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

[Korybut]

TL;DR

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.

 

[martinbn]

$z_1\otimes z_2 \mapsto (z_1z_2, z_1\bar{z}_2)$

 

[martinbn]

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

$\mathbb C \otimes \mathbb C = \mathbb C \otimes \mathbb R[T]/<T^2+1> = \mathbb C[T]/<T^2+1>=\mathbb C[T]/<(T-i)(T+i)> = \mathbb C \oplus \mathbb C$

 

[Korybut]

$z_1\otimes z_2 \mapsto (z_1z_2, z_1\bar{z}_2)$

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.

 

[martinbn]

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 $0$.

 

[Korybut]

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?

 

[Korybut]

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!

 

[martinbn]

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.

 

[martinbn]

ps By the way which book is that?

 

[Korybut]

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

 

[martinbn]

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.

 

[Korybut]

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)

 

[fresh_42]

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 $\mathbb{C}^1.$

 

[martinbn]

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.

 

[Sandra_Nair]

Here's a simple proof if you don't care about the explicit isomorphism:
\begin{align*}
\mathbb{C}\otimes_\mathbb{R}\mathbb{C}\cong \frac{\mathbb{R}[x]}{\langle x^2+1\rangle}\otimes_\mathbb{R}\mathbb{C}\cong \frac{\mathbb{C}[x]}{\langle x^2+1\rangle}\cong \frac{\mathbb{C}[x]}{\langle(x+i)(x-i)\rangle}\cong \frac{\mathbb{C}[x]}{\langle(x+i)\rangle}\times\frac{\mathbb{C}[x]}{\langle(x-i)\rangle}\cong \mathbb{C}\times\mathbb{C} = \mathbb{C}^{\oplus2}
\end{align*}

Here, we are identifying one of the two complex fields as an extension of the reals with a solution to the equation x^2+1=0. We do so by adjoining the indeterminate x to the reals, and quotienting out by the ideal (x^2+1). Once we are at that stage, we use the primary purpose of tensors: to extend the underlying scalar field. But in the complex field (which we get after tensoring with the second complex field), the formula x^2+1 factors into (x+i)(x-i). By the Chinese Remainder Theorem, the ring ought to split into a product, each with one factor in the quotient. Now each of the rings have i and -i living in C already, so this is nothing new. So each of the 2 rings must themselves be C. Thus, we get 2 copies of C.