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I Tensor Algebras and Graded Algebras - Cooperstein

  1. Apr 30, 2016 #1
    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:


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    ?temp_hash=f50d53053af31f1541d7434c8c2ec289.png




    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
     
  2. jcsd
  3. Apr 30, 2016 #2

    stevendaryl

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

    [itex]\mathcal{A}_k[/itex] and [itex]\mathcal{A}_l[/itex] are subsets of [itex]\mathcal{A}[/itex]. So I think that [itex]\mathcal{A}_k \mathcal{A}_l[/itex] just means the set of all elements [itex]x[/itex] such that [itex]x = y z[/itex], where [itex]y[/itex] is an element of [itex]\mathcal{A}_k[/itex] and [itex]z[/itex] is an element of [itex]\mathcal{A}_l[/itex].
     
  4. Apr 30, 2016 #3
    Thanks Steven ... appreciate your help ...
     
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