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Stream: learning: questions

Topic: Understanding Set -> Vect

Eric Forgy (Jan 28 2021 at 21:40):

$\mathbf{Set}$ has an initial object, i.e. the empty set $\emptyset$, and has a terminal object, i.e. the set with one element $1.$

The functor
${}$
$K[-]: \mathbf{Set\to Vect}_K$
${}$
sends the initial object $\emptyset$ to the vector space $0$ with one vector $0$, i.e.
${}$
$K[\emptyset] = 0.$
${}$
However, $K[-]$ sends the terminal object to the vector space $K,$ i.e.
${}$
$K[1] = K.$
${}$
I recently learned that $K$ is not the terminal object of $\mathbf{Vect}_K$ (something I should have known, but never really thought about).

Both $\mathbf{Set}$ and $\mathbf{Vect}_K$ are bicomplete, but apparently the functor $K[-]$ does not preserve limits (otherwise $K$ would be terminal in $\mathbf{Vect}_K$) :thinking:

I think the pushout of a span
${}$
$A\leftarrow K\rightarrow B$
${}$
in $\mathbf{Vect}_K$ is a cospan with the usual tensor product of $A$ and $B$ as apex, i.e.
${}$
$A\rightarrow A\otimes_K B\leftarrow B.$
${}$
If so, what would be the meaning of the pushout of
${}$
$A\leftarrow 0\rightarrow B$
${}$
? It should be something like
${}$
$A\rightarrow A\otimes_0 B\leftarrow B$
${}$
but I'm struggling to interpret what that would mean.

Can someone help? :pray:

Ralph Sarkis (Jan 28 2021 at 22:02):

Because there is a single morphism $0 \rightarrow V$ for any vector space $V$ (the trivial one), whatever you put in the corner with two well typed morphisms will make the square commute. In other words, this span does not put any restriction on the cospan (except for the endpoints) which means the colimit of this span is simply a universal cospan for $A$ and $B$, i.e.: a coproduct (direct sum of vector spaces).

Conclusion : $A\otimes_0B = A \oplus B$

Simon Burton (Jan 28 2021 at 23:31):

Pushouts in Vect are not tensor products, these are (quotients of) direct sums.. as Ralph is saying.

Eric Forgy (Jan 29 2021 at 00:54):

Thank you Ralph and thank you Simon :pray:

I see how the push out of span with $0$ as apex is the direct sum of vector spaces.

I got it in my head that composition of cospans in $\mathbf{Vect}_K$ corresponded to tensor product of bimodules :thinking:

Now I need to figure out where I went astray :thinking:

John Baez (Jan 29 2021 at 02:50):

Eric Forgy said:

Both $\mathsf{Set}$ and $\mathsf{Vect}_K$ are bicomplete, but apparently the functor $K[-]$ does not preserve limits (otherwise $K$ would be terminal in $\mathbf{Vect}_K$) :thinking:

The functor $K[-]$ is a left adjoint: it's left adjoint to the forgetful functor $\mathsf{Vect}_K \to \mathsf{Set}$.

Left adjoints preserve all colimits. So in particular $K[-]$ preserves the initial object, and coproducts, and pushouts.

But it's very rare for them to preserve limits. And indeed $K[-]$ does not preserves the terminal object, or products, or pullbacks.

Right adjoints, on the other hand, preserve limits. So the forgetful functor $\mathsf{Vect}_K \to \mathsf{Set}$ works the other way around: it preserves terminal objects, and products, and pullbacks, but not initial objects, or coproducts, or pushouts.

All this stuff should become second nature as you get used to category theory.

John Baez (Jan 29 2021 at 02:55):

Eric Forgy said:

I think the pushout of a span
${}$
$A\leftarrow K\rightarrow B$
${}$
in $\mathbf{Vect}_K$ is a cospan with the usual tensor product of $A$ and $B$ as apex, i.e.
${}$
$A\rightarrow A\otimes_K B\leftarrow B$

You've already been informed of this, but: no.

Tensor products in $\mathsf{Vect}_K$ are not pushouts in $\mathsf{Vect}_K$; they are not any sort of limit or colimit in $\mathsf{Vect}_K$. They are something new.

what would be the meaning of the pushout of
${}$
$A\leftarrow 0\rightarrow B$ ?

