# Exact forms on the 2 torus

1. Apr 30, 2010

### lavinia

I am having trouble visualizing when a 2 form is exact and have a specific case that I am struggling with at the moment. Any help is welcome.

Take an oriented 2 torus and divide it ,using parallel circles, into an even number of tube shaped regions.

In each tube, assign a 2-form that fades to zero at its bounding circles and require the following:

- these two forms fit together along the tube boundaries to give a global 2 form

- Each form has the opposite orientation from the 2 forms in its adjacent regions

- The integral of the induced global 2 form is zero.

- None of the forms are identically zero in any tube.

This form is exact. But how do I picture the one form that it is the exterior derivative of?

2. Apr 30, 2010

### AntideSitter

Hello Lavinia,

I think the best way to think about your two-form would be in terms of its 'dual', by which I mean

$$e^{ij} \omega_{ij}$$

where $$\omega_{ij}$$ the components of the two-form in a local coordinate frame, and $$e^{ij}$$ is the completely anti-symmetric matrix. This is a scalar function, and it is what you integrate to get the integral of $$\omega$$ over the torus.

$$\int_{T^2} \omega = \sum_{patches} \int_{\mathbb{R}^2} d^2 xe^{ij} \omega_{ij}$$

When the orientation changes between adjacent tubes, the sign of this 'dual' flips. So the when you integrate over your even number of tubes, the contribution adjacent tubes cancel each other, and the integral is zero.

You can think of this 'dual' as the single $$dx \wedge dy$$ component of the exterior derivative of the one-form you need. Call the one-form $$v$$. Then$$d v = \omega$$, i.e.:

$$dv = (\frac{\partial v^x}{\partial y} - \frac{\partial v^y}{\partial x}) dx \wedge dy = e^{ij} \omega_{ij} dx \wedge dy$$

up to a factor. So the curl appearing in the middle expression is our dual, and changes sign between adjacent tubes. $$v$$, when viewed as a vector field, will have arrows that point around each each tube. Their magnitude increases as you go length-wise along the tube. When you cross to the next tube they will start decreasing in size again. This will alternate between the tubes. You can check that the curl of this sort of vector field changes sign appropriately between tubes.

3. May 1, 2010

### lavinia

thank you. I was able to figure this out yesterday as well.

I just integrated along the x direction(the direction the coordinate direction that connects the boundary circles. Since the form changes sign form tube to tube, these integrals all fit together across the entire torus with the right choice of integration constants. As you said, you get a form that points in the y direction and oscillates in magnitude, increasing on one tube then decreasing on the next.

Last edited: May 1, 2010