\titlea{\S31}{Stably Conservative Reduction of Quadratic Forms}
\leftrunningtitle{Chapter IV. Specialization with Respect to Quadratic Places}
%\leftrunningtitle{Specialization of Quadratic and Symmetric
%Bilinear Forms}
\rightrunningtitle{\S31 Stably Conservative Reduction of Quadratic Forms}

The splitting theory of a quadratically specialized form, developed above, required the assumption that the characteristic is different from two. We would like to establish such a theory also for characteristic two, generalizing our  theory in Chapter~I, from \S5 onwards, and in Chapter~II.

Thence we will consider quadratic forms instead of bilinear forms. In the process we will endeavour to stay as close as possible to the specialization concept of ``almost good reduction'', developed in \S28--30.
Nevertheless important new problems will occur, which all arise from the lack of obedience of many forms and the existence of quasilinear parts.
An obvious, immediate generalization of the concept of ``almost good reduction'' will not suffice. 

We first recall some definitions from \S 7 and \S 11.

Let  $\lambda \colon K \to L \cup \infty$ be a place between fields of arbitrary characteristic, ${\nam o} = {\nam
o}_{\lambda}$, ${\nam m} = {\nam m}_{\lambda}$, $k = {\nam o}/{\nam
m}$. The place $\lambda$ yields a field embedding 
$\overline{\lambda} \colon k \to L$.

We choose a subgroup $T$ of $Q(K)$ with $Q(K) = T \times
Q({\nam o})$. A quadratic space $(E, q) = E$ over $K$
is called {\it weakly obedient with respect to} $\lambda$ if it has a decomposition
$$E  =  \raise5pt\hbox{$\newbot\limits_{\alpha \in T}$} \; F_{\alpha}, \quad F_{\alpha}  \cong  \langle \alpha \rangle \otimes
E_{\alpha} \leqno(1)$$
in which every $E_{\alpha}$ is a quadratic space over
$K$ of the form
$$E_{\alpha}  =  K \otimes_{{\nam o}} M_{\alpha}  =  K M_{\alpha}, \leqno(2)$$
for some quadratic module  $M_{\alpha}$ over
${\nam o}$, which is free of finite rank, and for which the quadratic module
$M_{\alpha} / {\nam m} M_{\alpha}$ over $k$ is nondegenerate.
($M_{\alpha}$ is then called ``reduced nondegenerate''.) We also called  (1) a 
{\it weakly $\lambda$-modular decomposition} of $E$. We described spaces of the form (2)  as having  {\it fair reduction} (FR),  and denoted the space
$M_{\alpha}/ {\nam m} M_{\alpha}
\otimes_{\overline{\lambda}} L$ by $\lambda_* (E_{\alpha})$. We
saw in \S 11 that for every $\alpha$ the Witt class $\{
\lambda_* (E_{\alpha}) \}$ is  determined solely by $E$, $\lambda$ and $\alpha$.

We should keep in mind that the quasilinear part  $QL (M_{\alpha}/
{\nam m} M_{\alpha})$ of $M_{\alpha} / {\nam m} M_{\alpha}$
has to be anisotropic,  but that $QL (\lambda_* (E_{\alpha}))$ 
could be isotropic. $\{$We defined Witt equivalence over fields also for
forms with isotropic quasilinear part, cf.  \S
6.$\}$

If  $E$ actually has the decomposition (1), (2), in which every $M_{\alpha}$
is a nondegenerate quadratic  ${\nam o}$-module, then we said that
$E$ is {\it obedient} with respect to $\lambda$, and further that 
$E_{\alpha}$ has {\it good reduction} (GR) with respect to $\Lambda$. 
Finally we then called  (1) a {\it $\lambda$-modular
decomposition} of $E$.

Assume now that  $\lambda$ is enlarged to a quadratic place
$\Lambda = (\lambda, H, \chi) \colon K \to L \cup \infty$. We choose
direct decompositions $H = H_0 \times Q ({\nam o})$, $Q(K) =
S \times H$. The group $T$ from above is now $S \times H_0 = SH_0$.

We switch to the language of quadratic forms instead of quadratic modules, and thus use $\varphi,
\varphi_{\alpha}$ etc. instead of $E, E_{\alpha}$ etc.  A ``form'' will now always be a 
quadratic form.

Let $\varphi$ be a nondegenerate form over $K$.

