Ordered Index Sets for Direct Sums & Products of Modules: Explained by B&K

[Math Amateur]

I am reading An Introduction to Rings and Modules With K-Theory in View by A.J. Berrick and M.E. Keating (B&K).

In Chapter2: Direct Sums and Short Exact Sequences in Sections 2.1.14 and 2.1.15 B&K deal with ordered index sets in the context of direct sums and products of modules.

Although it seems a quite simple concept at first glance, I find the motivation for, the meaning and the usefulness of the concept of ordered index set so bewildering I cannot really formulate a meaningful question ... ... but if anyone can begin to explain the motivation for, the meaning and the usefulness of the concept in the context of direct sums and products of modules, I would much appreciate it ... ...

B&K explain ordered index sets and their use in constructing the direct sum or product of copies of a module as follows:https://www.physicsforums.com/attachments/3361
View attachment 3362
View attachment 3363To try to explain my puzzlement with this notion ... ... I had previously thought that when we talk (without considering ordering) of a direct product say:

\(\displaystyle \prod^{k}_{i = 1} M_i \)

that we write the elements as:

\(\displaystyle (m_1, m_2, \ ... \ ... \ m_k) \)

... ... obviously, arbitrarily choosing a particular order ... ... and, indeed, we do not care about the order since all orderings give isomorphic products ... so why do we need ordered sets ... ? If we really accept there is NO ordering then how do we write down the elements ... ... ?

In the infinite cases (using unordered index sets) B&K's definitions of direct product and direct sum are as follows:View attachment 3364

Given that we are dealing with an unordered set, how do we interpret \(\displaystyle (\ell_i)\) - which looks like a sequence - ... ... well, presumably as an unordered collection ... is that right?Can someone clarify the above for me ... Peter
[Note: maybe I am overthinking it and it is quite simple ... but what is the point of it all? ... ... what do we gain from making such a distinction in this context between unordered and ordered index sets? ... what are the benefits from such a distinction? ... ... ]

 

[Evgeny.Makarov]

In Chapter2: Direct Sums and Short Exact Sequences in Sections 2.1.14 and 2.1.15 B&K deal with ordered index sets in the context of direct sums and products of modules.

Although it seems a quite simple concept at first glance, I find the motivation for, the meaning and the usefulness of the concept of ordered index set so bewildering I cannot really formulate a meaningful question ... ... but if anyone can begin to explain the motivation for, the meaning and the usefulness of the concept in the context of direct sums and products of modules, I would much appreciate it

This is a quite candid admission by the authors! (Smile) This must be an awesome book!

The cartesian product $\prod_I L_i$ consists of elements $(\ell_i)$ where each $\ell_i\in L_i$. This expression $(\ell_i)$ by definition means a function $f:I\to\bigcup_{i\in I} L_i$ such that $f(i)\in L_i$ for all $i\in I$. So, a direct sum consists of all such functions.

The inconvenience here is that there is no way to refer to a particular component of $(\ell_i)$ without specifying its index $i$. Imagine that you have a dossier for each of your employees. Thus, you have a function that maps people to folders. In order to tell the secretary to give you a certain folder, you need to identify the corresponding person, either by name, by picture or by pointing at him/her.

If $I$ is well-ordered, then it is order isomorphic to an ordinal, which means that elements of $I$ can be identified with ordinals. In the simplest case, when $I$ is finite or has order type $\omega$, elements of $I$ can be identified with natural numbers. Suppose that $g:\Bbb N\to I$ is such identification. Then instead of $f:I\to\bigcup_{i\in I} L_i$ you can consider $f\circ g:\Bbb N\to\bigcup_{i\in I} L_i$. The result is that each component of $(\ell_i)$ can be identified simply by a number. If you order your employees, say, alphabetically by name, then you can simply ask the secretary to give you the fifth folder.

 

[Math Amateur]

This is a quite candid admission by the authors! (Smile) This must be an awesome book!

The cartesian product $\prod_I L_i$ consists of elements $(\ell_i)$ where each $\ell_i\in L_i$. This expression $(\ell_i)$ by definition means a function $f:I\to\bigcup_{i\in I} L_i$ such that $f(i)\in L_i$ for all $i\in I$. So, a direct sum consists of all such functions.

