1547671870-The_Ricci_Flow__Chow

2. ETERNAL SOLUTIONS 27

So the Gauss curvature K is given by

2 r.p" ( s)

(2.10) K = D 1 (e2, e1) = - <.p (s).

Recalling equation (2.3), Remark 2.1, and Remark 2.2, we see that g con-
stitutes a steady gradient Ricci soliton if and only if there exists a function

f such that

(2.11) Kg= V'V' f.

Our assumptions are that K > 0 and that f is a radial function f = f ( s).

This implies in particular that f is a convex function. Recalling (2.9), we

observe that

and

1 r.p' ( s)

Y'e 2 e2 = W2 (e2) · e1 = - r.p(s) e1.

Since g = t5 with respect to the frame {e1, e2}, it follows that (2.11) holds if

and only if both

and

i.p^1 ( S) f I ( S)

K = e2 (e2f) - (V' e 2 e2) f = <.p (s)

hold, hence by (2.10) if and only if

(2.12) - 1.p" ( s) = !" ( s) = 1.p' ( s) !' ( s)

<.p (s) <.p (s) ·

Solving the separable ODE f" / J' = 1.p' / <.p implied by the second equality in

(2.12) implies that

(2.13) J' (s) = a<.p (s)

for some constant a. We cannot have a = 0 or else g would be fl.at. Since

<.p > 0, this forces J' to have a sign. Since f is convex, this sign must

be positive; so we may set a = 2a^2 for some a I 0. Then substituting

J' = 2a^2 <.p into the first equality in (2.12) yields the ODE

1.p" + 2a^2 r.pr.p' = 0.

Integrating this gives

r.p' + a2<.p2 = b

for some constant b. In order to g to extend to a smooth metric at s = 0,

we must have

<.p (0) = 0 and <p^1 (0) = 1

by Lemma 2.10 (below). This is possible if and only if b = 1. So we solve

the linear ODE r.p' + a^2 r.p^2 = 1 to obtain

1

<.p (s) = - tanh (as).

a