PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10

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Pullback of a riemannian metric is riemannian exactly for immersions

Statement

$F^{*} h$ is Riemannian if and only if $F$ is an immersion. In general it is positive semidefinite, with radical $\ker d F_{p}$ at $p$ .

Facts & Assumptions

Given: A smooth map $F : M \to N$ and a Riemannian metric $h$ .

[F1]

Pullback of a riemannian metric as a tensor: For smooth $F : M \to N$ and a Riemannian metric $h$ on $N$ , its pullback tensor is $(F^{*} h)_{p} (v, w) = h_{F (p)} (d F_{p} v, d F_{p} w)$ . This is def-pullback-of-a-covariant-tensor-field for the tensor in def-riemannian-metric-and-riemannian-manifold. It is always symmetric and positive semidefinite; the name does not assert positive definiteness. Smoothness and the precise immersion criterion are established next.

[F2]

Pullback of covariant tensors is smooth and functorial: If $F : M \to N$ is smooth and $T$ is a smooth covariant tensor field on $N$ , then $F^{*} T$ is a smooth covariant tensor field on $M$ . Moreover, $({id}_{M})^{*} T = T, (G \circ F)^{*} T = F^{*} (G^{*} T)$ for every composable smooth map $G$ .

[F3]

Immersions, submersions, and constant-rank maps: Let $F : M^{m} \to N^{n}$ be a smooth map. - $F$ is an immersion at $p \in M$ when $d F_{p}$ is injective. - $F$ is a submersion at $p \in M$ when $d F_{p}$ is surjective. - $F$ has constant rank $r$ on $A \subseteq M$ when ${rank}_{p} F = r$ for every $p \in A$ (def-rank-of-a-smooth-map-at-a-point). The map is an immersion or submersion without qualification when the corresponding pointwise condition holds at every point of $M$ .

Proof

technique · direct

1.1F1F2F3given

Tensor pullback is smooth, and $(F^{*} h)_{p} (v, v) = h_{F (p)} (d F_{p} v, d F_{p} v) \geq 0$ , with equality exactly when $d F_{p} v = 0$ . If $d F_{p}$ is injective at every point, the value is positive for every nonzero $v$ , so the pullback is Riemannian.

2.1F1F3step 1.1∎

Conversely, positive definiteness forces $d F_{p} v = 0$ to imply $v = 0$ , hence $F$ is an immersion. A vector in $\ker d F_{p}$ pairs to zero with every vector; if it is in the radical, its pairing with itself is zero, so the preceding equality forces it into $\ker d F_{p}$ . This proves the radical assertion as well.

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Chapter 13, pp.328–332 and 341–342.

Depends on

Dependency tree · 2 levels

Sources

John M. Lee, Introduction to Smooth Manifolds, second edition