The inconvenience here is that there is no way to refer to a particular component of $(\ell_i)$ without specifying its index $i$. Imagine that you have a dossier for each of your employees. Thus, you have a function that maps people to folders. In order to tell the secretary to give you a certain folder, you need to identify the corresponding person, either by name, by picture or by pointing at him/her.

If $I$ is well-ordered, then it is order isomorphic to an ordinal, which means that elements of $I$ can be identified with ordinals. In the simplest case, when $I$ is finite or has order type $\omega$, elements of $I$ can be identified with natural numbers. Suppose that $g:\Bbb N\to I$ is such identification. Then instead of $f:I\to\bigcup_{i\in I} L_i$ you can consider $f\circ g:\Bbb N\to\bigcup_{i\in I} L_i$. The result is that each component of $(\ell_i)$ can be identified simply by a number. If you order your employees, say, alphabetically by name, then you can simply ask the secretary to give you the fifth folder.

Thanks Evgeny ... Yes, the authors seem quite puzzled, don't they ...

Indeed, when explaining direct sums and products the authors do not explain or use your function $f:I\to\bigcup_{i\in I} L_i$ ... are you taking us into the wilds of category theory :) ... ... do you know of a book that gives more detail on what you have written plus some examples of how this works ... ?

Thanks again for your help ... still reflecting on what you have written ...

Peter

 

[Deveno]

The simple notation $(\ell_i)$ sort of "disguises" what it IS.

You are cautioned NOT to think of this as a SEQUENCE (that is: $i \in \Bbb N^+$), although this is ONE possibility.

Rather, $(\ell_i)$ indicates we have a FUNCTION:

$\displaystyle f: I \to L = \bigcup_{i \in I} L_i$ such that $f(i) \in L_i$.

Often (the most commonly-encountered case), all the "components" (the $L_i$) are in fact, the same set. For example, this is the case for $\Bbb R^n$, where a vector:

$v = (v_1,v_2,\dots,v_n)$ or $n$-tuple is actually a function:

$f:\{1,2,\dots,n\} \to \Bbb R$. For example, in $\Bbb R^3$, the triple: (1,-4,2) is the function:

f(1) = 1
f(2) = -4
f(3) = 2

We typically use the ordering of the natural numbers to induce a standard order on $\{1,2,3\}$ which might be loosely described as:

"first, second, third".

The ordering is implicitly referenced when we say "-4 is the second coordinate of (1,-4,2)".

However, you have to remember that $I$ (the indexing set) can be VERY LARGE (like the real numbers). If I ask you to bring me "the 3,067-th real number", I have very little confidence I will actually get the real number I am thinking about.

It may not be obvious, but this process "breaks down" for some larger sets. While the axiom of choice says any set has a well-order, actually explicitly DESCRIBING such an order for large sets is more of an abstraction, than a reality. If you get "infinite enough" you start to get into mathematics "woo-woo" land, where concrete examples are difficult to even describe, much less illustrate.

 

[Math Amateur]

The simple notation $(\ell_i)$ sort of "disguises" what it IS.

You are cautioned NOT to think of this as a SEQUENCE (that is: $i \in \Bbb N^+$), although this is ONE possibility.

Rather, $(\ell_i)$ indicates we have a FUNCTION:

$\displaystyle f: I \to L = \bigcup_{i \in I} L_i$ such that $f(i) \in L_i$.

Often (the most commonly-encountered case), all the "components" (the $L_i$) are in fact, the same set. For example, this is the case for $\Bbb R^n$, where a vector:

$v = (v_1,v_2,\dots,v_n)$ or $n$-tuple is actually a function:

$f:\{1,2,\dots,n\} \to \Bbb R$. For example, in $\Bbb R^3$, the triple: (1,-4,2) is the function:

f(1) = 1
f(2) = -4
f(3) = 2

We typically use the ordering of the natural numbers to induce a standard order on $\{1,2,3\}$ which might be loosely described as:

"first, second, third".

The ordering is implicitly referenced when we say "-4 is the second coordinate of (1,-4,2)".

However, you have to remember that $I$ (the indexing set) can be VERY LARGE (like the real numbers). If I ask you to bring me "the 3,067-th real number", I have very little confidence I will actually get the real number I am thinking about.

