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Stream: learning: questions

Topic: Module with finite bases of different sizes?

David Tanzer (Feb 29 2024 at 07:07):

Can a module have two bases that are finite, but with different sizes? If so what's a basic example?

Mike Shulman (Feb 29 2024 at 07:17):

Are you perhaps looking for the invariant basis number property of a ring?

David Tanzer (Feb 29 2024 at 14:26):

Yes, that hits the nail on the head. Thanks!

David Egolf (Feb 29 2024 at 18:08):

To give a specific example of a module with bases of different finite size:

Let $V$ be a vector space over $F$ with a countably infinite basis $B = \{ b_0, b_1, \dots \}$. Let $\mathcal{L}(V)$ be the ring of linear operators on $V$. Observe that $\mathcal{L}(V)$ is not commutative, as composition of functions is not commutative.

The ring $\mathcal{L}(V)$ is an $\mathcal{L}(V)$-module and as such, the identity map $i$ forms a basis for $\mathcal{L}(V)$. However, it is also possible to construct a basis for $\mathcal{L}(V)$ of any desired finite size $n$.

These quotes are from page 129 of "Advanced Linear Algebra" by Roman.

David Egolf (Feb 29 2024 at 18:08):

The author goes on to construct a basis for $\mathcal{L}(V)$ that has two elements. They do this by defining two linear operators $\beta_1:V \to V$ and $\beta_2: V \to V$. They are defined by: $\beta_1(b_{2k}) =b_k$ and $\beta_1(b_{2k+1})=0$; and $\beta_2(b_{2k}) =0$ and $\beta_2(b_{2k+1})=b_k$.

David Tanzer (Feb 29 2024 at 18:52):

Nice, thanks

Madeleine Birchfield (Feb 29 2024 at 20:01):

David Tanzer said:

Can a module have two bases that are finite, but with different sizes? If so what's a basic example?

I believe the trivial ring $0$, regarded as a module over $0$, has a basis of cardinality $n$ for every natural number $n$, since $0^n$ is isomorphic to $0$ for all natural numbers $n$.

Brendan Murphy (Feb 29 2024 at 20:09):

I don't see how you're getting a basis of cardinality n, given that 0^n has cardinality 1

Brendan Murphy (Feb 29 2024 at 20:09):

I would think every module over 0 has a unique basis, the empty set

Brendan Murphy (Feb 29 2024 at 20:10):

But I'm not confident in that assertion. It's such a degenerate case

Brendan Murphy (Feb 29 2024 at 20:16):

I guess the issue is that over the zero ring indexed bases behave very differently than over other rings, because {0} is linearly independent. What I said isn't right though, there are two (non indexed) bases for any module: the empty basis and the one element basis

Morgan Rogers (he/him) (Feb 29 2024 at 20:26):

Nope, what you said before was right @Brendan Murphy . The definition of linear dependence is the existence of non-zero coefficients weighting a sum to zero, so $\{0\}$ is always linearly dependent (and hence not a basis)

Brendan Murphy (Feb 29 2024 at 20:27):

But there are no nonzero coefficients!

Morgan Rogers (he/him) (Feb 29 2024 at 20:28):

Oh good point!

Morgan Rogers (he/him) (Feb 29 2024 at 20:29):

Ha so if you allow bases to be multisets then you do get arbitrarily large cardinality after all

Mike Shulman (Feb 29 2024 at 20:36):

The invariant basis number property is stated using the categorically-correct notion of "basis" for a module $M$, namely a set $X$ and an isomorphism from $M$ to the free module generated by $X$. In this sense, the zero ring does not have the IBN, since the free module on any set is isomorphic to the free module on any other set. But it's confusing because over any nonzero ring, any set $X$ injects into the free module it generates, and thus a basis for a module can be identified with a particular subset of that module.

David Tanzer (Mar 01 2024 at 04:02):

The wikipedia article on the IBN property gives a nice example of a non-IBN ring. For a ring $R$, let $R'$ be the ring of countably infinite matrices with entries in $R$ and columns with finite support. View $R'$ as a left $R'$ module over itself.

