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• CommentRowNumber1.
• CommentAuthorTodd_Trimble
• CommentTimeMar 8th 2012
• (edited Mar 8th 2012)

This comment is directed to MO user ’grok’, who posted a question here to which I gave an answer. Of course, anyone else should feel to chime in; I should very much like to know it if someone spots an error in the answer I gave.

Hi grok. I took a look at the link you provided, and the first question I have is this. To keep things simple for the moment, let’s suppose $C$ and $D$ are finite-dimensional cocommutative coalgebras. You start with a concrete description of the measuring cocommutative coalgebra $M(C, D)$ as sitting inside $Sym(Hom(C, D)^\ast)^\circ$ (I’ve replaced your $T$ by $Sym$, which is what I think you intended but didn’t edit in), and thus in a larger ambient space of commutative formal power series

$Sym(Hom(C, D)^\ast)^\ast \cong (\sum_{n \geq 0} S^n(Hom(C, D)^\ast))^\ast \cong \prod_{n \geq 0} S^n(Hom(C, D)^\ast)^\ast \cong \prod_{n \geq 0} S^n(Hom(C, D))$

($S^n$ being the $n^{th}$ symmetric power functor) in variables given by linear maps $\phi: C \to D$. That certainly accords with my understanding. (By the way, I should mention that I have never looked at Sweedler’s book, and I should further admit that I don’t count myself as an expert in this area, although I’ve done some thinking about the category of cocommutative coalgebras and wrote a lot of what is in the nLab on this subject, which still isn’t much yet.)

You then seem to be invoking a map

$\prod_{n \geq 0} S^n(Hom(C, D)) \to \prod_{n \geq 0} Hom(S^n(C), S^n(D)) = grHom(Sym(C), Sym(D))$

where you take a product over all $n$ of maps $S^n(Hom(C, D)) \to Hom(S^n(C), S^n(D))$. My first question (which may be a stupid one) is: why are these maps well-defined? For example, take $n=2$. Typical elements in $S^2(Hom(C, D))$ will be given by (linear combinations of) expressions $\phi_1 \otimes \phi_2$, but modulo an equivalence $\phi_1 \otimes \phi_2 \sim \phi_2 \otimes \phi_1$. I don’t see how the map described as

$c_1 \otimes c_2 \mapsto \phi_1(c_1) \otimes \phi_2(c_2)$

respects such equivalences.

• CommentRowNumber2.
• CommentAuthorgrok
• CommentTimeMar 13th 2012

Dear Todd:

I possibly got confused in trying to fit everything in commutative algebra. However, I think I really want to look first at $T(Hom(C,D)^*)^\circ$, the noncommutative power series in $Hom(C,D)$. There is then a well-defined map into $U:=grHom(T(C),T(D))$. Now $U$ is far too big, so we extract smaller and smaller subspaces till we get close to what we want. First step is to restrict to those graded homomorphisms that map $\langle c\otimes c'-c'\otimes c\rangle$ to $0$, namely that induce a graded homomorphism $S(C)\to S(D)$. Second and third steps are to restrict to graded homomorphisms that interlace the counit and coaugmentation. This is what I did in the note I posted on my website; and hopefully it does give the measuring coalgebra. In all cases, it seems to be the group-like coalgebra on Map(X,Y) when $C=k X$ and $D=k Y$.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeMar 13th 2012
• (edited Mar 13th 2012)

Hi grok –

I can’t tell how closely you examined my answer at MO. Is there any part of the reasoning that you think is wrong, or that needs clarification?

I will see whether I can sort out what you’re saying here, but it might help things along if you can state what you think the universal property of measuring coalgebra ought to be. From my point of view, the correct notion of measuring coalgebra $M(C, D)$ satisfies a universal property, namely that for the category of cocommutative coalgebras, we have a natural isomorphism

$CocommCoalg(E \otimes_k C, D) \cong CocommCoalg(E, M(C, D))$

(which, in the language I am accustomed to, means that cocommutative coalgebras form a closed monoidal category, in fact a cartesian closed category since $\otimes_k$ gives the cartesian product for cocommutative coalgebras).

If you are working with some other notion of measuring coalgebra, it would save us time if we both knew that in advance.

(Note: when I write $C(c, d)$ for a category $C$, I mean the hom-set $\hom_C(c, d)$.)

• CommentRowNumber4.
• CommentAuthorgrok
• CommentTimeMar 14th 2012

Dear Todd –

I should have clarified that. I examined very closely your answer on MO, and can’t find anything wrong there. I like very much the point of view of converting everything to Hom(Alg_fd,Set). I also agree that the measuring algebra should have the universal property of an internal hom.