It may not be obvious, but this process "breaks down" for some larger sets. While the axiom of choice says any set has a well-order, actually explicitly DESCRIBING such an order for large sets is more of an abstraction, than a reality. If you get "infinite enough" you start to get into mathematics "woo-woo" land, where concrete examples are difficult to even describe, much less illustrate.

Thanks Evgeny and Deveno ... appreciate your help ...

I can see difficulties in choosing elements elements from uncountably infinite sets ... ...

When choosing elements from an arbitrary family of indexed sets there are, I think, difficulties associated with an uncountably infinite index? ... ... is that so?

Are there any difficulties at all when we have a finite index, but where some of the sets are uncountably infinite; - such as \(\displaystyle \mathbb{R}\) ... I suspect not, even though one can never list or sequence the members of the set ... ... is that correct?We are actually dealing with the Axiom of Choice and all the surrounding issues ... ... BUT ... aren't these issues solved for both totally ordered sets and unordered sets by acceptance of the Axiom of Choice? ...

Why are B&K making a big distinction in the section of text displayed above, between unordered and ordered sets ...

Indeed, B&K write ... ... (for example)

"A module M is the internal direct sum of an ordered set \(\displaystyle \{ M_\lambda \ | \lambda \in \Lambda \}\) of submodules of \(\displaystyle M\) if it is already the internal sum of of the corresponding unordered set \(\displaystyle \{ M_i \ | i \in I \}\). ... ... "

So what is the problem or issue here ... is it simply that we can regard any direct sum as the direct sum of an ordered set ...

Mind you that prompts the question ... do direct sums exist for sets of submodules where the set cannot be totally ordered ... ... I do not think that B&K answer this question ... ... but you (Deveno) are implying above that the index set needs to be totally or well-ordered ...

I note in passing that B&K refer to "ordered sequences" in the section of text below:https://www.physicsforums.com/attachments/3376

Given that, in the text directly above, B&K are talking about ordered sequences \(\displaystyle (\ell_\lambda)\) for infinite ordered collections of sets that means that with the previous definitions which did not specify an order for the index, we have to think of "unordered sequences"!

Peter

 

[Deveno]

The difference is subtle, but it does matter.

Suppose I tell Bob and Alice to draw a pair of perpendicular axes (you should recognize this as a pair of mutually orthogonal idempotents, consisting of the projections onto either axis) and to label one $x$ and one $y$.

Alice may choose the vertical axis to be $x$, while Bob chooses the horizontal axis. The mathematics they do, and the graphs they draw, will look very SIMILAR, but they will not be IDENTICAL, nor even "close to identical". Bob thinks all of Alice's pictures are "backwards", even when he rotates then a quarter-turn so their axes line up.

We can write: $\Bbb R^2 = \Bbb R \oplus \Bbb R$, but since both copies of $\Bbb R$ are the same, it's difficult to be PRECISE about specifying a a change along one axis only. In physical situations, "mixing up the order" of the axes (the summands of a direct sum of one-dimensional linear subspaces) can fail to preserve ORIENTATION, leading to "sign errors".

 

[Math Amateur]

The difference is subtle, but it does matter.

Suppose I tell Bob and Alice to draw a pair of perpendicular axes (you should recognize this as a pair of mutually orthogonal idempotents, consisting of the projections onto either axis) and to label one $x$ and one $y$.

Alice may choose the vertical axis to be $x$, while Bob chooses the horizontal axis. The mathematics they do, and the graphs they draw, will look very SIMILAR, but they will not be IDENTICAL, nor even "close to identical". Bob thinks all of Alice's pictures are "backwards", even when he rotates then a quarter-turn so their axes line up.

We can write: $\Bbb R^2 = \Bbb R \oplus \Bbb R$, but since both copies of $\Bbb R$ are the same, it's difficult to be PRECISE about specifying a a change along one axis only. In physical situations, "mixing up the order" of the axes (the summands of a direct sum of one-dimensional linear subspaces) can fail to preserve ORIENTATION, leading to "sign errors".

Hi Deveno and Evgeny,

Just a note related to some thinking and reading ...