\begdefinition{1.} We say that  $\varphi$ has {\it fair reduction}\Idx{fair reduction (w.r.t. a Q-place)@fair reduction (w.r.t. a $Q$-place)}
(= FR)\Idx{111fr2@FR (w.r.t. a $Q$-place)}
 with respect to $\Lambda$, or that $\varphi$ is {\it weakly
$\Lambda$-unimodular},\Idx{weakly l2-unimodular@weakly $\Lambda$-unimodular} 
if $\varphi$ has a decomposition
$$\varphi  \cong \; \raise5pt\hbox{$\newbot\limits_{\alpha \in H_0}$} \alpha
\otimes \varphi_{\alpha}  = 
\raise5pt\hbox{$\newbot\limits_{\alpha \in H_{\alpha}}$} \alpha
\varphi_{\alpha} \leqno(3)$$
in which every  $\varphi_{\alpha}$ has FR with respect to $\lambda$, and then set
$$\Lambda_* (\varphi) \colon  =  \raise5pt\hbox{$\newbot\limits_{\alpha \in
H_{0}}$} \chi (\alpha) \lambda_* (\varphi_{\alpha}). \leqno(4)$$
If there actually exists a decomposition (3) in which all $\varphi_{\alpha}$ have GR
with respect to $\lambda$, then we say that $\varphi$ has GR with respect to
$\Lambda$, or that $\varphi$ is $\Lambda$-{\it unimodular}.\Idx{l2-unimodular@$\Lambda$-unimodular} 
\enddefinition

As in \S 28 we see that  $\Lambda_*
(\varphi)$ is determined by  $\varphi$ up to isometry, and also that it is independent of the choice of the complement  $H_0$ of $Q ({\nam o})$ in $H$, 
if we utilize  the weak specialization  $\lambda_W$ from \S 11 instead of the map
$\lambda_W$ from \S3.\fonote{In order to be able to repeat the conclusion of \S28, it is important that for our definition of Witt equivalence over fields (\S6, Def. 10),  the statement 
``If $E \sim F, \, \dim E = \dim F$, then $E \cong F$''
also holds for degenerate spaces.}  $\{$In the case of good reduction, the operator
$\lambda_W$ from \S 7 suffices.$\}$

It goes without saying that  $\varphi$ must be weakly obedient (resp. obedient) with respect to  $\lambda$ in order for  $\varphi$ to have FR (resp. GR) with respect to
$\Lambda$. It can happen that $\Lambda_* (\varphi)$ is degenerate;
for fair reduction even in the case where $\lambda$ is the canonical place 
$K \to {\nam o}/{\nam m} \cup \infty$ associated to ${\nam
o}$.

\begremarks{}
\smallskip	
\setitemindent{$(2)$}
\item{$(1)$} If  $\varphi, \psi$ have FR (resp. GR) with respect to $\Lambda$, and if 
$\varphi$ is strictly regular, then  $\varphi \perp \psi$  has FR (resp. GR)
with respect to $\Lambda$ and
$$\Lambda_* (\varphi \perp \psi)  \cong  \Lambda_* (\varphi) \perp \Lambda_*
(\psi), \leqno(5)$$
which can be seen easily.
\smallskip

\item{$(2)$} If $\sigma$ is a bilinear form which has GR with respect to $\Lambda$, $\varphi$ is a quadratic form which has  FR (resp. GR) with respect to $\Lambda$, and {\it if
$\Lambda_* (\varphi)$ is strictly regular}, then $\sigma \otimes
\varphi$ has FR (resp. GR) with respect to $\Lambda$ and
$$\Lambda_* (\sigma \otimes \varphi)  \cong  \Lambda_* (\sigma) \otimes
\Lambda_* (\varphi), \leqno(6)$$
which can also be verified easily. Note that $\Lambda_* (\sigma)$ is determined by $\Lambda,
\sigma$ only up to stable isometry. Nevertheless this formula (with true isometry) is correct. 
\smallskip

\item{$(3)$} It is important that $\Lambda_* (\varphi)$ is strictly regular in formula~(6), at any rate when  $\Lambda_* (\sigma)$ is  isotropic. This follows already from the following observation. Let $\Char L = 2$ and let $a, b, c \in
L^*$. Then
$$\langle a, a \rangle \otimes [c]  \cong  [ac, ac]  \cong  [ac, 0],$$
and likewise
$\langle b, b \rangle \otimes [c] \cong [bc, 0]$.  Furthermore
$$\langle a, a \rangle  \cong  \bigg< {a \atop 1} \; {1 \atop 0} \bigg> \approx
\bigg< {b \atop 1} \; {1 \atop 0} \bigg>  \cong  \langle
b, b \rangle,$$
but in general  $[ac, 0] \not\cong [bc, 0]$. \qed
\endremarks