However, I started with the construction of the measuring algebra given by Fox, which describes it as a subalgebra of the free coalgebra $U$ on $Hom(C,D)$. Taking finite-dimensional group-like $C,D$, say with group-like bases $\{c_i\}$ and $\{d_j\}$, I get a description of $U$ as an appropriate set of non-commutative power series in $\{e^i_j\}$, and plodding through the axioms seems again to indicate that $M(C,D)$ is group-like on the set of maps from $\{c_i\}$ to $\{d_j\}$. My formulation as a subset in $grHom(TC,TD)$ was just intended to simplify the notation.

So, in summary: what worries me is that the constructive approach from Fox gives me the answer I want, but I have more faith in the universal property characterization, which gives me the answer I don’t want.

As you’ll have guessed, I’m not an expert on coalgebras :)

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeMar 14th 2012

Dear grok –

Thank you for your candor, and for taking the time to look carefully at my argument. I have only recently cast my eyes at Fox’s paper, but I haven’t read it closely.

If it’s all right with you, I might contact Tom Fox myself. (He’s not someone I know well, but I’ve met him a few times at McGill and at category theory conferences.) I don’t have to say anything about who or where this problem comes from. But if you’re not comfortable with my doing that, then I would like to know (and you can contact me more privately at my gmail dot com address, where you should type topological dot musings before the at sign). I mention this possibility because I am guessing both of us would consider a firm answer from him as definitive.

I don’t regard myself as an expert on coalgebras either, and I haven’t tried to work out where a problem might be in your calculations (beyond the comment I made in #1), but are you reasonably confident that your construction gives you a cocommutative coalgebra?

And one more question: can I ask why you want the answer to be one way and not the other? Is it because it would enable some argument to go through more easily, or something like that? In my own experience, it’s not too common that a product-preserving functor (in this case, $k(-): Set \to CocommCoalg$) between cartesian closed categories, “in the wild” as it were, also preserves the internal hom up to isomorphism (it happens sometimes to be sure, but it’s somewhat attention-grabbing when it does, at least for me). So I was wondering why it’s important to you that it turn out this way here.

• CommentRowNumber6.
• CommentAuthorTodd_Trimble
• CommentTimeMar 15th 2012

(For anyone following this thread, it turns out I was being an idiot. Something ’grok’ said in private email made the scales fall from my eyes, and it turns out his original surmise was right, in complete generality. My MO answer has been updated to reflect this state of affairs, and I consider this a happy ending to the story.)

• CommentRowNumber7.
• CommentAuthorgrok
• CommentTimeMar 16th 2012

Dear Todd – Sorry to harrass you, but it’s not the end of the story yet. Here are more comments and questions:

CLAIM. Similar arguments work for Lie-type coalgebras. To wit: given a Lie algebra $g$, it has a universal enveloping algebra $U(g)$, which is a commutative coalgebra. It is represented by the functor

$A\mapsto CoAlg(A^*,U(g)) = Lin(A^*,g).$

Then one represents $M(U(c),U(d))$ by $A\mapsto Lin(A^*\otimes U(c),d)=Lin(A^*,Lin(U(c),d))=CoAlg(A^*,U(Lin(U(c),d))$. We have proven

For any Lie algebras $c,d$ we have $M(U(c),U(d))=U(Lin(U(c),d))$.

Do you agree with the proof?

Now come some more questions. I appreciate the value of working categorically, i.e. element-freely. However, I also want more out of $M(C,D)$. Assume that $D$ is a Hopf algebra. Then $M(C,D)$ should also be a Hopf algebra. Seeing $M(C,D)$ as sitting inside the free coalgebra on $Hom(C,D)$, this is obvious, because $Hom(C,D)$ is a Hopf algebra for convolution. How do I see the Hopf algebra structure on $Lex(Alg_{fd},Set)$?

Assume then that $C$ is a Hopf algebra. Then I would like to see that $M(C,D)$ is a Hopf $C$-module.

Finally, I would like to see the evaluation map $M(C,D) \to Hom(C,D)$. What is its translation in $Lex(Alg_{fd},Set)$?

• CommentRowNumber8.
• CommentAuthorTodd_Trimble
• CommentTimeMar 18th 2012
• (edited Mar 18th 2012)

Sorry to harrass you, but it’s not the end of the story yet.

No problem! I just meant “end of story” in deciding that your MO surmise that $k(-): Set \to CocommCoalg$ preserves exponentials was correct after all, contrary to what I initially thought. I know that the surmise is embedded in a larger story. :-)

Here are some reactions.

