Mixing
Witt extension Theorem Stronger Version
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This is Theorem 1.5.5 from Scharlau's book Quadratic and Hermitian Forms (Witt's theorem).
Let $(V, b)$ be a regular symmetric bilinear space.
Let $W$ be a subspace of $V$ and $sigma colon W to V$ an isometry. Then there exists
an isometry $tau colon V to V$ which extends $sigma $ i.e $tau$ restricted to $W$ is $sigma $. $tau$ is product of reflections which extends to $sigma$.
How to prove above result?
I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic. Also I want some explanation of totally isotropic subspace. What is step where we use induction in totally isotropic subspace and how we use it? What are reflection in second case? Please someone help me out.
linear-algebra bilinear-form
edited Jul 22 at 18:52
Martin Sleziak
43.5k6113259
asked Jul 18 at 18:08
sscool
198116
add a comment |Â
up vote
5
down vote
favorite
This is Theorem 1.5.5 from Scharlau's book Quadratic and Hermitian Forms (Witt's theorem).
Let $(V, b)$ be a regular symmetric bilinear space.
Let $W$ be a subspace of $V$ and $sigma colon W to V$ an isometry. Then there exists
an isometry $tau colon V to V$ which extends $sigma $ i.e $tau$ restricted to $W$ is $sigma $. $tau$ is product of reflections which extends to $sigma$.
How to prove above result?
I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic. Also I want some explanation of totally isotropic subspace. What is step where we use induction in totally isotropic subspace and how we use it? What are reflection in second case? Please someone help me out.
linear-algebra bilinear-form
edited Jul 22 at 18:52
Martin Sleziak
43.5k6113259
asked Jul 18 at 18:08
sscool
198116
@FedericoFallucca please take a look at question and link for the book is given in question ? Please look over it if possible for you and if possible then you please explain me for case of totally isotropic subspace and how can we use induction in that proof?
– Ninja hatori
Jul 22 at 15:13
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
This is Theorem 1.5.5 from Scharlau's book Quadratic and Hermitian Forms (Witt's theorem).
Let $(V, b)$ be a regular symmetric bilinear space.
Let $W$ be a subspace of $V$ and $sigma colon W to V$ an isometry. Then there exists
an isometry $tau colon V to V$ which extends $sigma $ i.e $tau$ restricted to $W$ is $sigma $. $tau$ is product of reflections which extends to $sigma$.
How to prove above result?
I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic. Also I want some explanation of totally isotropic subspace. What is step where we use induction in totally isotropic subspace and how we use it? What are reflection in second case? Please someone help me out.
linear-algebra bilinear-form
edited Jul 22 at 18:52
Martin Sleziak
43.5k6113259
asked Jul 18 at 18:08
sscool
198116
This is Theorem 1.5.5 from Scharlau's book Quadratic and Hermitian Forms (Witt's theorem).
Let $(V, b)$ be a regular symmetric bilinear space.
Let $W$ be a subspace of $V$ and $sigma colon W to V$ an isometry. Then there exists
an isometry $tau colon V to V$ which extends $sigma $ i.e $tau$ restricted to $W$ is $sigma $. $tau$ is product of reflections which extends to $sigma$.
How to prove above result?
I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic. Also I want some explanation of totally isotropic subspace. What is step where we use induction in totally isotropic subspace and how we use it? What are reflection in second case? Please someone help me out.
linear-algebra bilinear-form
edited Jul 22 at 18:52
Martin Sleziak
43.5k6113259
asked Jul 18 at 18:08
sscool
198116
edited Jul 22 at 18:52
Martin Sleziak
43.5k6113259
edited Jul 22 at 18:52
Martin Sleziak
43.5k6113259
edited Jul 22 at 18:52
Martin Sleziak
43.5k6113259
43.5k6113259
asked Jul 18 at 18:08
sscool
198116
asked Jul 18 at 18:08
sscool
198116
asked Jul 18 at 18:08
sscool
198116
198116
@FedericoFallucca please take a look at question and link for the book is given in question ? Please look over it if possible for you and if possible then you please explain me for case of totally isotropic subspace and how can we use induction in that proof?
– Ninja hatori
Jul 22 at 15:13
add a comment |Â
@FedericoFallucca please take a look at question and link for the book is given in question ? Please look over it if possible for you and if possible then you please explain me for case of totally isotropic subspace and how can we use induction in that proof?
