Tensor Algebras and Graded Algebras - Cooperstein - Theorem 10.11 and Defn 10.7

  • #1
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I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

I am focused on Section 10.3 The Tensor Algebra ... ...

I need help in order to get an understanding of an aspect of Example 10.11 and Definition 10.7 in Section 10.3 ...

The relevant text in Cooperstein is as follows:View attachment 5565
View attachment 5566My questions related to the above text from Cooperstein are simple and related ... they are as follows:Question 1

In the above text from Cooperstein ... at the start of the proof of Theorem 10.11 we read the following:

" ... ... Define a map \(\displaystyle S^k \ : \ V^k \longrightarrow \mathcal{A}\) by

\(\displaystyle S^k (v_1, \ ... \ ... \ , v_k ) = S(v_1) S(v_2) \ ... \ ... \ S(v_k)\)

... ... ... "


My question is ... what is the form and nature of the multiplication involved between the elements in the product \(\displaystyle S(v_1) S(v_2) \ ... \ ... \ S(v_k)\) ... ... ?
Question 2

In the above text from Cooperstein in Definition 10.7 we read the following:

" ... ... An algebra \(\displaystyle \mathcal{A}\) is said to be \(\displaystyle \mathbb{Z}\)-graded if it is the internal direct sum of subspaces \(\displaystyle \mathcal{A}_k , k \in \mathbb{Z}\) such that

\(\displaystyle \mathcal{A}_k \mathcal{A}_l \subset \mathcal{A}_{k + l}\)

... ... ... "My question is ... what is the form and nature of the multiplication involved between the elements in the product \(\displaystyle \mathcal{A}_k \mathcal{A}_l\) ... ... ?Hope someone can help ...

Peter
 
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  • #2
Peter said:
I am reading Bruce N. Coopersteins book: Advanced Linear Algebra (Second Edition) ... ...

I am focused on Section 10.3 The Tensor Algebra ... ...

I need help in order to get an understanding of an aspect of Example 10.11 and Definition 10.7 in Section 10.3 ...

The relevant text in Cooperstein is as follows:
My questions related to the above text from Cooperstein are simple and related ... they are as follows:Question 1

In the above text from Cooperstein ... at the start of the proof of Theorem 10.11 we read the following:

" ... ... Define a map \(\displaystyle S^k \ : \ V^k \longrightarrow \mathcal{A}\) by

\(\displaystyle S^k (v_1, \ ... \ ... \ , v_k ) = S(v_1) S(v_2) \ ... \ ... \ S(v_k)\)

... ... ... "


My question is ... what is the form and nature of the multiplication involved between the elements in the product \(\displaystyle S(v_1) S(v_2) \ ... \ ... \ S(v_k)\) ... ... ?

It is the product in the algebra $\mathcal{A}$ (recall an algebra has a "dual nature" as both an $R$-module *and* a ring, such that the ring multiplication is *compatible* with the module addition, AND the ring-action of $R$-this is why we require $R$-bilinearity. In addition, we require that $R$ commute with $\mathcal{A}$-that is, that $R \subseteq Z(\mathcal{A})$, the center of $\mathcal{A}$).
Question 2

In the above text from Cooperstein in Definition 10.7 we read the following:

" ... ... An algebra \(\displaystyle \mathcal{A}\) is said to be \(\displaystyle \mathbb{Z}\)-graded if it is the internal direct sum of subspaces \(\displaystyle \mathcal{A}_k , k \in \mathbb{Z}\) such that

\(\displaystyle \mathcal{A}_k \mathcal{A}_l \subset \mathcal{A}_{k + l}\)

... ... ... "My question is ... what is the form and nature of the multiplication involved between the elements in the product \(\displaystyle \mathcal{A}_k \mathcal{A}_l\) ... ... ?Hope someone can help ...

Peter

The set $\mathcal{A}_k\mathcal{A}_l = \{a \in \mathcal{A}: a = a_ka_l, a_k \in \mathcal{A}_k,a_l \in \mathcal{A}_l\}$ (it's just a "product set").
 
  • #3
Deveno said:
It is the product in the algebra $\mathcal{A}$ (recall an algebra has a "dual nature" as both an $R$-module *and* a ring, such that the ring multiplication is *compatible* with the module addition, AND the ring-action of $R$-this is why we require $R$-bilinearity. In addition, we require that $R$ commute with $\mathcal{A}$-that is, that $R \subseteq Z(\mathcal{A})$, the center of $\mathcal{A}$).The set $\mathcal{A}_k\mathcal{A}_l = \{a \in \mathcal{A}: a = a_ka_l, a_k \in \mathcal{A}_k,a_l \in \mathcal{A}_l\}$ (it's just a "product set").
Thanks Deveno ... appreciate the clarification ... most helpful ...

Peter
 

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