It's just $A \oplus B$. Pushing out over the initial object is the same as taking the coproduct, always!

John Baez (Jan 29 2021 at 02:59):

Eric Forgy said:

I got it in my head that composition of cospans in $\mathbf{Vect}_K$ corresponded to tensor product of bimodules

No. I told you all this wonderful stuff about how composing spans in a cartesian monoidal category $(\mathsf{C}, \times)$ is the same as tensor product of bimodules in the cocartesian monoidal category $(\mathsf{C}^{\rm op}, \times)$, and now you're applying that to $(\mathsf{Vect}_K , \otimes)$, where the tensor product is not cocartesian. So you're getting nonsense.

John Baez (Jan 29 2021 at 03:01):

(Remember, in $(\mathsf{C}^{\rm op}, \times)$ I'm using $\times$ to refer to the product in $\mathsf{C}$, which is the coproduct in $\mathsf{C}^{\rm op}$.)

John Baez (Jan 29 2021 at 03:04):

The only way to get out of the mess you got yourself in is to back out - straight back.

Tensor product of vector spaces or modules of algebras is not any sort of pushout or pullback. I told you the way of thinking that actually works, which is to treat spans in $\mathsf{Set}$ as bimodules in $\mathsf{Set}^{\rm op}$ and then map those bimodules to bimodules in $(\mathsf{Vect}, \otimes)$ via the monoidal functor $K[-]$, no longer treating them as spans or cospans.

John Baez (Jan 29 2021 at 03:07):

I guess if you tell someone how to cook a souffle they have to ruin a few to see why you described it so carefully.

Christoph Thies (Jan 29 2021 at 03:08):

John Baez said:

The only way to get out of the mess you got yourself in is to back out - straight back.

I keep saying this to my children but they won't listen either.

John Baez (Jan 29 2021 at 03:13):

I made the same mistake Eric just made, once upon a time. The interplay between products and coproducts and tensor products is confusing at first.

Christoph Thies (Jan 29 2021 at 03:15):

I was joking and not actually referring to his understanding. The concepts are certainly confusing for me.

John Baez (Jan 29 2021 at 03:16):

Yup. I tease Eric rather mercilessly, but I start defending him when other people join in teasing him.

Eric Forgy (Jan 29 2021 at 04:28):

According to Wikipedia:
${}$

$\mathbf{Vect}$, the category of vector spaces over a given field, can be made cocartesian monoidal with the "tensor product" given by the direct sum of vector spaces and the trivial vector space as unit.

${}$
$\mathbf{Vect}$ is a weird category where
${}$
$A\times B\cong A\sqcup B\cong A\oplus B.$
${}$
It has pushouts and an initial object so it is finitely cocomplete. It has pullbacks and a terminal object (which is the same as initial object $0$), so is also finitely complete.

Tensor product is a quotient of the product, which I believe means it is also a quotient of direct sum so I think it all actually works out.

I think it is true that the pushout
${}$
$A \otimes_0 B= A\oplus B$
${}$
and the pushout
${}$
$A\otimes_K B$
${}$
is a quotient of $A\oplus B$ in $\mathbf{Vect}.$
${}$
John Baez said:

Eric Forgy said:

I got it in my head that composition of cospans in $\mathbf{Vect}_K$ corresponded to tensor product of bimodules

${}$
No. I told you all this wonderful stuff about how composing spans in a cartesian monoidal category $(\mathsf{C}, \times)$ is the same as tensor product of bimodules in the cocartesian monoidal category $(\mathsf{C}^{\rm op}, \times)$, and now you're applying that to $(\mathsf{Vect}_K , \otimes)$, where the tensor product is not cocartesian. So you're getting nonsense.

${}$
Right. You told me all this wonderful stuff :raised_hands:

But since $(\mathbf{Vect},\oplus)$ is cocartesian with pushout, then $(\mathbf{Vect}^\mathsf{op},\oplus)$ is cartesian with pullback so I thought we have
${}$
$\mathbf{Cospan(Vect,\oplus)\cong Span(Vect^\mathsf{op},\oplus) \cong Bim(Vect,\oplus)}.$

John Baez (Jan 29 2021 at 04:59):

I think it is true that the pushout
${}$
$A \otimes_0 B= A\oplus B$

I don't know what $\otimes_0$ means - what do you mean by that?

and the pushout
${}$
$A\otimes_K B$
${}$
is a quotient of $A\oplus B$ in $\mathbf{Vect}.$

No:

1) I told you earlier tonight that the tensor product $A \otimes_K B$ is not any sort of pushout or pullback involving $A$ and $B$. Any thoughts along these lines are doomed. That's why I said "back straight out".