So $\{I\}$ is a basis for $R'$.

Then define f(m) = (even columns of m, odd columns of m); $f$ is an isomorphism from $R'$ to $(R')^2$. Turning this into a basis construction...

Let $R'_1$ be the subspace of matrices with zeros in the even columns, and $R'_2$ be the matrices with zeroes in the odd columns. Let $\sigma_1: R' \to R_1'$ be the injection inserting zeros into the even columns: $\sigma_1(c_0, c_1, c_2, ...) = (0, c_0, 0, c_1, 0, c_2, ...)$, and $\sigma_2: R' \to R_2'$ be the injection inserting zeros into the odd columns.

Let $I_1 = \sigma_1(I), I_2 = \sigma_2(I)$. Then $\{I_1, I_2\}$ is also a basis for $R'$.

David Tanzer (Mar 01 2024 at 04:03):

For $v \in R'$, its coordinates are $(v I_1, v I_2)$.

John Baez (Mar 01 2024 at 07:18):

Nice! Wikipedia points out that in this case we get $(R')^2 \cong R'$ as left $R'$-modules and thus

$(R')^n \cong (R')^m$

as left $R'$-modules for all $m, n \ge 1$.

John Baez (Mar 01 2024 at 07:19):

But intriguingly they mention there are examples of rings $R$ where

$R^n \cong R^m$

as left $R$-modules for some $m, n \ge 1$ with $m \ne n$, but not all!

John Baez (Mar 01 2024 at 07:23):

This sounds like a much more exotic and subtle phenomenon. They say that "Leavitt algebras" give examples, but I have no idea what a Leavitt algebra is, and they only point to this:

• Abrams, Gene; Ánh, P. N. (2002), "Some ultramatricial algebras which arise as intersections of Leavitt algebras", J. Algebra Appl. 1 (4): 357–363.

Mike Shulman (Mar 01 2024 at 16:06):

I wonder which partitions of the natural numbers are obtainable as the set of sets of ranks of free modules that coincide for some ring.

David Tanzer (Mar 01 2024 at 16:51):

Re: Leavitt algebras, here's a reference on Leavitt path algebras, which generalize Leavitt algebras:

• Leavitt path algebra, Wikipedia.

a Leavitt path algebra is a universal algebra constructed from a directed graph. Leavitt path algebras generalize Leavitt algebras and may be considered as algebraic analogues of graph C*-algebras.

For a fixed field $K$, and directed graph $E$, the article describes some construction of $L_K(E)$, the Leavitt path algebra for $E$, which is a $K$-algebra. There is a table showing certain cases which have been worked out.

• For the graph with one vertex and zero edges, the $K$-algebra is $K$ itself.
• For the graph with one vertex and one self-edge, you get the Laurent polynomials $K[x, x^{-1}]$.
• For a linear graph with $n-1$ edges on $n$ verticies, you get the $n \times n$ matrices with entries in $K$.
• For the flower-like graph with one vertex and $n$ self loops, you get the "Leavitt algebra" $L_K(n)$. I haven't found a direct reference on Leavitt algebras, but presumably this defines them.

There are some correspondences stated between graph and algebraic properties, including:

• Finite and acyclic graph corresponds to finite-dimensional Leavitt path algebra
• Finite vertex set corresponds to $L_K(E)$ being unital

The construction itself looks intricate, haven't grokked it yet. Also not clear how this could be brought to bear on the point John quoted from the article on the IBN property.

David Tanzer (Mar 01 2024 at 17:00):

• R. Hazrat, K. Rangaswamy, Leavitt algebras versus Leavitt path algebras

John Baez (Mar 01 2024 at 17:22):

David Tanzer said:

Re: Leavitt algebras, here's a reference on Leavitt path algebras, which generalize Leavitt algebras:

For a fixed field $K$, and directed graph $E$, the article describes some construction of $L_K(E)$, the Leavitt path algebra for $E$, which is a $K$-algebra. There is a table showing certain cases which have been worked out.