– Ninja hatori
Jul 22 at 15:13
@FedericoFallucca please take a look at question and link for the book is given in question ? Please look over it if possible for you and if possible then you please explain me for case of totally isotropic subspace and how can we use induction in that proof?
– Ninja hatori
Jul 22 at 15:13
@FedericoFallucca please take a look at question and link for the book is given in question ? Please look over it if possible for you and if possible then you please explain me for case of totally isotropic subspace and how can we use induction in that proof?
– Ninja hatori
Jul 22 at 15:13
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
DISCLAIMER: Take this answer with a grain of salt. This topic is new to me and basically I mostly rely only on the stuff I've learned here on the site when answering a few other questions from the same book.
5.3 Theorem. (Witt) Let $(V,b)$ be a regular symmetric bilinear space. Let $W$ be a subspace of $V$ and $sigma colon Wto V$ an isometry. Then there exists and isometry $Sigmacolon Vto V$ which extends $sigma$, that is $Sigma|_W=sigma$.
5.5 Theorem. Under the hypotheses of 5.3 there exists a product of reflections $Sigma$ which extends $sigma$.
You have asked about one part of this proof (and you have also mentioned in chat that basically you're after a more detailed breakdown of this proof):
I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic.
The part of proof you asked above:
We now proceed by induction on $m=m(W):=dim W+2dim(Wcap W^bot)$. (In this way we can avoid some technical steps in the classical proofs.)
Since I do not know the classical proofs, I cannot comment much on this. Probably somebody who is more familiar with theory of quadratic forms can say more, but it seems that goal of the question is mainly to understand this proof.
For $m=0$ we have $W=0$ and can take $Sigma=id$.
I guess that the case $W=0$ can be considered trivial.
We assume now $m>0$ and consider first the case that $W$ is not totally isotropic. By 5.6 there is an anisotropic vector $xin W$.
This is just an application of Lemma 5.6, not much to say here.
Let $W_1=x^botcap W$ so that $W=xKoplus W_1$.
From Lemma 3.4 we know $V=xK oplus x^bot$. By taking intersection with $W$ we get $W=xK oplus (x^botcap W)$.
In fact we know more than that - we know that $V= xK perp x^bot$ and $W= xKperp W_1$.
Since $m(W_1)=m(W)-1$ we can apply the induction hypothesis to $sigma_1=sigma|_W$.
We have $dim(W)=1+dim(W_1)$, hence $dim(W_1)=dim(W)-1$.
At the same time, we have $W_1cap W_1^bot=Wcap W^bot$. Indeed
beginalign*
Wcap W^bot
&= (xK oplus W_1) cap (xK oplus W_1)^bot \
&= (xK oplus W_1) cap x^bot cap W_1^bot \
&= (xK cap x^bot cap W_1^bot)oplus (W_1 cap x^bot cap W_1^bot) \
&= W_1 cap x^bot cap W_1^bot \
&= W_1 cap W_1^bot
endalign*
There is a product of reflections $Sigma_1$ extending $sigma_1$.
This is just application of the induction hypothesis. (Notice that a restriction of an isometry is again an isometry. So the induction hypothesis can be applied.)
The vectors $x$ and $y:=(Sigma_1^-1sigma)x$ lie in $W_1^bot$.
The fact that $xin W_1^bot$ follows simply from $W_1subseteq x^bot$.
Since $Sigma_1$ extends $sigma_1$, any element of the space $W_1$ is invariant under $Sigma_1^-1sigma$. I.e., $Sigma_1^-1sigma(w)=Sigma_1^-1sigma_1(w)=w$ for any $win W_1$.
Now using the fact that $Sigma_1^-1sigma$ is an isometry and any isometry maps orthogonal element to an orthogonal element, we get from $xin W_1^bot$ that $yin W_1^bot$.
By the first step of the proof there is a product of (at most 2) reflections of the form $tau_z$, $zin W_1^bot$ which maps $x$ to $y$.
This is application of the fact proved in the first paragraph of the proof (the case $W=xK$, $x$ anisotropic).
These reflections act as an identity on $W_1$.
The first part of Lemma 5.2 says that $tau_z$ acts as identity on $z^bot$.
