Mixing
What is this 2D division algebra?
Clash Royale CLAN TAG#URR8PPP
up vote
19
down vote
favorite
Consider the set $A$ of 2-tuples of real values $(a,b)$, equipped with an addition defined as
$$ (a,b) + (c,d) = (a+c,b+d)$$
and multiplication defined as
$$ (a,b) times (c,d) = (ac+bd,ad-bc).$$
What is this weird little thing?
This algebra has some nice properties. For instance:
on both sides multiplication distributes across addition, because the multiplication is bilinear
it is a division algebra, as there are no zero divisors
a subset is isomorphic to the reals, $(a,0) leftrightarrow a$
has a positive definite quadratic form
$$ (a,b)times(a,b) = (a^2+b^2,0) $$identity on left, $(1,0)times(a,b) = (a,b)$
But it has some weird properties:
there is no identity for multiplication on the right
$z=(0,1)$ anti-commutes with the subset noted above as being isomorphic to the reals
$$ (a,0)times z + z times (a,0) = (0,0)$$the multiplication is not associative
the multiplication is not even power associative, as seen with $z=(0,1)$,
$$ ztimes(ztimes z) = -(ztimes z)times z$$
- the center is trivial, as only $(0,0)$ commutes with all elements.
So I'm not sure on the terminology, but this can also be viewed as "extending" the Reals with an exotic sqrt of 1,
$$ z^2 = 1,$$
which anti-commutes with multiplication of reals,
$$forall a in mathbbR : az+za=0.$$
Then the set $A = a+bz : a,b in mathbbR$.
After playing with it a bit I realized it can also be viewed as taking the complex numbers and defining the operation:
$$x times y = x^* y$$
Which means this is also like taking a 1D complex Hilbert space, and treating the inner product as if it is a multiplication because in this case the scalar and vector are the same dimension.
This bizarre little thing is simple enough that I assume it has been studied before. Does it have a name?
Also, regardless if it has a name, I'd like to know the proper terminology for describing this.
Because of the relation to complex numbers, would mathematicians consider it "just the complex numbers" since the operations can be represented with complex numbers?
It at least isn't isomorphic to the complex numbers, correct?
Would you consider this a 2D real division algebra distinct from the complex numbers?
Since the structure was defined in terms of operations on the reals, and the elements are a tuple of reals, it feels like this would be some-object "over the Reals". Maybe a left semimodule over the Reals. Or does the phrase "over the reals" require that the Reals commute with everything?
Similarly, if you object to my use of terminology in the discussion of the properties, I'd appreciate if you could point that out and suggest more reasonable terminology with explanation.
abstract-algebra division-algebras
asked Aug 25 at 18:47
PPenguin
31018
add a comment |Â
up vote
19
down vote
favorite
Consider the set $A$ of 2-tuples of real values $(a,b)$, equipped with an addition defined as
$$ (a,b) + (c,d) = (a+c,b+d)$$
and multiplication defined as
$$ (a,b) times (c,d) = (ac+bd,ad-bc).$$
What is this weird little thing?
This algebra has some nice properties. For instance:
on both sides multiplication distributes across addition, because the multiplication is bilinear
it is a division algebra, as there are no zero divisors
a subset is isomorphic to the reals, $(a,0) leftrightarrow a$
has a positive definite quadratic form
$$ (a,b)times(a,b) = (a^2+b^2,0) $$identity on left, $(1,0)times(a,b) = (a,b)$
But it has some weird properties:
there is no identity for multiplication on the right
$z=(0,1)$ anti-commutes with the subset noted above as being isomorphic to the reals
$$ (a,0)times z + z times (a,0) = (0,0)$$the multiplication is not associative
the multiplication is not even power associative, as seen with $z=(0,1)$,
$$ ztimes(ztimes z) = -(ztimes z)times z$$
- the center is trivial, as only $(0,0)$ commutes with all elements.
So I'm not sure on the terminology, but this can also be viewed as "extending" the Reals with an exotic sqrt of 1,
$$ z^2 = 1,$$
which anti-commutes with multiplication of reals,
$$forall a in mathbbR : az+za=0.$$
Then the set $A = a+bz : a,b in mathbbR$.
