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About symmetric inner product space

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0

$begingroup$

How lattice $L = V otimes _k k[x] $ looks like? $L$ is free over $k[x]$ because $V$ is free over $k$ ; why this statement is true? Also I am trying to prove every unimodular $k[x]$ lattice has form $V otimes _k k[x]$ how this fact is related to orthogonal basis ? How this statment looks like in matrix version?

linear-algebra commutative-algebra modules projective-module

share|cite|improve this question

edited Mar 28 at 21:22

maths student

asked Mar 28 at 15:13

maths studentmaths student

6321521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm aware of the context since I've answered two of your previous questions on this topic, but you should probably add the context (and perhaps these links) into the body of this question so that it doesn't get closed.
  $endgroup$
  – jgon
  Mar 28 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, $V$ is a subset of $L$ via the map $vmapsto votimes 1$.
  $endgroup$
  – jgon
  Mar 28 at 15:30
  

  
  
  
  

add a comment | 

0

$begingroup$

How lattice $L = V otimes _k k[x] $ looks like? $L$ is free over $k[x]$ because $V$ is free over $k$ ; why this statement is true? Also I am trying to prove every unimodular $k[x]$ lattice has form $V otimes _k k[x]$ how this fact is related to orthogonal basis ? How this statment looks like in matrix version?

linear-algebra commutative-algebra modules projective-module

share|cite|improve this question

edited Mar 28 at 21:22

maths student

asked Mar 28 at 15:13

maths studentmaths student

6321521

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm aware of the context since I've answered two of your previous questions on this topic, but you should probably add the context (and perhaps these links) into the body of this question so that it doesn't get closed.
  $endgroup$
  – jgon
  Mar 28 at 15:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, $V$ is a subset of $L$ via the map $vmapsto votimes 1$.
  $endgroup$
  – jgon
  Mar 28 at 15:30
  

  
  
  
  

add a comment | 

$begingroup$

How lattice $L = V otimes _k k[x] $ looks like? $L$ is free over $k[x]$ because $V$ is free over $k$ ; why this statement is true? Also I am trying to prove every unimodular $k[x]$ lattice has form $V otimes _k k[x]$ how this fact is related to orthogonal basis ? How this statment looks like in matrix version?

linear-algebra commutative-algebra modules projective-module

share|cite|improve this question

edited Mar 28 at 21:22

maths student

asked Mar 28 at 15:13

maths studentmaths student

6321521

$endgroup$

share|cite|improve this question

edited Mar 28 at 21:22

maths student

asked Mar 28 at 15:13

maths studentmaths student

6321521

share|cite|improve this question

edited Mar 28 at 21:22

maths student

asked Mar 28 at 15:13

maths studentmaths student

6321521

asked Mar 28 at 15:13

maths studentmaths student

6321521

asked Mar 28 at 15:13

maths studentmaths student

6321521

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $V$ be a vector space over $k$ equipped with a nondegenerate bilinear form $B$. Then $L=Votimes_k k[x]$ is free and $B$ extends to a nondegenerate bilinear form $tildeB$ on $L$ with the same determinant as $B$ (in particular it is an element of $k^times$).

Proof.

$L$ is free because by choosing a basis $v_1,ldots,v_n$ we get that
$$Vcong k^n$$
for some $n$, so
$$L=Votimes_k k[x] cong k^notimes_k k[x] simeq k[x]^n.$$

Under this isomorphism the basis vectors for $L$ are $ell_i=v_iotimes 1$.

Now $B$ defines a bilinear form on $L$ by $tildeB(votimes p,wotimes q) = B(v,w)pq$.

Then $$tildeB(ell_i,ell_j)=tildeB(v_iotimes 1,v_jotimes_1)=B(v_i,v_j).$$
Thus $det(tildeB(ell_i,ell_j))=det(B(v_i,v_j))$.

Every unimodular $k[x]$-lattice has this form

Edit: My prior "proof" had an error. This is what's proven in the Theorem in the OP.

Edit 2:

To show how the theorem in the question proves every unimodular lattice has this form we do the following.

Let $ell_1,ldots,ell_n$ be the orthogonal $k[t]$ basis given by the theorem. Define $V=sum_i c_iell_i : c_iin k$ and define $B_0$ on $V$ by restricting $B$ to $V$. This is where we use orthogonality! The restriction is well-defined because the $ell_i$ are orthogonal, so $B(ell_i,ell_j)=c_idelta_ij$ for some constants $c_iin k^times$.

Then define $Votimes_k k[t] to L$ by $ell_iotimes p mapsto pell_i$. Since the $ell_i$ formed a $k[t]$ basis for $L$, this map is an isomorphism of $k[t]$ modules.

