# What are some explicit examples of nontrivial gradient almost Ricci solitons with harmonic curvature?

A Riemannian manifold $$(M, g)$$ is said to be an almost Ricci soliton if there exists a complete vector field $$X \in \Gamma(TM)$$ and a smooth function $$\lambda: M \to \mathbb{R}$$ such that $$\operatorname{Ric} + \frac{1}{2}\mathscr{L}_{X} g = \lambda g$$ When this vector field is the gradient of a smooth function $$f: M \to \mathbb{R}$$, we say $$M$$ is a gradient almost Ricci soliton, and this equation becomes: $$\operatorname{Ric} + \operatorname{Hess}(f) = \lambda g$$ Obviously, any Einstein manifold is a Ricci soliton and hence an almost Ricci soliton (gradient as well, trivially), so these are trivial examples.
If $$M$$ satisfies: $$\operatorname{div}({\operatorname{Rm}}) = 0$$ we then say $$M$$ has harmonic curvature (notice this happens if and only if $$M$$ has harmonic Weyl curvature and constant scalar curvature. I think that part of some work I've been doing with some other people shows that any gradient almost Ricci soliton with harmonic curvature satisfies the property that for any $$p \in M$$, there is a neighborhood $$U_p \ni p$$ such that $$U_p$$ has constant sectional curvature (and is therefore necessarily Einstein) (EDIT AT NOVEMBER 27: this supposes the dimension is $$\geq 4$$. Also, I've come to realize since the initial writing of this post that the Einstein examples might not be exhaustive).

As a sanity check, I'm looking for some explicit examples of nontrivial (i.e, not Einstein and with nonconstant $$\lambda$$) gradient almost Ricci solitons (preferably of dimension $$\geq 5$$) with harmonic curvature. Can anyone here provide some examples? I'd appreciate any help. Thanks in advance!

• I'm curious why you define the condition of 'harmonic curvature' in one sentence and then don't refer to again. Is this a stray sentence that should be deleted or did you intend to add this as a hypothesis somewhere and forgot? Commented Nov 2, 2022 at 14:33
• @RobertBryant sorry, I indeed forgot to add it as a hypothesis. I'll fix it, thanks for pointing it out. Commented Nov 2, 2022 at 16:41
• I thought a little bit about this and did a few caculations. In dimension 3, at least, it is not true that a gradient almost Ricci soliton with harmonic curvature is locally Einstein. There is a $3$-parameter family of mutually non-isometric, nontrivial examples $(g,f,\lambda)$ in dimension $3$ with the function $\lambda$ and the sectional curvatures not being constant. I haven't looked at higher dimensions, but I don't see why it would necessarily fail there. Commented Nov 27, 2022 at 14:06
• @RobertBryant thanks a lot! However, I must apologize again for not having included the hypothesis of dimension $\geq 4$ in my original post when I mentioned "part of some work that I've been doing with some other people...". But I am indeed very interested in knowing the explicit example of this $3$-parameter family you mentioned, I would very much appreciate it if you shared it. Commented Nov 27, 2022 at 18:43
• OK. As I suspected, the three-dimensional examples easily generalize to all higher dimensions as a 3-parameter family of non-trivial gradient almost Ricci solitons with harmonic curvature. They are not completely explicit, though, because each equivalence class of solutions corresponds to an integral curve of a vector field in $\mathbb{R}^3$. That vector field depends on a parameter: the (constant) scalar curvature $S=n(n{-}1)c$ of the metric $g$. I don't know how to integrate the vector field in elementary terms, but in the case $c=0$ phase portrait methods give good qualitative information. Commented Nov 27, 2022 at 21:41

I'm revising my answer to shorten it, since there is a much simpler way to describe these solutions more fully.

Let $$(N^n,h)$$ be a metric of constant sectional curvature $$k$$ and consider the quadratic form $$g = \frac{\mathrm{d}u^2}{k-a\,u^2+ b\,u^{1-n}} + u^2\,h$$ on $$M^{n+1} = \mathbb{R}^+\times N$$, where $$a$$ and $$b$$ are constants and $$u>0$$ is the coordinate on $$\mathbb{R}^+$$. If $$I\subset\mathbb{R}^+$$ is an interval on the $$u$$-line on which $$k-a\,u^2+ b\,u^{1-n} >0$$, then $$g$$ is a Riemannian metric on $$I\times N$$ that is conformally flat and has constant scalar curvature $$S = n(n{+}1)a$$. Hence it has harmonic curvature. The Ricci curvature is $$\mathrm{Ric}(g) = \bigl(n\,a - \tfrac{1}{2}\,b\,u^{-n-1}\bigr)\,g + \frac{(n^2{-}1)b\,\mathrm{d}u^2}{2\bigl(b\,u^2+k\,u^{n+1}-a\,u^{n+3}\bigr)},$$ so $$g$$ is Einstein if and only if $$b=0$$.