After playing with it a bit I realized it can also be viewed as taking the complex numbers and defining the operation:
$$x times y = x^* y$$
Which means this is also like taking a 1D complex Hilbert space, and treating the inner product as if it is a multiplication because in this case the scalar and vector are the same dimension.
This bizarre little thing is simple enough that I assume it has been studied before. Does it have a name?
Also, regardless if it has a name, I'd like to know the proper terminology for describing this.
Because of the relation to complex numbers, would mathematicians consider it "just the complex numbers" since the operations can be represented with complex numbers?
It at least isn't isomorphic to the complex numbers, correct?
Would you consider this a 2D real division algebra distinct from the complex numbers?
Since the structure was defined in terms of operations on the reals, and the elements are a tuple of reals, it feels like this would be some-object "over the Reals". Maybe a left semimodule over the Reals. Or does the phrase "over the reals" require that the Reals commute with everything?
Similarly, if you object to my use of terminology in the discussion of the properties, I'd appreciate if you could point that out and suggest more reasonable terminology with explanation.
abstract-algebra division-algebras
asked Aug 25 at 18:47
PPenguin
31018
Don't you mean $(a,b)times (a,-b)=(a^2+b^2,0)$?
– Frpzzd
Aug 25 at 18:51
3
Use informative titles... ones that will help future scholars find the question from similar questions. "What is this weird thing" is useless in this regard.
– David G. Stork
Aug 25 at 19:19
3
I don't think this has any standard name. A close variation is split-complex numbers (en.wikipedia.org/wiki/Split-complex_number).
– Alon Amit
Aug 25 at 19:23
add a comment |Â
up vote
19
down vote
favorite
up vote
19
down vote
favorite
Consider the set $A$ of 2-tuples of real values $(a,b)$, equipped with an addition defined as
$$ (a,b) + (c,d) = (a+c,b+d)$$
and multiplication defined as
$$ (a,b) times (c,d) = (ac+bd,ad-bc).$$
What is this weird little thing?
This algebra has some nice properties. For instance:
on both sides multiplication distributes across addition, because the multiplication is bilinear
it is a division algebra, as there are no zero divisors
a subset is isomorphic to the reals, $(a,0) leftrightarrow a$
has a positive definite quadratic form
$$ (a,b)times(a,b) = (a^2+b^2,0) $$identity on left, $(1,0)times(a,b) = (a,b)$
But it has some weird properties:
there is no identity for multiplication on the right
$z=(0,1)$ anti-commutes with the subset noted above as being isomorphic to the reals
$$ (a,0)times z + z times (a,0) = (0,0)$$the multiplication is not associative
the multiplication is not even power associative, as seen with $z=(0,1)$,
$$ ztimes(ztimes z) = -(ztimes z)times z$$
- the center is trivial, as only $(0,0)$ commutes with all elements.
So I'm not sure on the terminology, but this can also be viewed as "extending" the Reals with an exotic sqrt of 1,
$$ z^2 = 1,$$
which anti-commutes with multiplication of reals,
$$forall a in mathbbR : az+za=0.$$
Then the set $A = a+bz : a,b in mathbbR$.
After playing with it a bit I realized it can also be viewed as taking the complex numbers and defining the operation:
$$x times y = x^* y$$
Which means this is also like taking a 1D complex Hilbert space, and treating the inner product as if it is a multiplication because in this case the scalar and vector are the same dimension.
This bizarre little thing is simple enough that I assume it has been studied before. Does it have a name?
Also, regardless if it has a name, I'd like to know the proper terminology for describing this.
Because of the relation to complex numbers, would mathematicians consider it "just the complex numbers" since the operations can be represented with complex numbers?
It at least isn't isomorphic to the complex numbers, correct?
Would you consider this a 2D real division algebra distinct from the complex numbers?
Since the structure was defined in terms of operations on the reals, and the elements are a tuple of reals, it feels like this would be some-object "over the Reals". Maybe a left semimodule over the Reals. Or does the phrase "over the reals" require that the Reals commute with everything?