Then $Sigma:=Sigma_1Sigma_2$ is the map we want to find:
$$Sigma_1Sigma_2(xalpha+x_1)
=Sigma_1(yalpha+W_1)
=sigma(yalpha)+sigma x_1
=sigma(xalpha+x_1).$$
This is a direct computation using $Sigma_2x=y$ and $Sigma_1|_W_1=Sigma_2|_W_1=id_W_1$.
answered Jul 19 at 11:57
Martin Sleziak
43.5k6113259
@ MartinSleziak I have certain doubt in above proof; where they use induction in the above answer and I also saw proof for totally isotropic space in that proof also they use induction hypothesis in last statement what kind of induction they talk about? I also don't see explicit form of product of reflections for the case of totally isotropic subspace ? I understand whole idea but don't get where they form reflections in proof?
– Ninja hatori
Jul 19 at 19:30
@MartinSleziak I understand the whole proof for totally isotropic space but in the last statement they say by using induction hypothesis you can prove final result what is meaning of that?
– sscool
Jul 20 at 6:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
DISCLAIMER: Take this answer with a grain of salt. This topic is new to me and basically I mostly rely only on the stuff I've learned here on the site when answering a few other questions from the same book.
5.3 Theorem. (Witt) Let $(V,b)$ be a regular symmetric bilinear space. Let $W$ be a subspace of $V$ and $sigma colon Wto V$ an isometry. Then there exists and isometry $Sigmacolon Vto V$ which extends $sigma$, that is $Sigma|_W=sigma$.
5.5 Theorem. Under the hypotheses of 5.3 there exists a product of reflections $Sigma$ which extends $sigma$.
You have asked about one part of this proof (and you have also mentioned in chat that basically you're after a more detailed breakdown of this proof):
I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic.
The part of proof you asked above:
We now proceed by induction on $m=m(W):=dim W+2dim(Wcap W^bot)$. (In this way we can avoid some technical steps in the classical proofs.)
Since I do not know the classical proofs, I cannot comment much on this. Probably somebody who is more familiar with theory of quadratic forms can say more, but it seems that goal of the question is mainly to understand this proof.
For $m=0$ we have $W=0$ and can take $Sigma=id$.
I guess that the case $W=0$ can be considered trivial.
We assume now $m>0$ and consider first the case that $W$ is not totally isotropic. By 5.6 there is an anisotropic vector $xin W$.
This is just an application of Lemma 5.6, not much to say here.
Let $W_1=x^botcap W$ so that $W=xKoplus W_1$.
From Lemma 3.4 we know $V=xK oplus x^bot$. By taking intersection with $W$ we get $W=xK oplus (x^botcap W)$.
In fact we know more than that - we know that $V= xK perp x^bot$ and $W= xKperp W_1$.
Since $m(W_1)=m(W)-1$ we can apply the induction hypothesis to $sigma_1=sigma|_W$.
We have $dim(W)=1+dim(W_1)$, hence $dim(W_1)=dim(W)-1$.
At the same time, we have $W_1cap W_1^bot=Wcap W^bot$. Indeed
beginalign*
Wcap W^bot
&= (xK oplus W_1) cap (xK oplus W_1)^bot \
&= (xK oplus W_1) cap x^bot cap W_1^bot \
&= (xK cap x^bot cap W_1^bot)oplus (W_1 cap x^bot cap W_1^bot) \
&= W_1 cap x^bot cap W_1^bot \
&= W_1 cap W_1^bot
endalign*
There is a product of reflections $Sigma_1$ extending $sigma_1$.
This is just application of the induction hypothesis. (Notice that a restriction of an isometry is again an isometry. So the induction hypothesis can be applied.)
The vectors $x$ and $y:=(Sigma_1^-1sigma)x$ lie in $W_1^bot$.
The fact that $xin W_1^bot$ follows simply from $W_1subseteq x^bot$.
Since $Sigma_1$ extends $sigma_1$, any element of the space $W_1$ is invariant under $Sigma_1^-1sigma$. I.e., $Sigma_1^-1sigma(w)=Sigma_1^-1sigma_1(w)=w$ for any $win W_1$.
Now using the fact that $Sigma_1^-1sigma$ is an isometry and any isometry maps orthogonal element to an orthogonal element, we get from $xin W_1^bot$ that $yin W_1^bot$.
By the first step of the proof there is a product of (at most 2) reflections of the form $tau_z$, $zin W_1^bot$ which maps $x$ to $y$.
