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Is there a name for monoidal categories $(\mathscr V, \otimes, I)$ such that $\otimes$ has a left adjoint $(\ell, r) : \mathscr V \to \mathscr V^2$? Have they been studied anywhere? What are some interesting examples?

A couple of remarks: when $I : 1 \to \mathscr V$ has a left adjoint, then $\mathscr V$ is semicartesian, i.e. the unit is terminal. When $\otimes$ has a left adjoint, which is furthermore the diagonal $\Delta : \mathscr V \to \mathscr V^2$, then $\mathscr V$ has binary products.


I'll unwrap the definition here to make the structure more explicit. Let $(\mathscr V, \otimes, I)$ be a monoidal category. $\otimes$ has a left adjoint if we have the following.

  • endofunctors $\ell : \mathscr V \to \mathscr V$ and $r : \mathscr V \to \mathscr V$;
  • for every pair of morphisms $f : \ell(X) \to Y$ and $g : r(X) \to Z$, a morphism $\{f, g\} : X \to Y \otimes Z$;
  • for every morphism $h : X \to Y \otimes Z$, morphisms $h_\ell : \ell(X) \to Y$ and $h_r : r(X) \to Z$,

such that, for all $x : X' \to X$, $y : Y \to Y'$ and $z : Z \to Z'$, we have $$y \otimes z \circ \{ f, g \} \circ x = \{ y \circ f \circ \ell(x), z \circ g \circ r(x) \}$$ $$\{ h_\ell, h_r \} = h$$ $$\{ f, g \}_\ell = f$$ $$\{ f, g \}_r = g$$

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  • $\begingroup$ Usually tensor product is right exact, not left exact. Are you sure you want to be asking for a left adjoint here and not a right adjoint? $\endgroup$ – Noah Snyder Jan 3 at 20:28
  • $\begingroup$ @NoahSnyder: the concept is intended to be a weakening of the notion of cartesian category (where the cartesian product is right adjoint to the diagonal). However, one could just as well ask for a right adjoint instead, which would correspond to the cocartesian setting. I'm happy to know of references for either. $\endgroup$ – varkor Jan 3 at 20:36
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    $\begingroup$ @QiaochuYuan: Ah, right. In my setting, one looks at the Deligne-Kelly tensor product of the two categories rather than their Cartesian product, and so the functor out of that is also right exact. $\endgroup$ – Noah Snyder Jan 3 at 20:55
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    $\begingroup$ Reminds me of a tangentially related question. $\endgroup$ – Tim Campion Jan 3 at 21:12
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    $\begingroup$ Er -- rather, that other question is not-so-tangentially related -- it's the dual of the special case where the monoidal structure is cocartesian. $\endgroup$ – Tim Campion Jan 5 at 12:29
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Just to clean up the $\epsilon$ of room left after Qiaochu's answer -- we can get rid of the extra hypotheses. I'll write $I$ for the monoidal unit and $1$ for the terminal object.

Assume that $(\ell,r) \dashv \otimes$. Then the natural isomorphisms $A \cong I \otimes A \cong A \otimes I$ give rise, by adjunction, to maps $\ell A \to I$ and $r A \to I$, natural in $A$. We also have a unit map $A \to (\ell A) \otimes (r A)$, natural in $A$. Tensoring and composing, we get a map $A \to (\ell A) \otimes (r A) \to I \otimes I \cong I$, natural in $A$. That is, we have a cocone (with vertex $I$) on the identity functor for $V$. It follows that in the idempotent completion $\tilde V$ of $V$, there is a terminal object (which must be a retract of $I$).

Now, the idempotent completion $\tilde V$ again has a monoidal structure $\tilde \otimes$ with a left adjoint $(\tilde \ell, \tilde r)$. So the first part of Qiaochu's Eckmann-Hilton argument can be run in $\tilde V$: $I = I \otimes I = (I \times 1) \otimes (1 \times I) = (I \otimes 1) \times (1 \otimes I) = 1 \times 1 = 1$ (in the third expression, the products exist trivially, and in the fourth the product exists because $\otimes$ preserves products). That is, we must have $I_{\tilde V} = 1_{\tilde V}$. But $I_{\tilde V}$ is the image of $I_V$ in $\tilde V$, and the inclusion into the idempotent completion reflects terminal objects. Therefore $V$ has a terminal object, and $1_V = I_V$.

Then, as observed in the comments above, the second part of Qiaochu's Eckmann-Hilton argument can be run in $V$: $A \otimes B = (A \times 1) \otimes (1 \times B) = (A \otimes 1) \times (1 \otimes B) = A \times B$ (in the second expression, the products exist trivially, and in the third the product exists because $\otimes$ preserves products). That is, binary products exist in $V$ and agree with $\otimes$. In fact, the identity functor is an oplax monoidal functor from $(V,\otimes)$ to $(V,\times)$, which the argument shows is actually strong monoidal. Thus $(V,\otimes) \simeq (V,\times)$ as monoidal categories.

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  • $\begingroup$ What a beautiful argument and elimination of the extra hypotheses! I wish I could accept both answers! $\endgroup$ – varkor Jan 3 at 23:12
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If $\otimes : V \times V \to V$ has a left adjoint and $V$ has finite products then $\otimes$ preserves them in the sense that the natural map

$$(X \times Y) \otimes (Z \times W) \to (X \otimes Z) \times (Y \otimes W)$$

is an isomorphism. By a monoidal-categorical version of the Eckmann-Hilton argument it seems to me that this implies that $\otimes$ is the product. Explicitly, if we let $1_{\times}$ denote the terminal object and $1_{\otimes}$ denote the monoidal unit then we get isomorphisms

$$1_{\otimes} \cong 1_{\otimes} \otimes 1_{\otimes} \cong (1_{\otimes} \times 1_{\times}) \otimes (1_{\times} \times 1_{\otimes}) \cong (1_{\otimes} \otimes 1_{\times}) \times (1_{\times} \otimes 1_{\otimes}) \cong 1_{\times} \times 1_{\times} \cong 1_{\times}$$

so $1_{\otimes} \cong 1_{\times}$ (and this isomorphism is unique if it exists so we don't even need to worry all that much about naturality). Now we can drop the outrageous subscripts and just refer to $1$. This gives a natural isomorphism

$$X \otimes Y \cong (X \times 1) \otimes (1 \times Y) \cong (X \otimes 1) \times (1 \otimes Y) \cong X \times Y$$

for any $X, Y$. Actually I'm not sure if this argument shows that the associator and unitor of $\otimes$ match up with the associator and unitor of the product but I'd guess a more elaborate version of this argument does.

I don't know if it's possible that $V$ doesn't have finite products. (There was previously an argument here involving Day convolution but Tim has pointed out gaps in it in the comments.)

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    $\begingroup$ Nice! In the last paragraph, what's not clear to me is whether $\hat \otimes$ inherits a left adjoint from $\otimes$ (so that the same argument can really be run with $\hat \otimes$). One tweak which might work better would be to work with the functor $Fam$ which freely completes under finite products. I believe that $Fam$ is 2-functorial (so that it preserves the adjunction) and preserves finite products (so that $Fam(\otimes)$ is a monoidal structure on $Fam(V)$ analogous to Day convolution). But the embedding $V \to Fam(V)$ doesn't preserve finite products, so perhaps I'm spouting nonsense. $\endgroup$ – Tim Campion Jan 3 at 21:37
  • $\begingroup$ Cannot really catch it but some kind of category of binary trees could provide a counterexample. I mean, $\ell$ and $r$ being left and right branch and $\otimes$ pruning (or how is it called) $\endgroup$ – მამუკა ჯიბლაძე Jan 3 at 21:41
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    $\begingroup$ @QiaochuYuan I'm not sure I follow the argument that the map $(X \times Y) \hat \otimes (Z \times W) \to ( X \hat \otimes Z) \times (Y \hat \otimes Z)$ is an isomorphism on $V$ -- the domain doesn't really simplify if finite products aren't known to exist in $V$. However, assuming that $1_\times$ exists, the isomorphism $X \otimes Y \cong X \times Y$ that you construct above doesn't require us to know that $X \times Y$ exists a priori -- only that $X \times 1$ and $1 \times Y$ do (which they do!). So maybe we don't need to think about Day convolution at all! $\endgroup$ – Tim Campion Jan 3 at 21:56
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    $\begingroup$ I think this could be used to prove a version of the no-cloning theorem. If there was a way to clone quantum information then you could use it to define such an adjoint on $(\mathrm{Vect}_\mathbb{C},\otimes)$, with the $l$ and $r$ of the question being identities. Then $\otimes$ is a product, contradiction. $\endgroup$ – Oscar Cunningham Jan 4 at 17:19
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    $\begingroup$ @QiaochuYuan: thank you! This is a very good answer (along with the argument via Day convolution that was here previously). It's tricky to know which answer to accept, because the process of developing the proof was very much a collaboration: I wish I could accept both answers. I've decided to accept Tim's answer, simply because it contains the full proof that $\mathscr V$ has finite products, but I appreciate that this answer already contains many of the key ideas. Sorry, and thank you again! $\endgroup$ – varkor Jan 4 at 23:59

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