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I Free Abelian Groups ... Aluffi Proposition 5.6

  1. May 13, 2016 #1
    I am reading Paolo Aluffi's book: Algebra: Chapter 0 ... ...

    I am currently focussed on Section 5.4 Free Abelian Groups ... ...

    I need help with an aspect of Aluffi's preamble to introduce Proposition 5.6 ...

    Proposition 5.6 and its preamble reads as follows:


    ?temp_hash=437ca4dd8ac7eefa509f557e37154b23.png




    In the above text from Aluffi's book we find the following:

    " ... ... For ##H = \mathbb{Z}## there is a natural function ##j \ : \ A \longrightarrow \mathbb{Z}^{ \oplus A }## , obtained by sending ##a \in A## to the function ##j_a \ : \ A \longrightarrow \mathbb{Z}## ... ... "

    My problem is in (precisely and rigorously) understanding the claim that ##j## sends ##a \in A## to the function ##j_a \ : \ A \longrightarrow \mathbb{Z}## ... ...

    The function ##j_a## is actually a set of ordered pairs no two of which have the same member ... BUT ... ##j## does not (exactly anyway) seem to send ##a \in A## to this function ...


    If we pretend for a moment that A is a countable ordered set ... then we can say that what ##j## seems to do is send ##a \in A## to the image-set of ##j_a##, namely

    ##( \ ... \ ... \ ,0,0, 0, \ ... \ ... \ 0,1,0 \ ... \ ... \ ... \ \ ,0,0, 0, \ ... \ ... )##

    where the ##1## is in the ##a##'th position ...

    So then ##j## seems to send ##a \in A## to the image-set of ##j_a## and not to ##j_a## itself ... ...


    (... ... not sure how to put this argument for the case where the set ##A## is uncountable and not ordered ... ... )

    Given my analysis ... how do we justify or make sense of Aluffi's claim that " there is a natural function ##j \ : \ A \longrightarrow \mathbb{Z}^{ \oplus A }## , obtained by sending ##a \in A## to the function ##j_a \ : \ A \longrightarrow \mathbb{Z}## "

    Hope someone can critique my analysis and clarify the issue to which I refer ...

    Help will be much appreciated ...

    Peter


    *** EDIT ***

    Just another concern over possibly missing something in fully understanding Aluffi's text above ... he introduces the general case with a general abelian group ##H## ... ... and then defines ##H^{ \oplus A}## ... ... but never uses ##H## ... he just puts it equal to ##\mathbb{Z}## ... if you are just going to put ##H = \mathbb{Z}## ... ... why bother with ##H## ... does anyone have an idea what Aluffi is doing ... ... ??? ... ... he does a similar thing when explaining free modules ... ... am I missing something ... ... ??? ... ...




    ======================================================

    To give Physics Forum members reading this post a sense of the approach and notation of Aluffi to free Abelian groups I am here providing Aluffi's introduction to free Abelian groups up to and including Proposition 5.6 ... as follows:


    ?temp_hash=437ca4dd8ac7eefa509f557e37154b23.png
    ?temp_hash=437ca4dd8ac7eefa509f557e37154b23.png
    ?temp_hash=437ca4dd8ac7eefa509f557e37154b23.png
     

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    Last edited: May 13, 2016
  2. jcsd
  3. May 13, 2016 #2

    fresh_42

    Staff: Mentor

    What do you think about the following definitions:
    $$j = \bigoplus_{a \in A}^{} j_a \text{ and } j_a = \bigoplus_{b \in A}^{} \delta_{ab} \cdot a $$
     
  4. May 13, 2016 #3

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    Unless you are really having fun reading that book, I would suggest it is not going to do you a lot good. If you want to learn some algebra, you might try Michael Artin's book, Algebra,.
     
