangians, we have $S\mu\big{|}_{L}=0$ and
$S\mu^{\prime}\big{|}_{L^{\prime}}=0$, so $S\mu\big{|}_{N}\colon N\rightarrow
SQ$ and $S\mu^{\prime}\big{|}_{N^{\prime}}\colon N^{\prime}\rightarrow SQ$ are
surjective (because $\underline{\mu}$ and $\underline{\mu}^{\prime}$ are
full). By [N2, Lemma 2.21 a)] there is an $m\geq 0$ and an isomorphism
$h\colon N\oplus\mathbb{Z}^{m}\rightarrow N^{\prime}\oplus\mathbb{Z}^{m}$ such
that $(S\mu\big{|}_{N}+0)=(S\mu^{\prime}\big{|}_{N^{\prime}}+0)\circ h\colon
N\oplus\mathbb{Z}^{m}\rightarrow SQ$.

Let $\tilde{X}=X\oplus\mathbb{Z}^{2m}$, $\tilde{L}=L\oplus\mathbb{Z}^{m}$ and
$\tilde{N}=N\oplus\mathbb{Z}^{m}$, so that
$\tilde{X}=\tilde{L}\oplus\tilde{N}$ and if
$H_{-}(\mathbb{Z}^{m})=(\mathbb{Z}^{m}\oplus\mathbb{Z}^{m},\lambda_{H},\mu_{H})$
(where each copy of $\mathbb{Z}^{m}$ is a lagrangian), then
$\underline{\mu}\oplus
H_{-}(\mathbb{Z}^{m})=(\tilde{X},\lambda\oplus\lambda_{H},\mu\oplus\mu_{H})$
and $\tilde{L}$ is a lagrangian in $\underline{\mu}\oplus
H_{-}(\mathbb{Z}^{m})$. Let $l=k+m$ and suppose that
$e_{k+1},\ldots,e_{l},f_{k+1},\ldots,f_{l}$ is a symplectic basis of
$H_{-}(\mathbb{Z}^{m})$ (with $\left<e_{k+1},\ldots,e_{l}\right>$ and
$\left<f_{k+1},\ldots,f_{l}\right>$ being the standard lagrangians). Then
$\tilde{L}=\left<e_{1},\ldots,e_{l}\right>$,
$\tilde{N}=\left<f_{1},\ldots,f_{l}\right>$ and
$e_{1},\ldots,e_{l},f_{1},\ldots,f_{l}$ is a symplectic basis of
$(\tilde{X},\lambda\oplus\lambda_{H})$. Similarly let
$\tilde{X}^{\prime}=X^{\prime}\oplus\mathbb{Z}^{2m}$,
$\tilde{L}^{\prime}=L^{\prime}\oplus\mathbb{Z}^{m}$ and
$\tilde{N}^{\prime}=N^{\prime}\oplus\mathbb{Z}^{m}$ so that
$\underline{\mu}^{\prime}\oplus
H_{-}(\mathbb{Z}^{m})=(\tilde{X}^{\prime},\lambda^{\prime}\oplus\lambda_{H},\mu^{\prime}\oplus\mu_{H})$.

