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What is the curvature form $\Omega$ associated with the Levi-Civita connection $\nabla^{\text{L.C.}}$ for the $n$-sphere $S^n$ with respect to the standard metric, i.e. what is $\Omega=d\theta+\frac{1}{2}[\theta,\theta]$ for $\theta$ the connection form?

Thanks in advance!

Sergio Charles
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  • With respect to what metric? I'm assuming you're looking for the standard metric? – YoungMath Aug 14 '18 at 21:48
  • @YoungMath Yes. – Sergio Charles Aug 14 '18 at 21:53
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    Can give you a glimpse of what you have to compute if you like. First, you need to choose linearly indep. local sections $s_1,\dots,s_n$ of the tangent bundle. It would be suitable to choose coords in a neighborhood and take the induced sections. Then, you can calculate $\nabla$ by using the Koszul formula. From that you get $\theta$, which is simply given by $\nabla_{s_k} s_i = \sum^n_{j=1} \theta_i^j(s_k) s_j$. Could be a messy calculation if you use stereographical coordinates, for instance. But the cases $n=1,2,3$ seem to be viable. – YoungMath Aug 14 '18 at 22:29
  • @YoungMath Question: How does the Koszul formula give us $\nabla$? Couldn't we instead just compute $\nabla_{X}Y=\nabla_{X^i\frac{\partial}{\partial X^i}}Y^j\frac{\partial}{\partial Y^j}=X^iY^j\nabla_{i}Y_j=X^iY^j(\partial_iY_j-\Gamma^{t}_{ij}Y_t)$ where the Christoffel symbol of second kind is given by the metric on $S^n$ (see here). – Sergio Charles Aug 15 '18 at 17:01
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    That's another valid approach. There is just more than one way to compute stuff like that. But in the end, all of them are equivalent. – YoungMath Aug 15 '18 at 19:17

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Your question is a bit ambiguous, but here's how I choose to interpret it. Choose any local orthonormal moving frame $e_1,\dots,e_n$ on $S^n$. Indeed, if $e_0$ is the position vector, the matrix formed by $e_0,e_1,\dots,e_n$ gives us a mapping to $O(n+1)$. Thinking of $e_A$ ($A=0,\dots,n$) as maps to $\Bbb R^{n+1}$, we write $de_A = \sum\limits_{B=0}^n \omega_{AB}e_B$, we'll have $\omega_{0j} = \omega_j$ ($j=1,\dots,n$) the dual coframe to the original moving frame and $(\omega_{ij})$ will be the connection matrix of $1$-forms.

Now $d(de_A) = 0$ gives us $d\omega_{AB} = \sum\limits_C \omega_{AC}\wedge\omega_{CB}$. In particular, the entries of the curvature matrix of $2$-forms are $$\Omega_{ij} = d\omega_{ij} - \sum_{k=1}^n\omega_{ik}\wedge\omega_{kj} = \sum_{C=0}^n \omega_{iC}\wedge\omega_{Cj} - \sum_{k=1}^n\omega_{ik}\wedge\omega_{kj} = \omega_{i0}\wedge\omega_{0j} = -\omega_i\wedge\omega_j.$$ This tells us, in particular, that every sectional curvature is $1$.

(There are always sign issues with the structure equations depending on whether we think of $e_A$ as column vectors or row vectors, i.e., depending on whether we have $O(n+1)$ acting on the left or on the right on the orthonormal frame bundle. I'll leave you to sort this out ...)

Ted Shifrin
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  • Thanks @Ted Shifrin! I would like to explicitly calculate the first Chern class of the $n$-sphere (in a similar fashion to the way that you did for $S^2$) via $c_1(S^n)=\frac{i}{2\pi}\text{Tr}(\Omega)$. – Sergio Charles Aug 16 '18 at 17:14
  • That doesn't even make sense, @Multivariablecalculus. You need a complex vector bundle to talk about Chern classes. – Ted Shifrin Aug 16 '18 at 17:16
  • I thought $TS^n$ can be identified with the affine hyperquadric $Q=\left{(z_0,\dots,z_n)\in\mathbb{C}^{n+1}\big|\sum_{i=0}^nz_i^2=1\right}$, so it would be the same as $c_1(Q)$? @TedShifrin – Sergio Charles Aug 16 '18 at 17:17
  • That, too, doesn't make sense. By $c_1$ of a complex manifold one means $c_1$ of its tangent bundle. So that won't work. But the diffeomorphism between $TS^n$ and $Q$ is not in any sense a bundle mapping, is it? – Ted Shifrin Aug 16 '18 at 17:29
  • So why is $c_1(TS^2)$ allowed? @TedShifrin – Sergio Charles Aug 16 '18 at 17:30
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    Because $S^2 = \Bbb CP^1$ is a complex manifold. When $n$ is odd, $S^n$ cannot be an almost complex manifold, and it's known that it's also false for all even $n$ except $n=6$; whether $S^6$ can be a complex manifold is still an open question. Regardless, the Riemannian structure of $S^n$, as this question addressed, is not going to be relevant as it was for $S^2$. – Ted Shifrin Aug 16 '18 at 17:32
  • I see, thank you. @TedShifrin – Sergio Charles Aug 16 '18 at 17:34
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    On real vector bundles, like the tangent bundle, you can just calculate the Pontryagin classes which are given by the even Chern classes alternating signs, roughly speaking. But even though the absolute Pontryagin class (regarding as homology class) is constant 1 on $S^n$. – YoungMath Aug 16 '18 at 21:25
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    You have to complexify the bundle, re the first sentence, The second sentence is nonsense, @YoungMath. – Ted Shifrin Aug 17 '18 at 01:18
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    Right, it's the complexified bundle. Hm, no, I'm pretty sure it's not nonsense. Consider the normal bundle of $S^n \subset \mathbb{R}^{n+1}$ denoted by $N$ which is trivial (see here). Then we have $1=p(T\mathbb{R}^{n+1} \vert _{S^n})=p(TS^n \oplus N)=p(TS^n) \cdot p(N)=p(TS^n)$ since trivial vector bundles have total Chern (Pontryagin) class 1. All conceiving as cohomology classes not as pure differential forms. Otherwise the statement would be false. – YoungMath Aug 17 '18 at 10:35