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I'm learning "An Introduction to Exterior Differential Systems" by Gregor Weingart. I have some confusions regarding comodules:

  1. In section 3, the author defines a comodule: enter image description here

  2. Later, the "space of generator" of a comodule is defined: enter image description here

Here is my confusion:

  1. It seems that the standard definition of comodule is a vector space $X$ with a linear map $\rho: X \to X \otimes C$ with coassociativity condition, where $C$ is a coalgebra. How is the definition in this article consistent with the standard one?
  2. What's the meaning of this "space of generators"?

The author didn't give explanations in the article, and I can't find revelent references, so I have to ask here.

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The given definition seems to be the unwrapping of the definition of a comodule over the coalgebra $C:=\mathrm{Sym}T^{*}$ (in the category of graded vector spaces), presented in slightly modified way. I write 'seems' since I don't have access to the mentioned book, so some definitions may need to be compared.

The coproduct of the symmetric coalgebra $C$ is defined on the homogeneous component $T^{*}$ of degree one as $f\mapsto1\otimes f+f\otimes1\in C\otimes C$. One may extend it to a morphism of graded algebras $\Delta:C\to C\otimes C$ that gives a coassociative coproduct on $C$. The counit \varepsilon is the projection onto the zeroth component $\mathrm{Sym}^{0}T^{*}\cong\Bbbk$.

A comodule of $C$ (in the category of graded vector spaces) is a graded vector space $\mathcal{A}$ with a (homogeneous) morphism $\rho:\mathcal{A}\to\mathcal{A}\otimes C$ such that $(\mathrm{id}_{\mathcal{A}}\otimes\Delta)\circ\rho=(\rho\otimes\mathrm{id}_{C})\circ\rho$ and $(\mathrm{id}_{\mathcal{A}}\otimes\varepsilon)\circ\rho=\mathrm{id}_{\mathcal{A}}$. To compare the above definition with the explicit one quoted in the question, define the maps $\rho_{k}\overset{\textrm{def}}{=}(\mathrm{id}_{\mathcal{A}}\otimes\mathrm{pr}_{k})\circ\rho$ for all $k\in\mathbb{N}$ where $\mathrm{pr}_{k}:C\to C^{k}$ is the projection. By counitality, $\rho_{0}(a)=a\otimes1$. We claim that $\rho_{1}$ uniquely determines $\rho_{k}$ for all $k>1$. Indeed, applying $\mathrm{id}_{A}\otimes\mathrm{pr}_{k}\otimes\mathrm{pr}_{l}$ on the above equation gives$$\Big(\mathrm{id}_{\mathcal{A}}\otimes\big((\mathrm{pr}_{k}\otimes\mathrm{pr}_{l})\circ\Delta\big)\Big)\circ\rho_{k+l}=(\rho_{k}\otimes\mathrm{id})\circ\rho_{l}$$for all $k,l\geq0$. Note that $(\mathrm{pr}_{k}\otimes\mathrm{pr}_{l})\circ\Delta$ is left-invertible with one-sided inverse ${k+l \choose k}^{-1}\cdot m$ where $m:C\otimes C\to C$ is the multiplication. Therefore, $\rho_{k}$ and $\rho_{l}$ determine $\rho_{k+l}$ for all $k,l\geq 0$. It means that any (homogeneous) comodule structure map $\rho:\mathcal{A}\to\mathcal{A}\otimes\mathrm{Sym}T^{*}$ is uniquely determined by the (homogeneous) $\rho_{1}:\mathcal{A}\to\mathcal{A}\otimes T^{*}$ and hence by the associated bilinear map $\partial:T\times\mathcal{A}\to\mathcal{A}$ (of degree -1 for all fixed $t\in T$) given as $(t,a)\mapsto\rho_{1}(a)(t)$.

In fact, one can prove that the range of this assigment $\rho\mapsto\partial$ consists of all the $\partial$'s satisfying $\partial(t_{1},\partial(t_{2},a))=\partial(t_{2},\partial(t_{1},a))$ for all $t_{1},t_{2}\in T$ and $a\in\mathcal{A}$. The equation stands for the fact that $\mathrm{Sym}T^{*}$ is cocommutative.

Regarding the second question, the space of generators can also be found in the literature under the name coinvariants of a comodule, defined as $\{a\in\mathcal{A}\ |\ \rho(a)=a\otimes1_{C}\}$. The intuitive interpretation of the term may depend on the context. For example, if the coalgebra is $\mathcal{O}(G)$, the regular functions on an algebraic group (with its standard Hopf-algebra structure) then the coinvariants of a comodule $\mathcal{O}(X)$ for some homogeneous space $X$ are the $G$-invariant functions.

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  • $\begingroup$ Thank you for answering. Here's what I found following your answer. In a basis $\{t_1,\ldots,t_n\}$ of $T$ and the dual basis $dt_1,\ldots,dt_n$ of $T^*$, $\rho_k:A \to A \otimes \mathrm{Sym}^k T^*$ looks like $a \mapsto \sum_{i_1,\ldots,i_k}\partial^{i_1\ldots i_k}a\otimes dt_{i_1}\ldots dt_{i_k}$, which means the homogeneous components of $\rho(a)$ are all non-zero, so $\rho$ is actually a map from $A$ to $A \otimes \prod_k\mathrm{Sym}^k T^*$, which is strictly larger than $A \otimes \mathrm{Sym}T^*=A \otimes \bigoplus_k\mathrm{Sym}^k T^*$. This seems not right. Could you give more details? $\endgroup$
    – AaronS
    Commented Mar 29, 2017 at 15:37
  • $\begingroup$ And regarding the space of generators, I still don't understand. The coinvariants are about $1_C$, how could they be defined in terms of $0_C$? $\endgroup$
    – AaronS
    Commented Mar 29, 2017 at 15:39
  • $\begingroup$ The map $\rho$ is defined to have values in $A\otimes \mathrm{Sym}T^* = A\otimes \oplus_k \mathrm{Sym}^kT^*$ and not in the product, since that would indeed cause problems. Using the notation derivative-notation, all high enough derivatives of $a$ must vanish. $\endgroup$
    – SzM
    Commented Mar 30, 2017 at 5:43
  • $\begingroup$ With the given decomposition $\rho(a)=\rho_0(a) + \rho_1(a) + \dots$ where $\rho_0(a)=a \otimes 1_C$, the assumption $\rho(a)=a \otimes 1_C$ implies $\rho_1(a)=0$. In fact, the converse is also true: if $\rho_1(a)=0$ then -- by the given description of higher $\rho_k$'s using lower ones -- we have $\rho_k(a)=0$ for all k>1 as well. $\endgroup$
    – SzM
    Commented Mar 30, 2017 at 5:49
  • $\begingroup$ You are right, it's indeed the space of coinvariants. But how do they "generate" the comodule? $\endgroup$
    – AaronS
    Commented Mar 31, 2017 at 7:57

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