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I was wondering if anyone had a simple example of a manifold which only has Ricci curvature in one direction ie. such that the Ricci tensor only has one non-zero component.

Intuitively, I would expect such a thing to be possible somehow just from the definition of Ric but could not think of an example.

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Yes, this is possible. More precisely: In dimension $\geq 3$ there are metrics for which the Ricci curvature has rank $1$, that is, which can be written (locally, around any point) as $\phi \alpha \otimes \alpha$ for some smooth function $\phi$ and $1$-form $\alpha$. (In dimension $2$ the Ricci tensor is a (smooth) multiple of the metric, so this behavior is possible only in dimension $\geq 3$.)

Here's one way to construct an example. The warped product of the flat metric $\bar g := dx_1^2 + \cdots + dx_{n - 1}^2$ on $\Bbb R^{n - 1}$ with the flat metric $dy^2$ on $\Bbb R$ via $f(y)$ is $$g := \bar g \times_{f(y)} dy^2 = f(y) \bar g + dy^2 .$$

Computing gives that the Ricci curvature of $g$ is $$\operatorname{Ric} = - \frac{2 f''(y) f(y) + (n - 1) f'(y)^2}{4 f(y)} \bar g - \frac{(n - 1)(2 f''(y) f(y) - f'(y)^2)}{4 f(y)^2} dy^2$$

Demanding that the coefficient of $\bar g$ vanish defines a second-order differential equation in $f$ with general solution $$f(y) = \left[\frac{n - 1}{2} (A y + B)\right]^{2 / (n - 1)},$$ and substituting in $g$ gives (now restricting the metric to where $A y + B > 0$) that $$\operatorname{Ric} = \frac{(n - 2) A^2}{(n - 1) (A y + B)} dy^2 .$$ So, if we take $A \neq 0$, then for $n > 2$, $\operatorname{Ric}$ has rank $1$.

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  • $\begingroup$ I have two questions: would the final expression be neater if you assumed $f^2$ in the warped product? Also, I don't quite follow why you get one differential equation, if you assume the coefficient of g bar is 0 and the coefficient of $dy^2$ is 0, does that not give you a system of two differential equations to solve? $\endgroup$ – Tom Mar 20 at 20:12
  • $\begingroup$ ie. to solve for $f$, would you not need to assume the coefficients of both metrics are $0$ and that would then give you a pair of coupled differential equations? $\endgroup$ – Tom Mar 20 at 20:47
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    $\begingroup$ If you replaced $f$ with $f^2$ then the solution would be replaced with its square root. This would change the exponent in the formula for $f$ in the answer, but I don't think that's much of an improvement. And no---if we ask for both coefficients to vanish, that's asking $\operatorname{Ric}$ to be zero (and for $n > 2$ this imposes exactly $A =0$), but here we're looking for a metric for which $\operatorname{Ric}$ has rank $1$. So, by demanding that the coefficient of $\bar g$ in $\operatorname{Ric}$ vanish, we're looking for a metric for which $\operatorname{Ric}=\phi\,dy^2$ for some $\phi$. $\endgroup$ – Travis Willse Mar 20 at 20:54
  • $\begingroup$ I'm not sure if I have explained myself properly. Basically, take a manifold $(M,g)$ with a scalar curvature (call it $R$), I am looking for an example of a manifold such that one of the components of Ric is equal to the value of $R$ at every point, or maybe on most points if that is not possible, and Ric is $0$ for all the other components. In a sense this must mean you are solving two equations since you want the g bar coefficient to vanish but you want the coefficient for $dy^2$ to equal $R$. $\endgroup$ – Tom Mar 20 at 22:00
  • $\begingroup$ Strictly speaking one can only talk about "components of $\operatorname{Ric}$" once you choose additional information like a local frame. (This is why I refer to "rank $1$", which is a coordinate-independent condition, in my answer.) In any case, is your problem that you have a specific scalar function $R$ and you want to find a metric $g$ and coordinates $(x_1, \ldots, x_{n - 1}, y)$ such that $\operatorname{Ric} g = R \,dy^2$? $\endgroup$ – Travis Willse Mar 20 at 22:13

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