It is represented by the functor $A \mapsto CoAlg(A^\ast, U(g))=Lin(A^\ast, g)$.

In general, it would be an interesting and worthwhile exercise to collect a bunch of examples of cocommutative coalgebras and see which left exact functors they correspond to, and put the results somewhere on the Lab. Here’s what I get in this case.

One version of the Poincaré-Birkhoff-Witt theorem says that as a coalgebra, the universal enveloping algebra $U(g)$ is isomorphic to $Sym(g) = \sum_{n \geq 0} g^{\otimes n}/n!$, where the division by $n!$ indicates quotienting by an action of the symmetric group (written this way to bring out an analogy with the exponential function $\exp(z) = \sum_{n \geq 0} z^n/n!$). So one might as well see what $Sym(g)$ looks like as a left exact functor.

The isomorphism $Coalg(A^\ast, U(g)) \cong Lin(A^\ast, g)$ doesn’t look quite right to me, because we ought to have instead $Coalg(A^\ast, Cofree(g)) \cong Lin(A^\ast, g)$ (cofree cocommutative coalgebra on the underlying vector space of $g$), and $Cofree(g)$ is much larger than $Sym(g)$.

My own calculations lead me to

$Coalg(A^\ast, Sym(V)) \cong Vect(Nil(A)^\ast, V)$

for any vector space $V$, where $Nil(A)$ is the ideal of nilpotent elements of $A$. This can be approached in stages. If $V$ is 1-dimensional, then $Sym(V)$ is a space of polynomials $k[x]$ with its usual deconcatenation coalgebra structure. This is a filtered colimit of finite-dimensional subcoalgebras $Span\{1, \ldots, x^{n-1}\}$, each with a algebra dual given by $k[x]/(x^n)$, where algebra maps $k[x]/(x^n) \to A$ pick out elements $a$ such that $a^n = 0$. We have

$\array{ Coalg(A^\ast, Sym(k)) & = & Coalg(A^\ast, colim_n (k[x]/(x^n))^\ast) \\ & \cong & colim_n Coalg(A^\ast, (k[x]/(x^n))^\ast) \\ & \cong & Nil(A) \\ & \cong & Vect(Nil(A)^\ast, k) }$

where the second line holds because $Coalg(A^\ast, -)$ preserves filtered colimits since $A$ is finite-dimensional. For $V$ finite-dimensional, $V \cong k \oplus \ldots \oplus k$, we have

$Sym(V) = Sym(k) \otimes \ldots \otimes Sym(k)$

where the right side is a cartesian product of cocommutative coalgebras. So $Coalg(A^\ast, Sym(k \oplus \ldots \oplus k)) \cong Nil(A) \times \ldots \times Nil(A)$, which we can rewrite as

$Vect(V^\ast, Nil(A)) \cong Vect(Nil(A)^\ast, V)$

by naturality considerations. Finally, for $V$ arbitrary, we can write $V$ as a filtered colimit of finite-dimensional spaces $V_i$. Since $Sym(-): Vect \to Coalg$ and $Coalg(A^\ast, -)$ both preserve filtered colimits, we easily calculate

$Coalg(A^\ast, Sym(V)) \cong Vect(Nil(A)^\ast, V)$

in general.

Now come some more questions. I appreciate the value of working categorically, i.e. element-freely. However, I also want more out of $M(C,D)$. Assume that $D$ is a Hopf algebra. Then $M(C,D)$ should also be a Hopf algebra. Seeing $M(C,D)$ as sitting inside the free coalgebra on $Hom(C,D)$, this is obvious, because $Hom(C,D)$ is a Hopf algebra for convolution. How do I see the Hopf algebra structure on $Lex(Alg fd, Set)$?

If we are working with cocommutative coalgebras throughout, then a very conceptual argument can be given as follows. (If you do not wish to restrict to the cocommutative setting, let me know.)

Cocommutative Hopf algebras are precisely group objects in the category of cocommutative coalgebras. Since group objects are definable in any category with cartesian products, and since the functor $(-)^C = M(C, -)$ preserves products (because $(-)^C$ is a right adjoint, by cartesian closure), it must take group objects to group objects.

