I am looking for example of steady Ricci soliton with indefinite or nonpositive Ricci curvature. Any help will be appreciated. Thanks!
I have found doing some calculation that the metric:
$$g(t)=\frac{dx^2+dy^2}{e^{4t}x^2y^2}$$
satisfies $\frac{dg(t)}{dt}=2Ric(t)$
Where
$$\frac{dg(t)}{dt}=\frac{4e^{4t}(dx^2+dy^2)}{(e^{4t}x^2y^2)^2}\ {\rm and} \ Ric(t)=\frac{2e^{4t}( dx^2+dy^2)}{(e^{4t}x^2y^2)^2}$$
this should be a solution to the Ricci flow on $R^2$.
Furthermore, with some other calculation, I found that this metric $g(t)$ is the only solution within the family: $g_{\lambda}(t)=\frac{dx^2+dy^2}{e^{4\lambda t}x^{2 \lambda}y^{2 \lambda}}$, in fact, the only solution to the Ricci flow is only for $\lambda=1$.
Said this, my analysis has been to see the behavior in time for $g(t)$ and $Ric(t)$, when $t$ tends to $\infty$ and $ \infty$, but I want that the metric $g(t)$ remain positive.
a) Then for $t \rightarrow \infty$, to ensure that $g(t)$ is positive, I fixed $x=0$ and $y=0$ and I get $g(t)$ tends to $\infty$ and $Ric(t)$ tends to $2$
b) While for $t \rightarrow  \infty$ I get $g(t)$ tends to $0$ and $Ric(t)$ tends again to $2$.
For $t=0$,
I get $g_0= \frac{dx^2+dy^2}{1x^2y^2}$ and with a simple calculations, I found that the 1parameter family of diffeomorfism is $\phi_t=(e^{2t}x, e^{2t}y)$.
Considering $r^2=(x^2+y^2)$, the Scalar curvature, for $t=0$, is $R=\frac{4}{1r^2}$.
Now If I make an analysis of singularities (always for $(1r^2)>0$), I found that if $r$ tends to $1$, $g_0$ tends to $\infty$ and $R$ tends $ \infty$.
This soliton should be a steady soliton, it has negative curvature and it is not bunded from below.

$\begingroup$ There is a small error in your post, the given solution is not on R^2, it is only defined on the unit disk or a subset thereof. It also doesn't make sense to take t to infinity (what is the underlying space?). Usually steady solitons are defined for all time, it is not the case here since the metric is not complete. Any complete steady soliton has nonnegative scalar curvature. (theorem of BingLong Chen) $\endgroup$ Dec 5 '21 at 5:36