MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Hi friends,

Let me ask you about connexions having regular singularities. So imagine $X$ is some smooth algebraic variety over a subfield $k \subset \mathbb{C}$ and you have a locally free $\mathcal{O}_X$-module $\mathcal{E}$ with a flat connection $\nabla: \mathcal{E} \to \mathcal{E} \otimes \Omega^1$. This defines a complex $DR(\mathcal{E}, \nabla)$ and you can consider

$H^n(X, (\mathcal{E}, \nabla)):=\mathbb{H}^n(X, DR(\mathcal{E}, \nabla)).$

On the other hand, there is an associated analytic connection $(\mathcal{E}^{an}, \nabla^{an})$ on $X^{an}$ and the corresponding cohomology

$H^n(X^{an}, (\mathcal{E}^{an}, \nabla^{an})).$

There is of course a morphism (*)

$H^n(X,(\mathcal{E}, \nabla)) \otimes_k \mathbb{C} \to H^n(X^{an}, (\mathcal{E}^{an}, \nabla^{an}))$.

When $(\mathcal{E}, \nabla)$ is the trivial connection $(\mathcal{O}_X, d)$, then you just get algebraic and analytic de Rham cohomology, which are isomorphic by Grothendieck's theorem. In general, easy examples show that the above morphism is not an isomorphism.

The condition needed to guarantee that it (*) is an isomorphism is that the connection has regular singularities, meaning that there exists a logarithmic extension

$$ \nabla^{ext}: \mathcal{E}^{ext} \longrightarrow \mathcal{E}^{ext} \otimes \Omega^1(\log D) $$ to a good compactification $\bar{X}$ including $X$ as the complement of a normal crossings divisor.

Then Deligne proved that if $(\mathcal{E}, \nabla)$ has regular singularities, there is an isomorphism

$H^n(X,(\mathcal{E}, \nabla)) \otimes_k \mathbb{C} \stackrel{\sim}{\longrightarrow} H^n(X^{an}, (\mathcal{E}^{an}, \nabla^{an}))$.

I'm trying to understand what does it mean for a very simple connection, constructed on the trivial bundle on $\mathbb{G}_{m, k}$. To give a connection on $\mathcal{O}$ is the same as giving a differential $1$-form, so assume

$\nabla(1)=\alpha \frac{dt}{t}$

for some element $\alpha \in k^\times$. I would say that this connection has regular singularities because $\frac{dt}{t}$ has simple poles at $0$ and $\infty$, so $\mathcal{O}$ of $\mathbb{P}^1$ with $\nabla(1)=\alpha \frac{dt}{t} \in \Omega^1(\log (0+\infty))$ is a logarithmic extension.

My question is: how does one see that this two cohomologies are isomorphic in this particular case? And what is a description of the local system $\ker(\nabla^{an})$ underlying the connection?


share|cite|improve this question

We can see what $\ker(\nabla^\mathrm{an})$ will be by the following reasoning (which can be made a bit more precise):

$\nabla(f)=0\Leftrightarrow \frac{df}{f}=-\alpha\frac{dt}{t}\Leftrightarrow \log(f)=-\alpha\log(t)+c\Leftrightarrow f=c_0t^{-\alpha}$. So local sections of $\ker(\nabla^\mathrm{an})$ will then just be multiples of some fixed branch of the function $t\mapsto t^{-\alpha}$.

Global sections of this will exist iff $\alpha\in\mathbb{Z}$, and so in this case we can see that $H^0(\mathbb{G}_m^\mathrm{an},(\mathcal{O}^\mathrm{an},\nabla^\mathrm{an}))$ will be zero if $\alpha\notin\mathbb{Z} $ and $\mathbb{C}\cdot t^{-\alpha}$ otherwise.

On the other hand, algebraic de Rham cohomology will just be the cohomology of the complex of $k[t,t^{-1}]$-modules

$0\rightarrow k[t,t^{-1}]\overset{f\mapsto f'+\frac{f\alpha}{t}}{\rightarrow} k[t,t^{-1}]\rightarrow0$

Thus elements of $H^0(\mathbb{G}_{m,k},(\mathcal{O},\nabla))$ will be (multiples of) global algebraic branches of $t\mapsto t^{-\alpha}$. Again, if $\alpha\in\mathbb{Z}$, then this $H^0$ is just $k\cdot t^{-\alpha}$ and if $\alpha\notin\mathbb{Z}$ it will be zero.

I'm not sure how to see 'directly' that the two $H^1$'s will be the same though.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.