It appears that instead of considering the set of all functions

$f:I\to\bigcup_{i\in I} L_i$

to define a direct product of an arbitrary family of modules, one can use projections, which seems more natural in the setting of modules ... ... indeed Paul Bland in Rings and Their Modules does this by describing the direct product of an indexed family of modules \(\displaystyle \{ M_\alpha \}_\Delta\)as the R-module \(\displaystyle \prod_\Delta M_\alpha \) together with the family of canonical projections ... ... as below ...
https://www.physicsforums.com/attachments/3377
In doing this I am taking it that Bland is being general, and indeed, obviating the need (in the definition and in most - all I think - of the following theorems) to distinguish as B&K do between ordered and unordered indexes ...

Following this, Bland gives a formal definition of a direct product of the family of modules in terms of a universal mapping property ... ... which seems to involve projections ... ... ... BUT ... ... ... presumably none of this obviates the need for the Axiom of Choice ... ... but where is it used in Bland's approach ... ... Any comments ... ... ?The relationship of Bland's approach and the Axiom of Choice seems to be directly covered in Chapter 1 of Bert Mendelson's book "Introduction to Topology" as follows:https://www.physicsforums.com/attachments/3378
https://www.physicsforums.com/attachments/3379
Any comments on Mendelson's analysis ... ... ?I do have one simple/minor technical/notational question as follows:

What is \(\displaystyle x_\alpha\) exactly ?? ... and how does it differ from \(\displaystyle x( \alpha )\) ... they seem the same to me ... so why do we need to construct a function \(\displaystyle x\) (?) by setting them equal ?

Can someone please explain this point ... it is preventing me from understanding Mendelson's analysis ...

Would appreciate some help ...

Peter

 

[Deveno]

This is no accident, the UMP of the projection maps determine a product (up to a unique isomorphism). This works is ANY category, but it is conceivable in SOME categories that no actual object and pair of maps satisfies this UMP. If, for any indexed family of objects at least one example of a prodcut DOES exist, then the category is said to possesses products.

So, for example, sets, groups, abelian groups, rings, $R$-modules, vector spaces, and topological spaces all possesses products.

There is no "real difference" between $\alpha(i)$ and $\alpha_i$, that's why it's called an INDEX set, the domain of the function $\alpha$ is the set we take the indices ("subscripts") from. Most of the time, the indexing set isn't all that interesting, and direct reference to it is suppressed.

The construction $\alpha: I \to \bigcup_i L_i$ is used mainly to show such a thing EXISTS, something guaranteed for an arbitrary set $I$ by the axiom of choice. For most "products" in a category (particularly those categories that are "sets with structure") the underlying set of a product is the cartesian product of the factors of the product.

This for many categories $\mathcal{C}$ defines a "forgetful functor":

$U: \mathcal{C} \to \mathbf{Set}$ such that:

$\displaystyle U\left(\prod_{i \in I} A_i\right) = \prod_{i \in I} U(A_i)$

(Caveat: not all categories are "sets with structure", and not all categories that possesses products admit such a forgetful (or "underlying set") functor).

When working with indexed sets (and thus products, or co-products), it is often desirable to have a specified "ordering" of the indices. There may be a "preferred" way we wish to embed the copies of each identical summand into a direct sum, for example, as we may be using some sort of inductive process to prove something. This is the case, for example, when one row-reduces a matrix into row-reduced echelon form, what one is doing is simultaneously finding a basis for the row-space and column-space of a matrix, ordering the rows and columns by the natural ordering of the natural numbers allows us to make linear independence of the rows/columns "transparent" (which is why it's called REDUCED).

"Permuting" the row/columns of said matrix doesn't change the linear independence, but putting a matrix in rref allows one to often make an easy inductive argument on the size of a square matrix. Because "ordered set" is more restrictive than "set", it often allows us to say more, give more explicit results, or (more importantly) define a definite ALGORITHM for achieving some aim.

On a basic level: the axiom of choice is equivalent to the statement:

The cartesian product of an indexed family of non-empty sets is non-empty.

In any structure that must possesses an identity element for a binary operation, such a structure cannot be formed from a non-empty set, so for the indexed product of a family of (structure)s to be a (structure) itself, the axiom of choice has to hold. Remember, zero is usually the "bad guy" so typically needs to be considered as a "limiting case".

In algebra, one does well to examine EVERY theorem in the following light:

a) Is it true for the simplest, or smallest, example possible?
b) Is it true for the most complicated, or largest, example possible?