Unique multiplicative quadratic form on quaternion algebras
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I want to prove, that the only multiplicative quadratic form $Q$ (so $Q(xy)=Q(x)Q(y) forall x,y$) on a quaternion algebra $Big(dfrac{a,b}{F}Big)$ is the norm $mathrm{Nr}$, which is isometric to the form $langle 1,-a,-b,abrangle$ in the orthogonal basis $1,i,j,ij$.
One easily sees, that $Q(1)=Q(1^2)=Q(1)^2Rightarrow Q(1)=1$ and similarly $Q(i)^2=a^2$, $Q(j)^2=b^2$ and therfore $Q(ij)=(ab)^2$. But how do we show $Q(i)=-a$ and $Q(j)=-b$? Or is it even necessary?
It seems like I'm missing something obvious.
norm quadratic-forms quaternions algebras
asked Jan 24 at 16:08
BananenPateBananenPate
284
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add a comment |
$begingroup$
I want to prove, that the only multiplicative quadratic form $Q$ (so $Q(xy)=Q(x)Q(y) forall x,y$) on a quaternion algebra $Big(dfrac{a,b}{F}Big)$ is the norm $mathrm{Nr}$, which is isometric to the form $langle 1,-a,-b,abrangle$ in the orthogonal basis $1,i,j,ij$.
One easily sees, that $Q(1)=Q(1^2)=Q(1)^2Rightarrow Q(1)=1$ and similarly $Q(i)^2=a^2$, $Q(j)^2=b^2$ and therfore $Q(ij)=(ab)^2$. But how do we show $Q(i)=-a$ and $Q(j)=-b$? Or is it even necessary?
It seems like I'm missing something obvious.
norm quadratic-forms quaternions algebras
asked Jan 24 at 16:08
BananenPateBananenPate
284
$endgroup$
$begingroup$
A couple of things to keep in mind: (1) your argument that $Q(1) = 1$ is only valid if you assume $Q(1) ne 0$, and (2) in general, a quadratic form is not determined by its values on a basis.
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– Kimball
Jan 24 at 22:06
add a comment |
$begingroup$
I want to prove, that the only multiplicative quadratic form $Q$ (so $Q(xy)=Q(x)Q(y) forall x,y$) on a quaternion algebra $Big(dfrac{a,b}{F}Big)$ is the norm $mathrm{Nr}$, which is isometric to the form $langle 1,-a,-b,abrangle$ in the orthogonal basis $1,i,j,ij$.
One easily sees, that $Q(1)=Q(1^2)=Q(1)^2Rightarrow Q(1)=1$ and similarly $Q(i)^2=a^2$, $Q(j)^2=b^2$ and therfore $Q(ij)=(ab)^2$. But how do we show $Q(i)=-a$ and $Q(j)=-b$? Or is it even necessary?
It seems like I'm missing something obvious.
norm quadratic-forms quaternions algebras
asked Jan 24 at 16:08
BananenPateBananenPate
284
$endgroup$
I want to prove, that the only multiplicative quadratic form $Q$ (so $Q(xy)=Q(x)Q(y) forall x,y$) on a quaternion algebra $Big(dfrac{a,b}{F}Big)$ is the norm $mathrm{Nr}$, which is isometric to the form $langle 1,-a,-b,abrangle$ in the orthogonal basis $1,i,j,ij$.
One easily sees, that $Q(1)=Q(1^2)=Q(1)^2Rightarrow Q(1)=1$ and similarly $Q(i)^2=a^2$, $Q(j)^2=b^2$ and therfore $Q(ij)=(ab)^2$. But how do we show $Q(i)=-a$ and $Q(j)=-b$? Or is it even necessary?
It seems like I'm missing something obvious.
norm quadratic-forms quaternions algebras
norm quadratic-forms quaternions algebras
asked Jan 24 at 16:08
BananenPateBananenPate
284
asked Jan 24 at 16:08
BananenPateBananenPate
284
edited Jan 24 at 18:14
BananenPate
asked Jan 24 at 16:08
BananenPateBananenPate
284
asked Jan 24 at 16:08
BananenPateBananenPate
284
asked Jan 24 at 16:08
BananenPateBananenPate
284
284
$begingroup$
A couple of things to keep in mind: (1) your argument that $Q(1) = 1$ is only valid if you assume $Q(1) ne 0$, and (2) in general, a quadratic form is not determined by its values on a basis.
$endgroup$
– Kimball
Jan 24 at 22:06
add a comment |
$begingroup$
A couple of things to keep in mind: (1) your argument that $Q(1) = 1$ is only valid if you assume $Q(1) ne 0$, and (2) in general, a quadratic form is not determined by its values on a basis.
$endgroup$
– Kimball
Jan 24 at 22:06
$begingroup$
A couple of things to keep in mind: (1) your argument that $Q(1) = 1$ is only valid if you assume $Q(1) ne 0$, and (2) in general, a quadratic form is not determined by its values on a basis.
$endgroup$
– Kimball
Jan 24 at 22:06
$begingroup$
A couple of things to keep in mind: (1) your argument that $Q(1) = 1$ is only valid if you assume $Q(1) ne 0$, and (2) in general, a quadratic form is not determined by its values on a basis.
$endgroup$
– Kimball
Jan 24 at 22:06
add a comment |
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A couple of things to keep in mind: (1) your argument that $Q(1) = 1$ is only valid if you assume $Q(1) ne 0$, and (2) in general, a quadratic form is not determined by its values on a basis.
$endgroup$
– Kimball
Jan 24 at 22:06