Similarly, if you object to my use of terminology in the discussion of the properties, I'd appreciate if you could point that out and suggest more reasonable terminology with explanation.
abstract-algebra division-algebras
asked Aug 25 at 18:47
PPenguin
31018
Consider the set $A$ of 2-tuples of real values $(a,b)$, equipped with an addition defined as
$$ (a,b) + (c,d) = (a+c,b+d)$$
and multiplication defined as
$$ (a,b) times (c,d) = (ac+bd,ad-bc).$$
What is this weird little thing?
This algebra has some nice properties. For instance:
on both sides multiplication distributes across addition, because the multiplication is bilinear
it is a division algebra, as there are no zero divisors
a subset is isomorphic to the reals, $(a,0) leftrightarrow a$
has a positive definite quadratic form
$$ (a,b)times(a,b) = (a^2+b^2,0) $$identity on left, $(1,0)times(a,b) = (a,b)$
But it has some weird properties:
there is no identity for multiplication on the right
$z=(0,1)$ anti-commutes with the subset noted above as being isomorphic to the reals
$$ (a,0)times z + z times (a,0) = (0,0)$$the multiplication is not associative
the multiplication is not even power associative, as seen with $z=(0,1)$,
$$ ztimes(ztimes z) = -(ztimes z)times z$$
- the center is trivial, as only $(0,0)$ commutes with all elements.
So I'm not sure on the terminology, but this can also be viewed as "extending" the Reals with an exotic sqrt of 1,
$$ z^2 = 1,$$
which anti-commutes with multiplication of reals,
$$forall a in mathbbR : az+za=0.$$
Then the set $A = a+bz : a,b in mathbbR$.
After playing with it a bit I realized it can also be viewed as taking the complex numbers and defining the operation:
$$x times y = x^* y$$
Which means this is also like taking a 1D complex Hilbert space, and treating the inner product as if it is a multiplication because in this case the scalar and vector are the same dimension.
This bizarre little thing is simple enough that I assume it has been studied before. Does it have a name?
Also, regardless if it has a name, I'd like to know the proper terminology for describing this.
Because of the relation to complex numbers, would mathematicians consider it "just the complex numbers" since the operations can be represented with complex numbers?
It at least isn't isomorphic to the complex numbers, correct?
Would you consider this a 2D real division algebra distinct from the complex numbers?
Since the structure was defined in terms of operations on the reals, and the elements are a tuple of reals, it feels like this would be some-object "over the Reals". Maybe a left semimodule over the Reals. Or does the phrase "over the reals" require that the Reals commute with everything?
Similarly, if you object to my use of terminology in the discussion of the properties, I'd appreciate if you could point that out and suggest more reasonable terminology with explanation.
abstract-algebra division-algebras
asked Aug 25 at 18:47
PPenguin
31018
edited Aug 25 at 19:23
asked Aug 25 at 18:47
PPenguin
31018
asked Aug 25 at 18:47
PPenguin
31018
asked Aug 25 at 18:47
PPenguin
31018
31018
Don't you mean $(a,b)times (a,-b)=(a^2+b^2,0)$?
– Frpzzd
Aug 25 at 18:51
3
Use informative titles... ones that will help future scholars find the question from similar questions. "What is this weird thing" is useless in this regard.
– David G. Stork
Aug 25 at 19:19
3
I don't think this has any standard name. A close variation is split-complex numbers (en.wikipedia.org/wiki/Split-complex_number).
– Alon Amit
Aug 25 at 19:23
add a comment |Â
Don't you mean $(a,b)times (a,-b)=(a^2+b^2,0)$?
– Frpzzd
Aug 25 at 18:51
3
Use informative titles... ones that will help future scholars find the question from similar questions. "What is this weird thing" is useless in this regard.
– David G. Stork
Aug 25 at 19:19
3
I don't think this has any standard name. A close variation is split-complex numbers (en.wikipedia.org/wiki/Split-complex_number).
– Alon Amit
Aug 25 at 19:23
Don't you mean $(a,b)times (a,-b)=(a^2+b^2,0)$?
– Frpzzd
Aug 25 at 18:51
Don't you mean $(a,b)times (a,-b)=(a^2+b^2,0)$?
– Frpzzd
Aug 25 at 18:51
3
3
Use informative titles... ones that will help future scholars find the question from similar questions. "What is this weird thing" is useless in this regard.