2) In particular, taking $A = B = K^4$, $A \otimes_K B \cong K^{16}$ has dimension 16 so there's no way it can be a quotient of $A \oplus B \cong K^4$, which is a lower-dimensional vector space. So, it's not true that $A\otimes_K B$ is a quotient of $A\oplus B$ in $\mathbf{Vect}.$ That's just not how tensor products work.

John Baez (Jan 29 2021 at 05:06):

Eric Forgy said:

... since $(\mathbf{Vect},\oplus)$ is cocartesian with pushout, then $(\mathbf{Vect}^\mathsf{op},\oplus)$ is cartesian with pullback so I thought we have
${}$
$\mathbf{Cospan(Vect,\oplus)\cong Span(Vect^\mathsf{op},\oplus) \cong Bim(Vect,\oplus)}.$

John Baez (Jan 29 2021 at 05:07):

Yes, that's true.

John Baez (Jan 29 2021 at 05:10):

But note, none of this has anything to do with the usual tensor product $\otimes$ in $\mathsf{Vect}$.

John Baez (Jan 29 2021 at 05:13):

So this is false:

the pushout
${}$
$A\otimes_K B$
${}$
is a quotient of $A\oplus B$ in $\mathbf{Vect}$

and this is just meaningless by the standards of normal mathematicians:

I think it is true that the pushout
${}$
$A \otimes_0 B= A\oplus B$

since none of us knows what you mean by $\otimes_0$.

John Baez (Jan 29 2021 at 05:14):

In both cases you're trying to connect $\otimes$ to colimits in a way that just doesn't have any chance of working.

John Baez (Jan 29 2021 at 05:15):

If you just try examples you'll quickly be disabused of such notions.

Eric Forgy (Jan 29 2021 at 05:34):

Thanks John :sweat_smile:

For sure, I've gone astray somewhere. I know that. It is just a little difficult for me to pinpoint where.

In my first comment, I said what $A\otimes_0 B$ was, but not very clearly. It is the object obtained from pushing out the span $A\leftarrow 0\rightarrow B,$ which we know is coproduct. I used the tensor symbol because of the Wikipedia quote that said it was
${}$

"tensor product" given by direct sum

${}$
but $\otimes_0 = \oplus.$

If
${}$
$A = B = K^4 = K\oplus K\oplus K\oplus K$
${}$
then
${}$
$A\oplus B = K^8.$
${}$
I know that.

So the pushout of the span $A\leftarrow K\rightarrow B$ should be $A\oplus_K B.$

There are only so many operations that can happen in $\mathbf{Vect}$ so I would like to better understand how the tensor product relates to all this. I need to get to a point where I can understand a relationship (if it exists) between spans and bimodules in a way that involves actual tensor product or that beautiful knowledge I painfully acquired really isn't helpful to me, which would be a shame.

John Baez (Jan 29 2021 at 05:45):

Eric Forgy said:

In my first comment, I said what $A\otimes_0 B$ was, but not very clearly. It is the object obtained from pushing out the span $A\leftarrow 0\rightarrow B,$ which we know is coproduct

Yeah, the object obtained from pushing out the span $A\leftarrow 0\rightarrow B,$ is the coproduct of $A$ and $B$.

In $\mathsf{Vect}$ the coproduct is the direct sum, usually denoted $\oplus$. So call it $A \oplus B$ if you actually want people to understand you.

Or call it $A + B$ if you want to show off the fact that you know it's the coproduct.

Or, if you want to show off the fact that it's "the object obtained from pushing out the span $A\leftarrow 0\rightarrow B$", you can call it $A +_0 B$. Only a crazed category theorist would ever do this, but we'd still understand you.

The notation $A \otimes_0 B$ is basically just undefined crap. (Actually I can make up a meaning for it, but it's not at all useful here.)

I used the tensor symbol because of the Wikipedia quote that said it was
${}$

"tensor product" given by direct sum

Yeah, but they put "tensor product" in quotes because they meant it's just a monoidal structure on $\mathsf{Vect}$ - the quotes basically mean "haha, just kidding, don't write it as $\otimes$ unless you want people to think you're confused".