This is application of the fact proved in the first paragraph of the proof (the case $W=xK$, $x$ anisotropic).
These reflections act as an identity on $W_1$.
The first part of Lemma 5.2 says that $tau_z$ acts as identity on $z^bot$.
Then $Sigma:=Sigma_1Sigma_2$ is the map we want to find:
$$Sigma_1Sigma_2(xalpha+x_1)
=Sigma_1(yalpha+W_1)
=sigma(yalpha)+sigma x_1
=sigma(xalpha+x_1).$$
This is a direct computation using $Sigma_2x=y$ and $Sigma_1|_W_1=Sigma_2|_W_1=id_W_1$.
answered Jul 19 at 11:57
Martin Sleziak
43.5k6113259
@ MartinSleziak I have certain doubt in above proof; where they use induction in the above answer and I also saw proof for totally isotropic space in that proof also they use induction hypothesis in last statement what kind of induction they talk about? I also don't see explicit form of product of reflections for the case of totally isotropic subspace ? I understand whole idea but don't get where they form reflections in proof?
– Ninja hatori
Jul 19 at 19:30
@MartinSleziak I understand the whole proof for totally isotropic space but in the last statement they say by using induction hypothesis you can prove final result what is meaning of that?
– sscool
Jul 20 at 6:16
add a comment |Â
up vote
5
down vote
accepted
DISCLAIMER: Take this answer with a grain of salt. This topic is new to me and basically I mostly rely only on the stuff I've learned here on the site when answering a few other questions from the same book.
5.3 Theorem. (Witt) Let $(V,b)$ be a regular symmetric bilinear space. Let $W$ be a subspace of $V$ and $sigma colon Wto V$ an isometry. Then there exists and isometry $Sigmacolon Vto V$ which extends $sigma$, that is $Sigma|_W=sigma$.
5.5 Theorem. Under the hypotheses of 5.3 there exists a product of reflections $Sigma$ which extends $sigma$.
You have asked about one part of this proof (and you have also mentioned in chat that basically you're after a more detailed breakdown of this proof):
I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic.
The part of proof you asked above:
We now proceed by induction on $m=m(W):=dim W+2dim(Wcap W^bot)$. (In this way we can avoid some technical steps in the classical proofs.)
Since I do not know the classical proofs, I cannot comment much on this. Probably somebody who is more familiar with theory of quadratic forms can say more, but it seems that goal of the question is mainly to understand this proof.
For $m=0$ we have $W=0$ and can take $Sigma=id$.
I guess that the case $W=0$ can be considered trivial.
We assume now $m>0$ and consider first the case that $W$ is not totally isotropic. By 5.6 there is an anisotropic vector $xin W$.
This is just an application of Lemma 5.6, not much to say here.
Let $W_1=x^botcap W$ so that $W=xKoplus W_1$.
From Lemma 3.4 we know $V=xK oplus x^bot$. By taking intersection with $W$ we get $W=xK oplus (x^botcap W)$.
In fact we know more than that - we know that $V= xK perp x^bot$ and $W= xKperp W_1$.
Since $m(W_1)=m(W)-1$ we can apply the induction hypothesis to $sigma_1=sigma|_W$.
We have $dim(W)=1+dim(W_1)$, hence $dim(W_1)=dim(W)-1$.
At the same time, we have $W_1cap W_1^bot=Wcap W^bot$. Indeed
beginalign*
Wcap W^bot
&= (xK oplus W_1) cap (xK oplus W_1)^bot \
&= (xK oplus W_1) cap x^bot cap W_1^bot \
&= (xK cap x^bot cap W_1^bot)oplus (W_1 cap x^bot cap W_1^bot) \
&= W_1 cap x^bot cap W_1^bot \
&= W_1 cap W_1^bot
endalign*
There is a product of reflections $Sigma_1$ extending $sigma_1$.
This is just application of the induction hypothesis. (Notice that a restriction of an isometry is again an isometry. So the induction hypothesis can be applied.)
The vectors $x$ and $y:=(Sigma_1^-1sigma)x$ lie in $W_1^bot$.
The fact that $xin W_1^bot$ follows simply from $W_1subseteq x^bot$.