The map $h$ can be regarded as an isomorphism
$\tilde{N}\rightarrow\tilde{N}^{\prime}$ with
$S(\mu\oplus\mu_{H}\big{|}_{\tilde{N}})=S(\mu^{\prime}\oplus\mu_{H}\big{|}_{\tilde{N}^{\prime}})\circ
h\colon\tilde{N}\rightarrow SQ$. For every $1\leq i\leq l$ let
$f^{\prime\prime}_{i}=h(f_{i})$, then
$f^{\prime\prime}_{1},\ldots,f^{\prime\prime}_{l}$ is a basis of
$\tilde{N}^{\prime}$. Since the adjoint of $\lambda^{\prime}\oplus\lambda_{H}$
determines an isomorphism $\tilde{L}^{\prime}\cong(\tilde{N}^{\prime})^{*}$,
we get a basis $e^{\prime\prime}_{1},\ldots,e^{\prime\prime}_{l}$ of
$\tilde{L}^{\prime}$ such that
$e^{\prime\prime}_{1},\ldots,e^{\prime\prime}_{l},f^{\prime\prime}_{1},\ldots,f^{\prime\prime}_{l}$
is a symplectic basis of
$(\tilde{X}^{\prime},\lambda^{\prime}\oplus\lambda_{H})$. For every $1\leq
i\leq l$ we have
$S(\mu\oplus\mu_{H})(f_{i})=S(\mu^{\prime}\oplus\mu_{H})(f^{\prime\prime}_{i})\in
SQ=Q_{e}/\mathrm{Im}({\rm p})$, so
$(\mu^{\prime}\oplus\mu_{H})(f^{\prime\prime}_{i})\in Q_{e}$ is either
$(\mu\oplus\mu_{H})(f_{i})$ or $(\mu\oplus\mu_{H})(f_{i})+{\rm p}(1)$. We
define $f^{\prime\prime\prime}_{i}:=f^{\prime\prime}_{i}$ in the case where
$(\mu^{\prime}\oplus\mu_{H})(f^{\prime\prime}_{i})=(\mu\oplus\mu_{H})(f_{i})$
and $f^{\prime\prime\prime}_{i}:=f^{\prime\prime}_{i}+e^{\prime\prime}_{i}$
otherwise.

Then
$e^{\prime\prime}_{1},\ldots,e^{\prime\prime}_{l},f^{\prime\prime\prime}_{1},\ldots,f^{\prime\prime\prime}_{l}$
is again a symplectic basis of
$(\tilde{X}^{\prime},\lambda^{\prime}\oplus\lambda_{H})$. Moreover, we have
the equalities
$(\mu\oplus\mu_{H})(e_{i})=0=(\mu^{\prime}\oplus\mu_{H})(e^{\prime\prime}_{i})$
and
$(\mu\oplus\mu_{H})(f_{i})=(\mu^{\prime}\oplus\mu_{H})(f^{\prime\prime\prime}_{i})$
by the definition of $f^{\prime\prime\prime}_{i}$. Therefore the isomorphism
$I\colon\tilde{X}\rightarrow\tilde{X}^{\prime}$ given by
$I(e_{i})=e^{\prime\prime}_{i}$ and $I(f_{i})=f^{\prime\prime\prime}_{i}$ is
an isometry between $\underline{\mu}\oplus H_{-}(\mathbb{Z}^{m})$ and
$\underline{\mu}^{\prime}\oplus H_{-}(\mathbb{Z}^{m})$. ∎

Recall from Definition 2.26 that a $Q$-form $(X,\lambda,\mu)$ is called full
if its linearisation $S\mu\colon X\to SQ$ is surjective.

######  Lemma 6.5.

For any form parameter $Q$ there exists a full metabolic $Q$-form.

###### Proof.

Let $Q=(Q_{e},{\rm h},{\rm p})$ and suppose that $q_{1},q_{2},\ldots,q_{k}$ is
a generating system for $Q_{e}$. Consider the triple

$\left(\mathbb{Z}^{k}\oplus\mathbb{Z}^{k},\left[\begin{array}[]{cc}0&I_{k}\\\
\epsilon_{Q}I_{k}&D\end{array}\right],(0,\ldots,0,q_{1},\ldots,q_{k})^{\rm
t}\right),$

where $I_{k}$ is the $k\times k$ identity matrix, $D$ is a $k\times k$
diagonal matrix with entries ${\rm h}(q_{1}),\ldots,{\rm h}(q_{k})$ on the
diagonal and $(\cdot)^{\rm t}$ denotes the transpose. By Remark 2.20 this data
defines a $Q$-form, which is metabolic because the matrix is nonsingular and
$\mathbb{Z}^{k}\oplus 0$ a lagrangian. It is also full, because if $\pi\colon
Q_{e}\rightarrow SQ$ denotes the quotient homomorphism, then
$\pi(q_{1}),\ldots,\pi(q_{k})$ is a generating system for $SQ$. ∎