Then all we need to check is that the isomorphism sends $tildeB_0$ to $B$. We do this by observing that
$$tildeB_0(ell_iotimes p,ell_jotimes q) = B_0(ell_i,ell_j)pq=B(ell_i,ell_j)pq=B(pell_i,qell_j),$$
as desired.

share|cite|improve this answer

edited Mar 28 at 20:48

answered Mar 28 at 15:36

jgonjgon

16.4k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Referring to this theorem sir ; I have edited it
  $endgroup$
  – maths student
  Mar 28 at 17:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsstudent I'm a bit confused as to what your question is about the theorem.
  $endgroup$
  – jgon
  Mar 28 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  my question is how above theorem yields $L = V otimes _k k[x] $
  $endgroup$
  – maths student
  Mar 28 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsstudent Take the orthogonal $k[x]$-basis that is constructed by the theorem and define $V$ to be the $k$-span of that basis.
  $endgroup$
  – jgon
  Mar 28 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what is the meaning of k span of that basis can you explain that?
  $endgroup$
  – maths student
  Mar 28 at 17:59
  

  
  
  
  

 | 
show 10 more comments

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1

$begingroup$

Let $V$ be a vector space over $k$ equipped with a nondegenerate bilinear form $B$. Then $L=Votimes_k k[x]$ is free and $B$ extends to a nondegenerate bilinear form $tildeB$ on $L$ with the same determinant as $B$ (in particular it is an element of $k^times$).

Proof.

$L$ is free because by choosing a basis $v_1,ldots,v_n$ we get that
$$Vcong k^n$$
for some $n$, so
$$L=Votimes_k k[x] cong k^notimes_k k[x] simeq k[x]^n.$$

Under this isomorphism the basis vectors for $L$ are $ell_i=v_iotimes 1$.

Now $B$ defines a bilinear form on $L$ by $tildeB(votimes p,wotimes q) = B(v,w)pq$.

Then $$tildeB(ell_i,ell_j)=tildeB(v_iotimes 1,v_jotimes_1)=B(v_i,v_j).$$
Thus $det(tildeB(ell_i,ell_j))=det(B(v_i,v_j))$.

Every unimodular $k[x]$-lattice has this form

Edit: My prior "proof" had an error. This is what's proven in the Theorem in the OP.

Edit 2:

To show how the theorem in the question proves every unimodular lattice has this form we do the following.

Let $ell_1,ldots,ell_n$ be the orthogonal $k[t]$ basis given by the theorem. Define $V=sum_i c_iell_i : c_iin k$ and define $B_0$ on $V$ by restricting $B$ to $V$. This is where we use orthogonality! The restriction is well-defined because the $ell_i$ are orthogonal, so $B(ell_i,ell_j)=c_idelta_ij$ for some constants $c_iin k^times$.

Then define $Votimes_k k[t] to L$ by $ell_iotimes p mapsto pell_i$. Since the $ell_i$ formed a $k[t]$ basis for $L$, this map is an isomorphism of $k[t]$ modules.

Then all we need to check is that the isomorphism sends $tildeB_0$ to $B$. We do this by observing that
$$tildeB_0(ell_iotimes p,ell_jotimes q) = B_0(ell_i,ell_j)pq=B(ell_i,ell_j)pq=B(pell_i,qell_j),$$
as desired.

share|cite|improve this answer

edited Mar 28 at 20:48

answered Mar 28 at 15:36

jgonjgon

16.4k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Referring to this theorem sir ; I have edited it
  $endgroup$
  – maths student
  Mar 28 at 17:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsstudent I'm a bit confused as to what your question is about the theorem.
  $endgroup$
  – jgon
  Mar 28 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  my question is how above theorem yields $L = V otimes _k k[x] $
  $endgroup$
  – maths student
  Mar 28 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsstudent Take the orthogonal $k[x]$-basis that is constructed by the theorem and define $V$ to be the $k$-span of that basis.
  $endgroup$
  – jgon
  Mar 28 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what is the meaning of k span of that basis can you explain that?
  $endgroup$
  – maths student
  Mar 28 at 17:59
  

  
  
  
  

 | 
show 10 more comments

1

$begingroup$

Let $V$ be a vector space over $k$ equipped with a nondegenerate bilinear form $B$. Then $L=Votimes_k k[x]$ is free and $B$ extends to a nondegenerate bilinear form $tildeB$ on $L$ with the same determinant as $B$ (in particular it is an element of $k^times$).

Proof.

$L$ is free because by choosing a basis $v_1,ldots,v_n$ we get that
$$Vcong k^n$$
for some $n$, so
$$L=Votimes_k k[x] cong k^notimes_k k[x] simeq k[x]^n.$$

Under this isomorphism the basis vectors for $L$ are $ell_i=v_iotimes 1$.

Now $B$ defines a bilinear form on $L$ by $tildeB(votimes p,wotimes q) = B(v,w)pq$.

