# vector field on the torus [closed]

Is the following statement true?

Let $$T$$ be diffeomorphic to the solid torus. Let $$v$$ be a vector field such that $$v$$ and $$curl(v)$$ are both tangent to $$\partial T$$ everywhere and $$|v|$$ is constant on $$\partial T$$. Then $$v$$ is zero vector field.

• What is $curl(v)$? What is $|v|$? – abx Apr 4 at 4:52
• the metric is euclidean and $curl(v)= \nabla \times u$ – J. Doee Apr 4 at 12:19
• the question does not seem to be well written. Since you made no hypothesis at all for $v$ in the interior of $T$, but only on the boundary, you cannot conclude in the interior either. For example, any smooth vector field identically zero on a small neighborhood of $\partial T$ but not identically zero in $T$ is a counterexample. – Gael Meigniez Apr 5 at 11:07

## 1 Answer

The answer is no.

The vector field $$\nabla\times \vec{v}$$ is tangent to $$\partial T$$ iff for any region $$\Sigma\subset\partial T$$ bounded by the Jordan curve $$\partial\Sigma$$ the flux integral $$\iint_{\Sigma}(\nabla\times \vec{v}).\vec{n}\,{\rm{d}}S$$ is zero; this is because the tangency to $$\partial T$$ means orthogonality to the unit normal vector $$\vec{n}$$ at any point of $$\partial T$$. Stokes' Theorem now implies that $$\oint_{\,\partial\Sigma}\vec{v}.{\rm{d}}\vec{r}$$ is zero. Thus the line integral of $$\vec{v}$$ along any simple closed curve that encloses a region in $$\partial T$$ (i.e. is a boundary) must be zero. This provides a reformulation of the condition on $$\nabla\times \vec{v}$$: Denoting the differential $$1$$-form corresponding to $$\vec{v}$$ by $$\omega$$, the $$2$$-form $${\rm{d}}\omega|_{\partial T}=d\left(\omega|_{\partial T}\right)$$ vanishes.

Now I construct a counter-example: Write the solid torus $$T$$ as $$\overline{\Bbb{D}}\times\Bbb{S}^1$$ where $$\overline{\Bbb{D}}$$ with $$\Bbb{S}^1$$ and are the unit closed disk and the unit circle in $$\Bbb{R}^2$$. Equip $$T$$ with the Euclidean norm induced from $$\Bbb{R}^2\times\Bbb{R}^2$$. The vector field $$-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}$$ on $$\overline{\Bbb{D}}$$ could be thought of as a vector field on $$\overline{\Bbb{D}}\times\Bbb{S}^1$$ which is independent of the third coordinate. I claim that it works as $$\vec{v}$$. On $$\partial T=\Bbb{S}^1\times\Bbb{S}^1$$ the vector field $$\vec{v}$$ is given by $$-y\frac{\partial}{\partial x}+x\frac{\partial}{\partial y}|_{\Bbb{S}^1}=\frac{\partial }{\partial \theta}$$
which is tangent to the first component and is of length one. It remains to check the tangency condition imposed on $$\nabla\times \vec{v}$$. The $$1$$-form $$\omega$$ corresponding to $$\vec{v}$$ is the pullback of the $$1$$-form $$-y{\rm{d}}x+x{\rm{d}}y$$ on $$\overline{\Bbb{D}}$$ via the projection $$p_1:T=\overline{\Bbb{D}}\times\Bbb{S}^1\rightarrow\overline{\Bbb{D}}$$. Restricted to the boundary $$\partial T=\partial\overline{\Bbb{D}}\times\Bbb{S}^1=\Bbb{S}^1\times\Bbb{S}^1$$, $$\omega|_{\partial T}$$ is the pullback of the closed $$1$$-form $${\rm{d}}\theta$$ via $$p_1:\partial T=\Bbb{S}^1\times\Bbb{S}^1\rightarrow\Bbb{S}^1$$. Consequently, $$\omega|_{\partial T}$$ is closed as well.

• Thank-you very much for your answer. If I understand correctly, your counterexample is the unit vector field in the poloidal direction. What if one furthermore requires that the flow of the vector field generate $T$, namely that evolving a crosssection of $T$ by the flow of $v$ sweeps out all of $T$? An example of such a field would be the toroidal vector field, $e_\phi$ whose curl is in the $\hat{z}$ direction and is not tangent to the torus except on two circles. – J. Doee Apr 4 at 18:02