Since $Sigma_1$ extends $sigma_1$, any element of the space $W_1$ is invariant under $Sigma_1^-1sigma$. I.e., $Sigma_1^-1sigma(w)=Sigma_1^-1sigma_1(w)=w$ for any $win W_1$.
Now using the fact that $Sigma_1^-1sigma$ is an isometry and any isometry maps orthogonal element to an orthogonal element, we get from $xin W_1^bot$ that $yin W_1^bot$.
By the first step of the proof there is a product of (at most 2) reflections of the form $tau_z$, $zin W_1^bot$ which maps $x$ to $y$.
This is application of the fact proved in the first paragraph of the proof (the case $W=xK$, $x$ anisotropic).
These reflections act as an identity on $W_1$.
The first part of Lemma 5.2 says that $tau_z$ acts as identity on $z^bot$.
Then $Sigma:=Sigma_1Sigma_2$ is the map we want to find:
$$Sigma_1Sigma_2(xalpha+x_1)
=Sigma_1(yalpha+W_1)
=sigma(yalpha)+sigma x_1
=sigma(xalpha+x_1).$$
This is a direct computation using $Sigma_2x=y$ and $Sigma_1|_W_1=Sigma_2|_W_1=id_W_1$.
answered Jul 19 at 11:57
Martin Sleziak
43.5k6113259
@ MartinSleziak I have certain doubt in above proof; where they use induction in the above answer and I also saw proof for totally isotropic space in that proof also they use induction hypothesis in last statement what kind of induction they talk about? I also don't see explicit form of product of reflections for the case of totally isotropic subspace ? I understand whole idea but don't get where they form reflections in proof?
– Ninja hatori
Jul 19 at 19:30
@MartinSleziak I understand the whole proof for totally isotropic space but in the last statement they say by using induction hypothesis you can prove final result what is meaning of that?
– sscool
Jul 20 at 6:16
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
DISCLAIMER: Take this answer with a grain of salt. This topic is new to me and basically I mostly rely only on the stuff I've learned here on the site when answering a few other questions from the same book.
5.3 Theorem. (Witt) Let $(V,b)$ be a regular symmetric bilinear space. Let $W$ be a subspace of $V$ and $sigma colon Wto V$ an isometry. Then there exists and isometry $Sigmacolon Vto V$ which extends $sigma$, that is $Sigma|_W=sigma$.
5.5 Theorem. Under the hypotheses of 5.3 there exists a product of reflections $Sigma$ which extends $sigma$.
You have asked about one part of this proof (and you have also mentioned in chat that basically you're after a more detailed breakdown of this proof):
I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic.
The part of proof you asked above:
We now proceed by induction on $m=m(W):=dim W+2dim(Wcap W^bot)$. (In this way we can avoid some technical steps in the classical proofs.)
Since I do not know the classical proofs, I cannot comment much on this. Probably somebody who is more familiar with theory of quadratic forms can say more, but it seems that goal of the question is mainly to understand this proof.
For $m=0$ we have $W=0$ and can take $Sigma=id$.
I guess that the case $W=0$ can be considered trivial.
We assume now $m>0$ and consider first the case that $W$ is not totally isotropic. By 5.6 there is an anisotropic vector $xin W$.
This is just an application of Lemma 5.6, not much to say here.
Let $W_1=x^botcap W$ so that $W=xKoplus W_1$.
From Lemma 3.4 we know $V=xK oplus x^bot$. By taking intersection with $W$ we get $W=xK oplus (x^botcap W)$.
In fact we know more than that - we know that $V= xK perp x^bot$ and $W= xKperp W_1$.
Since $m(W_1)=m(W)-1$ we can apply the induction hypothesis to $sigma_1=sigma|_W$.
We have $dim(W)=1+dim(W_1)$, hence $dim(W_1)=dim(W)-1$.
At the same time, we have $W_1cap W_1^bot=Wcap W^bot$. Indeed
beginalign*
Wcap W^bot
&= (xK oplus W_1) cap (xK oplus W_1)^bot \
&= (xK oplus W_1) cap x^bot cap W_1^bot \
&= (xK cap x^bot cap W_1^bot)oplus (W_1 cap x^bot cap W_1^bot) \
&= W_1 cap x^bot cap W_1^bot \
&= W_1 cap W_1^bot
endalign*
There is a product of reflections $Sigma_1$ extending $sigma_1$.
This is just application of the induction hypothesis. (Notice that a restriction of an isometry is again an isometry. So the induction hypothesis can be applied.)
The vectors $x$ and $y:=(Sigma_1^-1sigma)x$ lie in $W_1^bot$.
The fact that $xin W_1^bot$ follows simply from $W_1subseteq x^bot$.
Since $Sigma_1$ extends $sigma_1$, any element of the space $W_1$ is invariant under $Sigma_1^-1sigma$. I.e., $Sigma_1^-1sigma(w)=Sigma_1^-1sigma_1(w)=w$ for any $win W_1$.
Now using the fact that $Sigma_1^-1sigma$ is an isometry and any isometry maps orthogonal element to an orthogonal element, we get from $xin W_1^bot$ that $yin W_1^bot$.
By the first step of the proof there is a product of (at most 2) reflections of the form $tau_z$, $zin W_1^bot$ which maps $x$ to $y$.
This is application of the fact proved in the first paragraph of the proof (the case $W=xK$, $x$ anisotropic).
These reflections act as an identity on $W_1$.
The first part of Lemma 5.2 says that $tau_z$ acts as identity on $z^bot$.
Then $Sigma:=Sigma_1Sigma_2$ is the map we want to find:
$$Sigma_1Sigma_2(xalpha+x_1)
=Sigma_1(yalpha+W_1)
=sigma(yalpha)+sigma x_1
=sigma(xalpha+x_1).$$
This is a direct computation using $Sigma_2x=y$ and $Sigma_1|_W_1=Sigma_2|_W_1=id_W_1$.
answered Jul 19 at 11:57
Martin Sleziak
43.5k6113259
DISCLAIMER: Take this answer with a grain of salt. This topic is new to me and basically I mostly rely only on the stuff I've learned here on the site when answering a few other questions from the same book.
5.3 Theorem. (Witt) Let $(V,b)$ be a regular symmetric bilinear space. Let $W$ be a subspace of $V$ and $sigma colon Wto V$ an isometry. Then there exists and isometry $Sigmacolon Vto V$ which extends $sigma$, that is $Sigma|_W=sigma$.
5.5 Theorem. Under the hypotheses of 5.3 there exists a product of reflections $Sigma$ which extends $sigma$.
You have asked about one part of this proof (and you have also mentioned in chat that basically you're after a more detailed breakdown of this proof):
I don't understand what book suggest in second paragraph when $W$ is not totally isotropic they prove above result by induction. Can someone please explain the case for which $W$ is not totally isotropic.
The part of proof you asked above:
We now proceed by induction on $m=m(W):=dim W+2dim(Wcap W^bot)$. (In this way we can avoid some technical steps in the classical proofs.)
Since I do not know the classical proofs, I cannot comment much on this. Probably somebody who is more familiar with theory of quadratic forms can say more, but it seems that goal of the question is mainly to understand this proof.
For $m=0$ we have $W=0$ and can take $Sigma=id$.
I guess that the case $W=0$ can be considered trivial.
We assume now $m>0$ and consider first the case that $W$ is not totally isotropic. By 5.6 there is an anisotropic vector $xin W$.
This is just an application of Lemma 5.6, not much to say here.
Let $W_1=x^botcap W$ so that $W=xKoplus W_1$.
From Lemma 3.4 we know $V=xK oplus x^bot$. By taking intersection with $W$ we get $W=xK oplus (x^botcap W)$.
In fact we know more than that - we know that $V= xK perp x^bot$ and $W= xKperp W_1$.
Since $m(W_1)=m(W)-1$ we can apply the induction hypothesis to $sigma_1=sigma|_W$.
We have $dim(W)=1+dim(W_1)$, hence $dim(W_1)=dim(W)-1$.
At the same time, we have $W_1cap W_1^bot=Wcap W^bot$. Indeed
beginalign*
Wcap W^bot
&= (xK oplus W_1) cap (xK oplus W_1)^bot \
&= (xK oplus W_1) cap x^bot cap W_1^bot \
&= (xK cap x^bot cap W_1^bot)oplus (W_1 cap x^bot cap W_1^bot) \
&= W_1 cap x^bot cap W_1^bot \
&= W_1 cap W_1^bot
endalign*
There is a product of reflections $Sigma_1$ extending $sigma_1$.
This is just application of the induction hypothesis. (Notice that a restriction of an isometry is again an isometry. So the induction hypothesis can be applied.)