###### Alternative proof of Lemma 6.2.

Metabolic forms have even rank, and so
$\textup{rk}(\mathrm{Ker}(\pi))\subseteq 2\mathbb{Z}$. Moreover, we have
$\textup{rk}([H_{\epsilon_{Q}}(\mathbb{Z}^{m}),\underline{0}])=2m$, and so
$\textup{rk}\colon\mathrm{Ker}(\pi)\to 2\mathbb{Z}$ is surjective. Finally
assume that $\underline{\mu_{0}}$ and $\underline{\mu_{1}}$ are metabolic
$Q$-forms of equal rank, and consider
$\big{(}\\{\underline{\mu_{0}}\\},\\{\underline{\mu_{1}}\\}\big{)}\in\mathrm{Ker}(\textup{rk}\big{|}_{\mathrm{Ker}(\pi)})$.
By Lemma 6.5 there is a full metabolic $Q$-form $\underline{\mu^{\prime}}$,
and by applying Lemma 6.4 to
$\underline{\mu_{0}}\oplus\underline{\mu^{\prime}}$ and
$\underline{\mu_{1}}\oplus\underline{\mu^{\prime}}$ we get that
$\underline{\mu_{0}}\oplus\underline{\mu^{\prime}}\oplus
H_{\epsilon_{Q}}(\mathbb{Z}^{m})\cong\underline{\mu_{1}}\oplus\underline{\mu^{\prime}}\oplus
H_{\epsilon_{Q}}(\mathbb{Z}^{m})$ for some $m$. This shows that
$\big{(}\\{\underline{\mu_{0}}\\},\\{\underline{\mu_{1}}\\}\big{)}=0$, and
therefore $\textup{rk}\big{|}_{\mathrm{Ker}(\pi)}\colon\mathrm{Ker}(\pi)\to
2\mathbb{Z}$ is an isomorphism. ∎

###  6.2. Absorbing $Q$-forms

In this subsection we characterise absorbing $Q$-forms.

######  Definition 6.6.

  1. a)

Let $\underline{\mu}$ and $\underline{\eta}$ be nonsingular $Q$-forms. We say
that $\underline{\mu}$ absorbs $\underline{\eta}$ if $\underline{\eta}$ can be
embedded in
$k\underline{\mu}=\underline{\mu}\oplus\ldots\oplus\underline{\mu}$ for some
$k\geq 1$.

  2. b)

A nonsingular $Q$-form $\underline{\mu}$ is called absorbing, if it absorbs
all nonsingular $Q$-forms.

######  Theorem 6.7.

A nonsingular $Q$-form is absorbing if and only if it is indefinite and full.

###### Proof.

A definite form cannot absorb any indefinite form, and a non-full form cannot
absorb any full form, so the conditions are necessary.

For the other direction first note that $\underline{\mu}$ absorbs
$\underline{\eta}\oplus\underline{\eta}^{\prime}$ if and only if it absorbs
both $\underline{\eta}$ and $\underline{\eta}^{\prime}$. As we saw before
Definition 2.31, $\underline{\eta}\oplus(-\underline{\eta})$ is metabolic for
every nonsingular $\underline{\eta}$. Moreover, it elementary to prove that
every metabolic form can be written as a direct sum of rank-$2$ metabolic
forms, and therefore a $Q$-form is absorbing if and only if it absorbs every
rank-$2$ metabolic form.

Now let $\underline{\mu}=(X,\lambda,\mu)$ be a full indefinite nonsingular
form over $Q=(Q_{e},{\rm h},{\rm p})$, and let $\pi\colon Q_{e}\rightarrow SQ$
denote the projection. Let $\underline{\eta}$ be a rank-$2$ metabolic form, it
is isomorphic to

$\left(\mathbb{Z}^{2}\left<e,f\right>,\left[\begin{array}[]{cc}0&1\\\
\epsilon_{Q}&a\end{array}\right],\left(\begin{array}[]{c}0\\\
q\end{array}\right)\right)$

for some $q\in Q_{e}$ and $a={\rm h}(q)\in\mathbb{Z}$. We will show that
$\underline{\eta}$ embeds into $3\underline{\mu}$.

