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Stream: theory: category theory

Topic: basis

Jean-Baptiste Vienney (Oct 24 2022 at 20:30):

In the category of vector spaces or of modules over a commutative ring, I could define a basis of an object $E$ as a family of maps $(x_{i}:R \rightarrow E)_{I \in I}$ such that for every object $F$ and for every family $(y_{i}:R \rightarrow F)_{i \in I}$ there exists a unique map $f:E \rightarrow F$ such that for every $i \in I$, $x_{i};f = y_{i}$. I can define such a basis of an object in every monoidal category. Do you think we can prove a few things about basis in such a generality? For instance, it is maybe possible to prove that objects admit a unique dimension under some conditions. Maybe in a symmetric monoidal category with biproducts? Or does a known categorical concept similar to this notion exist? If we have biproducts indexed on $I$, I guess it's equivalent to give a basis of $E$ or an isomorphism $E \cong \underset{i \in I}{\bigoplus} R^{i}$ (where $R$ is the monoidal unit).

Fabrizio Genovese (Oct 24 2022 at 21:09):

In general monoidal cats I'd expect to be able to say very little, as many algebraic structures for which the concept of basis makes little sense form monoidal cats (e.g. semirings, lattices, ...) I'd frame the question from a different angle: What properties do vectors spaces/modules over a commutative rings that make the concept of basis 'nice'? Then I'd try to abstract these properties to the categorical setting to pinpoint the definitions I need to make things work

Jean-Baptiste Vienney (Oct 24 2022 at 21:31):

We could look at the symmetric monoidal categories with biproducts. Let's write $1$ the monoidal unit. In this case we know that $\mathcal{C}[1,1]$ is a commutative semiring and that every $\mathcal{C}[A,B]$ is a $\mathcal{C}[1,1]$-semimodule. Thus, let's write $R := \mathcal{C}[1,1]$ and define the basis as before. In the category $Rel$, the monoidal unit is the one element set $*$. We find $Rel[*,*] \cong \mathbb{B}=\{0,1\}$, the boolean algebra with only false and true, addition given by or and multiplication equal to and. Clearly this $R$ looks a better choice than the monoidal unit. Do you first know about basis of semimodules over a commutative semiring? I don't even know if the notion of dimension works here, maybe not. Without the negatives, it looks harder. I would like a notion of dimension as a nice property... but I would prefer it to be a property rather than a part of the definition.

John Baez (Oct 24 2022 at 21:46):

The concept of basis doesn't even work well for modules over a commutative ring, unless the module is free, which essentially means it has a basis.

Jean-Baptiste Vienney (Oct 24 2022 at 21:46):

Yes, okay but I can define the notion of basis and say free module = module with a basis. Maybe we can do this in other cases.

Jean-Baptiste Vienney (Oct 24 2022 at 21:49):

No, it's maybe better to take $R$ equal to the monoidal unit. In the case of $Rel$, then every set admits a basis given by... the element of the set, I think

Jean-Baptiste Vienney (Oct 24 2022 at 21:50):

So there seems to be a dimension equal to the cardinality of the set.

John Baez (Oct 24 2022 at 21:51):

Let $R\mathsf{Mod}$ be the category of semimodules over the rig R. There's a forgetful functor $U: R\mathsf{Mod} \to \mathsf{Set}$, and this has a left adjoint $F : \mathsf{Set} \to R\mathsf{Mod}$. Say an R-module is free if it's of the form $FS$ for some set $S$. There's an inclusion $S \to UFS$ and elements in the image form a 'basis' of $FS$.

John Baez (Oct 24 2022 at 21:53):

You can describe a morphism between free semimodules in terms of a matrix with entries in $R$ indexed by their bases, unless I'm confused.

It works best for finitely generated free modules, just as in case where $R$ is a field, since finite coproducts are actually biproducts, so any $S \times T$ matrix of elements of $R$ determines (and is determined by) an $R$-semimodule homomorphism from $FT$ to $FS$.

Jean-Baptiste Vienney (Oct 24 2022 at 21:56):

Do we have the notion of dimension in $RMod$?

