Tensor Algebras and Graded Algebras - Cooperstein

[Math Amateur]

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 $S^k \ : \ V^k \longrightarrow \mathcal{A}$ by

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

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

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

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

$\mathcal{A}_k \mathcal{A}_l \subset \mathcal{A}_{k + l}$

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

Peter

 

[stevendaryl]

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

Well, the original assumption is that S is a linear map from the vector space V to the algebra \mathcal{A}. So S(v) is an element of the algebra \mathcal{A}, whatever that is--it's left unspecified. But every algebra over a field \mathbb{F} has its own notion of multiplication and addition. So you're just using that notion of multiplication in interpreting S(v_1) S(v_2)

My question is ... what is the form and nature of the multiplication involved between the elements in the product $\mathcal{A}_k \mathcal{A}_l$ ... ... ?

\mathcal{A}_k and \mathcal{A}_l are subsets of \mathcal{A}. So I think that \mathcal{A}_k \mathcal{A}_l just means the set of all elements x such that x = y z, where y is an element of \mathcal{A}_k and z is an element of \mathcal{A}_l.

 

[Math Amateur]

Thanks Steven ... appreciate your help ...