If  $\varphi$ is weakly obedient with respect to $\lambda$ and if
$$\varphi  =  \raise5pt\hbox{$\newbot\limits_{(s, \beta) \in S \times
H_0}$} s \beta \, \varphi_{s, \beta} \leqno(1)'$$
is a weakly  $\lambda$-unimodular decomposition of $\varphi$, in which thus every
form $\varphi_{s, \beta}$ has FR with respect to $\lambda$, then we can combine 
the forms $\varphi_{s, \beta}$, for fixed $s$,
in a form
$$\varphi_{s} \colon  =  \raise5pt\hbox{$\newbot\limits_{\beta \in
H_0}$} \beta \, \varphi_{s, \beta} \leqno(5)$$
which is weakly  $\Lambda$-unimodular. 
This results in a coarser decomposition
$$\varphi  =  \raise5pt\hbox{$\newbot\limits_{s \in S}$} s \,
\varphi_{s}.\leqno(6)$$
We call such a decomposition  (6) a {\it weakly 
$\Lambda$-modular decomposition}\Idx{weakly l2-modular decomposition@weakly $\Lambda$-modular decomposition} 
of $\varphi$. If $\varphi$ is actually
obedient with respect to $\lambda$, then we have such decompositions (5), (6),
in which the  $\varphi_{s, \beta}$ have GR with respect to $\lambda$ and the
$\varphi_{s}$ have GR with respect to $\Lambda$, and then call (6) a
$\Lambda$-{\it modular decomposition} of $\varphi$.\Idx{l2-modular decomposition@$\Lambda$-modular decomposition}


\begdefinition{2.} (a) We say that  $\varphi$ has {\it almost fair
reduction} (= AFR)\Idx{almost fair reduction (w.r.t. a Q-place)@almost fair reduction (w.r.t. a $Q$-place)}\Idx{111afr@AFR (w.r.t. a $Q$-place)} 
with respect to $\Lambda$ if $\varphi$ is weakly obedient
with respect to $\lambda$, and if in a  $\Lambda$-modular decomposition (6)
of $\varphi$ the specializations $\Lambda_* (\varphi_{s})$, with $s
\in S \setminus \{ 1 \}$, are all hyperbolic.

(b) We say that $\varphi$ has {\it almost good reduction} (= AGR)\Idx{almost good reduction (w.r.t. a Q-place)@almost good reduction (w.r.t. a $Q$-place)}\Idx{111agr2@AGR (w.r.t. a $Q$-place)}
with respect to $\Lambda$ if there is a $\Lambda$-modular decomposition
(6) in which 
$\Lambda_* (\varphi_{s})$ is  hyperbolic for each $s \in S \setminus \{ 1
\}$.

Note that these properties are independent of the choice of decomposition
(6)  since every $\Lambda_* (\varphi_{s})$ is up to Witt equivalence uniquely determined by  $\varphi$, $\Lambda$ and $s$.

As in \S28 we see that by almost fair reduction the form
$$\Lambda_* (\varphi) \colon  =  \Lambda_* (\varphi_1) \perp \frac{\dim \varphi
- \dim \varphi_1}{2} \times H$$
is determined up to isometry (instead of only Witt equivalence)
by $\Lambda$ and $\varphi$ alone.  We call  $\Lambda_*
(\varphi)$ the {\it specialization of $\varphi$ with respect to} $\Lambda$.\Idx{specialization (w.r.t. a Q-place)@specialization (w.r.t. a $Q$-place)}\Idx{111lsfi3@$\Lambda_* (\varphi)$}
\enddefinition


Next there follow two theorems which can be proved in exactly the same way as the analogous statements in \S 28 and \S 29.

