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Stream: learning: questions

Topic: is Ab tensored over Grp?

Emily (Oct 18 2021 at 18:03):

Is there a functor

$$-_{1}\odot-_{2}\colon\mathsf{Grp}\times\mathsf{Ab}\to\mathsf{Ab}$$

characterised by a natural bijection

$$\mathbf{Hom}_{\mathsf{Ab}}(G\odot A,B) \cong \mathbf{Hom}_{\mathsf{Grp}}(G,\mathbf{Hom}_{\mathsf{Ab}}(B,C))$$

for $G$ a group and $A$ and $B$ abelian groups, making $\mathsf{Ab}$ into a $\mathsf{Grp}$-tensored category?

Emily (Oct 18 2021 at 18:03):

For what it's worth, $\mathsf{Ab}$ does have $\mathsf{Grp}$-cotensors: $G\pitchfork A$ is the group consisting of

• The set $\mathrm{Hom}_{\mathsf{Grp}}(G,A)$;
• The pointwise product $(f,g)\mapsto f*g$ with

$$[f*g](a)\overset{\mathrm{def}}{=}f(a)g(a)$$

• The unit $\Delta_{1_{A}}\colon G\to A$ given by $\Delta_{1_{A}}(g)\overset{\mathrm{def}}{=}1_{A}$.

This is because for $f*g$ to be a group homomorphism we only need $A$ to be abelian:

$$\begin{aligned} [f*g](ab) &= f(ab)g(ab)\\ &= f(a)f(b)g(a)g(b)\\ &= f(a)g(a)f(b)g(b)\\ &= [f*g](a)[f*g](b). \end{aligned}$$

Reid Barton (Oct 18 2021 at 18:08):

Yes, it's $G \odot A = G_{\mathrm{ab}} \otimes A$.

Reid Barton (Oct 18 2021 at 18:09):

Because $\mathbf{Hom}_{\mathsf{Grp}}(G,\mathbf{Hom}_{\mathsf{Ab}}(B,C))$ is also $\mathbf{Hom}_{\mathsf{Ab}}(G_{\mathrm{ab}},\mathbf{Hom}_{\mathsf{Ab}}(B,C))$.

Reid Barton (Oct 18 2021 at 18:09):

I guess I'm thinking of both sides as sets really, but the abelian group structures are also the same.

Emily (Oct 18 2021 at 18:12):

Ah, perfect! Thanks, Reid!

John Baez (Oct 19 2021 at 02:48):

If $\mathsf{Grp}$ is tensored over $(\mathsf{Ab},\otimes)$ shouldn't tensoring any group $G$ with $\mathbb{Z}$ give $G$? $\mathbb{Z}$ is the unit for the tensor product in $(\mathsf{Ab},\otimes)$, and I thought the definition of "tensored over" requires that tensoring with the unit object is the identity functor (up to natural isomorphism).

But Reid seems to be saying $G \odot \mathbb{Z}$ is the abelianization of $G$, not $G$.

Reid Barton (Oct 19 2021 at 09:19):

It's the other way around: Ab is tensored over Grp.

Reid Barton (Oct 19 2021 at 09:21):

Another fun fact: This functor $\odot$ preserves colimits in each argument, and it's the universal example of a functor $\mathsf{Grp}\times\mathsf{Ab}\to C$ that does so. That means it is the functor witnessing the fact that "a group object in the category of abelian groups is an abelian group".

John Baez (Oct 19 2021 at 12:07):

Okay, whew!

John Baez (Oct 19 2021 at 12:14):

Is the category of p-torsion abelian groups tensored over abelian groups? Can we define a tensor product of an abelian group $A$ and a p-torsion abelian group $T$ to be $A_p \otimes T$, where $A_p$ is the p-torsion part of $A$?

John Baez (Oct 19 2021 at 12:14):

(Here $\otimes$ is the usual tensor product of abelian groups.)

Reid Barton (Oct 19 2021 at 12:18):

Yes--or you can also define it to be $A \otimes T$, which will turn out to be $p$-torsion.

