I'm trying to find an oriented atlas on the circle $S^1$, i.e.,

I want to find an atlas for $S^1$ such that for any two overlapping charts $(U,s)$ and $(V,t)$ of the atlas, the derivative $d s/d t$ is everywhere positive on $U\cap V$.

I can define one atlas $\mathfrak U = \{(U_i,\phi_i)\}_{i=1}^4$ for $S^1$ by letting $U_1,U_2$ be the upper and lower semicircles, respectively, and defining $\phi_i(x,y) = x$ for $i = 1,2$, and letting $U_3,U_4$ be the right and left semicircles, respectively, and defining $\phi_i(x,y) = y$ for $i = 3,4$.

However, I think this atlas is not oriented, since for example taking the overlapping charts $(U_1,x)$ and $(U_3, y)$ of $\mathfrak U$, we have $y = \sqrt{1 - x^2}$ and $$ \frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}} \text, $$ which is not positive on $U_1\cap U_3$.

How can I modify $\mathfrak U$ to fix this issue? For example, should I redefine $\phi_3(x,y) = -y$, and then keep on checking the other seven derivatives? (I have to check the pairs $dy/dx$ and $dx/dy$ for each quarter-circle.)


1 Answer 1


The given idea (about reversing charts) will work, but you'll have to change signs of two charts, as two of the four are oriented clockwise and two are anticlockwise.

Note that you can cover a circle (or indeed, $\Bbb S^n$ for any $n$) using just two charts (though not two graph charts, as in the question): Indeed, the stereographic projection $$(x, y) \mapsto \frac{y}{x + 1}$$ is a homeomorphism $\Bbb S^1 - \{(-1, 0)\} \to \Bbb R$ and the domain omits just one point of the circle. (This is the map that maps a point $z$ to the slope of the line through $z$ and $(-1, 0)$.)


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