As everyone knows, the steady gradient Ricci soliton is a triple $(M,g,f)$ such that $$ Ric+\nabla^2 f =0 $$ Then, how to normalize it to $$ ~~~~~~~~~~~~~~~~~~~Ric=\nabla^2 f ~~~~~~~~~~~~~~~~~~~~~~~~~~(1)\\ ~~~~~~~~~~~~~~~~~~~R+|\nabla f|^2 =1 ~~~~~~~~~~~~~~~~~~~~(2) $$ I know how to get $(1)$, just let $\tilde f = -f$. But I don't know how to get $(2)$. This problem origin An optimal volume growth estimate for noncollapsed steady gradient Ricci solitons. Pictures below also from it.
2026-03-25 17:37:32.1774460252
Normalize of steady gradient Ricci soliton
82 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
1
There are 1 best solutions below
Related Questions in PARTIAL-DIFFERENTIAL-EQUATIONS
- PDE Separation of Variables Generality
- Partial Derivative vs Total Derivative: Function depending Implicitly and Explicitly on Variable
- Transition from theory of PDEs to applied analysis and industrial problems and models with PDEs
- Harmonic Functions are Analytic Evan’s Proof
- If $A$ generates the $C_0$-semigroup $\{T_t;t\ge0\}$, then $Au=f \Rightarrow u=-\int_0^\infty T_t f dt$?
- Regular surfaces with boundary and $C^1$ domains
- How might we express a second order PDE as a system of first order PDE's?
- Inhomogeneous biharmonic equation on $\mathbb{R}^d$
- PDE: Determine the region above the $x$-axis for which there is a classical solution.
- Division in differential equations when the dividing function is equal to $0$
Related Questions in RIEMANNIAN-GEOMETRY
- What is the correct formula for the Ricci curvature of a warped manifold?
- How to show that extension of linear connection commutes with contraction.
- geodesic of infinite length without self-intersections
- Levi-Civita-connection of an embedded submanifold is induced by the orthogonal projection of the Levi-Civita-connection of the original manifold
- Geodesically convex neighborhoods
- The induced Riemannian metric is not smooth on the diagonal
- Intrinsic vs. Extrinsic notions of Harmonic maps.
- Equivalence of different "balls" in Riemannian manifold.
- Why is the index of a harmonic map finite?
- A closed manifold of negative Ricci curvature has no conformal vector fields
Related Questions in RICCI-FLOW
- Ricci soliton solves Ricci flow
- Can Ricci flow develop singularity if metric is bounded?
- Ricci flow of the Torus
- Ricci Flow Matlab Code / Algorithm
- Cylinder to sphere rule - Ricci Flow
- Short-time existence of Ricci flow
- Normalized Ricci flow
- Calculation using contracted Binachi identity
- Gram-Schmidt with curvature
- Curvature of a homogenous manifold.
Related Questions in SOLITON-THEORY
- Qualitative solutions of KDV
- Topological solitons in general dimension
- Integrals of motion remain constant (proof)
- Are there Soliton Solutions for Maxwell's Equations?
- Non-linear waves and solitons: verifying solution for Nonlinear Schrödinger Equation
- N soliton solution KdV equation
- Good book on solitons focusing mainly on analysis of PDE
- What is a soliton?
- Literature about solitons and Hirota derivatives
- How dispersive term in KdV PDE causes smoothing
Trending Questions
- Induction on the number of equations
- How to convince a math teacher of this simple and obvious fact?
- Find $E[XY|Y+Z=1 ]$
- Refuting the Anti-Cantor Cranks
- What are imaginary numbers?
- Determine the adjoint of $\tilde Q(x)$ for $\tilde Q(x)u:=(Qu)(x)$ where $Q:U→L^2(Ω,ℝ^d$ is a Hilbert-Schmidt operator and $U$ is a Hilbert space
- Why does this innovative method of subtraction from a third grader always work?
- How do we know that the number $1$ is not equal to the number $-1$?
- What are the Implications of having VΩ as a model for a theory?
- Defining a Galois Field based on primitive element versus polynomial?
- Can't find the relationship between two columns of numbers. Please Help
- Is computer science a branch of mathematics?
- Is there a bijection of $\mathbb{R}^n$ with itself such that the forward map is connected but the inverse is not?
- Identification of a quadrilateral as a trapezoid, rectangle, or square
- Generator of inertia group in function field extension
Popular # Hahtags
second-order-logic
numerical-methods
puzzle
logic
probability
number-theory
winding-number
real-analysis
integration
calculus
complex-analysis
sequences-and-series
proof-writing
set-theory
functions
homotopy-theory
elementary-number-theory
ordinary-differential-equations
circles
derivatives
game-theory
definite-integrals
elementary-set-theory
limits
multivariable-calculus
geometry
algebraic-number-theory
proof-verification
partial-derivative
algebra-precalculus
Popular Questions
- What is the integral of 1/x?
- How many squares actually ARE in this picture? Is this a trick question with no right answer?
- Is a matrix multiplied with its transpose something special?
- What is the difference between independent and mutually exclusive events?
- Visually stunning math concepts which are easy to explain
- taylor series of $\ln(1+x)$?
- How to tell if a set of vectors spans a space?
- Calculus question taking derivative to find horizontal tangent line
- How to determine if a function is one-to-one?
- Determine if vectors are linearly independent
- What does it mean to have a determinant equal to zero?
- Is this Batman equation for real?
- How to find perpendicular vector to another vector?
- How to find mean and median from histogram
- How many sides does a circle have?



The basic idea is the following: The assumption $\mathrm{Ric} + \nabla^2f=0$ implies that there is a constant $c$ such that $R + \lvert \nabla f \rvert^2 = c$ (apply 2nd Bianchi). A maximum principle argument implies that $R \geq 0$, and hence $c > 0$ unless $f$ is constant. In the case when $c>0$, note that under the homothetic scaling $\hat g = \lambda^2g$, it holds that $$ \mathrm{Ric}_{\hat g} + \nabla_{\hat g}^2f = 0 \quad\text{and}\quad R_{\hat g} + \lvert \nabla_{\hat g} f \rvert_{\hat g}^2 = \lambda^{-2}c . $$ In particular, choosing $\lambda$ appropriately makes $c=1$.
How to derive the existence of c
We compute that \begin{align*} 0 & = \nabla^k ( R_{jk} + f_{jk} ) \\ & = \frac{1}{2}\nabla_jR + f_{kj}{}^k \\ & = \frac{1}{2}\nabla_jR + f_k{}^k{}_j + R_j^kf_k \\ & = \frac{1}{2}\nabla_j ( R + 2\Delta f) - f_j^k f_k \\ & = \frac{1}{2}\nabla_j ( R + 2\Delta f - f_k f^k) \\ & = -\frac{1}{2}\nabla_j (R + \lvert \nabla f \rvert^2) , \end{align*} where the first line is the GRS assumption, the second is 2nd Bianchi, the third is the Ricci identity, the fourth is the GRS assumption and grouping terms, the fifth is the product rule, and the sixth is that the trace of the GRS assumption gives $\Delta f = -R$. If your manifold is connected, you conclude that $R + \lvert\nabla f\rvert^2$ is constant.