John Baez (Oct 24 2022 at 21:58):

That's a somewhat fuzzy question. Maybe you're asking: given an $R$-module that's isomorphic to $FS$ and also $FS'$, must we have $S \cong S'$? I.e., given two bases of a free module, must they have the same cardinality?

Jean-Baptiste Vienney (Oct 24 2022 at 21:59):

Yes that's it

John Baez (Oct 24 2022 at 22:02):

Okay! Already for rings $R$, the answer to my question is "sometimes yes, sometimes no". If the answer to my question is yes for all finite $S, S'$ then we say the ring has the invariant basis number property. Here are some examples of rings that do or do not have this property:

• Wikipedia, Invariant basis number.

Jean-Baptiste Vienney (Oct 24 2022 at 22:03):

It looks like we could use the same word for semirings!

John Baez (Oct 24 2022 at 22:04):

I don't know much about this stuff, so I'll just quote them:

• every commutative ring has the invariant basis number property except the 1-element ring.

• every division ring (= ring where every nonzero element has a two-sided inverse) has the invariant basis number property.

Jean-Baptiste Vienney (Oct 24 2022 at 22:06):

That's a good question to look whether commutative semirings have the invariant basis number...

Jean-Baptiste Vienney (Oct 24 2022 at 22:07):

I guess it must be known

Jean-Baptiste Vienney (Oct 24 2022 at 22:07):

Probably in the books on semirings

John Baez (Oct 24 2022 at 22:07):

Yes, that's a good algebra question! Since I don't know the proof for commutative rings I won't try to guess whether it works for semirings.

John Baez (Oct 24 2022 at 22:08):

Are there books about semimodules of semirings?

John Baez (Oct 24 2022 at 22:08):

There should be, but I haven't seen them.

Jean-Baptiste Vienney (Oct 24 2022 at 22:10):

In "Semirings and their Applications" by Golan, there are sections on semimodules

Jean-Baptiste Vienney (Oct 24 2022 at 22:10):

But I've found the answer in a paper

Jean-Baptiste Vienney (Oct 24 2022 at 22:11):

Corollary 5.2 of https://arxiv.org/pdf/1711.05163.pdf

Jean-Baptiste Vienney (Oct 24 2022 at 22:11):

"Corollary 5.2. Division semirings and commutative semirings satisfy IBN."

John Baez (Oct 24 2022 at 22:12):

Nice! So both the examples I mentioned carry over to semirings!

Jean-Baptiste Vienney (Oct 24 2022 at 22:12):

Yes, exactly!

Jean-Baptiste Vienney (Oct 24 2022 at 22:13):

I don't know a lot about representation theory. Could it make sense to speak of a basis of a finite-dimensional representation of a group?

John Baez (Oct 24 2022 at 22:14):

Since a representation of a group is a representation on a vector space, if you said this people would think you meant a basis of that vector space.

John Baez (Oct 24 2022 at 22:14):

Lemma 5.1 in that paper is surprising to me.

Jean-Baptiste Vienney (Oct 24 2022 at 22:16):

That's look a very weak condition to have a morphism from $S$ to $T$ for transmitting the IBN from $T$ to $S$ indeed...

Jean-Baptiste Vienney (Oct 24 2022 at 22:19):

So if we have a terminal object, if it has the IBN, every object also, but $0$ is a terminal object in the category of semimodules over any semiring and has the IBN so every semimodules also??? Is the lemma or me wrong?

Jean-Baptiste Vienney (Oct 24 2022 at 22:23):

But I don't know what $R^{0}$ could mean. On wikipedia they require the IBN to be $\ge 1$

John Baez (Oct 24 2022 at 22:27):

Why are you talking about the terminal object in the category of semimodules?

John Baez (Oct 24 2022 at 22:27):

They're saying if $\phi : S \to T$ is a homomorphism of semirings then if $T$ has the invariant basis number property then so does $S$.

John Baez (Oct 24 2022 at 22:28):

So, if the terminal semiring had the invariant basis number property, all semirings would have it. But it doesn't!

John Baez (Oct 24 2022 at 22:30):

Indeed the terminal ring, the 1-element semiring, is about the furthest you could imagine from having the invariant basis number property, because its 1-element semimodule is free on any number of generators!

Jean-Baptiste Vienney (Oct 24 2022 at 22:36):

Oh sorry, I confused the semirings and the modules. Okay, I understand what you say.

Jean-Baptiste Vienney (Oct 24 2022 at 22:42):

We could say that a monoidal category with unit $I$ and biproducts has the IBN if all objects isomorphic to $I^{\oplus n}$ for some $n \ge 0$ (we put $I^{\oplus 0} = 0$) are like this for only one $n$.

Jean-Baptiste Vienney (Oct 24 2022 at 22:43):

So the category of sets and relations has the IBN as well as semimodules over a commutative semiring or division semiring

Jean-Baptiste Vienney (Oct 24 2022 at 22:46):

I guess it should work for representations of algebraic things and vector bundles over spatial things in such semimodules if it makes sense

John Baez (Oct 24 2022 at 22:49):

Btw, notice that a relation between finite sets $S$ and $T$ is the same as an $R$-module homomorphism between $FS$ and $FT$, where $R$ is the 2-element rig $\{F,T\}$ and $F$ is the free $R$-module functor.

Jean-Baptiste Vienney (Oct 24 2022 at 22:54):

I wasn't sure if $Rel$ has the IBN finally, I was thinking to $Set$. But so it must work because $\{F,T\}$ is commutative

Jean-Baptiste Vienney (Oct 24 2022 at 22:55):

I'm not sure what it is an iso in $Rel$. And the biproducts are different also. Oh sorry there isn't biproducts in $Set$

Jean-Baptiste Vienney (Oct 24 2022 at 22:59):

Oh okay, isos in $Rel$ are just bijections so the IBN is just the number of elements of the set

Martti Karvonen (Oct 25 2022 at 02:25):

I wonder how this works for infinite-dimensional (semi)modules over (semi)rings. If a $R$-module $M$ is free over an infinite set $X$ with $\left\vert X\right\vert>\left\vert R\right\vert$, then the size of a basis of $M$ is unique by cardinality reasons. However, I don't know what could happen when $\left\vert X\right\vert\leq \left\vert R\right\vert$. Is asking for uniqueness of size of basis in this case a stronger property than just the invariant basis number property?

Oscar Cunningham (Oct 25 2022 at 10:19):

The definition of adjunction in terms of universal morphisms (https://en.wikipedia.org/wiki/Adjoint_functors#Definition_via_universal_morphisms) works objectwise. So you can say what it means to have a basis even if you don't have an actual adjoint to the forgetful functor. Given a monoidal category $\mathcal{C}$ and an object $C\in\mathcal{C}$ a basis of $C$ is a set $X$ and a function $e:X\to\mathcal{C}(1,C)$ such that for any $D\in\mathcal{C}$ and any morphism $f:X\to\mathcal{C}(1,D)$ there exists a unique morphism $g:C\to D$ such that $f = \mathcal{C}(1,g)\circ e$.

Oscar Cunningham (Oct 25 2022 at 10:59):

Oh, which is what Jean-Baptiste said in the first place.

John Baez (Oct 25 2022 at 11:31):

That sounds like a tricky question that people may have thought about for rings, since people have studied lots of subtle questions for modules of rings.

Jean-Baptiste Vienney (Oct 25 2022 at 12:31):

It works a little bit for a basis of a topological space but that's not very satisfying. Consider the category of open sets of a topological space $X$ where morphisms $U \rightarrow V$ are inclusions. A family $(U_{i})_{I \in I}$ of open sets is a basis of $X$ iff for every open set $V$, $\exists J \subseteq I$ and morphisms $(U_{j} \rightarrow V)_{j \in J}$ which form a basis of $V$ in the categorical sense ie. for every family $(U_{j} \rightarrow W)_{j \in J}$, there exists $V \rightarrow W$ (such that $U_{j} \rightarrow V \rightarrow W = U_{j} \rightarrow W$ but that's automatic).