and(1.57)- CONSTRUCTING THE BRYANT STEADY SOLITON
x^2 -xy+n-1
w"=------
w(n -1) (X^2 -yin (X - CYY^2 ))
wY^2
= O(-Y^3 ) = O(-r-^312 )
as r---+ oo.25A moving frames calculation like that mentioned above^10 shows that the
sectional curvatures of the Bryant solitons are
1-(w')^2 w"
(1.58) v1 = w 2 , V2 = ---;;;,where v1 is the curvature for planes tangent to the spheres, and v 2 for planes
tangent to the radial direction. In particular, 2v1 is the eigenvalue of the
curvature operator corresponding to the eigenspace
El ~ {CY /\ ,8 : CY) ,8 E A^1 ( sn x { r}) }
and 2v2 is the eigenvalue corresponding to the eigenspace
E2 ~ {CY /\ dr : CY E A^1 ( sn x { r})}.
From the asymptotic behavior of w and its first two derivatives, we obtain
v1 = O(r-^1 ) and v2 = O(r-^2 ). So, the sectional curvatures decay inverse
linearly in terms of the distance from the origin. Like the cigar, the curvature
is positive:LEMMA 1.37 (Curvature of Bryant soliton). The curvature operator ofthe Bryant soliton is strictly positive (i.e., v1 > 0 and v2 > 0) away from the
origin, and these curvatures have a positive limit at the origin. Moreover,v1 = O(r-^1 ) and v2 = O(r-^2 ).
PROOF. The positivity of v1 for r > 0 follows from the fact that w' = x
is strictly less than 1. To show that v2 is positive for r > 0, we will show
that w" is negative. From (1.57), it suffices to show that x^2 - xy + n - 1 is
negative, or equivalently that X^2 - foX + Y^2 is negative. Linearizing the
system (1.48) about the saddle point shows that x^2 - xy + n - 1 is initially
negative when the trajectory emerges. Moreover,
.!!:._(X2 - Vn + y2) = X2(X2 + y2 - 1) + (X2 - l)(X2 - vnX + y2)
dsshows that if X^2 - fo + Y^2 were ever equal to zero in the middle of the
trajectory, it would have a negative derivative. So, w" is negative for all
r > 0.
At the origin, all sectional curvatures are equal. The scalar curvature R
satisfies R +IV fl^2 = C, and we know C must be positive because
R = n(n - l)v1 + 2n v2
(^10) See, for example, the solution to Exercise 1.188 in [111].