Then $$tildeB(ell_i,ell_j)=tildeB(v_iotimes 1,v_jotimes_1)=B(v_i,v_j).$$
Thus $det(tildeB(ell_i,ell_j))=det(B(v_i,v_j))$.

Every unimodular $k[x]$-lattice has this form

Edit: My prior "proof" had an error. This is what's proven in the Theorem in the OP.

Edit 2:

To show how the theorem in the question proves every unimodular lattice has this form we do the following.

Let $ell_1,ldots,ell_n$ be the orthogonal $k[t]$ basis given by the theorem. Define $V=sum_i c_iell_i : c_iin k$ and define $B_0$ on $V$ by restricting $B$ to $V$. This is where we use orthogonality! The restriction is well-defined because the $ell_i$ are orthogonal, so $B(ell_i,ell_j)=c_idelta_ij$ for some constants $c_iin k^times$.

Then define $Votimes_k k[t] to L$ by $ell_iotimes p mapsto pell_i$. Since the $ell_i$ formed a $k[t]$ basis for $L$, this map is an isomorphism of $k[t]$ modules.

Then all we need to check is that the isomorphism sends $tildeB_0$ to $B$. We do this by observing that
$$tildeB_0(ell_iotimes p,ell_jotimes q) = B_0(ell_i,ell_j)pq=B(ell_i,ell_j)pq=B(pell_i,qell_j),$$
as desired.

share|cite|improve this answer

edited Mar 28 at 20:48

answered Mar 28 at 15:36

jgonjgon

16.4k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Referring to this theorem sir ; I have edited it
  $endgroup$
  – maths student
  Mar 28 at 17:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsstudent I'm a bit confused as to what your question is about the theorem.
  $endgroup$
  – jgon
  Mar 28 at 17:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  my question is how above theorem yields $L = V otimes _k k[x] $
  $endgroup$
  – maths student
  Mar 28 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mathsstudent Take the orthogonal $k[x]$-basis that is constructed by the theorem and define $V$ to be the $k$-span of that basis.
  $endgroup$
  – jgon
  Mar 28 at 17:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what is the meaning of k span of that basis can you explain that?
  $endgroup$
  – maths student
  Mar 28 at 17:59
  

  
  
  
  

 | 
show 10 more comments

$begingroup$

Let $V$ be a vector space over $k$ equipped with a nondegenerate bilinear form $B$. Then $L=Votimes_k k[x]$ is free and $B$ extends to a nondegenerate bilinear form $tildeB$ on $L$ with the same determinant as $B$ (in particular it is an element of $k^times$).

Proof.

$L$ is free because by choosing a basis $v_1,ldots,v_n$ we get that
$$Vcong k^n$$
for some $n$, so
$$L=Votimes_k k[x] cong k^notimes_k k[x] simeq k[x]^n.$$

Under this isomorphism the basis vectors for $L$ are $ell_i=v_iotimes 1$.

Now $B$ defines a bilinear form on $L$ by $tildeB(votimes p,wotimes q) = B(v,w)pq$.

Then $$tildeB(ell_i,ell_j)=tildeB(v_iotimes 1,v_jotimes_1)=B(v_i,v_j).$$
Thus $det(tildeB(ell_i,ell_j))=det(B(v_i,v_j))$.

Every unimodular $k[x]$-lattice has this form

Edit: My prior "proof" had an error. This is what's proven in the Theorem in the OP.

Edit 2:

To show how the theorem in the question proves every unimodular lattice has this form we do the following.

Let $ell_1,ldots,ell_n$ be the orthogonal $k[t]$ basis given by the theorem. Define $V=sum_i c_iell_i : c_iin k$ and define $B_0$ on $V$ by restricting $B$ to $V$. This is where we use orthogonality! The restriction is well-defined because the $ell_i$ are orthogonal, so $B(ell_i,ell_j)=c_idelta_ij$ for some constants $c_iin k^times$.

Then define $Votimes_k k[t] to L$ by $ell_iotimes p mapsto pell_i$. Since the $ell_i$ formed a $k[t]$ basis for $L$, this map is an isomorphism of $k[t]$ modules.

Then all we need to check is that the isomorphism sends $tildeB_0$ to $B$. We do this by observing that
$$tildeB_0(ell_iotimes p,ell_jotimes q) = B_0(ell_i,ell_j)pq=B(ell_i,ell_j)pq=B(pell_i,qell_j),$$
as desired.

share|cite|improve this answer

edited Mar 28 at 20:48

answered Mar 28 at 15:36

jgonjgon

16.4k32143

$endgroup$

share|cite|improve this answer

edited Mar 28 at 20:48

answered Mar 28 at 15:36

jgonjgon

16.4k32143

answered Mar 28 at 15:36

jgonjgon

16.4k32143

answered Mar 28 at 15:36

jgonjgon

16.4k32143