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I have a very silly question. While showing that $S^1$ is oriented we use two stereographic projection from the north and south pole. I have the atlas and everything. However, I just could not figure out how to obtain the Jacobian which idefine as follows: $\frac{\partial x^i}{\partial y^j}$. I need to show that the determinant of this Jacobian matrix is positive but I did not get how to write this matrix.

I try to prove the fact by using Proposition 7.1 in the following link.

Lilith
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3 Answers3

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One way to prove that $S^1$ is orientable is by recognizing it as the zero set of $f(x,y)=x^2+y^2-1$ and so $\nabla f$ is a continuous (even smooth) normal vector field defined globally on $S^1$.

lhf
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  • My question is not how to prove it, it is actually how to obtain this Jacobian matrix in terms of chart coordinates. It is may be a problem of linear algebra or calculus than differential geometry. – Lilith Dec 11 '13 at 00:19
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The circumference $\mathbf{S}^1$ is an orientable manifold. One standard atlas for $\mathbf{S}^1$ is given by the stereographic projections, defined as $\pi_+ :\mathbf{S}^1-N\to \mathbb{R}$ and $\pi_-:\mathbf{S}^1-S\to\mathbb{R}$:

\begin{equation} \pi_+(x,y)= \frac{x}{1-y}, \qquad \pi_-(x,y)=\frac{x}{1+y}, \end{equation}

with inverses \begin{equation} \pi_+^{-1}(t)= \left(\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right), \; \pi_-^{-1}(t)=\left(\frac{2t}{t^2+1},\frac{1-t^2}{t^2+1}\right), \end{equation}

where $N=(0,1)$ and $S=(0,-1)$ are the north and south poles, respectively. For instance, $\pi_+$ maps a point $p$ in $S^1$ to $\mathbb{R}$ by computing the intersection of the line by $N$ and $p$ with the horizontal line $y=0$. The other projection $\pi_-$ works analogously with $S$. To check the orientability of $S^1$ the transition matrix has to be taken into analysis. Such matrix is the Jacobian of the transition function between two charts of an atlas. In this case it is given by $\pi_-\circ\pi_+^{-1}:\mathbb{R}\to\mathbb{R}$ or in the other direction $\pi_+\circ\pi_-^{-1}:\mathbb{R}\to\mathbb{R}$. In the both cases

\begin{equation} \pi_-\circ\pi_+^{-1} (t) = \frac{1}{t} = \pi_+\circ\pi_-^{-1} (t) \end{equation}

and its Jacobian, the transition matrix, is $-\frac{1}{t^2}<0$, hence it seems that the circumference is non orientable. However, by changing, for instance the chart $\pi_-$ with $-\pi_-$ the sign reverses and the Jacobian becomes positive, and thus $\mathbf{S}^1$ is orientable. Such construction is possible inductively for all the spheres $\mathbf{S}^n$ by generalising the stereographic projection.

Marco
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We have to obtain an $\Bbb R\to\Bbb R$ mapping, by applying stereographic projections first from the North pole, then from the South pole. If you draw it, you can recognise several right triangles and hence some equal angles and similarity of triangles, that would lead to $x\mapsto 4/x$ (unless I miscalculated), if $S^1$ is placed on the unit circle.

The $1\times 1$ Jacobian is now its differentiate, which is $-4/x^2$, actually it seems negative...

This is because we implicitly oriented both our tangent lines from left to right. So, if we consider e.g. the upper line to be scaled from right to left, then the same picture will yield to the function $x\mapsto -4/x$, and this has indeed positive differential.

Berci
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  • I did not get where $4/x$ comes from actually. What I obtain is from the north pole $(x,y) \mapsto (\frac{x}{1-y})$ and from the south pole $(x,y)\mapsto (\frac{x}{1+y})$. There is this theorem: https://people.maths.ox.ac.uk/hitchin/hitchinnotes/Differentiable_manifolds/Chapter_3.pdf Proposition 7.1, I'm trying to prove the fact by using this proposition. – Lilith Dec 11 '13 at 00:36
  • Yes. DRAW IT, follow the first $3$ lines of the answer. Anyway, what are the inverses of these functions $\Bbb R\to S^1$? – Berci Dec 11 '13 at 00:43
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    As Berci said, the issue is the orientation on the charts that you are using with north and south stereographic projection. If you want your positive orientation on the circle to be counterclockwise, then standard stereographic projection from the north pole is orientation preserving, whereas standard stereographic projection from the south pole is orientation reversing. The Jacobian of the transition function will have a negative determinant because of this.

    If you don't insist on using stereographic projection (and just want an exercise in following the transition maps), you might be . .

     – THW Dec 11 '13 at 15:36
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    . . .better served by taking your second chart to be a counterclockwise rotation of your first chart. – THW Dec 11 '13 at 15:37