Reid Barton (Oct 19 2021 at 12:21):

An abelian group $A$ is p-torsion if and only if the map $\mathbb{Z} \otimes A \to A$ sending $1 \otimes a$ to $a$ descends (necessarily uniquely) to $\mathbb{Z}/p \otimes A$. In that case, the resulting action $\mathbb{Z}/p \otimes A \to A$ is the inverse of the map $A \to \mathbb{Z}/p \otimes A$ sending $a$ to $[1] \otimes a$. So, $A$ is p-torsion if and only if this latter map is an isomorphism, and then we can canonically identify $\mathbb{Z}/p \otimes A$ with $A$.

Reid Barton (Oct 19 2021 at 12:22):

(Also, $\mathbb{Z}/p \otimes A$ is the p-torsion $A_p$.)

Reid Barton (Oct 19 2021 at 12:22):

Oh wait!

Reid Barton (Oct 19 2021 at 12:22):

No, the tensor product should be given by $(A/pA) \otimes T$.

Reid Barton (Oct 19 2021 at 12:22):

Er now I'm totally confused, sorry.

Reid Barton (Oct 19 2021 at 12:25):

OK, I unconfused myself.

Reid Barton (Oct 19 2021 at 12:28):

I guess you can define the symbol "$A \otimes T$" to mean whatever you want, but if you want it to satisfy these properties:

• $- \otimes T$ preserves colimits.
• $\mathbb{Z}/p \otimes T = T$ (for $T$ a $p$-torsion group).

then you can't define it using the formula $A_p \otimes T$, where $A_p$ is the p-torsion part, by which I assume you mean the kernel of multiplication by $p$. Instead, you should use the formula $A/pA \otimes T$.

Reid Barton (Oct 19 2021 at 12:29):

Because let's consider what $\mathbb{Z} \otimes T$ should be. The p-torsion part of $\mathbb{Z}$ is 0, so the $A_p \otimes T$ formula would give 0. But there is a coequalizer $\mathbb{Z} \rightrightarrows \mathbb{Z} \to \mathbb{Z}/p$ (where the maps are $p$ and $0$), and applying $- \otimes T$ to this, we would deduce that $\mathbb{Z}/p \otimes T$ is zero as well.

Zhen Lin Low (Oct 19 2021 at 12:31):

If you have a reflective subcategory of Ab with additive reflector, the reflector preserves tensors (and weighted colimits more generally). You can use that to figure out what the tensor is.

Reid Barton (Oct 19 2021 at 12:33):

What's going on is this. Let's write $R$ for the ring $\mathbb{Z}/p$. The category of $R$-modules is actually a full subcategory of abelian groups, which I think is equivalent to $R$ being a solid ring ($m : R \otimes R \to R$ is an isomorphism). So the tensor product of a $\mathbb{Z}$-module and an $R$-module is an $R$-module. We can also calculate that tensor product $A \otimes T$ by replacing $A$ by $R \otimes A$ = the free $R$-module on $A$ first, if we like--in this case that's $A/pA$.

John Baez (Oct 19 2021 at 13:17):

Thanks! I wanted to look at this example to get a better feeling for things. But I screwed up and tried $A_p$ instead of $A/pA$ as my way of "p-torsion-ifying" the abelian group $A$.

Mike Shulman (Oct 19 2021 at 16:01):

More generally, if $C$ is any category and $D$ is a reflective subcategory of $C$ with a closed monoidal structure, then $D$ has this sort of "tensors" over $C$. If $L$ is the reflector and $c\in C$, $d_1,d_2\in D$, then $D(Lc \otimes d_1, d_2) \cong D(Lc, [d_1,d_2]) \cong C(c,[d_1,d_2])$.

However, the name "tensor" (or the more modern name "copower") is not really appropriate here, because $D$ is not enriched over $C$, and indeed that question doesn't even make sense to ask because $C$ may not be a monoidal category. In particular, in the original question, $\mathsf{Gp}$ is not a monoidal category in any relevant way, and so $\mathsf{Ab}$ can't be enriched over it.

