Rotational symmetry of self-similar solutions to the Ricci flow

Abstract

Let (M,g) be a three-dimensional steady gradient Ricci soliton which is non-flat and κ-noncollapsed. We prove that (M,g) is isometric to the Bryant soliton up to scaling. This solves a problem mentioned in Perelman’s first paper.

Appendix: The eigenvalues of some elliptic operators on S 2

In this section, we collect some well-known results concerning the eigenvalues of certain elliptic operators on S ². In the following, $g_{S^{2}}$ will denote the standard metric on S ² with constant Gaussian curvature 1.

Proposition A.1

Let σ be a one-form on S ² satisfying

$$\varDelta _{S^2} \sigma+ \mu \sigma= 0, $$

where $\varDelta _{S^{2}}$ denotes the rough Laplacian and μ∈(−∞,1) is a constant. Then σ=0.

Proof

We can find a real-valued function α and a two-form ω such that σ=dα+d ^∗ ω. Using the Bochner formula for one-forms, we obtain

Consequently, the function $\varDelta _{S^{2}} \alpha+ (\mu+1) \alpha$ is constant, and the two-form $\varDelta _{S^{2}} \omega+ (\mu+1) \omega$ is a constant multiple of the volume form. Since μ+1<2, we conclude that α is constant and ω is a constant multiple of the volume form. Thus, σ=0, as claimed. □

Proposition A.2

Let χ be a symmetric (0,2)-tensor on S ² satisfying

$$\varDelta _{S^2} \chi- 4 \overset{\mathrm{o}}{\chi} + \mu \chi= 0, $$

where $\overset{\mathrm{o}}{\chi}$ denotes the trace-free part of χ and μ∈(−∞,2) is a constant. Then χ is a constant multiple of $g_{S^{2}}$.

Proof

The trace of χ satisfies

$$\varDelta _{S^2}(\operatorname {\mathrm {tr}}\chi) + \mu (\operatorname {\mathrm {tr}}\chi) = 0. $$

Since μ<2, we conclude that $\operatorname {\mathrm {tr}}\chi$ is constant. Moreover, the trace-free part of χ satisfies

$$\varDelta _{S^2} \overset{\mathrm{o}}{\chi} + (\mu- 4) \overset{\mathrm{o}}{\chi} = 0. $$

Since μ−4<0, it follows that $\overset{\mathrm{o}}{\chi} = 0$. Putting these facts together, the assertion follows. □