  5. May 13, 2016 #4

    fresh_42

    Staff: Mentor

    I agree. It seems Paolo Aluffi has a very categorial approach. This might have its right when (co-)homology theory is the goal. However, I've seen physical concepts in which the latter is needed. But I agree it's a rather dusty ground to start with algebra. I took my first steps with these books:
    https://www.amazon.com/s/ref=nb_sb_...n+algebra&sprefix=van+der+waer,stripbooks,384
     
    Last edited by a moderator: May 7, 2017
  6. May 13, 2016 #5

    Thanks mathwonk and fresh_42 ... but note that I wish to understand a category theory approach ... seems an interesting theoretical basis and language for algebra ... and possibly category theory has an interesting way to help in algebraic topology as well ... it is challenging abstract though ...

    Just thinking about your first post, fresh_42 ...

    Peter
     
    Last edited by a moderator: May 7, 2017
  7. May 13, 2016 #6

    Fresh_42 ... can you expand a bit on how to interpret your definitions ...

    Peter
     
  8. May 13, 2016 #7
    Hi Fresh_42 ... just some vague thoughts on your definitions ... what they may mean, that is ...

    Assume (in order to get an idea of the meaning of the definitions) that ##A## is an ordered and countable set ...

    ... then we have ##j \ : \ A \longrightarrow \mathbb{Z}^{ \oplus A}## can be written as follows:

    ##j = \bigoplus_{a \in A} \ j_a##

    ##= ( \ j_{a_1} \ , \ j_{a_2} \ , \ j_{a_3}, \ ... \ ... \ ...\ ... )##

    ... so we can write ... ...

    ##j(a_1) = ( \ j_{a_1} (a_1) \ , \ j_{a_2} (a_1) \ , \ j_{a_3} (a_1) , \ ... \ ... \ ...\ ... )##


    ... BUT ... how do we interpret ## j_{a_1} (a_1) \ , \ j_{a_2} (a_1) \ , \ j_{a_3} (a_1) , \ ... \ ... ## etc ...


    I am assuming that ## j_{a_1} (a_1) \ = 1 \ , \ j_{a_2} (a_1) \ = 0 \ , \ j_{a_3} (a_1) \ = 0 \ , \ ... \ ## etc

    But ... I am not sure how that comes out of your definition/formula for ##j_a## ... ...



    Can you please comment and clarify ...

    Peter
     
    Last edited: May 13, 2016
  9. May 13, 2016 #8

    mathwonk

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    Science Advisor
    Homework Helper

    please forgive me, but to be blunt, before learning a fancy way to express something, it helps to actually understand the topic. and you do not. so in my opinion you are handicapping yourself by pursuing this less than useful presentation of the material. long ago i read briefly paolo's book and wondered just who in the world he aimed it at. i would never recommend this book to any young person, not just you.

    your questions make it obvious that you are not understanding anything in this book. this is not a criticism of you by the way. so i suggest an experiment, try one of the other books we recommend and see for yourelf if they do not speak more clearly to you.
     
  10. May 13, 2016 #9

    fresh_42

    Staff: Mentor

    I should probably better have written ##j_a## without the plus because it's only a function ##j_a : A → ℤ##. Your dots behind the ##ℤ## above misleaded me together with Aluffi's tendency to denote different things with the same letter.

    ##j_a## is defined exactly as the Kronecker-delta(-function): ##j_a(b) = \delta_{ab}##, i.e. ##1## if ##a=b## and ##0## elsewhere. It is a set function which picks a special ##a## out of ##A## and shouts "Got it!" by returning a ##1##.

    ##j## is then the cartesian product of all ##j_a##, i.e. ##j = \oplus_{a \in A} j_a##, a function ##j: A → \{0,1\}^A ⊆ ℤ^{A}##. It is also a set function only that here you can pick any element of ##A##.
    If you want to define it as the general function ##j :A → F^{ab}(A)## then it would be ##j = \oplus_{a \in A}j_a \cdot a## and
    $$j(b) = \oplus_{a \in A}j_a(b) \cdot a = \oplus_{a \in A}\delta_{ab} \cdot a = 1 \cdot b = b \in F^{ab}(A)$$

    Unfortunately Aluffi uses both versions and denotes them equally by ##j##.
    In 5.4. it's the natural embedding ##j : A → F^{ab}(A)## where you have to multiply the ##j_a## with an ##a \in A## for ##1 \in ℤ## isn't an element of ##F^{ab}(A)## and there is no order either,
    while in Prop. 5.6. he omits these multiplications by ##a## and ends up in ##ℤ^A## instead. Regarding that he exactly wants to prove that ##F^{ab}(A) ≅ ℤ^{\oplus A}## it's rather sloppy.