Since $\underline{\mu}$ is indefinite, there is a primitive $x\in X$ such that
$\lambda(x,x)=0$, and since it is nonsingular, there is a $y\in X$ such that
$\lambda(x,y)=1$. Finally, $\underline{\mu}$ is full, so there is a $z\in X$
such that $S\mu(z)=\pi(q)-S\mu(y)$. If $Q$ is symmetric, then let
$\delta=\frac{1}{2}{\rm h}(q-\mu(y)-\mu(z))$, which is an integer, because
$\pi(q-\mu(y)-\mu(z))=\pi(q)-S\mu(y)-S\mu(z)=0$. It follows that
$q-\mu(y)-\mu(z)\in\mathrm{Ker}(\pi)=\mathrm{Im}({\rm p})$, and ${\rm
h}\circ{\rm p}=2\textup{Id}_{\mathbb{Z}}$. If $Q$ is anti-symmetric, then let
$\delta=0$ if $\mu(y)+\mu(z)=q$ and $\delta=1$ otherwise.

We will use the notation $(X^{3},\bar{\lambda},\bar{\mu})$ for
$3\underline{\mu}$. We define the homomorphism
$i\colon\mathbb{Z}^{2}\rightarrow X^{3}$ by

$i(e)=(x,-x,0)\quad\text{and}\quad i(f)=(y+\delta x,-\delta x,z).$

Then $\bar{\lambda}(i(e),i(e))=0$ and $\bar{\lambda}(i(e),i(f))=1$. We also
have

$\bar{\mu}(i(e))=\mu(x)+\mu(-x)=\mu(x)+\mu(-x)+{\rm
p}(\lambda(x,-x))=\mu(x-x)=0$

and $S\bar{\mu}(i(f))=S\mu(y)+S\mu(z)=\pi(q)$. If $Q$ is symmetric, then

$\bar{\lambda}(i(f),i(f))=\lambda(y,y)+2\delta+\lambda(z,z)={\rm
h}(\mu(y))+{\rm h}(q-\mu(y)-\mu(z))+{\rm h}(\mu(z))={\rm h}(q)=a.$

This means that ${\rm h}(\bar{\mu}(i(f)))={\rm h}(q)$ and
$\pi(\bar{\mu}(i(f)))=\pi(q)$, so by Lemma 2.10 $\bar{\mu}(i(f))=q$. If $Q$ is
anti-symmetric, then $\bar{\lambda}(i(f),i(f))=0=a$. Moreover,

$\bar{\mu}(i(f))=\mu(y+\delta x)+\mu(-\delta x)+\mu(z)=\mu(y)+\mu(\delta
x)+\delta{\rm p}(1)+\mu(-\delta x)+\mu(z)=\mu(y)+\mu(z)+\delta{\rm p}(1).$

The previously established equality $S\mu(y)+S\mu(z)=\pi(q)$ implies that
$\mu(y)+\mu(z)-q\in\mathrm{Ker}(\pi)=\mathrm{Im}({\rm p})$, so either
$\mu(y)+\mu(z)=q$ or $\mu(y)+\mu(z)=q+{\rm p}(1)$. By the definition of
$\delta$ we have $\bar{\mu}(i(f))=q$ in both cases.

Therefore $i$ is a morphism of $Q$-forms. Since $\underline{\eta}$ is
nonsingular, this implies that $i$ is injective. Therefore $\underline{\eta}$
embeds into $3\underline{\mu}$, and by our earlier observations this shows
that $\underline{\mu}$ is absorbing. ∎

##  7\. Witt groups and algebraic Poincaré cobordism

Extended quadratic forms and their Witt groups were previously defined by
Ranicki in [R1], where they are located in the larger context of the bordism
groups of algebraic Poincaré complexes with structure maps to chain bundles.
In this section we briefly recall this larger setting and relate Ranicki’s
algebraic Wu classes to the quasi-Wu classes of Definition 2.9.

As the objects and definitions of algebraic surgery are involved, we will not
go into them here in detail. Rather, we assume that the reader is familiar
with the theory of chain bundles $(B_{*},\beta)$ and algebraic Poincaré
complexes with $(B_{*},\beta)$-structure as developed by Weiss and Ranicki
[We, R1]; we use [R1] as a reference. We fix the ring $\mathbb{Z}$ with
trivial involution and let $B_{*}$ be a chain complex of finitely generated
free $\mathbb{Z}$-modules with $B_{i}=0$ for all $i<0$. A chain bundle
$(B_{*},\beta)$ is an algebraic model for a stable vector bundle, and more
generally a stable spherical fibration. Indeed, if $\xi$ is a spherical
fibration over a simply connected $CW$-complex $X$, then $\xi$ defines a chain
bundle $(C_{*}(X),\beta(\xi))$ over the cellular chain complex of $X$. An
$n$-dimensional $(B_{*},\beta)$-Poincaré complex is an algebraic analogue of
closed $n$-dimensional normal $\xi$-manifold and the $\beta$-twisted
$L$-groups, $L^{n}(B_{*},\beta)$, which are the algebraic bordism classes of
$n$-dimensional $(B_{*},\beta)$-Poincaré complexes, model the bordism group of
closed $n$-dimensional normal $\xi$-manifolds.

