# Does stereographic projection preserve or reverse orientation?

Let $S^n\subset\mathbb{R}^{n+1}$ denote the standard unit sphere with normal bundle $\nu$, let $N=(0,\dots,0,1)$ and $S=(0,\dots,0,-1)$. Then there are two sterographic projections $$\sigma_+\colon S^n-S\to\mathbb{R}^n$$ and $$\sigma_-\colon S^n-N\to\mathbb{R}^n$$ Both of these maps are homeomorphisms and they form an atlas for the standard smooth structure on $S^n$. I'm interested in how they should induce the standard orientation on $S^n$. (Here I orient $S^n$ by putting the standard orientation on $T S^n \oplus \nu\cong T\mathbb{R}^{n+1}|_{S^n}$ and orient $\nu$ by declaring that the outward-pointing direction is positive.)

Let $v\in S^{n-1}\subset S^n$ so that $\sigma_+^{-1}\circ \sigma_-(v)=v$, and let $B=\{b_1,\dots,b_n\}$ be a positively-oriented orthonormal basis for $T_vS^n$ where $b_1$ points along the great circle from $N$ to $S$. Geometrically, it seems like $D_v (\sigma_+^{-1}\circ \sigma_-) (b_i)=-b_1$ if $i= 1$ and $b_i$ otherwise, suggesting that the transition function is orientation reversing, and hence exactly one of $\sigma_+$ and $\sigma_-$ is orientation reversing. My question is which one reverses orientation and which preserves?

I haven't succeeded in computing anything in terms of the formulas for stereographic projection. The Jacobean matrix I get in the 2-dimensional case for $\sigma_+$ or $\sigma_-$ is $2\times 3$ so I don't know how I'm supposed to interpret its "determinant".

(I should point out that in general my experiences with differential geometry have been very, very, very bad, and I'm much more topologically minded. In particular my definition for an orientation of a vector bundle is a Thom class.)

• The Jacobian matrix you are getting seems like that because you are looking at a map $f: \mathbb{R}^3 \rightarrow \mathbb{R}^2$ by looking at the sphere inside $\mathbb{R}^3$, which you shouldn't. You should take the charts for your sphere to compute the determinant if you want to think like that, after all you are on a manifold. – Aloizio Macedo Feb 10 '16 at 15:49

It looks like, in all dimensions, $\sigma_+$ preserves orientation and $\sigma_-$ reverses it. Here is the case of $S^2$ explicitly worked out:

For a vector space $V$ of dimension $n$, an orientation will be an element of $$\{(b_1,\dots,b_n)\ |\ V=\langle b_1,\dots, b_n\rangle \}/GL^+(V) \cong \mathbb{Z}/2$$ and given oriented vector spaces $(V,o)$ and $(V',o')$, a linear isomorphism $L\colon V\to W$ will be orientation preserving if $L(o)=o'$, and orientation reversing if $L(o)=-o'$.

For the sake of clarity, we will write $\mathbb{R}^3=\langle e_1,e_2,e_3\rangle$ and $\mathbb{R}^2=\langle e_1',e_2'\rangle$, and we will give these vector spaces the natural orientations.

Explicitly, the stereographic projections are given by $$\sigma_+(x,y,z)=(\frac{x}{1+z},\frac{y}{1+z})\text{ and }\sigma_-(x,y,z)=(\frac{x}{1-z},\frac{y}{1-z})$$ for $(x,y,z)$ in the appropriate domains.

Consider specifically the point $e_1=(1,0,0)$, which is in the domains of both functions. An orientation of $T_{e_1}S^2\cong \langle e_2,e_3\rangle$ is an ordered basis with the property that concatenating with the outward-pointing normal vector $e_1$ gives a positively oriented basis for $\mathbb{R}^3$. (In this case it doesn't matter if the concatenation happens at the beginning or end, because they differ by an even number of transpositions.) The preferred orientation for $\mathbb{R}^3$ is $[e_1,e_2,e_3]$, so the orientation on $T_{e_1}S^2\cong \langle e_2,e_3\rangle$ is given by $[e_2,e_3]$.

Stereographic projection sends the standard great circle $S^{1}\subset S^2$ to the unit circle $S^{1}\subset\mathbb{R}^2=\langle e_1',e_2'\rangle$, and in particular $e_1$ is sent to $e_1'$. The tangent space $T_{e_1'}\mathbb{R}^2$ is naturally isomorphic to $\langle e_1',e_2'\rangle$, and the preferred orientation is $[e_1',e_2']$.

