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Consider the unit hypersphere in $\mathbb{R}^n$, i.e. with the Euclidean metric, using spherical coordinates.

The metric tensor is then: $$ g_{11}=1 $$ $$ g_{ij}=\delta_{ij}\prod_{k=1}^{i-1}\sin^2(\theta_k) $$ where $\delta_{ij}$ is the Kronecker delta. The inverse metric is then: $$ g^{11}=1 $$ $$ g^{ij}=\delta_{ij}\prod_{k=1}^{i-1}\csc^2(\theta_k) $$ Notice that if any of $\theta_k=0$, then there will be a problem with $\csc(\theta_k)$. But $\theta_k=0$ seems to be a perfectly reasonable coordinate to be on.

Questions:

  • (1) Why does the inverse metric fail to exist when $\theta_k=0$?
  • (2) How do I fix this so that I can use the inverse metric computationally in cases where $\theta_k$ are allowed to freely vary (and vanish) as they can for $g$?

I'm hoping I'm making a silly mistake :)

(Note: this is true for $n=3$ too)

user3658307
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    The singularity is not a mistake. You have to use more than one chart if you want to describe every point on a sphere. This means converting to another set of coordinates when you are near $\theta=0$. – Spencer Aug 12 '17 at 05:23
  • @user3658307: Your formulas appear to be awry: The factors inside the product don't depend on $k$, the index of multiplication. ;) That said: 1. Geographic coordinates "collapse" at the poles, so the corresponding metric components in those coordinates vanish. 2. In case it matters, inverting $(g_{ij})$ does not give another metric on the sphere. (But perhaps you're just computing index-raising and lowering?) It might be worth giving the problem you're actually attacking; there may be an easier strategy than you're taking. – Andrew D. Hwang Aug 12 '17 at 10:51
  • @AndrewD.Hwang Thanks, fixed the error :) The actual problem I'm attacking is computational, involving searches over the unit tangent bundle of a manifold. I don't want the cost of constantly changing coordinates or something. Do you have any suggestions? – user3658307 Aug 12 '17 at 11:49
  • @Spencer Thank you... I know you can't cover the sphere with one patch, but apparently it didn't cross my mind before I went to sleep. >_> – user3658307 Aug 12 '17 at 11:51
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    The unit tangent bundle of the $n$-sphere is the set of ordered orthonormal pairs in $\mathbf{R}^{n+1}$; if it's feasible, I'd be inclined to use ambient linear algebra rather than intrinsic coordinates. – Andrew D. Hwang Aug 12 '17 at 12:04
  • @AndrewD.Hwang Sorry, I meant to say I am looking at the unit tangent bundle of a different manifold $(M,g)$, which I wanted to characterize with coordinates $(x_i,\theta_j)$ i.e. spatial coordinates with an angular coordinates appended onto the end, since $UTM=T_1M$ should be $2n-1$ dimensional in this representation. This question arises by doing this in (flat) $\mathbb{R}^n$ (g=I), where $UTM$ is (I think) just $\mathbb{R}^n$ with $\mathbb{S}^n$ associated to every point, which I can nicely characterise with spherical coords. But I want to compute quantities over $UTM$ involving $g^{-1}$. – user3658307 Aug 12 '17 at 12:42

1 Answers1

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Notice that if $\theta_i=0$ for $i=1,\dots,n-1$, then your metric $g_{ij}$ fails to be Riemannian (it's not an inner product for $\theta_i=0$, because it's degenerate).

Hence, if your metric is not well-defined in some subset, you definitely cannot expect the inverse to be well-defined (in this case, it does not even exist!)

What you can do, however, is to change coordinates so that you cover the part of the manifold which is not covered by your first set of coordinates. The coordinates that induce your metric are the following: $$x_1={\cos{\theta_1}},$$

$$x_p={\cos{\theta_p}}{\prod_{m=1}^{p-1}}{\sin{\theta_{m}}},$$

$$x_n={\prod_{m=1}^{n-1}}{\sin{\theta_{m}}},$$

yet you can verify yourself that these are not one-to-one if one of the $\theta_i$ vanishes, so you can only have well-defined coordinates for the subset $$ \{(\theta_1,\dots,\theta_{n-1})\ |\ 0<\theta_i<\pi \textrm{ for }i\not=n-1,\ 0<\theta_{n-1}<2\pi\}. $$ If you seek another coordinate patch, you may for instance "flip" the sphere and build the coordinates in the exact same way, and then notice the relation between the two sets of coordinates (which will be the translation given by the "flip").

Alessio Di Lorenzo
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  • Thanks for your answer! Computationally, is there a way to use $g^{-1}$ to compute quantities ranging over the unit sphere, without many "checks" and coordinate changes? – user3658307 Aug 12 '17 at 12:34