The $\beta$-twisted $L$-groups are natural for maps of chain bundles, and by
definition, $L_{n}(\mathbb{Z})$, the usual quadratic $L$-group of the ring
$\mathbb{Z}$ (see e.g. [R2, Defns. 11.57 & 12.23]), is $L_{n}(0,0)$, where
$(0,0)$ denotes the unique chain bundle over the zero chain complex. It
follows that there is a natural map $L_{n}(\mathbb{Z})\to L^{n}(B_{*},\beta)$,
which Weiss [We, 2.21(iii) & (iv)] places into a long exact sequence of
abelian groups

$\dots\xrightarrow{}Q_{n+1}(B_{*},\beta)\xrightarrow{}L_{n}(\mathbb{Z})\xrightarrow{}L^{n}(B_{*},\beta)\xrightarrow{}Q_{n}(B_{*},\beta)\xrightarrow{}L_{n-1}(\mathbb{Z})\xrightarrow{}\dots~{},$
(8)

where $Q_{n}(B_{*},\beta)$ is the group of $n$-dimensional $\beta$-twisted
quadratic structures on $B_{*}$ defined in [R1, §6].

The $\beta$-twisted $L$-groups receive homomorphisms from certain Witt groups,
as we now explain. For each non-negative integer $q$, Ranicki defines a
$(-1)^{q}$-symmetric form parameter $Q_{(B_{*},\beta)}(q)$ [R1, §10], which
for brevity we denote by $Q_{\beta}(q)$. Moreover, given a nonsingular
$Q_{\beta}(q)$-form $(X,\lambda,\mu)$, Ranicki constructs a $2q$-dimensional
$(B_{*},\beta)$-Poincaré complex concentrated in dimension $q$ [R1, p. 250],
and proves that this construction induces a natural homomorphism

$\eta_{\beta}\colon W_{0}(Q_{\beta}(q))\to L^{2q}(B_{*},\beta).$

Below we formulate a conjecture which describes conditions under which the
homomorphism $\eta_{\beta}$ may be a split injection or an isomorphism. At all
events, to use $\eta_{\beta}$ it is important to identify the form parameter
$Q_{\beta}(q)$ by finding the quasi-Wu class of $Q_{\beta}(q)$, and we do this
now. By construction (see [R1, §10]) the linearisation of the form parameter
$Q_{\beta}(q)$ is given by

$SQ_{\beta}(q)=H_{q}(B_{*}).$

The chain bundle $\beta$ also defines algebraic Wu classes
$\widehat{v}_{i}(\beta)\in H^{i}(B_{*};\mathbb{Z}_{2})$ [R1, §4] such that if
$\beta=\beta(\xi)$ is the chain bundle defined by a spherical fibration then,
$\widehat{v}_{i}(\beta(\xi))=v_{i}(\xi)$, where $v_{i}(\xi)$ are the usual Wu
classes of $\xi$. By the Universal Coefficient Theorem, we have a split short
exact sequence

$0\to\textup{Ext}_{1}^{\mathbb{Z}}(H_{q-1}(B_{*}),\mathbb{Z}_{2})\to
H^{q}(B_{*};\mathbb{Z}_{2})\xrightarrow{\mathrm{ev}}\textup{Hom}(H_{q}(B_{*}),\mathbb{Z}_{2})\to
0.$

Recalling Definition 2.9, we note that the quasi-Wu classes of $Q_{\beta}(q)$
are such that if $q=2k$ is even then
$v_{Q_{\beta}(2k)}\in\textup{Hom}(H_{q}(B_{*}),\mathbb{Z}_{2})$ and if
$q=2k{+}1$ is odd, then either $v^{\prime}_{Q_{\beta}(2k+1)}=0$, or
$v^{\prime}_{Q_{\beta}(2k+1)}\neq 0$ is determined by the extension

