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 ...