In order to compute the derivative of $\sigma_{\pm}$ at $e_1$, consider the two paths $$\gamma_2(t)=(cos(t),sin(t),0)\text{ and }\gamma_3(t)=(cos(t),0,sin(t))$$ Then $\frac{d}{dt}\gamma_i(t)|_{t=0}=e_i$, i.e. the derivatives form a basis for the tangent space at $e_1$. Compute

$$\frac{d}{dt}\sigma_-\big(\gamma_2(t)\big)|_{t=0} = \frac{d}{dt}\big(cos(t),sin(t)\big)|_{t=0}=(0,1)=e_2'$$ and $$\frac{d}{dt}\sigma_-\big(\gamma_3(t)\big)|_{t=0} = \frac{d}{dt}\left(\frac{cos(t)}{1-sin(t)},0\right)|_{t=0}=(1,0)=e_1'$$ Hence the derivative of $\sigma_-$ at $e_1$ takes the positively oriented basis $[e_2,e_3]$ to $[e_2',e_1']=-[e_1', e_2']$, and so $\sigma_-$ reverses orientation at $e_1$. That it reverses orientation at all points in its domain should follow from continuity of $\sigma_-$ and connectedness of its domain.

Similarly, one computes that the derivative of $\sigma_+$ at $e_1$ takes the basis $[e_2,e_3]$ to $[e_2',-e_1']=[e_1’,e_2’]$, and so it preserves orientation.

A similar computation also works in higher dimensions, but I wrote out the case of $S^2$ because it still illustrates the key idea and computation. For $S^{2n+1}$ one needs to be careful about defining an oriented basis for a tangent space, because moving a normal vector from the end of an ordered basis to the beginning is an odd permutation; however, if the convention is that the normal vector is added at the end, then again we should have $\sigma_-$ orientation reversing and $\sigma_+$ orientation preserving.

$\def\ip {\, \lrcorner \, }$ I get a different result than the previous answer: $\sigma^-$ preserves orientation iff $n$ is even:

Consider $\Bbb S^{n-1}$ embedded in $\Bbb R^n$.The orientation form on $\Bbb R^n$ is $dx_1 \wedge \ldots \wedge dx_n$.

If $v(x)$ is a non-vanishing vector field on $\Bbb S^{n-1}$ then the induced orientation form on $\Bbb S^{n-1}$ is $v \ip (dx_1 \wedge \ldots dx_n)$. In particular the standard orientation on $\Bbb S^{n-1}$ is given by using the outward normal to the sphere.

If $p \in \Bbb R^n$, and we can identify $T_p\Bbb R^n$ with $\Bbb R^n$. If $p \in \Bbb S^{n-1}$ then of course the outward normal is just $p$. Hence the orientation form on $\Bbb S^{n-1}$ is $\hat \omega = p \ip (dx_1 \wedge \ldots \wedge dx_n)$.

The stereographic projection map $\sigma^- : \Bbb S^{n-1} \rightarrow \Bbb R^{n-1}$ is defined by:

$$(u_1, \ldots u_{n-1}) = \sigma^-(x_1, \ldots, x_n) = \left({ x_1, \ldots , x_{n-1} \over 1 - x_n} \right)$$

It should be clear that at $-e_n = (0, \ldots, 0, -1)$ we have:

$$d \sigma^-_{e_n}\left({\partial \over \partial x_i}\right) = {\partial \over \partial u_i} \text{ for }i < n\\ d \sigma^-_{e_n}\left({\partial \over \partial x_n}\right) = 0.$$

If we identify tangent vectors in $\Bbb R^{n-1}$ with $\Bbb R^{n-1}$, then set of tangent vectors $(e_1, \ldots, e_{n-1})$ is trivially mapped to $(e_1, \ldots, e_{n-1})$.

In $\Bbb R^{-1}$, $(e_1, \ldots, e_{n-1})$ is positively orientated, but on $\Bbb S^{n-1}$,

\begin{align}\hat \omega (e_1, \ldots, e_{n-1}) &= -e_n \ip (dx_1 \wedge \ldots \wedge dx_n)(e_1, \ldots, e_{n-1}) \\ &= (dx_1 \wedge \ldots \wedge dx_n) (-e_n, e_1, \ldots, e_{n-1})\\ &= (-1)^{n-1}(dx_n \wedge dx_1 \wedge \ldots \wedge dx_{n-1}) (-e_n, e_1, \ldots, e_{n-1})\\ &= (-1)^{n}(dx_n \wedge dx_1 \wedge \ldots \wedge dx_{n-1}) (e_n, e_1, \ldots, e_{n-1})\\ &= (-1)^{n} \\\end{align}

Thus at the point $-e_n$, $\sigma^-$ is orientation reversing iff $n$ is odd. Since $\sigma^-$ is diffeomorphism, it has to have the same orientation everywhere, so $\sigma^-$ preserves orientation iff $n$ is even.

Am I mistaken somehow?

• you say "If $\nu(x)$ is a non-vanishing vector field on $\mathbb{S}^{n−1}$ then the induced orientation form on $\mathbb{S}^{n−1}$ is $\nu \lrcorner (dx_1\wedge…dx_n)$", but what happens in the case where $n$ is odd, and there are no non-vanishing vector fields on $\mathbb{S}^{n−1}$? – William Jan 27 '19 at 18:59

Inversions in the plane and likewise stereographic projection from $\mathbb R^2$ (having an inversion component) reverse sense in projections of a closed loop of the surface.