$[v^{\prime}_{Q_{\beta}(2k+1)}]:=\bigl{[}0\to\mathbb{Z}_{2}\to(Q_{\beta}(2k{+}1))_{e}\to
H_{2k+1}(B_{*})\to 0\bigr{]}.$

######  Proposition 7.1.

The algebraic Wu classes $\widehat{v}_{*}(\beta)$ of a chain bundle
$(B_{*},\beta)$ determine the quasi-Wu classes of the form parameter
$Q_{\beta}(q)$ as follows:

  1. a)

If $q=2k$, then $v_{Q_{\beta}(2k)}=\mathrm{ev}(\widehat{v}_{2k}(\beta))$.

  2. b)

If $q=2k{+}1$ and $\mathrm{ev}(\widehat{v}_{2k+2}(\beta))\neq 0$ then then
$v^{\prime}_{Q_{\beta}(2k+1)}=0$.

  3. c)

If $q=2k{+}1$ and $\mathrm{ev}(\widehat{v}_{2k+2}(\beta))=0$, then
$\widehat{v}_{2k+2}(\beta)\in\textup{Ext}^{\mathbb{Z}}_{1}(H_{2k+1}(B_{*}),\mathbb{Z}_{2})$
and $[v^{\prime}_{Q_{\beta}(2k+1)}]=\widehat{v}_{2k+2}(\beta)$.

###### Proof.

a) This follows directly by inspecting the formula for ${\rm h}$ on [R1, p.
249] (it is denoted there by $s$), and the formula for $\widehat{v}_{2k}$ on
[R1, p. 232].

b) The braid [R1, p. 249] gives an exact sequence of abelian groups

$H_{2k+2}(B)\xrightarrow{\mathrm{ev}(\widehat{v}_{2k+2}(\beta))}(Q_{-})_{e}\xrightarrow{}Q_{\beta}(2k{+}1)_{e}\to
H_{2k+1}(B)\to 0,$

where $(Q_{-})_{e}\to Q_{\beta}(2k{+}1)$ is $v^{\prime}_{Q_{\beta}(2k{+}1)}$.
The exactness of this sequence proves Part b).

c) This follows by inspecting the formula on [R1, p. 249] for addition in
$Q_{\beta}(2k{+}1)$. ∎

With Proposition 7.1 and Theorem 1.5 we can compute the Witt group
$W_{0}(Q_{\beta}(q))$ from knowledge of $H_{q}(B)$ and the algebraic Wu
classes of $\beta$. To apply these computations to the $\beta$-twisted
$L$-groups and $\beta$-twisted $Q$-groups, we need more information about the
homomorphism $\eta_{\beta}$.

######  Conjecture 7.2.

Let $(B_{*},\beta)$ be a chain bundle and $q\geq 0$. If $H_{q-1}(B_{*})$ is
torsion free, then the natural homomorphism $\eta_{\beta}\colon
W_{0}(Q_{\beta}(q))\to L^{2q}(B_{*},\beta)$ is split injective. If, in
addition, $H_{i}(B_{*})=0$ for all $i<q$, then $\eta_{\beta}$ is an
isomorphism.

######  Remark 7.3.

One idea for constructing an inverse to $\eta_{\beta}$ goes as follows. In
[R1, §10] Ranicki proves that the $q$-th homology group of a $2q$-dimensional
$(B_{*},\beta)$-Poincaré complex supports a $Q_{\beta}(q)$-form. If the map
from $(B_{*},\beta)$-Poincaré complex to $B_{*}$ is $q$-connected, then since
$H_{q-1}(B_{*})$ is torsion free, Poincaré duality ensures that this
$Q_{\beta}(q)$-form is nonsingular, and we can take it’s Witt class a
candidate for the value of $\eta_{\beta}^{-1}$. If the map from the
$(B_{*},\beta)$-Poincaré complex to $B_{*}$ is not $q$-connected, we expect
that algebraic surgery below the middle dimension, e.g. [We, Proposition 5.4],
can be used to show both that the $(B_{*},\beta)$-Poincaré complex is
$(B_{*},\beta)$-cobordant to a $(B_{*},\beta)$-Poincaré complex-connected
complex with a $q$-connected map to $B_{*}$, and that the Witt class of the
$Q_{\beta}(q)$-forms obtained is well-defined.