However, if $C$ is a closed monoidal category and $D$ is a reflective subcategory closed under the tensor product and internal-hom, then $D$ does become enriched over $C$ and this gives its copowers. That's probably what's happening for $p$-torsion abelian groups.

John Baez (Oct 19 2021 at 18:51):

Mike Shulman said:

In particular, in the original question, $\mathsf{Gp}$ is not a monoidal category in any relevant way, and so $\mathsf{Ab}$ can't be enriched over it.

Maybe that's why I slipped and started talking about Grp being tensored over Ab rather than what was being discussed, Ab being tensored over Grp. When I heard "tensored over" and saw some tensor product symbols I assumed folks were talking about Grp being tensored over the monoidal category $(\mathsf{Ab}, \otimes)$. And it seems like it almost is, via

$A \odot G := A \otimes G_{\rm{ab}} ,$

except that the unit object doesn't act unitally!

Reid Barton (Oct 19 2021 at 20:11):

Mike Shulman said:

$\mathsf{Gp}$ is not a monoidal category in any relevant way

Right, of course. But if we consider the graded category (= family of categories) $(C_n)_{n \ge 0}$ with $C_n$ = $n$-fold group objects in Set, then the whole graded category has a "graded monoidal" structure involving functors $\otimes_{n,m} : C_n \times C_m \to C_{n+m}$, of which the functor $\odot$ under discussion is one component--actually many components, the functors $\otimes_{1,m}$ for any $m \ge 2$. So it does make sense to think of $\odot$ as a kind of tensor product, insofar as we also consider the components of an action of a graded algebra on a graded module as a kind of multiplication.

Emily (Oct 20 2021 at 00:27):

Is there some way to fit units, associators, and unitors into the 'graded closed monoidal category' $(\mathsf{Sets},\mathsf{Grp},\mathsf{Ab},\mathsf{Ab},\ldots)$?

Emily (Oct 20 2021 at 00:27):

On a related issue, the $\mathsf{Sets}$-tensoring of $\mathsf{Grp}$ leads to a kind of "tensor semiproduct"

$$\triangleleft\colon\mathsf{Grp}\times\mathsf{Grp}\to\mathsf{Grp}$$

on $\mathsf{Grp}$, obtained by precomposing it with the forgetful functor $\mathsf{Grp}\to\mathsf{Sets}$. (So we define

$$\begin{aligned} G\triangleleft H &= |H|\odot G\\ &\cong \coprod_{h\in H}G \end{aligned}$$

) Supposedly "monoids" with respect to $\triangleleft$ should recover near-rings, but it seems again we don't have units, associators, and unitors (at least if we refuse to work with 'lax-monoidal' categories)...

Emily (Oct 20 2021 at 00:27):

All this is to say that this is a somewhat weird situation: there are these 'almost co/tensors' between $\mathsf{Sets}$, $\mathsf{Grp}$, and $\mathsf{Ab}$, but we can't apply the usual categorical tools for capturing them...

Emily (Oct 20 2021 at 00:27):

This also only gets worse moving from sets to categories, as now e.g. $\mathsf{SymMonCats}$ is now "co/tensored" over four different categories: $\mathsf{Cats}$, $\mathsf{MonCats}$, $\mathsf{BrMonCats}$, and itself. Indeed, $\mathsf{Fun}^\otimes(\mathcal{C},\mathcal{D})$:

• Exists if $\mathcal{D}$ is a monoidal category;
• Has a monoidal structure if $\mathcal{D}$ is a braided monoidal category;
• Has a braided monoidal structure (which is also symmetric since "$\mathbb{E}_{\infty-1}=\mathbb{E}_{\infty}$") if $\mathcal{D}$ is a symmetric monoidal category.

Emily (Oct 20 2021 at 00:27):

On this last point, is there also a "tensor"

$$\odot\colon\mathsf{MonCats}\times\mathsf{BrMonCats}\to\mathsf{BrMonCats}$$

of braided monoidal categories by monoidal categories?