Eric Forgy (Jan 29 2021 at 05:56):

Btw, I have seen a few places that describe tensor product as pushout and it made sense to me, but maybe it is in a different context? For example:

https://math.stackexchange.com/questions/1548206/prove-that-tensor-product-is-pushout

[Edit: Yeah. It looks like I wanted $\mathbf{CRing}$ instead of $\mathbf{Vect}$ I guess :face_palm: ]

John Baez (Jan 29 2021 at 06:01):

There they are saying that if $A$ and $B$ are commutative algebras over some field $K$, the commutative algebra $A \otimes_K B$ is the pushout in $\mathsf{CommAlg}_K$ of a certain cospan

$A \leftarrow K \to B$

That's true. But this is a pushout in $\mathsf{CommAlg}_K$, not in $\mathsf{Vect}_K$, so it's a completely different story.

John Baez (Jan 29 2021 at 06:03):

By the way, for any field $K$ the monoidal category $(\mathsf{CommAlg}_K, \otimes_K)$ is cocartesian so some of the general stuff I said about cospans in cocartesian categories applies here. But $(\mathsf{Vect}_K, \otimes_K)$ is not cocartesian, and that's what you were working with.

Eric Forgy (Jan 29 2021 at 06:13):

I think I see where I confused myself.

Since I got one thing right, I'll copy it here:
${}$
$\mathbf{Cospan(Vect,\oplus)\cong Span(Vect^\mathsf{op},\oplus) \cong Bim(Vect,\oplus)}.$
${}$
Among other things, this means that every vector space is a monoid object and I know monoid objects in $\mathbf{Vect}$ are algebras. However, this is a monoid object in $(\mathbf{Vect},\oplus,0).$ It wasn't a huge step in my mind to get from there to commutative algebras, but I went astray (surprise!).

John Baez (Jan 29 2021 at 15:38):

Eric Forgy said:

... every vector space is a monoid object and I know monoid objects in $\mathsf{Vect}$ are algebras. However, this is a monoid object in $(\mathsf{Vect},\oplus,0).$

It's good to be precise to avoid confusing oneself:

1. Monoid objects in $(\mathsf{Vect}_K,\otimes,K)$ are algebras.

2. Monoid objects in $(\mathsf{Vect}_K,\oplus,0)$ are something else, and they turn out to be just vector spaces.

The monoidal structure matters!!!

If a random guy walks up to you and says "monoid object in $\mathsf{Vect}$", they almost always mean "monoid object in $(\mathsf{Vect}_K,\otimes,K)$". But you can also ask them to say which monoidal structure they're talking about.

Eric Forgy (Feb 01 2021 at 20:22):

For future reference, I think another source of confusion for me was an example of bicategory on the nLab.

If $(C,\otimes,1)$ is a monoidal category, there is a corresponding bicategory $\mathbf{B}C$ with
${}$

• One object $\bullet$
• For each object $c\in C$, a morphism $[c]$ with composition given by monoidal product $[c]\circ[c']:=[c\otimes c']$
• For each morphism $f:c\to c'$ in $C$, a 2-morphism $[f]:[c]\to[c'].$

${}$
For $(\mathbf{Vect},\otimes,K)$, the composition is $[S]\circ[T] := [S\otimes_K T].$ I want to relate this to spans / bimodules somehow :thinking:

Joe Moeller (Feb 01 2021 at 20:32):

If you take a bicategory, pick one object, and take all of its endomorphism 1-cells, and all 2-cells between those, its the same thing as a monoidal category, related to the fact you gave about $\mathbf B C$. So what do you get when you do this in the bicategory of bimodules?

Eric Forgy (Feb 01 2021 at 20:52):

Thanks Joe. This is precisely the situation I'm interested in. Super cool :blush:

In the case of $\mathbf{Bim}$, we'd pick a ring $A$ and morphisms would be $A$-bimodules and 2-morphisms would be $R$-bimodule homomorphisms.