To apply Conjecture 1.3 we need a further definition in the symmetric case.
This arises since Weiss’s sequence (8) involves the quadratic $L$-group
$L_{4k}(\mathbb{Z})=W_{0}(Q_{+})$, whereas our methods compare
$W_{0}(Q_{\beta}(2k))$ with the symmetric $L$-group
$W_{0}(Q^{+})=L^{4k}(\mathbb{Z})$. For $v\colon A\to\mathbb{Z}_{2}$ a
homomorphism, we recall $\Sigma(v)\subset\mathbb{Z}\oplus\Gamma(E)$ from
Definition 5.3. We make the following mod $8$ modification of $\Sigma(v)$.

######  Definition 7.4.

Let $\Sigma_{8}(v)\leq\mathbb{Z}/8\oplus\Gamma(A)$ denote the subgroup
generated by the following elements:

$(1,x\otimes 1),\quad(0,[k,x]\otimes 1-k\otimes
1),\quad\text{and}\quad(0,[k_{1},k_{2}]\otimes 1),$

where $x\in v^{-1}(1)$ and $k,k_{1},k_{2}\in\mathrm{Ker}(v)$.

The next conjecture would be a direct consequence of Conjecture 7.2,
Proposition 7.1 and Theorem 1.5. Recall that
$\iota_{1}\colon\mathbb{Z}_{2}\to\mathbb{Z}_{2}$ is the identity. For any
homomorphism $v^{\prime}\colon\mathbb{Z}_{2}\to A$, there is a canonical map
of homomorphisms $\iota_{1}\to v^{\prime}$, which induces a homomorphism
$\Lambda(\iota_{1})\to\Lambda(v^{\prime})$.

######  Conjecture 7.5.

Let $(B_{*},\beta)$ be a chain bundle with algebraic Wu class
$\widehat{v}_{2k}(\beta)$.
If $(B_{*},\beta)$ is $(2k{-}1)$-connected then

  1. a)

$L^{4k}(B_{*},\beta)\cong
W_{0}(Q_{\beta}(2k))\cong\Sigma\bigl{(}\mathrm{ev}(\widehat{v}_{2k}(\beta))\bigr{)}$;

  2. b)

$Q_{4k}(B_{*},\beta)\cong\Sigma_{8}\bigl{(}\mathrm{ev}(\widehat{v}_{2k}(\beta))\bigr{)}$.

If $(B_{*},\beta)$ is $(2k{-}2)$-connected and
$\mathrm{ev}(\widehat{v}_{2k}(\beta))=0$, then
$[v^{\prime}_{Q_{\beta}(2k{-}1)}]=\widehat{v}_{2k}(\beta)$ and

  1. c)

$L^{4k-2}(B_{*},\beta)\cong
W_{0}(Q_{\beta}(2k{-}1))\cong\Lambda(v^{\prime}_{Q_{\beta}(2k{-}1)})$;

  2. d)

$Q_{4k-2}(B_{*},\beta)\cong\textup{Coker}\bigl{(}\Lambda(\iota_{1})\to\Lambda(v^{\prime}_{Q_{\beta}(2k{-1})})\bigr{)}$.

If $(B_{*},\beta)$ is $(2k{-}2)$-connected and
$\mathrm{ev}(\widehat{v}_{2k}(\beta))\neq 0$ then $Q_{\beta}(2k{-1})\cong
Q^{-}\oplus H_{2k-1}(B_{*})$ and

  1. e)

$L^{4k-2}(B_{*},\beta)\cong Q_{4k-2}(B_{*},\beta)\cong
W_{0}(Q_{\beta}(2k{-}1))\cong\Lambda(H_{2k-1}(B_{*}))$.

######  Remark 7.6.

The conjectural computation of $\beta$-twisted $Q$-groups in Conjecture 7.5 is
consistent with the computations of Banagl and Ranicki [B-R, Proposition 52].

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