– David G. Stork
Aug 25 at 19:19
Use informative titles... ones that will help future scholars find the question from similar questions. "What is this weird thing" is useless in this regard.
– David G. Stork
Aug 25 at 19:19
3
3
I don't think this has any standard name. A close variation is split-complex numbers (en.wikipedia.org/wiki/Split-complex_number).
– Alon Amit
Aug 25 at 19:23
I don't think this has any standard name. A close variation is split-complex numbers (en.wikipedia.org/wiki/Split-complex_number).
– Alon Amit
Aug 25 at 19:23
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
24
down vote
accepted
The two-dimensional division algebras over the real numbers were completely classified by Burdujan in 1985. See for example section 2 of this article for a short review. Borrowing notation from the article, we have that every such algebra is isomorphic to $mathbbC_ST$, the complex numbers equipped with a modified multiplication $x circ y = S(x)cdot T(y)$, where $S, T$ are invertible $mathbbR$-linear transformations of $mathbbC$.
For the division algebra in your question, $S$ is complex conjugation and $T$ the identity.
answered Aug 25 at 19:45
pregunton
1,3991231
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
24
down vote
accepted
The two-dimensional division algebras over the real numbers were completely classified by Burdujan in 1985. See for example section 2 of this article for a short review. Borrowing notation from the article, we have that every such algebra is isomorphic to $mathbbC_ST$, the complex numbers equipped with a modified multiplication $x circ y = S(x)cdot T(y)$, where $S, T$ are invertible $mathbbR$-linear transformations of $mathbbC$.
For the division algebra in your question, $S$ is complex conjugation and $T$ the identity.
answered Aug 25 at 19:45
pregunton
1,3991231
add a comment |Â
up vote
24
down vote
accepted
The two-dimensional division algebras over the real numbers were completely classified by Burdujan in 1985. See for example section 2 of this article for a short review. Borrowing notation from the article, we have that every such algebra is isomorphic to $mathbbC_ST$, the complex numbers equipped with a modified multiplication $x circ y = S(x)cdot T(y)$, where $S, T$ are invertible $mathbbR$-linear transformations of $mathbbC$.
For the division algebra in your question, $S$ is complex conjugation and $T$ the identity.
answered Aug 25 at 19:45
pregunton
1,3991231
add a comment |Â
up vote
24
down vote
accepted
up vote
24
down vote
accepted
The two-dimensional division algebras over the real numbers were completely classified by Burdujan in 1985. See for example section 2 of this article for a short review. Borrowing notation from the article, we have that every such algebra is isomorphic to $mathbbC_ST$, the complex numbers equipped with a modified multiplication $x circ y = S(x)cdot T(y)$, where $S, T$ are invertible $mathbbR$-linear transformations of $mathbbC$.
For the division algebra in your question, $S$ is complex conjugation and $T$ the identity.
answered Aug 25 at 19:45
pregunton
1,3991231
The two-dimensional division algebras over the real numbers were completely classified by Burdujan in 1985. See for example section 2 of this article for a short review. Borrowing notation from the article, we have that every such algebra is isomorphic to $mathbbC_ST$, the complex numbers equipped with a modified multiplication $x circ y = S(x)cdot T(y)$, where $S, T$ are invertible $mathbbR$-linear transformations of $mathbbC$.
For the division algebra in your question, $S$ is complex conjugation and $T$ the identity.
answered Aug 25 at 19:45
pregunton
1,3991231
answered Aug 25 at 19:45
pregunton
1,3991231
answered Aug 25 at 19:45
pregunton
1,3991231
answered Aug 25 at 19:45
pregunton
1,3991231
1,3991231
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2894403%2fwhat-is-this-2d-division-algebra%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Don't you mean $(a,b)times (a,-b)=(a^2+b^2,0)$?
– Frpzzd
Aug 25 at 18:51
3
Use informative titles... ones that will help future scholars find the question from similar questions. "What is this weird thing" is useless in this regard.
– David G. Stork
Aug 25 at 19:19
3
I don't think this has any standard name. A close variation is split-complex numbers (en.wikipedia.org/wiki/Split-complex_number).
– Alon Amit
Aug 25 at 19:23