Composition of morphisms would be tensor product
${}$
$\Omega \circ \Omega := \Omega\otimes_R\Omega.$
${}$
If you choose a field $K$ (which is also a ring), then a $K$-bimodule is a vector space $V$ and
${}$
$V\circ V := V\otimes_K V.$

John Baez (Feb 01 2021 at 21:26):

Right, so in the case where we start with the bicategory

[rings, bimodules, bimodule homomorphisms]

and choose one object that happens to be a field $K$, the monoidal category we get from Joe's construction is our old friend $(\mathsf{Vect}_K, \otimes_K)$.

John Baez (Feb 01 2021 at 21:28):

Oh, actually that's not true!

John Baez (Feb 01 2021 at 21:29):

The problem is that a $(K,K)$-bimodule is not the same as a vector space over $K$.

Vector spaces over $K$ give some examples of $(K,K)$-bimodules, but not all.

John Baez (Feb 01 2021 at 21:30):

The point is that whenever our ring $R$ is commutative, any left $R$-module becomes a $(R,R)$-bimodule in a standard way, but we don't get all $(R,R)$-bimodules this way.

John Baez (Feb 01 2021 at 21:31):

A fun example is: figure out how to make the abelian group $\mathbb{C}$ into a $(\mathbb{C}, \mathbb{C})$-bimodule in a nonstandard way.

John Baez (Feb 01 2021 at 21:33):

So $\mathsf{Vect}_K$ is usually just a sub-monoidal category of Joe's monoidal category, when $K$ is a field.

John Baez (Feb 01 2021 at 21:33):

And Joe's construction is even more interesting when the ring we pick is noncommutative.

John Baez (Feb 01 2021 at 21:34):

But I don't think any of this helps Eric very much, except to give him yet another workout in categorical algebra. :muscle:

Eric Forgy (Feb 01 2021 at 21:34):

I feel so disappointed. My hero $(\mathbf{Vect}_K,\otimes,K)$ seems a lot less special now :sweat_smile:

John Baez (Feb 01 2021 at 21:38):

The point is that $\mathsf{Vect}_K$ is a category of modules of a ring, not a category of bimodules of a ring.

John Baez (Feb 01 2021 at 21:39):

I thought of my counterexample pretty quickly because I've thought about doing quantum mechanics with bimodules instead of modules...

Eric Forgy (Feb 01 2021 at 21:52):

A hunch suggests that the distinction might come down to zero objects. If a $K$-bimodule has a zero object, it is a vector space?

Eric Forgy (Feb 01 2021 at 21:53):

A $K$-bimodule has scalar multiplication, so if it also has addition, it is a vector space.

Eric Forgy (Feb 01 2021 at 21:56):

I think addition requires a zero object.

Eric Forgy (Feb 01 2021 at 21:57):

https://ncatlab.org/nlab/show/additive+category

John Baez (Feb 01 2021 at 22:11):

Eric Forgy said:

A hunch suggests that the distinction might come down to zero objects. If a $K$-bimodule has a zero object, it is a vector space?

It doesn't make sense to ask if a $K$-bimodule has a zero object. A category can have a zero object, not a bimodule.

Eric Forgy (Feb 01 2021 at 22:14):

I need an emoticon for "You know what I mean" :sweat_smile:

Eric Forgy (Feb 01 2021 at 22:16):

All I want to say is that if a $K$-bimodule has addition, it should also be a vector space.

John Baez (Feb 01 2021 at 22:19):

Well, that's at least meaningful, but it's false. All bimodules of rings have addition, by definition.

Eric Forgy (Feb 01 2021 at 22:21):

Is everything I thought I knew about vector spaces wrong? I thought scalar multiplication that distributes over addition is enough to have a vector space :sweat_smile:

Eric Forgy (Feb 01 2021 at 22:22):

(scalar -> field)

John Baez (Feb 01 2021 at 22:24):

Yes, that's true.

John Baez (Feb 01 2021 at 22:26):

A vector space is a left $K$-module.

Every $K$-bimodule gives a vector space by forgetting the right $K$-module structure and only remembering the left $K$-module structure.

What I'm saying is something else: every left $K$-module gives a $K$-bimodule by defining the right module structure to equal the bimodule structure. But, not every $K$-bimodule arises this way.

John Baez (Feb 01 2021 at 22:27):

So: there's a forgetfulful functor from $K$-bimodules to vector spaces, and also a functor going back the other way, but these do not form an equivalence.

