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Supposing a Riemannian manifold $\{M,g\}$, $\lambda g$ is also a Riemannian metric on $M$ with the constant scaling factor $\lambda>0$. The geodesics, Riemannian exponential map, Riemannian logarithm (the inverse of exponential map), and parallel transportation along the geodesic are all the same under $g$ and $\lambda g$.

Is this claim right? I think it is right. The following is my proof. The key is that Chritoffel symbols are the same.

Proof:

According to the following, Chritoffel symbols are the same: $$ \Gamma_{i j}^m=\frac{1}{2} \sum_k\left\{\frac{\partial}{\partial x_i} g_{j k}+\frac{\partial}{\partial x_j} g_{k i}-\frac{\partial}{\partial x_k} g_{i j}\right\} g^{k m} $$

According to the following two equations for geodesic and parallel transporation, we can see the equivalence of the operators under $g$ and $\lambda g$, including geodesics,exponential,logarithm,parallel transportation along a geodesic: $$ \frac{d^2 x_k}{d t^2}+\sum_{i, j} \Gamma_{i j}^k \frac{d x_i}{d t} \frac{d x_j}{d t}=0, \quad k=1, \ldots, n $$ $$ \frac{D V}{d t}=\sum_k\left\{\frac{d v^k}{d t}+\sum_{i, j} v^j \frac{d x_i}{d t} \Gamma_{i j}^k\right\} X_k=0 $$

gsoldier
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They are not really the same but related. For example, the exponential map gives the same trajectories (the geometric geodesics) but they are traversed at a different speed. Namely, the exponential map by definition would produce unit speed geodesics, but when the metric is modified by the factor $\lambda$ this changes the speed by the same factor. The effect on other "Riemannian operators" can be deduced from their explicit form in terms of the conformal factor. For example, on a surface the Laplace-Beltrami operator would appear in conformal coordinates $f^2(x,y)(dx^2+dy^2)$ as $\frac{1}{f^2}\Delta_0$ where $\Delta_0$ is the flat Laplacian. Modifying $f$ by the factor $\lambda$ would lead to a corresponding change in the Laplace-Beltrami operator.

Mikhail Katz
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  • For $p \in M$ and $v \in T_pM$, the exponential $Exp_p(v)$ is defined by $\gamma(1,p,v)$. I think $\gamma(1,p,v)$ have the same results under $g$ and $\lambda g$. Why did you say $Exp_p(v)$ differs under $g$ and $\lambda g$? – gsoldier Nov 08 '23 at 13:30
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    You are right. We are arguing about semantics. Consider the exponential map of the unit sphere. It maps the boundary of the ball of radius $\pi$ to a point. Now modify the metric by a factor. The boundary of the ball of radius $\pi$ is no longer mapped to a single point. But if you think of the tangent space in purely differential-geometric sense, the exponential map remains "the same". As far as the Laplace-Beltrami operator is concerned, it gets modified by the square of $\lambda$. I was responding to the title of your question which mentioned general "riemannian operators". – Mikhail Katz Nov 08 '23 at 13:42
  • Thanks so much. I understand now. Perhaps i did not explain my question clearly. So let me formalize my question further. For $p,q \in M$ and $v \in T_pM$, then $Exp_p(v)$, $Log_p(q)$, and $PT_{p \rightarrow q}(v)$ are the same under $g$ and $\lambda g$, where $PT$ is the parallel transportation along the geodesic connecting $p$ and $q$. In this sense, I think my claim is right, am I right please? – gsoldier Nov 08 '23 at 13:51
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    @gsoldier, I think so. The main point, as you note, is that the Gamma symbols are the same. You may want to be more explicit about the cancellation of $\lambda^2$ in the calculation of the Gamma symbols. – Mikhail Katz Nov 08 '23 at 13:54
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    This can be answered by writing out carefully the equations (algebraic, ODE, PDE) used to define everything and working out how each metric-dependent parameter scales if the metric is rescaled and how they scale if the coordinates are rescaled. This is a good exercise when learning Riemannian geometry and a useful tool when doing calculations and proofs. – Deane Nov 08 '23 at 14:47
  • Hi @Deane, did you mean to post your comment under the question? – Mikhail Katz Nov 08 '23 at 14:49
  • Thanks. I have written down the ODE of geodesics, and parallel transportation along geodesics. The Chritoffel symbols remain the same. So I think that Riemannian exponential/logarithmic map, parallel transportation remain the same. However, others involve the inner product will change, like geodesic distance, Ricci curvature. – gsoldier Nov 08 '23 at 15:42
  • The exponential map does not remain the same, even though the ODE remains unchanged. Note that the initial data for the ODE (at $t=0$) changes when the metric is rescaled. Also, you can check that, surprisingly, the Ricci tensor is invariant under rescaling of the metric but the Riemann tensor is not. It's worthwhile to figure out why this makes sense geometrically. – Deane Nov 08 '23 at 22:02
  • Thanks. I check that the Ricci curvature is invariant and Riemann tensor is not. But I do not think $Exp$ will change. For instance, For $p\in M$ and $v \in T_p M$, then $Exp_p(v)=\gamma(1,p,v)$. Since the geodesic $\gamma(t,p,v)$ remain unchanged, $Exp_p(v)$ also do not change. – gsoldier Nov 09 '23 at 08:09
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    @Deane The $(0,4)$ Riemann tensor is not invariant, but so is its $(1,3)$-version. This actually helped me to understand why Ricci is invariant – Didier Nov 09 '23 at 23:12
  • @Didier, good point! Thanks. – Deane Nov 10 '23 at 00:55
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    @Deane, if you have a vector of unit length for the original metric, it will get mapped to a point at distance 1 from $p$ under the exponential map at $p$. If we now multiply the metric by $\lambda^2$, the vector will have length $\lambda$, and so it is expected to be sent to the point at distance $\lambda$ from $p$ for the new metric, i.e., it should be the same (here I am ignoring problems with the injectivity radius). So why doesn't the exponential map stay the same? I am confused. – Mikhail Katz Nov 10 '23 at 11:39
  • @MikhailKatz, I'm sorry. I was doing this all in my head, got mixed up, and was being stubborn. I'll blame it on old age.. Now that I've written everything carefully, I see my error. – Deane Nov 10 '23 at 14:45