Mixing
Typo? $H$ is quaternion algebra and $C$ is complex number. $M_n(H)cong M_{2n}(C)$?
$begingroup$
Denote $H$ as quaternion algebra and $C$ as complex number.
Take any $ain H$. One has $a=(a_1+ia_2)+(a_3+ia_4)j=c_1+jc_2$ with $ij=k$. Denote $M_n(F)$ as the matrix entries in division algebra $F$.
The book says $M_n(H)to M_{2n}(C)$ is isomorphic. If this is true, then this must hold for $n=1$. So $Hcong M_{2}(C)$ as ring.
This homomorphism is given as the following. Using $c_1,c_2$ notation in first paragraph and labeling $xin M_{2n}(C)$ as $x_{11},x_{12},x_{21},x_{22}$, I have $x_{11}=c_1,x_{12}=c_2,x_{21}=-bar{c_2},x_{22}=bar{c_1}$ where $bar{x}$ denotes complex conjugation.
It is clear that this map is $R-$linear. So this had better to be isomorphic as $R-$vector space to start with even before talking about ring isomorphism.
Counting dimension over $R$ is $4times 2=8$ for $M_{2}(C)$ and $H$ is only $4$. This cannot be isomorphism.
$textbf{Q:}$ Is this a typo in the book? What is the correction? I could see it has to be identified to the image and this is clearly injection for sure.
Ref. D. Bump Lie Groups Chpt 5 Exercise 5.6
abstract-algebra group-theory geometry differential-geometry lie-groups
asked Jan 4 at 0:06
user45765user45765
2,6792722
$endgroup$
add a comment |
$begingroup$
Denote $H$ as quaternion algebra and $C$ as complex number.
Take any $ain H$. One has $a=(a_1+ia_2)+(a_3+ia_4)j=c_1+jc_2$ with $ij=k$. Denote $M_n(F)$ as the matrix entries in division algebra $F$.
The book says $M_n(H)to M_{2n}(C)$ is isomorphic. If this is true, then this must hold for $n=1$. So $Hcong M_{2}(C)$ as ring.
This homomorphism is given as the following. Using $c_1,c_2$ notation in first paragraph and labeling $xin M_{2n}(C)$ as $x_{11},x_{12},x_{21},x_{22}$, I have $x_{11}=c_1,x_{12}=c_2,x_{21}=-bar{c_2},x_{22}=bar{c_1}$ where $bar{x}$ denotes complex conjugation.
It is clear that this map is $R-$linear. So this had better to be isomorphic as $R-$vector space to start with even before talking about ring isomorphism.
Counting dimension over $R$ is $4times 2=8$ for $M_{2}(C)$ and $H$ is only $4$. This cannot be isomorphism.
$textbf{Q:}$ Is this a typo in the book? What is the correction? I could see it has to be identified to the image and this is clearly injection for sure.
Ref. D. Bump Lie Groups Chpt 5 Exercise 5.6
abstract-algebra group-theory geometry differential-geometry lie-groups
asked Jan 4 at 0:06
user45765user45765
2,6792722
$endgroup$
2
$begingroup$
It is indeed a typo or imprecision of language. The map is an isomorphism to_its_image. That is all that's asserted, and you already see that.
$endgroup$
– paul garrett
Jan 4 at 0:17
add a comment |
$begingroup$
Denote $H$ as quaternion algebra and $C$ as complex number.
Take any $ain H$. One has $a=(a_1+ia_2)+(a_3+ia_4)j=c_1+jc_2$ with $ij=k$. Denote $M_n(F)$ as the matrix entries in division algebra $F$.
The book says $M_n(H)to M_{2n}(C)$ is isomorphic. If this is true, then this must hold for $n=1$. So $Hcong M_{2}(C)$ as ring.
This homomorphism is given as the following. Using $c_1,c_2$ notation in first paragraph and labeling $xin M_{2n}(C)$ as $x_{11},x_{12},x_{21},x_{22}$, I have $x_{11}=c_1,x_{12}=c_2,x_{21}=-bar{c_2},x_{22}=bar{c_1}$ where $bar{x}$ denotes complex conjugation.
It is clear that this map is $R-$linear. So this had better to be isomorphic as $R-$vector space to start with even before talking about ring isomorphism.
Counting dimension over $R$ is $4times 2=8$ for $M_{2}(C)$ and $H$ is only $4$. This cannot be isomorphism.
$textbf{Q:}$ Is this a typo in the book? What is the correction? I could see it has to be identified to the image and this is clearly injection for sure.
Ref. D. Bump Lie Groups Chpt 5 Exercise 5.6
abstract-algebra group-theory geometry differential-geometry lie-groups
asked Jan 4 at 0:06
user45765user45765
2,6792722
$endgroup$
Denote $H$ as quaternion algebra and $C$ as complex number.
Take any $ain H$. One has $a=(a_1+ia_2)+(a_3+ia_4)j=c_1+jc_2$ with $ij=k$. Denote $M_n(F)$ as the matrix entries in division algebra $F$.
