Timeline for Calculating scalar curvature of a warped product $I \times N^n(k)$
Current License: CC BY-SA 4.0
14 events
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| Jan 14, 2023 at 17:43 | comment | added | Matheus Andrade | Ah, I see now how my confusion arose, there is no sign inconsistency in your comment. In order for $\Omega^{i}_j = \kappa \theta^i \wedge \theta^j$ in a manifold of constant sectional curvature $\kappa$, I thought your convention for the sectional curvature was that $R_{ijji}$ was the sectional curvature of the plane spanned by $e_{i}, e_j$, with the convention that $R_{ijk}^{\ell} e_{\ell} = R(e_i, e_j) e_k$. With your convention, the scalar curvature is indeed the expression you wrote, and with the one I cited, it is another, but ultimately we both get the same thing. Thanks again! | |
| Jan 14, 2023 at 14:55 | comment | added | Deane | Ah, there is how to choose the order of indices for the curvature tensor. I always want, with respect to an orthonormal frame, $R_{ijij}$ to be the sectional curvature of the plane spanned by $e_i, e_j$. To get this, my convention is $$R^i{}_{jkl}e_i = R(e_k,e_l)e_j$$ and $R_{ijkl} = g^{ip}R_{pjkl}$. This leads to the formula $$\Omega^i_j = \frac{1}{2}R^i_{jkl}\omega^k\wedge\omega^i,$$ which holds for any frame, not just orthonormal ones. | |
| Jan 14, 2023 at 14:36 | comment | added | Deane | Let$$g = d\rho^2 + (\sin\rho)^2\,d\theta^2.$$An orthonormal coframe is\begin{align*} \omega^1 &= d\rho\\\omega^2 &= \sin\rho\,d\theta\end{align*}Differentiating,\begin{align*}d\omega^1 &= 0\\d\omega^2 &= \cos\rho\,d\rho\wedge d\theta\\&= - \cos\rho\,d\theta\wedge\omega^1.\end{align*}which implies that$$\omega^2_1 = \cos\rho\,d\theta.$$Therefore,\begin{align*}\Omega^1_2&= d\omega^1_2\\&= d(-\cos\rho\,d\theta)\\&= \sin\rho\,d\rho\wedge d\theta\\&= \omega^1\wedge\omega^2.\end{align*}Since$$\Omega^i_j = \frac{1}{2}R_{ijkl}\omega^j\wedge\omega^k,$$we get$$R_{1212} = 1.$$ | |
| Jan 14, 2023 at 14:36 | comment | added | Deane | Thanks. Could you elaborate? Since I'm never completely sure about the sign conventions, I checked this using the standard metric on the $2$-sphere. | |
| Jan 14, 2023 at 3:57 | comment | added | Matheus Andrade | Just a minor comment, which is kind of irrelevant but may be useful for future readers: I think you used some inconsistent sign conventions (which ultimately canceled out). It appears that the definition you use for the curvature tensor is $[\nabla_X, \nabla_Y ] - \nabla_{[X, Y]}$, and in that case $R_{00} = \sum_{1 \leq i \leq n} R_{i00i}$, rather than $R_{00} = \sum_{1 \leq i \leq n} R_{i0i0}$, and similarly for the other components. | |
| Jan 13, 2023 at 21:06 | comment | added | Matheus Andrade | Thanks for the tips! I'll keep them in mind. | |
| Jan 13, 2023 at 20:48 | comment | added | Deane | Yes, this is an easy mistake to make. I try, whenever possible, to use a tensor in its "natural" state. This means in a form that is independent of the metric $g$. So the derivative of a function in its natural state is a $1$-form, and the gradient is a vector field whose definition requires the metric. I very rarely use the gradient of a function. Instead, as needed, I will use the metric tensor or its inverse to raise and lower indices. When I do calculations, I try to write everything out concretely and avoid having to remember anything, such as dependence on the metric, in my head. | |
| Jan 13, 2023 at 19:20 | comment | added | Matheus Andrade | I managed to find my mistake and see where my computations diverged from yours. In the end, it was very, very silly. In the back of my mind, I always thought that the gradient of a function does not depend on the metric of a manifold. In my particular case, this led me to assume that $\nabla u$ with respect to the metric $\mathrm{d}u^2$ was the same as $\nabla u$ with respect to the metric $f^{-2} \mathrm{d} u^2$. This IS NOT the case! While $g(\nabla f, X) = \mathrm{d} f(X)$ is independent of $g$, the vector field $\nabla f$ itself is not! This led me to miss a crucial factor of $f^2$. | |
| Jan 13, 2023 at 18:52 | comment | added | Matheus Andrade | Also, I still think the Ricci tensor in Robert's answer might be wrong. Comparing yours to his I don't think the two are the same. But I'll recheck my calculations. | |
| Jan 13, 2023 at 18:51 | comment | added | Matheus Andrade | First of all, thanks a lot for taking the time to write this absolutely wonderful answer! I have been trying to see where my mistake was and managed to track it down to my calculation of $R_{ii}$, but it'll still take me a while to see where my computations diverged from yours. I also very much appreciate all your comments, they are all very valuable and very helpful. | |
| Jan 13, 2023 at 18:50 | comment | added | Deane | I should mention that since you already know from Robert's answer what $\phi$ is, you do not need to solve the ODE as I've done above. You simply have to check that Robert's formula for $\phi$ satisfies the ODE above. | |
| Jan 13, 2023 at 18:49 | vote | accept | Matheus Andrade | ||
| Jan 13, 2023 at 18:40 | history | edited | Deane | CC BY-SA 4.0 | added 589 characters in body |
| Jan 13, 2023 at 18:32 | history | answered | Deane | CC BY-SA 4.0 |