The vectors $x$ and $y:=(Sigma_1^-1sigma)x$ lie in $W_1^bot$.
The fact that $xin W_1^bot$ follows simply from $W_1subseteq x^bot$.
Since $Sigma_1$ extends $sigma_1$, any element of the space $W_1$ is invariant under $Sigma_1^-1sigma$. I.e., $Sigma_1^-1sigma(w)=Sigma_1^-1sigma_1(w)=w$ for any $win W_1$.
Now using the fact that $Sigma_1^-1sigma$ is an isometry and any isometry maps orthogonal element to an orthogonal element, we get from $xin W_1^bot$ that $yin W_1^bot$.
By the first step of the proof there is a product of (at most 2) reflections of the form $tau_z$, $zin W_1^bot$ which maps $x$ to $y$.
This is application of the fact proved in the first paragraph of the proof (the case $W=xK$, $x$ anisotropic).
These reflections act as an identity on $W_1$.
The first part of Lemma 5.2 says that $tau_z$ acts as identity on $z^bot$.
Then $Sigma:=Sigma_1Sigma_2$ is the map we want to find:
$$Sigma_1Sigma_2(xalpha+x_1)
=Sigma_1(yalpha+W_1)
=sigma(yalpha)+sigma x_1
=sigma(xalpha+x_1).$$
This is a direct computation using $Sigma_2x=y$ and $Sigma_1|_W_1=Sigma_2|_W_1=id_W_1$.
answered Jul 19 at 11:57
Martin Sleziak
43.5k6113259
edited Jul 19 at 13:00
answered Jul 19 at 11:57
Martin Sleziak
43.5k6113259
answered Jul 19 at 11:57
Martin Sleziak
43.5k6113259
answered Jul 19 at 11:57
Martin Sleziak
43.5k6113259
43.5k6113259
@ MartinSleziak I have certain doubt in above proof; where they use induction in the above answer and I also saw proof for totally isotropic space in that proof also they use induction hypothesis in last statement what kind of induction they talk about? I also don't see explicit form of product of reflections for the case of totally isotropic subspace ? I understand whole idea but don't get where they form reflections in proof?
– Ninja hatori
Jul 19 at 19:30
@MartinSleziak I understand the whole proof for totally isotropic space but in the last statement they say by using induction hypothesis you can prove final result what is meaning of that?
– sscool
Jul 20 at 6:16
add a comment |Â
@ MartinSleziak I have certain doubt in above proof; where they use induction in the above answer and I also saw proof for totally isotropic space in that proof also they use induction hypothesis in last statement what kind of induction they talk about? I also don't see explicit form of product of reflections for the case of totally isotropic subspace ? I understand whole idea but don't get where they form reflections in proof?
– Ninja hatori
Jul 19 at 19:30
@MartinSleziak I understand the whole proof for totally isotropic space but in the last statement they say by using induction hypothesis you can prove final result what is meaning of that?
– sscool
Jul 20 at 6:16
@ MartinSleziak I have certain doubt in above proof; where they use induction in the above answer and I also saw proof for totally isotropic space in that proof also they use induction hypothesis in last statement what kind of induction they talk about? I also don't see explicit form of product of reflections for the case of totally isotropic subspace ? I understand whole idea but don't get where they form reflections in proof?
– Ninja hatori
Jul 19 at 19:30
@ MartinSleziak I have certain doubt in above proof; where they use induction in the above answer and I also saw proof for totally isotropic space in that proof also they use induction hypothesis in last statement what kind of induction they talk about? I also don't see explicit form of product of reflections for the case of totally isotropic subspace ? I understand whole idea but don't get where they form reflections in proof?
– Ninja hatori
Jul 19 at 19:30
@MartinSleziak I understand the whole proof for totally isotropic space but in the last statement they say by using induction hypothesis you can prove final result what is meaning of that?
– sscool
Jul 20 at 6:16
@MartinSleziak I understand the whole proof for totally isotropic space but in the last statement they say by using induction hypothesis you can prove final result what is meaning of that?
– sscool
Jul 20 at 6:16
add a comment |Â
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@FedericoFallucca please take a look at question and link for the book is given in question ? Please look over it if possible for you and if possible then you please explain me for case of totally isotropic subspace and how can we use induction in that proof?
– Ninja hatori
Jul 22 at 15:13