ag.algebraic geometry – Monodromy of a non flat connection

I’m asking the following question (i couldn’t find a similar question in the litterature or on the web). Let $mathcal{E}$ be a vector bundle of rank $n$ over a complex manifold $M$ and $D$ a divisor. Let $nabla$ be a (non flat) meromorphic connection on it, with pole over $D$.

  1. Although the connection is not flat, we can consider the locally constant sheaf $mathcal{E}^nabla$ over $Mbackslash D$ of flat sections, which is a subsheaf of $mathcal{E}|_{Msetminus D}$. Suppose the sections of $mathcal{E}^nabla$ have moderate growth around $D$ (for local basis of $mathcal{E}$). Then let $mathcal{L}$ the $mathcal{O}_M$-module defined by : $$mathcal{L}=(mathcal{O}_{Msetminus D} underset{mathcal{O}_M}{otimes} mathcal{E}^nabla) cap mathcal{E}$$
    Can we say that it is equivalent to a lattice for $nabla$ in the sense that $z_1 nabla_{frac{partial}{partial z_1}}(mathcal{L})subset mathcal{L}$ ?
  2. If 1., is true, suppose the following : for every $x in D$, there exist a local basis $Y^x_1,ldots,Y^x_n$ of $mathcal{E}$ in a neighbhorood of $x$ such that : $$nabla_{(Y^x_i)_i} = d + A$$ where the connection form $A$ has no logarithmic terms.

Can we say that $nabla$ has no monodromy in the sense that $mathcal{E}^nabla$ is a locally constant sheaf over $M$ ?

Thanks very much for trying to help me, best regards

Edit : In fact, i’m pretty sure that we can find a $mathcal{O}_M$-module $mathcal{L}$ which is loccally free, and is a submodule of $mathcal{E}$ such that $nabla$ is regular on $mathcal{L}$ : we know that a such module always exists by Riemann Hilbert on punctured disks , but since for $(X,U) in mathcal{E}^nabla(U)$, and for $(Z,V) in mathcal{L}(V)$ we have $Z|_V = f X$ with $fin mathcal{O}_{Msetminus D}(U)$ which has a moderate growth, and $X$ has moderate growth relatively to $mathcal{E}$, then $Z$ must have moderate growth relatively to $mathcal{E}$ so that $$mathcal{L}hookrightarrow mathcal{E}otimes mathcal{O}_M(-D)$$ Then we can consider $$mathcal{L}’=z_1^N mathcal{L}$$ for $N$ large enough for $mathcal{L}’$ to be in $mathcal{E}$ and it remains a lattice for $nabla$.

I think my problem is to know wether $mathcal{E}/mathcal{L}$ is locally free. If so, then i think it’s immediate that the monodromy can be read on the matrix form $A$ relatively to a (local) basis of $mathcal{E}$.

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vector bundles – Integrability condition for flat connections

I am reading Kobayashi’s book “Differential geometry of complex vector bundles”. More precisely, I am on section 2 of chapter 1, page 5.

Kobayashi is trying to prove that if $E$ is a vector bundle on some manifold $M$, with a flat connection $D$, then it admits a “flat structure” ${U,s_U}$ which consists on an open cover of $M$ and a local frame of $E$ such that the transition functions are locally constant.

In order to do this, he starts with some arbitrary local frame $s’$ and looks for functions $a:U rightarrow GL(r,mathbb{C})$ such that in the frame $s_U= s’ a$ the connection $1$-form is $0$.

Therefore, if $omega’$ is the connection $1$-form in the frame $s’$, what he is trying to do is solve the following equation for $a$

$$
omega’ a + da = 0.
$$

He claims that solutions exists since the “integrability condition” for this equation is obtained by differentiating it

$$
0=(domega’) a -omega’ wedge da = (domega’)a + (omega’ wedge omega’)a = Omega’ a,
$$

which is true since we assumed that the connection is flat.

My question is what does he mean by the “integrability condition”. Moreover, why is that the integrability condition for that equation? And, also, why can he use the fact $da=-omega’ a$ when computing it?

I think he might be using some form of the Frobenius theorem, since I know that it is what you use from a “global” point of view.

In any way, I want to know precisely in this context what he means by that “integrability condition”, maybe it is just something basic or standard that I am missing.

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ag.algebraic geometry – Any finite flat commutative group scheme of $p$-power order is etale if

This question is immediately related to Discriminant ideal in a member of Barsotti-Tate Group
dealing with Barsotti-Tate groups and here I
would like to clarify a proof presented by
Anonymous in the comments from linked thread which I not completly understand. Although meanwhile I found another proofs of the claim below I have a big interest on understanding this proof below.

Assume $G$ is a finite flat commutative group scheme
over a field $k$ of
order $p^N$. Assume $p$ prime and $p in k backslash {0}$,
equivalently invertible on the base.

Claim: Any finite flat commutative group scheme of $p$-power order
is etale if $p$
is invertible on the base.

Anonymous’ proof works as follows:
Firstly we reduce to case over a field (because a
finitely presented flat map is etale if it is
fibrewise etale). Since we assumed $G$ commutative
the multiplication by $p^N$ map $f_{p^N}: G to G$
is well defined and by Deligne’s theorem
$p^N$ kills $G$ since it’s the order of $G$.
That means that $f_{p^N}$ is the zero map: equivalently
it factorize over $Spec(k)$.

What now comes I not understand:

It is clamed that $f_{p^N}$ is unramified
“as the map on the tangent spaces is given
by $p^N$, which is invertible”.

Question I: why the induced by $f_{p^N}$
maps on tangent spaces is given
by $p^N$?

Question II: assume we understand Question 1.
Why this implies $G$ unramified?

when we know this we are done because unramified finite type schemes over a field are etale.

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Where’s a road of McMansions in New Zealand that opens up to green flat field and mountains?

When my family and we travelled to New Zealand, we were just driving around on a road in some suburb. We turned on a road of at least 40 town houses or houses packed in close together on postage-stamp-sized lots on a road. But we got surprised because this road ABRUPTLY separated the houses from the wilderness. The other side of road had just green grass flatland for kilometers until mountains. Anyone know where this is?

These pictures don’t show the abruptness, but I can’t find anything relevant.

Top. mid. Bottom.
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Online exhibition design: 3d user interface VS 2d (flat) interface

I was planning to design an online exhibition platform.

I went to have a look at existing exhibition platforms for reference and found that most of them have 3d interfaces.

I was triggered by the thought that is it a good idea to have a 3d interface as I was uncomfortable in navigation and have thoughts like a good flat design will work better.

So my question is can a flat interface be better than 3d interface.

What factors should I consider to find the trade-off between these two?