# Thread: Isomorphism of Algebras

1. ## Isomorphism of Algebras

Hello,

Here are two Algebras and i want to show that they are isomorphic:

1) $
\mathcal{L}(\bigotimes_{i\in \mathbb{N}} (E_i)) \
$

2) $\bigotimes_{i\in \mathbb{N}} \mathcal{L}(E_i)$

The algebra of the form $\mathcal{L} (E)$ is the set of all endomorphisms. And the $E_i$ are vector spaces.

I try to construct a isomorphism, but it failed all the time. Can you please help me?

I think that the Algebra-Multiplikation is composition in both cases.

I have defined this map, but i couldn't show that it is an isomorphism:
$\phi: \bigotimes_{i\in \mathbb{N}} \mathcal{L}(E_i) -> \bigotimes_{i\in \mathbb{N}} \mathcal{L}(E_i) , \phi (\otimes f_i):=f$ with f defined as: $f: \bigotimes E_i -> \bigotimes E_i, \; f(\otimes e_i):= \otimes f_i(e_i)$

Regards

2. Originally Posted by Bongo
Hello,

Here are two Algebras and i want to show that they are isomorphic:

1) $
\mathcal{L}(\bigotimes_{i\in \mathbb{N}} (E_i)) \
$

2) $\bigotimes_{i\in \mathbb{N}} \mathcal{L}(E_i)$

The algebra of the form $\mathcal{L} (E)$ is the set of all endomorphisms. And the $E_i$ are vector spaces.

I try to construct a isomorphism, but it failed all the time. Can you please help me?

I think that the Algebra-Multiplikation is composition in both cases.

I have defined this map, but i couldn't show that it is an isomorphism:
$\phi: \bigotimes_{i\in \mathbb{N}} \mathcal{L}(E_i) -> \bigotimes_{i\in \mathbb{N}} \mathcal{L}(E_i) , \phi (\otimes f_i):=f$ with f defined as: $f: \bigotimes E_i -> \bigotimes E_i, \; f(\otimes e_i):= \otimes f_i(e_i)$

Regards
I'll assume you're talking about the tensor product of algebras.

Ok, so you have the map $\displaystyle \phi:\bigotimes_{i\in\mathbb{N}}\text{End}\left(E_ i\right)\to\text{End}\left(\bigotimes_{i\in\mathbb {N}}E_i\right)$ by $\phi_{\otimes^i f_i}(\otimes^i e_i)=\otimes^i f_i(e_i)$. I agree that it's the right map. So, recall that an isomorphism of associative unital algebras is a linear isomorphism which is multiplicative and maps the multiplicative identity to the multiplicative identity (i.e. it preserves both the vector space and monoidal structures). It's clearly linear because you defined it to be so. In particular, you've defined $\phi$ (really you've defined it on more than a basis, but that's unimportant) on a basis of $\displaystyle \bigotimes_{i\in\mathbb{N}}\text{End}\left(E_i\rig ht)$ and of course are letting $\phi$ be the unique linear map. So the question is why it respects the multiplicative properties. It clearly maps the identity to the identity since the identity for $\displaystyle \bigotimes_{i\in\mathbb{N}}\text{End}\left(E_i\rig ht)$ is $\otimes^i\mathbf{1}_i$ where $\mathbf{1}_i$ is the identity map on $E_i$ and thus for any $\displaystyle \otimes^i e_i\in\bigotimes_{i\in\mathbb{N}}E_i$ we have that $\phi_{\otimes^i \mathbf{1}_i}(\otimes^i e_i)=\otimes^i \mathbf{1}_i(e_i)=\otimes^i e_i$ so that $\displaystyle \phi_{\otimes^i \mathbf{1}_i}=\mathbf{1}_{\otimes^i E_i}$. Can you do the multiplicative part?

3. How is the tensor product is defined for infinetly many vector spaces? (or algebras)

4. Hello,

Thank you for your help!

Here is my multiplication part:

It suffices to check the multiplication for decomposed elm.:

$
\phi((\otimes f_i)* (\otimes g_i))=\phi (\otimes (f_i * g_i))=\phi(\otimes f_i)*\phi(\otimes g_i),
$
since we have
$
\phi (\otimes (f_i * g_i))(\otimes e_i)=\otimes ((f_i *g_i)(e_i))=(\otimes(f_i))* \otimes (g_i (e_i))=\phi(\otimes f_i)*\phi(\otimes g_i)(\otimes e_i)
$

In fact, these are only the definitions. Is it correct?

Regards

5. Originally Posted by Bongo
Hello,

Thank you for your help!

Here is my multiplication part:

It suffices to check the multiplication for decomposed elm.:

$
\phi((\otimes f_i)* (\otimes g_i))=\phi (\otimes (f_i * g_i))=\phi(\otimes f_i)*\phi(\otimes g_i),
$
since we have
$
\phi (\otimes (f_i * g_i))(\otimes e_i)=\otimes ((f_i *g_i)(e_i))=(\otimes(f_i))* \otimes (g_i (e_i))=\phi(\otimes f_i)*\phi(\otimes g_i)(\otimes e_i)
$

In fact, these are only the definitions. Is it correct?

Regards
Right! It looks nasty, but the proof is just by definition.

6. Hello,

i have a little question. Why is our function bijective? suddenly i forgot! Can you remember me?
Furthermore, i'm not sure whether it suffices to show miltiplication property for decomposed elements. Is this true?

Regards