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For $\tilde{g}=e^{2\sigma }g$, the displayed correction is a symmetric tensor, so adding it to $\nabla$ gives a torsion-free connection. The two extra metric-compatibility pairings sum to $2d\sigma (X)g(Y,Z)$; this is exactly the derivative of the conformal factor. Uniqueness of the Levi-Civita connection proves the formula. It also yields the critical-point criterion for equal conformal connections.

The Levi-Civita connection is the unique connection on a vector bundle $TM$ that is torsion-free and metric-compatible. These conditions areExistence and uniqueness follow from the Koszul formula; equivalently, these are the defining properties that characterize the connection.

Define the correction tensorand set $D_{X}Y={\nabla }_{X}Y+A(X,Y)$. All terms of $A$ are $C^{\infty }(M)$-linear in both $X$ and $Y$, so $D$ is a connection. The symmetry $A(X,Y)=A(Y,X)$ preserves vanishing of the torsion tensor.

To check compatibility with $\tilde{g}$, computeOn the other hand,The cross terms cancel, so their sum is $2X(\sigma )g(Y,Z)$. Multiplying by $e^{2\sigma }$ proves $X(\tilde{g}(Y,Z))=\tilde{g}(D_{X}Y,Z)+\tilde{g}(Y,D_{X}Z)$. Hence $D$ is torsion-free and compatible with $\tilde{g}$, so uniqueness of the Levi-Civita connection givesThis is the Levi-Civita connection under conformal rescaling.

On a nonempty compact manifold without boundary, $\sigma$ attains a maximum at some $p$. The derivative of a smooth function vanishes at an interior extremum, so $(d\sigma {)}_p=0$ and $(V_{\sigma }{)}_p=0$. The correction tensor therefore vanishes at $p$, givingThis is equality for all vector fields at the same point, not an assertion that either connection vanishes there. More generally, the two connections agree at any critical point of $\sigma$. Conversely, if their correction vanishes for all $X,Y$ at a point of positive dimension, evaluating $g(A(v,v),v)=g(v,v)d\sigma (v)$ for every $v$ shows that the point is critical. This is the critical-point criterion for equal conformal connections.

The boundaryless hypothesis is the usual convention for “manifold” here. If manifolds with boundary were included, compactness alone would not suffice: on $[0,1]$ with $g=d{x}^2$ and $\sigma =x$, one has ${\tilde{\nabla }}_{{\partial }_x}{\partial }_x={\partial }_x$ everywhere while ${\nabla }_{{\partial }_x}{\partial }_x=0$. An extremum at the boundary need not be a critical point.