\begtheorem{1.} Let $\varphi$ and $\psi$ be quadratic forms over
$K$. Assume that $\varphi$ is strictly regular and has AGR with respect to $\Lambda$. Assume that $\psi$ has AFR $($resp. AGR$)$ with respect to $\Lambda$. Then
$\varphi \perp \psi$ has AFR $($resp. AGR$)$ with respect to $\Lambda$ and
$$\Lambda_* (\varphi \perp \psi)  \cong  \Lambda_* (\varphi) \perp \Lambda_*
(\psi).\eqno\square$$
\endtheorem

\begtheorem{2.} Let $\varphi$ be a quadratic form, weakly obedient $($resp. obedient$)$  with respect to  $\lambda \colon K \to L \cup
\infty$. As before, let  $\Lambda = (\lambda, \chi, H) \colon K \to L \cup \infty$
be a $Q$-place with first component $\lambda$. Let further $E \supset K$ be a
field extension and  $M \colon
E \to L \cup \infty$ a $Q$-place with  $M \vert K \supset
\Lambda$.
\smallskip	
\setitemindent{$(b)$}

\item{$(a)$} If $\varphi$ has AFR $($resp. AGR$)$ with respect to $\Lambda$, then
$\varphi \otimes E$ has AFR $($resp. AGR$)$ with respect to $M$, and $M_*
(\varphi \otimes E) = \Lambda_* (\varphi)$.
\smallskip

\item{$(b)$} If actually  $M \vert K = \Lambda$, and if $\varphi \otimes
E$ has  AFR $($resp. AGR$)$ with respect to $M$, then $\varphi$ has AFR $($resp. AGR$)$ with respect to $\Lambda$. \hfill $\square$
\endtheorem

However, can we establish, as in \S28, wether for two forms
$\varphi, \psi$ over $K$ where $\varphi$ and $\varphi
\perp \psi$ have AGR with respect to $\Lambda \colon K \to L \cup \infty$,
also $\psi$ has AGR with respect to $\Lambda$, even when $\varphi$
and $\psi$ are both strictly regular?

This is a debatable question since (at least for the author) there is no discernible reason why $\psi$ should be obedient with respect to the place
$\lambda$.

Thus we are forced to make a new turn and follow a different path. The fact that 
for almost fair reduction, or even almost good reduction, the quasilinear part
$QL (\Lambda_* (\varphi))$ could be isotropic, will be an obstacle. We want to exclude this possibility.

\begdefinition{3.} Let  $\Lambda \colon K \to L \cup \infty$ be a 
quadratic place and let $\varphi$ be a form over $K$. We say that
$\varphi$ has {\it conservative reduction} (= CR) {\it with respect to}\Idx{conservative reduction}\Idx{111cr@CR}
$\Lambda$ if $\varphi$ has AFR with respect to $\Lambda$ and $QL(\Lambda_*
(\varphi))$ is anisotropic. 
\enddefinition

If  $\varphi$ actually has   AGR with respect to $\Lambda$, then clearly
$QL(\Lambda_* (\varphi)) = \Lambda_* (QL(\varphi))$. Hence
$\varphi$ certainly has   CR in this case if $\varphi$ is in addition regular.
If $\varphi$ only has   AFR, then $QL (\Lambda_* (\varphi))$ may be isotropic, even if $\varphi$ is regular.

As a result of the following lemma we obtain a connection between conservative reduction with respect to $\Lambda$ and fair reduction with respect to an ordinary place. In the process we will be able to utilize results from \S11.

\beglemma{1.} Let $(K \hookrightarrow E, \mu)$ be a realization of
 $\Lambda$, i.e.,  $E \supset K$ is a field extension and $\mu
\colon E \to L \cup \infty$ is a place with $\hat{\mu} \vert K
\supset \Lambda$. $\{$Note that there exists a realization of $\Lambda$
with $E \supset K$ even purely transcendental, as demonstrated in \S $30$.$\}$ If $\varphi$ has conservative reduction with respect to
 $\Lambda$, then $\varphi \otimes E$ has fair reduction with respect to
$\mu$ and
$\mu_* (\varphi \otimes E) = \Lambda_* (\varphi)$.
\endlemma

\begproof{} We use Theorem 2$(b)$. Let  $\varphi = 
\raise5pt\hbox{$\newbot\limits_{s \in S}$} \langle s \rangle \otimes
\varphi_s$ be a weakly $\Lambda$-modular  decomposition of $\varphi$.
For every $s \in S$ we have $\langle s \rangle_E \in Q ({\nam
o}_{\mu})$. If $s \ne 1$, then $\varphi_s \otimes E$ has FR with respect to
$\mu$, and $\mu_* (\varphi_s \otimes E) = \Lambda_* (\varphi_s)$ is
hyperbolic. Furthermore, the anisotropy of  $QL (\Lambda_*
(\varphi)) = QL (\Lambda_* (\varphi_1))$ yields that $\varphi_1 \otimes
E$ has fair reduction with respect to
 $\mu$ and $\mu_* (\varphi_1 \otimes E)
= \Lambda_* (\varphi_1)$.  Hence $\varphi \otimes E$ has FR with respect to
$\mu$ and
$$\mu_* (\varphi \otimes E)  =  \Lambda_* (\varphi_1)  \perp  \frac{\dim \varphi -
\dim \varphi_1}{2} \times H  =  \Lambda_* (\varphi).\eqno\square$$
\endproof