Moreover, it is now easy to construct (by quadrature) a function $$f = f(u)$$ on $$I$$ such that $$\mathrm{Ric}(g) + \mathrm{Hess}_g(f) = \lambda\,g$$ for some function $$\lambda$$. When $$b\not=0$$, $$\lambda$$ will not be constant. Thus, this gives a completely explicit $$3$$-parameter family of non-trivial almost Ricci solitons with harmonic curvature.

If $$I = (r_1,r_2)$$ where $$r_2>r_1>0$$ are simple roots of $$k-a\,u^2+ b\,u^{1-n}=0$$, then the curve $$v^2 = k-a\,u^2+ b\,u^{1-n}$$ in the $$uv$$-plane has a smooth circle component $$C$$ between the lines $$u=r_1$$ and $$u=r_2$$. In this case, the metric $$g$$ extends to a smooth complete metric on $$C\times N$$ (assuming that $$(N,h)$$ is complete). In this way, one can construct many complete or compact examples of such metrics. However, when $$b\not=0$$, the functions $$f$$ and $$\lambda$$ will only be locally defined unless one passes to the simply-connected cover of $$C$$, so that $$M = \mathbb{R}\times N$$. On this covering space, $$f$$ (and $$\lambda$$) can be globally defined.

Remark: In dimension $$3$$, it turns out that every conformally flat metric $$(M^3,g)$$ with constant scalar curvature (i.e., every metric in dimension $$3$$ with harmonic curvature) that admits a 'Ricci potential', i.e., a function $$f$$ such that $$\mathrm{Ric}(g)+\mathrm{Hess}_g(f) = \lambda\,g$$ for some function $$\lambda$$, is locally of the above form for some $$(N^2,h)$$ with constant curvature. It was after I worked out this fact, via an exterior differential system analysis, that I realized that the above construction could be used to produce examples in any dimension.

In dimensions above $$3$$, it is not likely that every metric with harmonic curvature that admits a Ricci potential in the above sense is of the form given above, but I don't know a classification, even when $$n=4$$.

• Thanks very much! This confirmed a few suspicions of mine. When you have the time I'd love to see all the other examples you mentioned, but this is already pretty good :) Commented Nov 28, 2022 at 15:50
• I just saw your addendum. What a wonderful answer! Thanks a lot. This completely explicit construction goes above and beyond what I expected. Commented Dec 1, 2022 at 19:36
• Could you perhaps have made a typo when defining $g$? I calculated the scalar curvature of a metric $$g = \frac{\mathrm{d}u^2}{(f(u))^2} + u^2 h$$ and found it to be equal to $$\text{Scal}_{g} = \frac{n f'}{fu} - \frac{n f f'}{u} + \frac{n(n-1)}{u^2}\left(k - \frac{1}{f^2} \right)$$ Setting $f = \sqrt{k - au^2 + bu^{1-n}}$ as in your example, I found that $\text{Scal}_g$ is not constant. Commented Jan 12, 2023 at 13:35
• @MatheusAndrade: Hmmm. You are right that something is wrong with my answer, but I'm not sure what. I'm traveling now and don't have those calculations with me so I can't check them. I'll have a look at them when I get back home next week, check them over, and see what I can do about correcting the formulae. Commented Jan 13, 2023 at 7:29
• Alright, I'm looking forward to it, thanks for the reply! I think maybe $\text{Ric}(g)$ or the definition of $g$ is not right. Commented Jan 13, 2023 at 17:52

The Riemannian product $$\mathbb{R}^m \times S^n$$ is always of this type, where $$\mathbb{R}^m$$ is given the flat metric and $$S^n$$ the round metric of constant sectional curvature one. In this case $$\lambda=n-1$$ and you can take $$f(x,\theta) = \frac{n-1}{2}\lvert x\rvert^2$$.

• Thanks very much for the answer, but I would prefer examples where $\lambda$ is not constant. I'm sorry I didn't specify that, I'll edit my question. Commented Nov 2, 2022 at 0:05