##\mathbb{C}\oplus\mathbb{C}\cong\mathbb{C}\otimes\mathbb{C}##

  • #1
Korybut
64
3
TL;DR Summary
Algebra isomorphism
Hello!

Reading book o Clifford algebra authors claim that ##\mathbb{C}\oplus\mathbb{C}\cong\mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}## as algebras. Unfortunately proof is absent and provided hint is pretty misleading

As vector spaces they are obviously isomorphic since
##\dim_{\mathbb{R}} \mathbb{C}\oplus\mathbb{C}=\dim_{\mathbb{R}} \mathbb{C}\otimes_{\mathbb{R}}\mathbb{C}=4##.
Product in ##\mathbb{C}\otimes_{\mathbb{R}} \mathbb{C}## looks as follows
##(z_1\otimes z_2) (z_3\otimes z_4)=z_1 z_3 \otimes z_2 z_4##
Product in ##\mathbb{C}\oplus\mathbb{C}## looks as follows
##(z_1,z_2) (z_3,z_4)=(z_1 z_3,z_2 z_4)##
From that perspective it is quite tempting to identify ##z_1 \otimes z_2## with ##(z_1,z_2)## however ##z_1\otimes z_2## and ##\lambda z_1 \otimes \frac{1}{\lambda} z_2## (for some real ##\lambda##) are the same elements in ##\mathbb{C}\otimes \mathbb{C}## but they will be mapped to different elements of ##\mathbb{C}\oplus\mathbb{C}##. Obviously I need better map if this isomorphism indeed take place. But which one?

Many thanks in advance.
 
Physics news on Phys.org
  • #2
##z_1\otimes z_2 \mapsto (z_1z_2, z_1\bar{z}_2)##
 
  • Like
Likes Korybut
  • #4
martinbn said:
##z_1\otimes z_2 \mapsto (z_1z_2, z_1\bar{z}_2)##
I've seen this formula before and indeed this identification doesn't suffer from the problem I mentioned earlier however I don't get why this is isomorphism.
If I consider this map
##i: \mathbb{C}\otimes \mathbb{C}\rightarrow \mathbb{C}\oplus\mathbb{C}##
then I can look at element which are mapped to zero in ##\mathbb{C}\oplus\mathbb{C}## and there are way more than one.
 
  • #5
Korybut said:
I've seen this formula before and indeed this identification doesn't suffer from the problem I mentioned earlier however I don't get why this is isomorphism.
If I consider this map
##i: \mathbb{C}\otimes \mathbb{C}\rightarrow \mathbb{C}\oplus\mathbb{C}##
then I can look at element which are mapped to zero in ##\mathbb{C}\oplus\mathbb{C}## and there are way more than one.
How? If it is mapped to ##0##, then either ##z_1## or ##z_2## is ##0##, so ##z_1\otimes z_2## is also ##0##.
 
  • Like
Likes Korybut
  • #6
martinbn said:
This one shows up at the bottom of the page
https://www.physicsforums.com/threads/tensor-product-of-c-with-itself-over-r.675424/
but it should read as follow

##\mathbb C \otimes \mathbb C = \mathbb C \otimes \mathbb R[T]/<T^2+1> = \mathbb C[T]/<T^2+1>=\mathbb C[T]/<(T-i)(T+i)> = \mathbb C \oplus \mathbb C##
I wish someone can explain notation in this formula. What is ##T##? What does ##/<...> means?
 
  • #7
martinbn said:
How? If it is mapped to ##0##, then either ##z_1## or ##z_2## is ##0##, so ##z_1\otimes z_2## is also ##0##.
Sorry! My Bad! Indeed! Thanks for help!
 
  • #8
Korybut said:
I wish someone can explain notation in this formula. What is ##T##? What does ##/<...> means?
##T## is a variable, you are looking at polynomials. ##/<...>## means the quotient by the ideal.
 
  • #9
ps By the way which book is that?
 
  • #10
martinbn said:
ps By the way which book is that?
Lawson and Michelson "Spin Geometry". They suggested the following
##(1,0)\rightarrow \frac{1}{2}(1\otimes 1+i\otimes i)##
##(0,1)\rightarrow \frac{1}{2}(1\otimes 1 -i \otimes i)##
And I don't get how I proceed with the full proof with just that
 
  • #11
Korybut said:
Lawson and Michelson "Spin Geometry". They suggested the following
##(1,0)\rightarrow \frac{1}{2}(1\otimes 1+i\otimes i)##
##(0,1)\rightarrow \frac{1}{2}(1\otimes 1 -i \otimes i)##
And I don't get how I proceed with the full proof with just that
This is meant to be as ##\mathbb C##-algebras, where the structure on the tensor product is through the first argument. If you solve it for the inverse map it sends ##1\otimes 1## to ##(1,1)## and ##i\otimes i## to ##(1,-1)##. As real algebras it will map as follows

##1\otimes 1## to ##(1,1)##
##i\otimes 1## to ##(i,i)##
##i\otimes i## to ##(1,-1)##
##-1\otimes i## to ##(i,-i)##

which is the map ##a\otimes b \mapsto (a\bar{b}, ab)## in that basis.
 
  • Like
Likes Korybut
  • #12
martinbn said:
This is meant to be as ##\mathbb C##-algebras, where the structure on the tensor product is through the first argument.
Thanks for another clarification. I wish those lines were in the book)
 
  • #13
This is one of the cases that show why I dislike the tensor notation without mentioning the scalar field. We have ##\mathbb{C}\otimes_\mathbb{R} \mathbb{C}\cong \mathbb{C}^2.##

I first read it as ##\mathbb{C}\otimes_\mathbb{C} \mathbb{C}## which would be isomorphic to ##\mathbb{C}^1.##
 
  • #14
fresh_42 said:
This is one of the cases that show why I dislike the tensor notation without mentioning the scalar field. We have ##\mathbb{C}\otimes_\mathbb{R} \mathbb{C}\cong \mathbb{C}^2.##

I first read it as ##\mathbb{C}\otimes_\mathbb{C} \mathbb{C}## which would be isomorphic to ##\mathbb{C}^1.##
But he did write it indicating it is over the
reals. See the first post.
 
  • Like
Likes fresh_42

Similar threads

  • Linear and Abstract Algebra
Replies
7
Views
530
  • Linear and Abstract Algebra
Replies
18
Views
644
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
883
  • Linear and Abstract Algebra
Replies
21
Views
1K
Replies
14
Views
3K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
968
  • Linear and Abstract Algebra
Replies
11
Views
3K
Back
Top