In terms of $Lex(Alg_{fd}, Set)$, the desired Hopf algebra multiplication is a coalgebra map

$m: M(C, D) \otimes M(C, D) \to M(C, D)$

that induces a map

$Coalg(A^\ast, M(C, D)) \times Coalg(A^\ast, M(C, D)) \to Coalg(A^\ast, M(C, D))$

or in other words

$Coalg(A^\ast \otimes C, D) \times Coalg(A^\ast \otimes C, D) \to Coalg(A^\ast \otimes C, D)$

which we can rewrite again in the form

$Coalg(A^\ast \otimes C, D \otimes D) \to Coalg(A^\ast \otimes C, D)$

and this is gotten by applying $Coalg(A^\ast \otimes C, -)$ to the Hopf algebra multiplication $D \otimes D \to D$. There are ways to make this even more explicit-looking by manipulating units and counits of adjunctions, but this is the general idea.

Assume then that $C$ is a Hopf algebra. Then I would like to see that $M(C,D)$ is a Hopf $C$-module.

When you say “assume then”, does the ’then’ mean we are still assuming $D$ is a Hopf algebra?

Under the cocommutativity assumption, we can take advantage of the fact that $CocommCoalg$ is cartesian closed and think of $M(C, D)$ is an exponential $D^C$, and think of $\otimes$ as the categorical cartesian product $\times$. My best guess for what you want is that we first have a map

$C \otimes C \otimes D^C \stackrel{1 \otimes 1 \otimes \Delta}{\to} C \otimes C \otimes D^C \otimes D^C \cong C \otimes D^C \otimes C \otimes D^C \stackrel{ev \otimes ev}{\to} D \otimes D \stackrel{m}{\to} D$

and this $C \otimes C \otimes D^C \to D$ corresponds to a map $C \otimes D^C \to D^C$ which at least has the form you want. Does this look like the thing you’re after?

Finally, I would like to see the evaluation map $M(C,D) \to Hom(C,D)$. What is its translation in $Lex(Alg_{fd},Set)$?

Details to come.

• CommentRowNumber9.
• CommentAuthorTodd_Trimble
• CommentTimeMar 18th 2012

I wrote

My best guess for what you want is

and a stupid guess it was. If I were applying my brain better, I would have written something more like

$C \otimes D^C \stackrel{1 \otimes D^m}{\to} C \otimes D^{C \otimes C} \cong C \otimes (D^C)^C \stackrel{ev}{\to} D^C$

where the middle isomorphism is an instance of $Z^{X \otimes Y} \cong (Z^Y)^X$. We’re not using a Hopf algebra structure on $D$, just the coalgebra structure.

The idea behind the formula is to capture, in the language of cartesian closed categories, a mapping which in $Set$ would be interpreted as

$(c, f) \mapsto (c' \mapsto f(c c'))$

where $f$ is an element in $D^C$, i.e., a function $C \to D$. Alternatively, we might want something like

$(c, f) \mapsto (c' \mapsto f(c' c^{-1}))$

which would involve the antipode.

More later…

• CommentRowNumber10.
• CommentAuthorgrok
• CommentTimeMar 18th 2012

Thanks, I’m starting to get the feel now. Sorry about my too-quick post on u.e.a., you’re right that one must only look at nilpotent $A$.

I guess a more fundamental question I should have asked is: how does one get from the functor to the coalgebra? I.e., is there a map $CoAlg(A^*,C)\to C$? I.e., I have a functor that gives me, for every f.d. algebra $A$, a set of coalgebra morphisms. Can I construct a vector space $D$ and a linear map $D\to D\otimes D$ that turn $D$ into a coalgebra isomorphic to $C$? Somehow I feel that this is the wrong question to ask; but it would help if it were true.

Now, evaluation: ideally, it would be a map $CoAlg(A^*\otimes C,D) \times C\to D$. In view of the previous paragraph, let’s rather ask for a map $CoAlg(A^*\otimes C,D)\times C\to CoAlg(A^*,D)$. This map is easy to define: for a coalgebra morphism $\phi:A^*\otimes C\to D$ and $c\in C$, I just consider the morphism $A^*\to D$ given by $\phi(-\otimes c)$.

On the other hand, if I wanted a map $CoAlg(A^*\otimes C,D)\times CoAlg(A^*,C)\to CoAlg(A^*,D)$, I would send $(\phi,\psi)$ to the morphism $f\mapsto \phi(f')\psi(f'')$ in Sweedler notation for $f\in A^*$.

In summary, it seems that whatever I wrote about free coalgebras on $Hom(C,D)$ and appropriate subcoalgebras of $grHom(...)$ can be deleted, and everything I will ever need (except possibly putting my hand on a concrete element of a coalgebra) can be packaged into left exact functors.