\begtheorem{3.} Let $\varphi, \psi$ be quadratic forms over $K$,
both having CR with respect to $\Lambda \colon K \to L \cup \infty$. Assume that
$\varphi < \psi$. Then $\Lambda_* (\varphi) < \Lambda_* (\psi)$.
\endtheorem

\begproof{} We choose a place $\mu \colon E \to L \cup
\infty$ with $E \supset K$ and $\hat{\mu} \vert K \supset \Lambda$.
By the lemma $\varphi \otimes E$ and $\psi \otimes E$ both have
FR with respect to $\mu$ with $\Lambda_* (\varphi) = \mu_* (\varphi \otimes
E)$, $\Lambda_* (\psi) = \mu_* (\psi \otimes E)$. The form $\varphi
\otimes E$ is a subform of $\psi \otimes E$. By \S 11, Theorem 2
(Corollary~2)  $\mu_* (\varphi \otimes E)$ is a subform of
$\mu_* (\psi \otimes E)$. \qed
\endproof

Now we come to the key definition, which allows us to tame disobedient forms
at least a little bit.  

\begdefinition{4.} Let $\varphi$ be a quadratic form over $K$ and, as before,
$\Lambda \colon K \to L \cup \infty$ a quadratic place.
\smallskip	
\setitemindent{$(b)$}

\item{(a)} $\varphi$ has {\it stably conservative} reduction\Idx{stably conservative reduction}\Idx{111scr@SCR} 
($=$ SCR) with respect to
$\Lambda$ if there exists an $r \in {\Bbb N}_0$ such that $\varphi
\perp r \times H$ has conservative reduction with respect to $\Lambda$.
\smallskip

\item{(b)} In this case  $r \times H$ is a subform of
$\Lambda_* (\varphi \perp r \times H)$ by Theorem~3. We then denote by
$\Lambda_* (\varphi)$ the form over $L$, uniquely determined up to isometry, for which
$$\Lambda_* (\varphi \perp r \times H)  =  \Lambda_* (\varphi) \perp r \times
H,$$
and call $\Lambda_* (\varphi)$ the specialization of  $\varphi$ with respect to
$\Lambda$.
\enddefinition

\begremark{}  $\Lambda_* (\varphi)$ is independent of the choice of
$r$.  Namely, if $s \in {\Bbb N}$ is arbitrary then, in the situation of Definition~4$(a)$, 
$\varphi \perp (r + s) \times H =
(\varphi \perp r \times H) \perp s \times H$ also has conservative reduction
and
$$\Lambda_* (\varphi \perp (r \times s) \times H)  =  \Lambda_* (\varphi \perp r \times H)
\perp s \times H  =  \Lambda_* (\varphi) \perp (r + s)
\times H.$$
\endremark

\begdefinition{5.} Similarly we define: a quadratic form $\varphi$ over $K$ has {\it stably almost good reduction} ($=$
SAGR)\Idx{stably almost good reduction}\Idx{111sagr@SAGR} 
with respect to $\Lambda$ if there exists an $r \in {\Bbb N}_0$ such that
 $\varphi \perp r \times H$ has AGR with respect to $\Lambda$.
\enddefinition 

\begremark{} If $\varphi$ has SAGR, then $QL(\varphi)$ has good reduction
with respect to $\Lambda$. Such a form $\varphi$ has SCR
with respect to $\Lambda$ if and only if  $\Lambda_* (QL(\varphi))$ is anisotropic, and then  $\Lambda_* (QL(\varphi)) = QL (\Lambda_* (\varphi))$.
\endremark

Finally we can derive a full analogue of the additive statements in \S 28, Theorem~3 and \S 28, Theorem 4.