John Baez (Feb 01 2021 at 22:28):

I already said part of this:

John Baez said:

The point is that whenever our ring $R$ is commutative, any left $R$-module becomes a $(R,R)$-bimodule in a standard way, but we don't get all $(R,R)$-bimodules this way.

John Baez (Feb 01 2021 at 22:28):

"Having zero objects" or "having addition" has nothing to do with this stuff.

Eric Forgy (Feb 01 2021 at 22:29):

Ok

John Baez (Feb 01 2021 at 22:31):

You would understand this stuff better if you found two different ways to make $\mathbb{C}$, with its usual addition, into a $(\mathbb{C}, \mathbb{C})$-bimodule.

John Baez (Feb 01 2021 at 22:32):

There's the obvious way and the less obvious way.

Eric Forgy (Feb 01 2021 at 22:32):

John Baez said:

You would understand this stuff better if you found two different ways to make $\mathbb{C}$, with its usual addition, into a $(\mathbb{C}, \mathbb{C})$-bimodule.

Yeah. I was thinking about it :thinking:

John Baez (Feb 01 2021 at 22:32):

Okay. It doesn't make much sense to talk about this until after you do that....

John Baez (Feb 01 2021 at 22:33):

It's like talking about the difference between lizards and salamanders before you've ever seen them.

Eric Forgy (Feb 01 2021 at 22:35):

The obvious way, I think is just that $\mathbb{C}$ is already a $(\mathbb{C},\mathbb{C})$-bimodule with action given by $+$.

John Baez (Feb 01 2021 at 22:35):

Huh?

Eric Forgy (Feb 01 2021 at 22:38):

Well, $\mathbb{C}$ with action given by multiplication seems too obvious and you stressed it was an abelian group :sweat_smile:

John Baez (Feb 01 2021 at 22:38):

I think we're in deep trouble here.

John Baez (Feb 01 2021 at 22:38):

I think you don't know what a bimodule of a ring is.

John Baez (Feb 01 2021 at 22:39):

So I should just ask you "what's a bimodule of a ring?"

Eric Forgy (Feb 01 2021 at 22:43):

I'll try to answer without looking, which is dangerous :sweat_smile:

Recalling the puzzles, we start with monoid objects $R,S$ in a monoidal category $(C,\otimes,1)$. An $(R,S)$-bimodule is an object togoether with compatible left and right actions.

John Baez (Feb 01 2021 at 22:46):

That's pretty good, at least if you know what "compatible" means.

John Baez (Feb 01 2021 at 22:46):

But we were doing the classic case: bimodules of rings.

John Baez (Feb 01 2021 at 22:46):

What's the monoidal category that gives this classic case?

John Baez (Feb 01 2021 at 22:47):

(When you talk to algebraists and say "bimodule", you're referring to this classic case.)

Eric Forgy (Feb 01 2021 at 22:53):

I'd need to know what monoidal category has rings as monoid objects. Without looking, my guess would be $\mathbf{Ab},$ but I've never worked this out. "Working it out" would mean checking that a bunch of intuitive diagrams commute, which I can't remember off the top of my head, but I'm fairly certain they work out here.

John Baez (Feb 01 2021 at 22:57):

Good, yes, it's $\mathsf{Ab}$. So you need to know what the tensor product of abelian groups is, and then check you get a monoidal category this way, and check that monoid objects in here are exactly the same as rings, etc.

John Baez (Feb 01 2021 at 22:58):

So a bimodule in this context is an abelian group A with a left action of a ring R and a right action of a ring S, obeying the "compatibility" law (which is a version of the associative law).

John Baez (Feb 01 2021 at 23:00):

So my puzzle:

You would understand this stuff better if you found two different ways to make $\mathbb{C}$, with its usual addition, into a $(\mathbb{C}, \mathbb{C})$-bimodule.

says to treat $\mathbb{C}$ as an abelian group in the usual way, and then find two different ways to make it into a $(\mathbb{C}, \mathbb{C})$-bimodule, where these two $\mathbb{C}$ 's are being treated as a ring in the usual way.

John Baez (Feb 01 2021 at 23:01):

So start by telling me one way to do this. Pick the way that's most likely to work, since you were saying some crazy stuff before.

