# Abstract Nonsense

## Product of Modules

Point of Post: In this post we discuss the product of modules, including their characterizations via univeral mapping properties.

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Motivation

We now consider the product of modules, which as always, is just endowing the Cartesian product of a set of modules with operations that turn the resulting (set) product into a module. That said, while we have mentioned before that products of things can be characterized via certain universal mapping properties (e.g. for rings and groups) here we shall actually start with thinking of products in terms of these universal mapping properties and then define the “natural product”  only to prove existence of such modules. Why? What precisely is the point of doing? Well, we all have an intuitive idea about what products are, we have seen them as objects for much of our mathematical careers. That said, we understand them intuitively in a concrete “I can see them” sense. What we are now more currently interested is understanding them in a “how do they act” sense–what makes “products” products from the view of mappings. Well, this is precisely what the universal characterization of products tells us. Stated it says that “a product of the set $\left\{M_\alpha\right\}_{\alpha\in\mathcal{A}}$ of left $R$-modules is a left $R$-module $M$ together with a set of maps $f_\alpha:M\to M_\alpha$ with the following property: given any left $R$-module $N$ and maps $g_\alpha:N\to M_\alpha$ there exists a unique map $g:N\to M$ such that $f_\alpha\circ g=g_\alpha$.” Ok, so fine, but what is this big-long-horrible definition really telling us about this $M$? The existence of these $f_\alpha$‘s tell us that $M$ is “put together” in some way via the $M_\alpha$‘s. The fact that any map $N\to M$ is determined by its values on $M_\alpha$ (translated via the “put together maps”, $f_\alpha$) tells us that $M$ is put together in a fairly minimal way–i.e. this $M$ isn’t put together in such a way that there is much “room outside the $M_\alpha$s” for movement. That said, we see that the $M_\alpha$‘s are living inside of $M$ in a fairly faithful manner, in the sense that a functions values on each of the $M_\alpha$s are independent of one another–this tells intuitively that we didn’t squish the $M_\alpha$‘s, or that we didn’t underepresent any of the $M_\alpha$s at any stage of construction. Thus, with this in mind, it’s clear why we’d consider products of modules.

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November 11, 2011

## Projections Into the Group Algebra (Pt. II)

Point of Post: This post is a continuation of this one.

April 9, 2011

## Projections Into the Group Algebra (Pt. I)

Point of post: In this post we discuss the notion of the group algebra, in preparation for our eventual discussion about the representation of symmetric groups.

Motivation

We’ve seen in past posts that the group algebra is isomorphic in all the important ways to the direct sum of matrix algebras. We’ll use this fact to study projections in the group algebra which are functions generalizing the notion of projections on an endomorphism algebra. Namely, projections are elements of the group algebra which are idempotent under convolution. These shall prove to be very important when we attempt, at a later date, to classify the representations of the symmetric group.

April 9, 2011

## Projections (Pt. III)

Point of post: This is a continuation of this post.

January 13, 2011

## Projections (Pt. II)

Point of post: This is a continuation of this post.

January 13, 2011

## Projections (Pt. I)

Point of post: In this post we discuss the concepts of projections onto subspaces of a vector space.

Motivation

It makes sense that if the internal direct sum of subspaces of a vector space should have any significance in the overall theory, then so should the projections onto theses subspaces. The idea of a projection is that if $\mathscr{V}=\mathscr{U}\oplus\mathscr{W}$ with $\mathscr{U},\mathscr{W}\leqslant \mathscr{V}$ then for every $v\in\mathscr{V}$ the projection along $\mathscr{U}$ should be a function which takes in $v$ and gives back the $\mathscr{U}$ part.

January 13, 2011