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Combining metric compatibility with the torsion-free condition forces the Koszul formula:Nondegeneracy of $g$ proves uniqueness of the Levi-Civita connection. For existence, use the right side to define $2g({\nabla }_{X}Y,Z)$. Expanding brackets shows that this expression is $C^{\infty }$-linear in $Z$, so it defines a smooth one-form; the musical isomorphism gives the required vector field. The same expansion shows ${\nabla }_{aX}Y=a{\nabla }_{X}Y$, additivity, and ${\nabla }_X(aY)=X(a)Y+a{\nabla }_{X}Y$, establishing the connection rules. Subtracting the formulas with $X,Y$ exchanged gives ${\nabla }_{X}Y-{\nabla }_{Y}X=[X,Y]$. Adding the formulas pairing ${\nabla }_{X}Y$ with $Z$ and ${\nabla }_{X}Z$ with $Y$ gives metric compatibility. Thus this construction has both required properties.

In coordinate vector fields the brackets vanish. Consequently the Christoffel symbols and coordinate derivative areRepeated indices are summed. The displayed connection type in the source is best understood in its standard form on two vector fields; if the first input is an individual tangent vector at a point, the output lies in the tangent fiber at that point rather than in the space of global sections.

For the parallel metrics with a common Levi-Civita connection, let $\gamma$ be a piecewise smooth path from the point $x$ of equality to any $y$. Connected smooth manifolds admit such paths because coordinate balls are path connected. Parallel transport $P_{\gamma }$ for the common connection is invertible and preserves both metrics. ThereforeEvery pair of tangent vectors at $y$ arises this way, so $\boxed{\tilde{g}=g}$ everywhere.

Dropping the agreement at one point removes the conclusion. For any constant $c>0$, $\tilde{g}=cg$ has the same Christoffel symbols. Nor must the two metrics be proportional: on ${\mathbb{R}}^n$, $n\ge 2$, the constant metrics ${\sum }_i(d{x}^i{)}^2$ and $2(d{x}^1{)}^2+{\sum }_{i\ge 2}(d{x}^i{)}^2$ both have zero connection coefficients. In general write $\tilde{g}(u,v)=g(Au,v)$. Since both metrics are parallel, $0=({\nabla }_X\tilde{g})(Y,Z)=g(({\nabla }_{X}A)Y,Z)$, so $\nabla A=0$. Conversely, a positive $g$-self-adjoint parallel $A$ makes the same torsion-free connection compatible with $\tilde{g}$, proving equality of their Levi-Civita connections. Thus one-point agreement specifies $A=I$ and forces it everywhere; without it, nontrivial parallel choices can remain.