Eric Forgy (Feb 01 2021 at 23:29):

$\mathbb{C}$ is a monoid object in $\mathbf{Ab}.$

The most obvious way to make $\mathbb{C}$ into a $(\mathbb{C},\mathbb{C})$-bimodule is to define the left and right actions to be multiplication, i.e. $\mathbb{C}$ is already a $(\mathbb{C},\mathbb{C})$-bimodule just like any ring $R$ is an $(R,R)$-bimodule. In this case, compatible is simply associativity.

John Baez (Feb 01 2021 at 23:53):

Yes! This is what I meant by the "obvious" way to make $\mathbb{C}$ into a $(\mathbb{C}, \mathbb{C})$-bimodule, since as you mention, this way works for any ring.

John Baez (Feb 01 2021 at 23:54):

So the interesting part of my puzzle was to find another way to make $\mathbb{C}$ into a $(\mathbb{C}, \mathbb{C})$-bimodule, which is not isomorphic to the "obvious" one.

John Baez (Feb 01 2021 at 23:55):

As a hint: this other way would not work for $\mathbb{R}$; we really use the complex numbers here.

Eric Forgy (Feb 02 2021 at 00:16):

I'm drawing a blank :thinking:

John Baez (Feb 02 2021 at 01:44):

Well, think about what the complex has, as a ring, that the real number doesn't. That seems like a pretty robust approach!

John Baez (Feb 02 2021 at 01:44):

Note I say as a ring since this is a question about bimodules of rings. So all sorts of stuff connected to analysis is irrelevant.

Eric Forgy (Feb 02 2021 at 01:46):

Complex conjugates?

John Baez (Feb 02 2021 at 01:49):

Hmm, so try using those.

Eric Forgy (Feb 02 2021 at 01:52):

I'm just throwing darts, but maybe we could define left and right actions by multiplication by the complex conjugates?

Reid Barton (Feb 02 2021 at 01:58):

You're in the general vicinity of the dartboard. :darts:

Reid Barton (Feb 02 2021 at 01:59):

You get three darts right? So what can you try?

John Baez (Feb 02 2021 at 04:40):

Yes, Eric's suggestion was a bit too vague for me to tell if it's right or not... I think Reid was pointing out there are 3 things to try here.

Eric Forgy (Feb 02 2021 at 04:52):

I meant, left action
${}$
$z\otimes m\mapsto z^* m$
${}$
and right action
${}$
$m\otimes z\mapsto m z^*.$

John Baez (Feb 02 2021 at 04:56):

Sorry. That's a perfectly fine way to make $\mathbb{C}$ into a bimodule, but it's isomorphic to the usual way! See why?

John Baez (Feb 03 2021 at 03:20):

Okay, I'll say why. Your way to make $\mathbb{C}$ into a bimodule is isomorphic to the usual where the left action is

$z \otimes m \mapsto zm$

and the right action is

$m \otimes z \mapsto z m$

The isomorphism is complex conjugation: $m \mapsto m^\ast$ !

To see this, notice that if we apply your left action and then the isomorphism:

$z \otimes m \mapsto z^\ast m \mapsto (z^\ast m)^\ast = z m^\ast$

it's the same as applying the isomorphism to $m$ and then applying the usual left action:

$z \otimes m \mapsto z \otimes m^\ast \mapsto z m^\ast$

The same is true for the right action.

John Baez (Feb 03 2021 at 03:22):

So, to get a really new $(\mathbb{C}, \mathbb{C})$ bimodule structure on $\mathbb{C}$, we should do something like this: use the usual left action

$z \otimes m \mapsto z m$

but tweak the right action using complex conjugation:

$m \otimes z \mapsto m z^\ast$

John Baez (Feb 03 2021 at 03:25):

This is not isomorphic to the usual bimodule structure: the attempted isomorphism $m \mapsto m^\ast$ works for the right action but not the left, and the attempted isomorphism $m \mapsto m$ works for the left action but not the right.

John Baez (Feb 03 2021 at 03:25):

We could also have tweaked the left action using complex conjugation, and used the usual right action.

David Michael Roberts (Feb 05 2021 at 09:56):

For what it's worth, I edited that misleading Wikipedia sentence, so it doesn't say

can be made cocartesian monoidal with the "tensor product" given by the direct sum

anymore, but

can be made cocartesian monoidal with the monoidal product given by the direct sum.

It was clearly confusing to at least one person with a PhD in a quantitative science, so ...