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Fix a field $k$, denote by $k[x]$ the polynomial algebra. The category of finite-dimensional modules over $k[x]$ is precisely the category $\mathcal{C}$ consisting of pairs $(V, T_V: V \to V)$ of finite-dimensional vector spaces equipped with an endomorphism. Morphisms in this category are those linear maps $f: V \to W$ such that $f T_V = T_W f$. The category $\mathcal{C}$ is $k$-linear and abelian, every object is of finite length, and it has a natural fibre functor $\omega: \mathcal{C} \to \mathsf{Vect}_k$ by forgetting the endomorphism associated with a vector space.

Various things I have read say that the category $\mathcal{C}$ should be equivalent to the category of finitely-generated comodules over some coalgebra $B$, in a way which is compatible with the fiber functor $\omega$. But I am struggling to find any coalgebra which realises this. Do I have an assumption wrong? Perhaps something about generators of the category?


Addition: Nilpotent transformations. I can solve this problem if I require all endomorphisms to be nilpotent. Let $\mathcal{N}$ be the full subcategory of $\mathcal{C}$ where each endomorphism is nilpotent. Then $\mathcal{N}$ is the category of comodules over the following coalgebra $N$:

  • As a $k$-vector space, $N = \{x_0, x_1, x_2, \ldots\}$.
  • The counit is $\epsilon(x_i) = \delta_{i0}$.
  • The coproduct is $\Delta(x_n) = \sum_{i + j = n} x_i \otimes x_j$

This can be systematically derived by taking the dual coalgebras of the algebras $k[x]/x^n$ for increasing $n$, and then taking a limit. However, I have no idea how to do this for eigenvalues which are not zero.

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    $\begingroup$ "Various things I have read say that..." Could you be more precise? $\endgroup$ – Pedro Tamaroff Oct 22 '18 at 4:56
  • $\begingroup$ @PedroTamaroff I find it difficult to be more precise, since a lot of this is hearsay from other people. It should be a kind of underpowered Tannakian reconstruction (I don't want to consider it as a monoidal category). It seems like Theorem 2.4 on page 12 of these notes asserts this, but again I'm not entirely sure. $\endgroup$ – Joppy Oct 22 '18 at 5:13
  • $\begingroup$ I see. Just glossing over that paper, it seems that the the coalgebra will not be described too easily, unfortunately. Or perhaps it will, who knows. Your example of nilpotent actions is fine, though. $\endgroup$ – Pedro Tamaroff Oct 22 '18 at 5:43
  • $\begingroup$ It is true that your category is equivalent to the category of finite dimensional comodules over a coalgebra. This was (essentially) proven by Gabriel in the dual language of pseudocompact algebras (or if you prefer the direct language of coalgebras this is by Takeuchi). A nice summary is: [Raedschelders-Van den Bergh: The Tannaka-Krein formalism and (re)presentations of universal quantum groups]. The coalgebra is the continuous dual of the endomorphism of the fibre functor (or equivalently the coend of this functor). I have not worked out what it is explicitly. $\endgroup$ – Julian Kuelshammer Oct 22 '18 at 8:23
  • $\begingroup$ @JulianKuelshammer Do you know of a nice way of computing the endomorphism algebra of the fibre functor, perhaps even for the simpler case of nilpotent endomorphisms? $\endgroup$ – Joppy Oct 22 '18 at 8:26
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Here's one way of thinking about this. First, over a field $k$, the category of coalgebras is the Ind-category of the category of finite-dimensional coalgebras. Second, the category of finite-dimensional coalgebras is equivalent to the opposite of the category of finite-dimensional algebras, by taking linear duals. Hence we have

$$\text{Coalg}(k) \cong \text{Ind}(\text{Coalg}_f(k)) \cong \text{Ind}(\text{Alg}_f(k)^{op}) \cong \text{Pro}(\text{Alg}_f(k))^{op}$$

from which we conclude that the category of coalgebras is equivalent to the opposite of the category of profinite algebras; that is, formal cofiltered limits of finite-dimensional algebras. The correspondence comes again from taking linear duals.

Furthermore, this correspondence respects modules in the following way: the category of finite-dimensional comodules over a coalgebra $C$ is equivalent to the category of finite-dimensional modules over the corresponding dual profinite algebra (where "module" means "module over a finite quotient," or equivalently "continuous module"). Finally:

Observation: The category of finite-dimensional modules over an algebra $A$ is equivalent to the category of finite-dimensional modules over its profinite completion $\widehat{A}$.

Here the profinite completion of an algebra $A$ is the cofiltered limit over all finite-dimensional quotients $A/I$.

So the question remains: what is the profinite completion of $k[x]$? Every finite quotient takes the form $k[x]/f(x)$ for some monic polynomial $f(x)$, which has some factorization $f(x) = \prod_i f_i(x)^{m_i}$ into irreducibles. The computation of the resulting cofiltered limit splits up into an independent piece for each irreducible, and we end up getting

$$\widehat{k[x]} \cong \prod_f \lim_m k[x]/f(x)^m$$

where the product runs over all monic irreducibles. In the special case that $k$ is algebraically closed, these are all linear polynomials $f(x) = x - a$, and we get

$$\widehat{k[x]} \cong \prod_{a \in k} k[[x - a]].$$

Compare to the usual computation of the profinite completion of $\mathbb{Z}$ as the product $\prod_p \mathbb{Z}_p$ over the $p$-adics. The corresponding dual coalgebra is the direct sum of the dual coalgebra of each $k[[x - a]]$ which is the thing in Julian's answer.

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Let me assume $k$ to be algebraically closed. Otherwise the story is more complicated.

Then finite dimensional modules over $k[x]$ can be described by Jordan normal form. In particular, the indecomposable modules are given by Jordan blocks. Moreover it is easy to see that there are no homomorphisms between indecomposable modules corresponding to different eigenvalues. Thus, the category of finite dimensional modules over $k[x]$ decomposes into 'blocks', i.e. subcategories such that there are no homomorphisms between each other.

Every block is isomorphic to the block for eigenvalue $0$ (just subtract $\lambda\operatorname{id}$ from a Jordan block with eigenvalue $\lambda$ to obtain a nilpotent Jordan block). One can check that this defines an equivalence of categories.

You already computed that the coalgebra corresponding to $\mathcal{N}$ is the tensor coalgebra $k[x]$. The category of comodules over a direct sum of coalgebras is given precisely by the "union" of the categories of comodules of the blocks (i.e. each object is just given by a direct sum of objects in the blocks). Therefore the coalgebra corresponding to finite dimensional comodules over $k[x]$ is $\bigoplus_{\lambda\in k} k[x]$.

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  • $\begingroup$ Thanks for the answer! This makes a lot of sense. Is the story much more complicated if $k$ is not algebraically closed? The indexing set for the blocks should be the monic irreducible polynomials in $k[x]$ instead, and the blocks in this case might look different to $\mathcal{N}$. For example, do you know how I could find the coalgebra for a block corresponding to a complex eigenvalue of an $\mathbb{R}[x]$ module? (I use the term "eigenvalue" loosely, I know for example that in this block a simple object will be of dimension 2). $\endgroup$ – Joppy Oct 22 '18 at 11:51
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    $\begingroup$ @Joppy For $\mathbb{R}$, I think the corresponding coalgebra should be $\mathbb{C}[x]$ for the degree $2$ irreducible polynomials (instead of $\mathbb{R}[x]$ for the real eigenvalues). $\endgroup$ – Julian Kuelshammer Oct 22 '18 at 12:21

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