    Of course you could as well define the multiplication by an ##a \in A## as part of ##j_a## and omit it in the definition of ##j##. But in this case ##j_a## would be the natural embedding ##j_a : \{a\} → F^{ab}(A)## and not a mapping in ##ℤ##.

    The crucial part of the story in the book, however, is another. Although ##A## can have arbitrary many elements and the cartesian product will be respectively large, the elements of ##F^{ab}(A)## are all products ##a_1^{n_1} \dots a_N^{n_N}## of (arbitrary but) finite length ##N##.
    This is the difference in Aluffi's notation between ##ℤ^A## and ##ℤ^{\oplus A}## (see above with ##H## in the role of ##ℤ##).
     
    Last edited: May 14, 2016
  11. May 13, 2016 #10

    fresh_42

    Staff: Mentor

    This is something that makes me wonder, too. For a representation like this, full of categorial context, Aluffi is rather sloppy in his notations. A fact that doesn't work very well on category theory. And I seriously doubt that there are more than five lines needed to explain a free (Abelian) group in an algebra textbook.
     
  12. May 14, 2016 #11
     
  13. May 14, 2016 #12
    Hi mathwonk ... look I know you are essentially trying to help ... but I think when you write:

    " ... your questions make it obvious that you are not understanding anything in this book. ... "

    I think you are being somewhat negative, de-motivating and overly harsh in your judgement ...

    Peter
     
    Last edited: May 14, 2016
  14. May 14, 2016 #13

    Thanks Fresh_42 ... most helpful ... appreciate the help ...

    Peter
     
  15. May 14, 2016 #14

    fresh_42

    Staff: Mentor

    You're welcome!
     
  16. May 14, 2016 #15
    Hi Fresh_42, mathwonk,

    What do you think of Dummit and Foote: Abstract Algebra as an alternative to Aluffi ...

    Peter
     
  17. May 14, 2016 #16

    fresh_42

    Staff: Mentor

    A note on the diagram in 5.4.

    The free abelian group ##F^{ab}(A)## is basically the set of all words over an alphabet ##A## plus commutativity and formal inverse. (The neutral element can be seen as empty word.)
    It has no additional structure other than neutral element, (formal) inverse, associativity and commutativity. No rules like ##a^2=1## which reflections have, or ##a^3=1## for rotations of 120°.
    ##j## is the embedding of the alphabet in it, the one-letter-words.
    Now you can have any abelian group ##G## with an additional structure (e.g. rotations, reflections) and a mapping ##f : A → G##.
    Then there is a group homomorphism ##φ: F^{ab}(A) → G## which puts the structure upon ##F^{ab}(A)## and gets ##G##.
     
  18. May 14, 2016 #17

    Thanks fresh_42 ... yes ... do understand that ... but appreciate the further help ...

    Peter
     
  19. May 14, 2016 #18

    fresh_42

    Staff: Mentor

    I don't know it. Perhaps I got driven to van der Waerden because I knew the girl who typed the German version :smile:. But I never regretted it and I still use it to look things up which I have forgotten in detail. But it's more about groups, fields and numbers and practically nothing homological.
     
  20. May 14, 2016 #19
    Thanks anyway fresh_42 ... I did check van der Waerden at Amazon ... seems it is a classic ... but may be a bit dated ..

    Peter
     
  21. May 14, 2016 #20

    fresh_42

    Staff: Mentor

    It arose from lectures by Artin and Emmy Noether. What a pedigree!
     
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