• For the graph with one vertex and zero edges, the $K$-algebra is $K$ itself.
• For the graph with one vertex and one self-edge, you get the Laurent polynomials $K[x, x^{-1}]$.
• For a linear graph with $n-1$ edges on $n$ vertices, you get the $n \times n$ matrices with entries in $K$.

John Baez (Mar 01 2024 at 17:22):

In these three examples what we're getting is just the usual path algebra, so I'm wondering if "Leavitt path algebra" is a synonym for "path algebra". The idea is that given a directed graph you form an algebra whose basis consists of all directed paths in that graph. Given directed paths $\gamma$ and $\delta$, if $\delta$ starts where $\gamma$ ends we define the product $\gamma \delta$ to be the usual composite path: the path where you first go along $\gamma$ and then $\delta$. If $\delta$ doesn't start where $\gamma$ ends we set $\gamma \delta = 0$ because... what else can we do?

This completely describes multiplication in the path algebra, since to describe an algebra you just need to say how to multiply basis elements and check associativity.

John Baez (Mar 01 2024 at 17:51):

Okay, now I looked at the Wikipedia article and see that yes, the description there is indeed truly horrible!

John Baez (Mar 01 2024 at 17:51):

I can't even bear to understand it.

John Baez (Mar 01 2024 at 17:54):

However, my guess is that it's a bit like the graph algebra I described, but it's a $\ast$-algebra where for each path $\gamma$ we get another basis element $\gamma^\ast$ which we think of as the "reverse" path.... and they seem to include some extra relations as well.

James Deikun (Mar 01 2024 at 18:03):

There's an idempotent for each vertex, the product of the idempotents for two different vertices is 0, the product of going "backward then forward" is the idempotent for the vertex you started on, but if you go "forward then backward" you have to add up all the possible ways of doing that to get the same thing ... but the relation only holds when it "makes sense" (the sum is finite and nonempty).

James Deikun (Mar 01 2024 at 18:04):

(all the possible ways of doing it in one step anyways)

James Deikun (Mar 01 2024 at 18:08):

As a result, the basis elements are paths that go forward then backward, and where you "turn around" there has to be another choice besides going straight back where you came, even if you didn't take it. That's why you get things like the Laurent algebra and the matrix algebra as examples--the graphs for those examples are thin so you can only trivially go forward or back and never turn around in the middle. The proper Leavitt algebras are generally not thin though.

James Deikun (Mar 01 2024 at 18:13):

The ordinary Leavitt algebra $L_K(n)$ has $n$ generators $e_k$ each of which has a left-sided inverse and $\sum_k e^{\;}_k e^{*}_k = 1$. As for $e^{*}_i e^{\;}_j$ for $i \ne j$ it's $0$.

James Deikun (Mar 01 2024 at 18:21):

The $e^{\;}_k e^{*}_k$ form an orthogonal partition of unity which is probably what makes these algebras useful for describing modules with finite bases of different sizes.

David Tanzer (Mar 01 2024 at 19:34):

That's making more sense - thanks. I see that after the user-unfriendly definition, there is a nice redefinition:

...one can show that $L_{K}(E)=\operatorname {span} _{K}\{s_{\alpha }s_{\beta }^{*}:\alpha {\text{ and }}\beta {\text{ are paths in }}E\}.$

Xuanrui Qi (Mar 04 2024 at 09:22):

Note however, that all commutative rings have the IBN property, so say in commutative algebra people just take it for granted.

Madeleine Birchfield (Mar 04 2024 at 16:27):

Xuanrui Qi said:

Note however, that all commutative rings have the IBN property, so say in commutative algebra people just take it for granted.

All non-trivial commutative rings have the IBN property. Earlier in this thread there is a discussion on how the trivial ring doesn't have the IBN property.

Xuanrui Qi (Mar 05 2024 at 05:16):

Yeah, my bad.