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The standard sphere immersion and its normal line

Example

Let j:S2↪R3 be the standard unit sphere inclusion. Then (j,dj) is a formal immersion, and the normal bundle of the formal immersion is νdj=dj(TS2)⊥, the fibrewise orthogonal complement of the tangent planes of S2 in j∗TR3. The outward unit normal N(x)=x is a global nonvanishing section, so νdj≅ε1 is trivial, and the tangent-normal identity gives TS2⊕ε1  ≅  j∗TR3=ε3, the standard trivialization of the tangent bundle of S2 stabilised by one trivial line. The bundle isomorphism (v,a)↦v+ax therefore trivializes TS2⊕ε1; the inverse images of the standard basis vectors ei give the global frame x↦(ei−⟨ei,x⟩x,⟨ei,x⟩), i=1,2,3, while TS2 alone admits no nowhere-zero global section by A positive even sphere has no nowhere-zero tangent field; hence the extra normal line is essential and the splitting is not a triviality of TS2. This verifies the tangent-normal identity in the first nontrivial even-dimensional case and provides the normal line used in the sphere-eversion computation on the next page.

Facts & Assumptions

Given: The standard unit sphere inclusion j:S2↪R3 and the standard structures on TR3≅R3×R3 (The tangent bundle as a disjoint union).

[F1]

(j,dj) is a formal immersion whose fibres are injective, and the normal bundle of the formal immersion is νdj=dj(TS2)⊥, the orthogonal complement with respect to the standard metric (Formal immersion between smooth manifolds, Normal bundle of a formal immersion).

[L1]

The tangent-normal identity gives a smooth bundle isomorphism TS2⊕νdj≅j∗TR3 (Formal immersion gives the tangent normal-bundle identity), and j∗TR3≅ε3 is the trivial rank-three bundle (Smooth vector bundles, rank, fibres, and trivial bundles).

Verification

technique · direct
1.1F1givenalgebra

For x∈S2 the tangent space TxS2 is the orthogonal complement of the radial line Rx in R3, and djx is the inclusion TxS2↪R3. Hence the normal line of dj at x is spanned by x, and the outward unit normal N(x):=x is a global smooth nonvanishing section of νdj.

2.1L1step 1.1∎

A line bundle with a global nonvanishing section is trivial, so νdj≅ε1; the tangent-normal identity of [L1] then gives TS2⊕ε1≅j∗TR3=ε3, and the isomorphism (v,a)↦v+ax pulls back the standard basis to the three smooth sections (ei−⟨ei,x⟩x,⟨ei,x⟩), which are a global frame because their images are a basis in every fibre.

Remarks

The failure of a nowhere-zero section of TS2 is proved by the identity-to-antipodal homotopy obstruction in A positive even sphere has no nowhere-zero tangent field. The explicit normal-line verification above and that obstruction together distinguish stable triviality from triviality.

Depends on

Used by

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Dependency tree · two levels

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Sources