The book says $M_n(H)to M_{2n}(C)$ is isomorphic. If this is true, then this must hold for $n=1$. So $Hcong M_{2}(C)$ as ring.
This homomorphism is given as the following. Using $c_1,c_2$ notation in first paragraph and labeling $xin M_{2n}(C)$ as $x_{11},x_{12},x_{21},x_{22}$, I have $x_{11}=c_1,x_{12}=c_2,x_{21}=-bar{c_2},x_{22}=bar{c_1}$ where $bar{x}$ denotes complex conjugation.
It is clear that this map is $R-$linear. So this had better to be isomorphic as $R-$vector space to start with even before talking about ring isomorphism.
Counting dimension over $R$ is $4times 2=8$ for $M_{2}(C)$ and $H$ is only $4$. This cannot be isomorphism.
$textbf{Q:}$ Is this a typo in the book? What is the correction? I could see it has to be identified to the image and this is clearly injection for sure.
Ref. D. Bump Lie Groups Chpt 5 Exercise 5.6
abstract-algebra group-theory geometry differential-geometry lie-groups
abstract-algebra group-theory geometry differential-geometry lie-groups
asked Jan 4 at 0:06
user45765user45765
2,6792722
asked Jan 4 at 0:06
user45765user45765
2,6792722
asked Jan 4 at 0:06
user45765user45765
2,6792722
asked Jan 4 at 0:06
user45765user45765
2,6792722
asked Jan 4 at 0:06
user45765user45765
2,6792722
2,6792722
2
$begingroup$
It is indeed a typo or imprecision of language. The map is an isomorphism to_its_image. That is all that's asserted, and you already see that.
$endgroup$
– paul garrett
Jan 4 at 0:17
add a comment |
2
$begingroup$
It is indeed a typo or imprecision of language. The map is an isomorphism to_its_image. That is all that's asserted, and you already see that.
$endgroup$
– paul garrett
Jan 4 at 0:17
2
2
$begingroup$
It is indeed a typo or imprecision of language. The map is an isomorphism to_its_image. That is all that's asserted, and you already see that.
$endgroup$
– paul garrett
Jan 4 at 0:17
$begingroup$
It is indeed a typo or imprecision of language. The map is an isomorphism to_its_image. That is all that's asserted, and you already see that.
$endgroup$
– paul garrett
Jan 4 at 0:17
add a comment |
1 Answer
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$begingroup$
I am almost sure what is meant there is that it is an isomorphism to its image. Namely, the map is obviously not surjective. But you can check that it is an injective morphism of $Bbb R$-algebras. Indeed, the case $n=1$ gives a standard representation of the quaternions as sub-$Bbb R$-algebra of $M_2(Bbb C)$.
answered Jan 4 at 0:17
Torsten SchoenebergTorsten Schoeneberg
3,9112833
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$begingroup$
I am almost sure what is meant there is that it is an isomorphism to its image. Namely, the map is obviously not surjective. But you can check that it is an injective morphism of $Bbb R$-algebras. Indeed, the case $n=1$ gives a standard representation of the quaternions as sub-$Bbb R$-algebra of $M_2(Bbb C)$.
answered Jan 4 at 0:17
Torsten SchoenebergTorsten Schoeneberg
3,9112833
$endgroup$
add a comment |
$begingroup$
I am almost sure what is meant there is that it is an isomorphism to its image. Namely, the map is obviously not surjective. But you can check that it is an injective morphism of $Bbb R$-algebras. Indeed, the case $n=1$ gives a standard representation of the quaternions as sub-$Bbb R$-algebra of $M_2(Bbb C)$.
answered Jan 4 at 0:17
Torsten SchoenebergTorsten Schoeneberg
3,9112833
$endgroup$
add a comment |
$begingroup$
I am almost sure what is meant there is that it is an isomorphism to its image. Namely, the map is obviously not surjective. But you can check that it is an injective morphism of $Bbb R$-algebras. Indeed, the case $n=1$ gives a standard representation of the quaternions as sub-$Bbb R$-algebra of $M_2(Bbb C)$.
answered Jan 4 at 0:17
Torsten SchoenebergTorsten Schoeneberg
3,9112833
$endgroup$
I am almost sure what is meant there is that it is an isomorphism to its image. Namely, the map is obviously not surjective. But you can check that it is an injective morphism of $Bbb R$-algebras. Indeed, the case $n=1$ gives a standard representation of the quaternions as sub-$Bbb R$-algebra of $M_2(Bbb C)$.
answered Jan 4 at 0:17
Torsten SchoenebergTorsten Schoeneberg
3,9112833
answered Jan 4 at 0:17
Torsten SchoenebergTorsten Schoeneberg
3,9112833
answered Jan 4 at 0:17
Torsten SchoenebergTorsten Schoeneberg
3,9112833
answered Jan 4 at 0:17
Torsten SchoenebergTorsten Schoeneberg
3,9112833
3,9112833
add a comment |
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It is indeed a typo or imprecision of language. The map is an isomorphism to_its_image. That is all that's asserted, and you already see that.
$endgroup$
– paul garrett
Jan 4 at 0:17