\begtheorem{4.} Let  $\varphi$ and $\psi$ be quadratic forms over
 $K$. Assume that $\varphi$ is {\it strictly regular} and that it has SAGR with respect to $\Lambda \colon K \to L \cup \infty$.
\smallskip	
\setitemindent{$(b)$}

\item{$(a)$} If  $\psi$ has SCR with respect to $\Lambda$, then $\varphi \perp \psi$ has SCR with respect to $\Lambda$ and
$$\Lambda_* (\varphi \perp \psi)  \cong  \Lambda_* (\varphi) \perp
\Lambda_* (\psi).$$
\smallskip

\item{$(b)$} If $\varphi \perp \psi$ has SCR with respect to $\Lambda$, then $\psi$ has SCR with respect to  $\Lambda$.
\endtheorem

\begproof{} $(a)$ Choose $r \in {\Bbb N}_0$, $s \in {\Bbb N}_0$ such that
$\varphi \perp r \times H$ has AGR and $\psi \perp s \times
H$ has CR with respect to $\Lambda$. From Theorem~1 it follows immediately 
that $\varphi \perp \psi \perp (r + s) \times H$ has CR with respect to $\Lambda$ and that
$$\Lambda_* (\varphi) \perp r \times H \perp \Lambda_* (\psi) \perp s \times
H  =  \Lambda_* (\varphi \perp \psi \perp (r +s) \times
H).$$
Therefore,
$$\Lambda_* (\varphi) \perp \Lambda_* (\psi) =  \Lambda_* (\varphi \perp \psi).$$

$(b)$ We choose integers $r, t \in {\Bbb N}_0$ with $t \geq r$
and such that $\rho \colon = \varphi \perp r \times H$ has  AGR and that
$\tau \colon = (\varphi \perp \psi) \perp t \times H$ has AFR. We have 
$$\rho \perp \psi \perp (t-r) \times H  \cong  \tau.$$
Therefore,
$$\psi \perp (t - r + \dim \rho) \times H  \cong  (- \rho) \perp \tau.$$
By part $(a)$ of the Theorem,  $(- \rho) \perp
\tau$ has SCR with respect to $\Lambda$. Thus $\psi$ has SCR with respect to $\Lambda$.\qed
\endproof

Furthermore we can extend Theorem~3 and part of Theorem~2 to stably conservative reduction. 

\begtheorem{5.} Let $\varphi, \psi$ be quadratic forms over $K$ which have
SCR with respect to $\Lambda$. Assume that $\varphi < \psi$. Then
$\Lambda_* (\varphi) < \Lambda_* (\psi)$.
\endtheorem

\begproof{} We choose an  $r \in {\Bbb N}$ such that $\varphi
\perp r \times H$ and $\psi \perp r \times H$ have CR. By Theorem~3 we have
$$\Lambda_* (\varphi) \perp r \times H  =  \Lambda_* (\varphi \perp r \times H)
<  \Lambda_* (\psi \perp r \times H)  =  \Lambda_* (\psi)
\perp r \times H.$$
Hence, $\Lambda_* (\varphi) < \Lambda_* (\psi)$. \qed
\endproof

\begtheorem{6.} Let $K \subset E$ be a field extension and let
$\Lambda \colon K \to L \cup \infty$,\break 
$M \colon E \to L \cup \infty$
be quadratic places with $M \vert K \supset \Lambda$. Let
$\varphi$ be a quadratic form over $K$ which has SCR with respect to
$\Lambda$. Then $\varphi \otimes E$ has SCR with respect to $M$ and
$M_* (\varphi \otimes E) = \Lambda_* (\varphi)$.
\endtheorem

\begproof{} We choose an $r \in {\Bbb N}_0$ such that
$\varphi \perp r \times H$ has CR with respect to $\Lambda$. By Theorem~2$(a)$,
$\varphi_E \perp r \times H$ has AFR with respect to
$\Lambda$ and
$$M_* (\varphi_E \perp r \times H)  \cong  \Lambda_* (\varphi \perp r \times H)
 =  \Lambda_* (\varphi) \perp r \times H.$$
In particular,  $QL (M_* (\varphi_E \perp r \times H))$
is anisotropic. Thus $\varphi_E \perp r \times H$ has CR with respect to $M$. Therefore  $\varphi_E$ has SCR with respect to $\Lambda$ and
$$M_* (\varphi_E) \perp r \times H  \cong  \Lambda_* (\varphi) \perp r \times
H.$$
It follows that $M_* (\varphi_E) \cong \Lambda_* (\varphi)$. \qed
\endproof

