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Suppose we have a scheme $X$ over a field $L$ of characteristic $0$, and suppose that $S$ is an $L$-scheme. Then let $X_S:=X \times _L S$ denote the base change.

As in Le group fondamentale de la droite projective moins trois pointes (Deligne) we let $Un(X)$ be the category of unipotent vector bundles $\mathcal{V}$ on $X$ with flat connection $\nabla$, where by unipotent we mean that $X$ has a filtration (of finite length) by sub-bundles stabilised by the connection $\nabla$. The morphisms are morphisms of $O_X$-modules preserving the connection. Let $Vect_S$ be the category of vector bundles on $S$.

Given an $L$-points $b,x$ of $X$, there are $S$-points $b(S),x(S)$ of $X_S$. We fix $b$ as our base point and consider the following functor:

$$ e_b(S): Un(X_S) \rightarrow Vect_S;\;\; \mathcal{V} \mapsto b(S)^* \mathcal{V}$$

In particular, if $S=L$ then we are just taking the fibre of $\mathcal{V}$ at the point $b$.

I'm trying to understand how the above construction gives us a neutral Tannakian category by which we construct the de Rham fundamental group. So we require that $Un(X)$ be a rigid abelian tensor category with $e_b$ a fibre functor. The bit that's confusing me is the following: showing $e_b$ is a fibre functor.

I have looked at the paper referenced above, and it simply says that $e_b$ is a fibre functor. I have tried to show that $e_b$ is a fibre functor based on the definitions of everything presented. That is, I've tried to show that $e_b$ is an exact, $L$-linear and faithful tensor functor. My ideas we're as follows:

  1. Tensor functor: this follows from the definition of $b^*$ and the definition of tensor product of sheaves of modules

  2. Exact: We want to show that if we have an exact sequence $$ 0 \rightarrow \mathcal{U} \rightarrow \mathcal{V} \rightarrow \mathcal{W} \rightarrow 0$$ then we have an exact sequence $$ 0 \rightarrow b^*\mathcal{U} \rightarrow b^*\mathcal{V} \rightarrow b^*\mathcal{W} \rightarrow 0$$ We know that $b^*$ is right exact, so we just need to check that it is also left exact for these vector bundles. Now if we are working over $S=L$, then this is very easy by checking exactness on the stalks since $O_L(L)=L$. If $S \neq L$ I'm a bit more confused - $S$ will be covered by the spectra of some $L$-algebras $R$. At the level of stalks , if we pick a point $s$ we'll have something like $$ 0 \rightarrow \mathcal{U}_{b(S)(s)} \otimes_{O_{X_S,b(S)(s)}} O_{S,s} \rightarrow b^*\mathcal{V}_{b(S)(s)} \otimes_{O_{X_S,b(S)(s)}} O_{S,s} \rightarrow b^*\mathcal{W}_{b(S)(s)} \otimes_{O_{X_S,b(S)(s)}} O_{S,s} \rightarrow 0$$ So then my thinking is that since any $L$-algebra is a flat, we find that $O_{S,s}$ is flat and hence the sequence is exact at the level of stalks and hence exact. Is this correct?

  3. Faithful: here's where I'm really confused about things. We want to show that the morphism $$ Hom_{Un(X_S)}(\mathcal{V}, \mathcal{W}) \rightarrow Hom_{Vect_S}(b(S)^*\mathcal{V}, b(S)^*\mathcal{W})$$ is an injection. Now, if we let $S=L$, then already this is confusing to me. Suppose that the $L$-point $b$ corresponds to a point $P \in X$. Then to my mind, the above is saying that if $f,g: \mathcal{V} \rightarrow \mathcal{W}$ are such that $f_P=g_P$ then $f=g$. Now, obviously this must have something to do with the connections on the bundles $\mathcal{V}, \mathcal{W}$ but I'm not sure how to go forward here.

Any help would be greatly appreciated!

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The exactness of $e_b$ follows from the fact that a short exact sequence of vector bundles is locally split, so pull-back of short exact sequences of vector bundles is exact. Another way to think about it is that $\mathcal W$ is flat over $X_S$, and pull-back is a tensor product, so the Tor that could cause left exactness to fail actually vanishes.

The faithfulness uses the connection: certainly a map of vector bundles (with no extra structure) is not characterized by its value on a fibre. For example, suppose that a morphism of bundles with connection, that is compatible with the connection, vanishes on the fibre at $b$. Then uniqueness of solutions of first order linear ODE's with fixed initial conditions shows that this morphism has to be identically zero.

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  • $\begingroup$ So, the first part makes sense to me but I need to work through the details to be sure. $\endgroup$ – Aaron Nov 29 '16 at 11:17
  • $\begingroup$ As for the second part, do we mean uniqueness of formal solutions here, since we have algebraic connections? $\endgroup$ – Aaron Nov 29 '16 at 11:19
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    $\begingroup$ @Aaron: Yes, uniqueness of formal solutions, together with the fact that sections of a vector bundle over an irreducible variety embed into the formal completion at any chosen point, so that it's enough to check formal vanishing to deduce actual vanishing. $\endgroup$ – tracing Nov 29 '16 at 12:20

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