This question asks pictures or intuitions of Riemannian geometry on a compact surface.

Pull the connection 1 form on the tangent circle bundle down to the tangent bundle with a vector field with isolated singularities. Take its dual vector field under the Riemannian metric. Now there is a vector field that is dual to the connection. In fact there are many such forms, one for each vector field with isolated singularities.

My question is: What picture of the connection does this vector field or its flow give us?

Is there a particularly clear picture if the vector field is dual to a meromorphic 1 form?


1 Answer 1


Most of your question is about local Riemannian geometry on a surface $(M,g)$, and that part can be answered without too much trouble, so I'll do that. Whether you can get anything global out of it is another matter, and I'm somewhat doubtful, but maybe the following will be of some use.

First, let's avoid the singular points for a moment and work on the part of the surface where the vector field isn't zero. Let's start with a unit vector field $\mathbf{e}_1$ on an open set $U\subset M$. Let $\mathbf{e}_2$ be such that $(\mathbf{e}_1,\mathbf{e}_2)$ is an oriented $g$-orthonormal frame field on $U$.

Direct computation shows that the vector field your process associates to $\mathbf{e}_1$ is the vector field $[\mathbf{e}_1,\mathbf{e}_2]$, i.e., the Lie bracket of the two unit vector fields.

Let $\mathbf{v}$ be another vector field on $U$. One can write $\mathbf{v} = r\cos\theta\ \mathbf{e}_1+r\sin\theta\ \mathbf{e}_2$ and, on the open set where $r>0$ (so that $\theta$ is well-defined), and your process associates to $\mathbf{v}$ the vector field $$ \bigl[\cos\theta\ \mathbf{e}_1+\sin\theta\ \mathbf{e}_2, \ -\sin\theta\ \mathbf{e}_1+\cos\theta\ \mathbf{e}_2\bigr] = [\mathbf{e}_1,\mathbf{e}_2] - \nabla^g\theta, $$ where $\nabla^g\theta$ is the gradient of $\theta$ with respect to the metric $g$.

It is not clear to me that you can get much local geometric information from this formula, and I'm not sure what you are hoping to see globally, but maybe you have something in mind. Locally, what you see is that the vector fields you get this way all differ by gradient vector fields. If you think of this as a first-order differential operator from vector fields to vector fields, it's nonlinear, but if you think of the map from $C^\infty(M)$ to vector fields given by $\theta\mapsto [\mathbf{e}_1,\mathbf{e}_2] - \nabla^g\theta$, then this is obviously an affine operator, so, you can almost think of it as a linear operator.

Note that, if $\mathbf{v}$ has isolated zeros of nonzero index, then $\nabla^g\theta$ blows up at those points, so your process yields vector fields with unbounded lengths.

As for the case of a meromorphic $1$-form, I guess you mean a vector field $\mathbf{v}$ that is $g$-dual to the real part of a meromorphic $1$-form $\omega$. Otherwise, I'm not sure what you mean. In this case, you can see, without too much trouble, that, away from the poles and zeros (where it isn't defined) the flow of the resulting vector field, say $\mathbf{w}$, is $g$-area preserving. The converse is true also (at least locally); i.e., if the flow of the resulting vector field is $g$-area preserving, then, up to a (real, nonzero) multiple, it is dual to a $g$-harmonic $1$-form (i.e., one that is closed and co-closed), so that (again, locally) it is dual to the real part of a holomorphic $1$-form. Locally, at least, there is not anything more to say.

  • $\begingroup$ So the flows that are dual to the pull backs of the connection 1 form are Lie brackets of orthonormal frames. Given any vector field one obtains a flow by your construction - assuming that you have a conformal structure.Yes? How does this restrict the possible flows - if at all? I guess an equivalent way to ask this question is to start with a vector field and ask for another field that is parallel along it. $\endgroup$
    – marc
    Sep 5, 2011 at 13:23

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