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Stream: deprecated: mathematics

Topic: tensor product of modules

Matteo Capucci (he/him) (Feb 18 2022 at 23:37):

I'm realizing there's two tensor products of modules around! How can it be possible?

Let $R$ be a (not necessarily commutative) ring, and let $A$ be a right $R$-module and $B$ a left $R$-module.
One defines $A \otimes B$ to be the representing object for balanced maps, that is maps $f:A \times B \to C$ where $f(ar,b)=f(a,rb)$ ($C$ can be just a set). Then, for every such map, there is a map $\tilde f: A \otimes B \to C$. So $A \otimes B$ is basically $A \times B$ quotiented by the relations $(ar,b) \sim (a, rb)$.
Now suppose $A,B,C$ are $R$-bimodules. Consider $\mathrm{Hom}(B,C)$: it has an obvious bimodule structure given by pointwise scalar multiplication. It can be shown this is a closed structure for $R$-bimodules. What's the associated monoidal product? Now a map $f:A \to \mathrm{Hom}(B,C)$ is linear iff $f(ra) = rf(a)$ and $f(ar) = f(a)r$. You then define $A \otimes' B$ to be such that there is exactly one map uncurrying $f$ to give you a linear map $A \otimes' B \to C$. This map, by virtue of being linear, has the property $f(r a, b) = r f(a, b)$ and $f(a, br) = f(a, b)r$. Also by virtue of being adjoint, we know that $f(ra, b) = rf(a,b)=f(a,rb)$ and symmetrically on the right, so $f$ is actually a bilinear map.

So... we have two tensors: $\otimes$ represents balanced maps, and $\otimes'$ represents bilinear balanced maps...!
I always believed every balanced map of bimodules to be bilinear, but it's not the case: if $R$ is a non-commutative ring, then its product map $f = \cdot$ is balanced but not bilinear! In fact $a(rb) = (ar)b$ but not necessarily $a(rb) = (ab)r$.
Even if $R$ were commutative though, suppose $B$ is an $R$-bimodule with the property that $br \neq rb$. This can be achieved since not every bimodule, even if $R$ is commutative, arises as the 'symmetrization' of a left or right $R$-module (in other words: $R$-$Mod \to R$-$Mod$-$R$ is not essentially surjective). Then a balanced map $A \times B \to C$ would require only $f(ar,b)=f(a,rb)$ but a bilinear map would also require $f(ar,b)=f(a,br)$. But since $rb \neq br$, it is not necessary that the second condition hold!

Well, I'm quite confident I'm missing some very important bit, but I've never seen this spelled out clearly.
Are there really two tensor products of modules around?

Mike Shulman (Feb 18 2022 at 23:40):

The first one is not really a "tensor product of modules", as it doesn't pay any attention to the abelian group structure. It's more like a tensor product of sets with an action by the multiplicative monoid of R.

Matteo Capucci (he/him) (Feb 18 2022 at 23:41):

Right, I missed to mention the quotient by the other part of the linearity -- but that's the same for both balanced and bilinear maps.
Do you think it amounts to the difference between the two tensors?

Mike Shulman (Feb 18 2022 at 23:44):

In particular, your $C$ in the first case really does have to be an abelian group at least.

Matteo Capucci (he/him) (Feb 18 2022 at 23:44):

Indeed, but is that the culprit? :thinking:

Mike Shulman (Feb 18 2022 at 23:45):

With that modification, the two tensor products are the same; the second one just remembers more structure.

Mike Shulman (Feb 18 2022 at 23:46):

If $A$ is a right $R$-module and $B$ a left $R$-module, then $A\otimes_R B$ is an abelian group. Moreover, if $A$ is a left $S$-module and $B$ a right $T$-module (compatibly with their $R$-structures), for any rings $S$ and $T$, then this same abelian group $A\otimes_R B$ inherits an $S$-$T$-bimodule structure.

Matteo Capucci (he/him) (Feb 18 2022 at 23:49):

Mmh yeah I know that too... I dont' see how you would go on about proving $f(ar, b) = f(a, br)$ for a given balanced map using that fact though

Mike Shulman (Feb 18 2022 at 23:49):

Why would you want to prove that?

Matteo Capucci (he/him) (Feb 18 2022 at 23:51):

In order to prove that if $f : A \times B \to C$ is balanced (hence a map $f:A \otimes B \to C$) then it's also bilinear (hence a map $f:A \otimes' B \to C$)

Matteo Capucci (he/him) (Feb 18 2022 at 23:51):

I feel like I'm making a fool of myself but I really don't see how :sweat_smile:

Zhen Lin Low (Feb 18 2022 at 23:52):

If A and B are symmetric bimodules over a commutative ring you get that. If not, you don't.

Mike Shulman (Feb 18 2022 at 23:52):

A bilinear map doesn't have to satisfy that property.

Matteo Capucci (he/him) (Feb 18 2022 at 23:52):

Zhen Lin Low said:

If A and B are symmetric bimodules over a commutative ring you get that. If not, you don't.

That's what I concluded too

Mike Shulman (Feb 18 2022 at 23:52):

Bilinear means $f(ar,b) = f(a,rb)$ (which I guess you're calling "balanced"), $f$ respects addition in both variables, and $f(ra,b) = rf(a,b)$ and $f(a,br) = f(a,b)r$.

Matteo Capucci (he/him) (Feb 18 2022 at 23:53):

Indeed

Matteo Capucci (he/him) (Feb 18 2022 at 23:53):

Mmh I see

Matteo Capucci (he/him) (Feb 18 2022 at 23:54):

Matteo Capucci (he/him) said:

I feel like I'm making a fool of myself but I really don't see how :sweat_smile:

Here it is

Matteo Capucci (he/him) (Feb 18 2022 at 23:56):

Also by virtue of being adjoint, we know that $f(ra, b) = rf(a,b)=f(a,rb)$ and symmetrically on the right, so $f$ is actually a bilinear map.

Ok this is where I fooled myself

Matteo Capucci (he/him) (Feb 18 2022 at 23:57):

$rf(a,b)=f(a,rb)$ doesn't follow from adjointess

Matteo Capucci (he/him) (Feb 18 2022 at 23:57):

Thanks for putting up with me! :D

Mike Shulman (Feb 18 2022 at 23:58):

Sorry it took me a while to realize what you were actually asking. (-:

Zhen Lin Low (Feb 19 2022 at 00:04):

In this situation it would have been good to use three different rings, as Mike did. Then it would have been a type error to even write $r f (a, b) = f (a, r b)$.

John Baez (Feb 19 2022 at 01:37):

Matteo Capucci (he/him) said:

$rf(a,b)=f(a,rb)$ doesn't follow from adjointess

A good rule of thumb is that letters should not jump over each other unless there's a darn good reason, for example some commutativity or braiding. So when that $r$ jumped out of $f(a,rb)$ and landed out in front, you should have quizzed it on how it was allowed to do that.

Matteo Capucci (he/him) (Feb 19 2022 at 17:48):

Indeed

Matteo Capucci (he/him) (Feb 19 2022 at 17:50):

I've been think about this quite a lot in the context of actions of monoidal categories and I'm very strict at not commuting stuff that shouldn't... but I was also convinced one false theorem was true, and yesterday I realized (erroneously) it was because bilinearity actually allows $f(a,rb)=rf(a,b)$. So basically I wish-thought that property into existence and then got angry at it.

John Baez (Feb 20 2022 at 00:42):

:upside_down: