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It is often claimed as "obvious" that a pair of adjoint functors: $L\colon{\cal V}\to {\cal M}$ and $R\colon{\cal M}\to {\cal V}$ defines a cotriple $(\bot, \epsilon, \delta)$ and a monad. What is wrong with the following counterexample?

Let ${\cal V}$ be the category of vector spaces over $\mathbb R$ and ${\cal M}$ be the category of modules over a Lie algebra $\mathfrak g.$ Then (I believe) the forgetful functor $R:{\cal M}\to {\cal V}$ is right adjoint to $L\colon{\cal V}\to {\cal M},$ $L(V)=V\otimes \mathfrak g.$

The counit $\epsilon\colon M\otimes \mathfrak g\to M$ is given by the action of $\mathfrak g$ on $M.$ In Weibel's "Intro to Homological Algebra", the construction of a monad is based on the following identity: $\epsilon\circ (LR\epsilon)=\epsilon\circ(\epsilon LR)$. That means that $(m\cdot x)\cdot y=m\cdot (x\cdot y)$ for $x,y\in \mathfrak g.$ But for Lie algebra modules we have $(m\cdot x)\cdot y=m\cdot (x\cdot y)+(m\cdot y)\cdot x$!

Przemysław Scherwentke
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student
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    try to prove for yourself that adjoint functors give rise to a monad (or read the proof online or in Mac Lane's book) and then you'll see that 1) it is obvious; 2) the answer to your question. – Ittay Weiss Oct 26 '13 at 19:57
  • Dear Ittay, I tried and I arrived at the above counterexample. Unfortuntely, your comment does not contribute anything to my question. MacLane claims it is "obvious". – student Oct 26 '13 at 20:11
  • @student What should exactly be the action of $\mathfrak g$ on $M$? $M$ should be $\mathfrak g$ module or just a module? – Giorgio Mossa Oct 26 '13 at 20:20
  • $M$ is a $\mathfrak g$-module. That means that the equality on the bottom of my post holds. – student Oct 26 '13 at 20:25
  • When given an adjunction such as yours you always get a monad by composing $R \circ L$, not the other way around, i.e. you get a monad on $\mathcal{V}$. The unit of the monad is the unit of the adjunction, and the multiplication of the monad is given by $R(\varepsilon L)$, i.e $\mu_X = R(\varepsilon_{F(X)})$. You seem to want to define a monad on $\mathcal{M}$. – Aleš Bizjak Oct 26 '13 at 20:44
  • Why do you think $\epsilon\circ (LR\epsilon)=\epsilon\circ(\epsilon LR)$ should hold? If your are explicit about this people can probably pinpoint your confusion. In the monad from an adjunction the multiplication is $R \epsilon L$, so the monad laws you need to check will involve that and not naked $\epsilon$'s or the combination $LR\epsilon$. – Omar Antolín-Camarena Oct 27 '13 at 02:00
  • @Omar -- to make things specific I am interested in the property: $\epsilon(\epsilon \bot)=\epsilon(\bot\epsilon)$, where $\epsilon$ is the counit of adjunction (see eg. Exercise 8.6.2 in Weibel "Intro to Homological Algebra"). My example seems to contradict it. I believe this identity is necessary to construct a monad. That is how Weibel does it. Cf. my comment below. – student Oct 27 '13 at 04:16
  • @Ales -- please see my comment for Omar. – student Oct 27 '13 at 04:16
  • @OmarAntolín-Camarena that identity holds simply because $\varepsilon$ is natural. – Aleš Bizjak Oct 27 '13 at 06:38
  • "cotriple" is deprecated nowadays. Better use "comonad" instead. See CWM for terminology. – magma Oct 27 '13 at 12:18
  • @AlešBizjak The OP is interested in the comonad, not in the monad – magma Oct 27 '13 at 12:29
  • @magma Yes, I know that now but it was not clear what he's asking at the time I posted the comment. – Aleš Bizjak Oct 27 '13 at 13:08
  • @AlešBizjak: you're right of course, I don't know what I was thinking. – Omar Antolín-Camarena Oct 27 '13 at 17:19
  • Oh, sorry, I only posted the first comment because I believed you were right in saying that the identity was false! But now I'll ask something else: are you sure the free $\mathfrak{g}$-module on $V$ is $\mathfrak{g} \otimes V$? Because the definition of a $\mathfrak{g}$-module is rigged to match that of a module over the universal enveloping algebra $U(\mathfrak{g})$, I would expect the free module to be $U(\mathfrak{g}) \otimes V$. – Omar Antolín-Camarena Oct 27 '13 at 17:19

3 Answers3

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I've edited the answer to address the real problems of the question and to shorten it a little bit.

The data you have indicated cannot give an adjunction, the problem being in the supposed counit. The details follows.

The point is the the morphisms $\epsilon \colon M \otimes \mathfrak g \to M$ in order to be the (components of the) counit of the adjunction should be morphism in the category $\mathcal M$, i.e. they should be morphisms of lie algebras. And that's not possible here's why.

Ofcourse since the action of $\mathfrak g$ over a module $M$ is a bilinear map we can identify the action with a linear map $\epsilon_M \colon M \otimes \mathfrak g \to M$ which is the mapping such that $$\epsilon_M(m \otimes x) =m \cdot x$$ for $m \in M$ and $x \in \mathfrak g$, what we said until now tells that $\epsilon$ should satisfy the equality $$\epsilon_M( (m \otimes x) \cdot y) = \epsilon(m \otimes x) \cdot y$$ where the action $\cdot$ on the left is the action in the $\mathfrak g$-module $M \otimes \mathfrak g$ while in the action on the right is that in the module $M$.

Now the last equation could be rewritten as $$m \cdot [x,y] = (m \cdot x)\cdot y$$ that as you said in the question doesn't hold. So the $\epsilon$ is not a family of morphisms in $\mathcal M$ and so cannot be the counit of an a adjunction.

Just for completeness and to convince you that the property of adjoint functors that you stated, namely that the equality $\epsilon \circ LR(\epsilon) = \epsilon \circ \epsilon_{LR}$, holds I've written a proof below.

A little notation: $F \colon \mathcal X \to \mathcal A$ and $R \colon \mathcal A \to \mathcal X$ are the adjoint functors, $\varphi \colon \mathcal A(L(-),-) \cong \mathcal X(-,R(-))$ is the adjunction and $\epsilon \colon LR \Rightarrow 1_\mathcal{A}$ is the counit.

Now by the properties of adjunction we get that for every object $A \in \mathcal A$ $$\epsilon_A \circ \epsilon_{LR(A)}=\mathcal A(L(1_{R(M)}),\epsilon_A)\circ \varphi^{-1}(1_{RLR(A)})$$ $$\epsilon_A \circ \epsilon_{LR(A)} = \varphi^{-1} \circ \mathcal V(1_{R(A)},R(\epsilon_A))(1_{RLR(A)})$$ $$\epsilon_A \circ \epsilon_{LR(A)} = \varphi^{-1}(R(\epsilon_A))$$ and that $$\epsilon_A \circ LR(\epsilon_A) = \mathcal M(LR(\epsilon_A),1_{LR(A)}) \circ \varphi^{-1}(1_{R(A)})$$ $$\epsilon_M \circ LR(\epsilon_A) = \varphi^{-1}\circ \mathcal V(R(\epsilon_A),R(1_{LR(A)}))(1_{R(A)})$$ $$\epsilon_A \circ LR(\epsilon_A) = \varphi^{-1}(R(\epsilon_A))$$ this proves the so wished equality.

Giorgio Mossa
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  • @Gorgio: You are right about the set up of monad, however I am afraid that my contradiction remains.

    Details: my $\epsilon$ is the "counit of adjunction". What I am really interested is the cotriple $(\bot, \epsilon, \delta)$ induced by my adjoint functors. By definition of cotriple, $\epsilon$ should satisfy $\epsilon(\epsilon \bot)=\epsilon(\bot\epsilon)$, and my example seems to contradict that.

    "Intro to Homological Algebra" by Weibel and others use this identity for $\epsilon$ to prove "coherence conditions" for $\mu=R\epsilon L$ which you mention.

     – student Oct 27 '13 at 03:55
  • @student I think I've found your mistake take a look and let me know if it convinces you :) – Giorgio Mossa Oct 27 '13 at 10:24
  • @GiorgioMossa the OP is interested in the comonad – magma Oct 27 '13 at 12:48
  • @student Where exactly in Weibel are you looking at? – magma Oct 27 '13 at 12:49
  • @magma I dont understand your edit to Giorgio's response. Which "result" do you mean saying "Now that you apply ϵM to the result"? Also, I am using ⋅ for the Lie product in g or the action of g on V.You seem to be using it in some (undefined?) associative way(?), since you use [,] for Lie bracket. I dont understand your last equation and how it resolves the problem, since we need to have ϵ(ϵ⊥)=ϵ(⊥ϵ) which does not have any minus sign. – student Oct 27 '13 at 22:40
  • @student I've edited the answer, I hope I made more clear why the equation should hold in this case. If you're interested in seeing a proof of the general result for the comonad induced by the adjoint pair of functor I'll ask you to be a little patient, since I've to write it down. Usually to prove that $LR,\mu,\eta$ induce a monad I've always followed Maclane's direct approach :) – Giorgio Mossa Oct 27 '13 at 23:11
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    @student I just edited a small typo. No math or meaning. Anything else what edited by someone else. – magma Oct 27 '13 at 23:22
  • @student after some long and accurate (I hope) calculations I've finally found the problem. As apologize for being a little messy before I've also added the proof that $\epsilon \circ LR \epsilon = \epsilon \circ \epsilon_{LR}$ for $\epsilon$ a counit of an adjunction. – Giorgio Mossa Oct 28 '13 at 15:20
  • @Georgio Thanks for the proof! The only confusing part is that you seem to assume that $L$ and $R$ are adjoint, while now I am convinced that they are not! Does it mean that you disagree with the solution I posted? At what point? – student Oct 28 '13 at 23:28
  • @student Well if you are referring to the last proof then that's just the proof of the exercise in Weibel's book, i.e. that for two generic adjoint functor the equality involving the $\epsilon$ should holds. Instead I'm also convinced that the pair of functor that you have indicated are not adjoint, at least not with the counit you've indicated. – Giorgio Mossa Oct 29 '13 at 07:00
  • Now I see probably the answer is too long and confusing, I'll try to shorten it a little bit. – Giorgio Mossa Oct 29 '13 at 07:03
  • @student I've edited the answer. Let me know if now it's more clear. – Giorgio Mossa Oct 29 '13 at 07:23
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I would leave this as a comment, but see Mac Lane's "Categories for the Working Mathematician", second edition, p.138. He does not leave it at "it's obvious", but explains very clearly.

EDIT: Also, it took me a bit to notice this, but draw out the identity $\epsilon\circ LR\epsilon = \epsilon\circ \epsilon LR$ as a commutative diagram and observe that it's a naturality square for $\epsilon$. If your "counterexample" holds, $\epsilon$ isn't even a natural transformation, and thus not the counit of an adjunction.

Malice Vidrine
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  • Exactly! As a matter of fact, the same explanation appears in the first edition of CWM too. – magma Oct 27 '13 at 12:30
  • Nor would it have been appropriate to label it "obvious", since (as Mac Lane well knew) all this took some time to work out when it was still new (in the early 1960's). – user43208 Oct 27 '13 at 12:46
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I think I found the error: My $L$ and $R$ are not adjoint! For an associative algebra $A,$ over $\mathbb R$, the forgetful functor $R$ is right adjoint to $L: {\mathcal V}\to {\mathcal M}$ (= category of right $A$-modules) by sending for example $z\in M=Hom_{\mathbb R}({\mathbb R},M)$ to $r_z\in Hom_{A}(A,M),$ $r_z(a)=z\cdot a.$ This is an $A$-module homomorphisms because $r_z(ab)=r_z(a)b$. But that fails when $A$ is not associative!

Thank you all for your comments, especially @Omar who pointed in that direction!

student
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  • I don't think you were wrong. It should be a pair of adjoint functor, and I'm pretty much sure that your mapping is exactly the counit. – Giorgio Mossa Oct 27 '13 at 23:38
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    @GiorgioMossa: He was wrong, the left adjoint to the forgetful functor is $V \mapsto U(\mathfrak{g}) \otimes V$, so it cannot also be $V \mapsto \mathfrak{g} \otimes V$ (a left adjoint is uniquely determined up to natural isomorphism by the right adjoint). – Omar Antolín-Camarena Oct 28 '13 at 00:13
  • I'd just add that the category of $\mathfrak{g}$-modules is a category of left modules for an associative algebra, but that associative algebra is the universal enveloping algebra $U(\mathfrak{g})$ (not $\mathfrak{g}$, which, as you point out, is not associative). – Omar Antolín-Camarena Oct 28 '13 at 00:17