• CommentRowNumber11.
• CommentAuthorgrok
• CommentTimeMar 18th 2012

Here is an attempt to answer my own question. To $C$ a coalgebra, Todd associates the functor $CoAlg(-^*,C)$; namely, a bunch of sets $C_A$, one for each f.d. algebra $A$, and natural maps $C_A\to C_B$ for every algebra homomorphism $A\to B$. (if $f:A\to B$ and $\Phi$ is a coalgebra morphism $A^*\to C$, then $f(\Phi)(x) := \Phi(x\circ f)$ defines a coalgebra morphism $f(\Phi):B^*\to C$.)

Out of these $C_A$, one can probably reconstruct $C$ by taking the vector space spanned by $\bigsqcup_A C_A$, and quotient by all algebra homomorphisms.

• CommentRowNumber12.
• CommentAuthorTodd_Trimble
• CommentTimeMar 18th 2012

In summary, it seems that whatever I wrote about free coalgebras on $Hom(C, D)$ and appropriate subcoalgebras of $grHom(...)$ can be deleted, and everything I will ever need (except possibly putting my hand on a concrete element of a coalgebra) can be packaged into left exact functors.

Possibly that’s true, although there will certainly always be readers who are more comfortable with elementwise nuts-and-bolts descriptions. It’s good to feel at home in both worlds!

I guess a more fundamental question I should have asked is: how does one get from the functor to the coalgebra?

That is indeed the fundamental question, and I think this (i.e., the equivalence between coalgebras and left exact functors) is also at the root of your other question about expressing the evaluation homomorphism in the language of left exact functors.

There’s a big overarching categorical theory surrounding these issues, namely the theory of locally presentable categories, and its earlier predecessor, the theory of locally finitely presentable categories – key words here are “Gabriel-Ulmer duality”. Basically we are applying Gabriel-Ulmer duality to a special case.

A brief review will suffice here. A locally small category $C$ (which in the case at hand will be the category of coalgebras or of cocommutative coalgebras over a field $k$) is locally finitely presentable if

• The category is small-cocomplete;

• There is a small full subcategory of “finitely presentable objects” (sometimes called “compact objects”) $\mathcal{K}$ such that every object in $C$ is a filtered colimit of objects belonging to $\mathcal{K}$. An object $c$ of $C$ is finitely presentable if $\hom(c, -): C \to Set$ preserves filtered colimits.

(Note: there are various ways of defining locally finitely presentable category; what I have written is not literally what is in the nLab, but it is the one given by Adámek and Rosicky, Locally Presentable and Accessible Categories, which is a very good source.)

For $C$ the category of coalgebras, the finitely presentable objects are finite-dimensional coalgebras. There are of course only a small set of isomorphism classes of finite-dimensional coalgebras. The fundamental theorem of coalgebras says that every coalgebra is the union (which gives an example of a filtered colimit) of its finite-dimensional subcoalgebras, so the category of coalgebras is indeed locally finitely presentable.

In this situation, Gabriel-Ulmer duality guarantees that $C$ is equivalent to the category of left-exact functors $\mathcal{K}^{op} \to Set$, and that of course is what I’m using here. In the present circumstance, the equivalence

$Lex(Coalg_{fd}^{op}, Set) \to Coalg$

takes a left exact functor $F: Coalg_{fd}^{op} \to Set$ which – and here is a key fact – can be canonically represented as a filtered colimit of representables $\hom(-, c_i)$ (is a “flat presheaf” in the parlance), and sends this to the coalgebra given by the corresponding filtered colimit $colim_i c_i$ in $Coalg$. Needless to say, the $c_i$ are finitely presentable, meaning here finite-dimensional coalgebras.

The functor $Coalg \to Lex(Coalg_{fd}^{op}, Set)$ is just as you surmised: it sends a coalgebra $c$ to

$C(i-, c): Coalg_{fd}^{op} \to Set$

(where $i$ denotes the full inclusion functor $Coalg_{fd} \to Coalg$).

There are lots of wonderful facts about locally presentable categories (which of course includes locally finitely presentable categories). One is an adjoint functor theorem, to the effect that a colimit-preserving functor between locally presentable categories automatically has a right adjoint (NB: no solution set condition needed!).

This has the fairly immediate consequence that for any coalgebra $C$, the functor $C \otimes -: Coalg \to Coalg$ has a right adjoint. (Colimits in the category of coalgebras are computed just as they are at the level of their underlying vector spaces, and of course $C \otimes -$ preserves colimits in $Vect$.) This shows that the category of coalgebras is a closed monoidal category. And that the category of cocommutative coalgebras is a cartesian closed category (since the monoidal product $\otimes$ behaves as a categorical product there).

This gives a sketch of ingredients one would need to flesh out the evaluation homomorphism of coalgebras in terms of left exact functors.