Induced Metrics and Degenerate Restrictions

An immersion pulls the ambient metric back to a symmetric bilinear form on the submanifold. That form is a genuine pseudo-Riemannian metric exactly when the tangent space has no nonzero vector orthogonal to every tangent vector. A null hypersurface fails this test: its normal direction lies inside its tangent space, producing a one-dimensional kernel. Consequently there is no ordinary unit normal and no inverse intrinsic metric on the full tangent bundle.

Required background. Tensor Fields and Natural Operations supplies immersions, pullbacks, tangent spaces, and tensor restrictions.

Helpful background. Calculus and Linear Algebra for Relativity supplies kernels and orthogonal complements; Exterior Algebra and Differential Forms supplies normal covectors; Events, Intervals, and Minkowski Geometry supplies null directions in flat spacetime.

Pulling a metric onto a submanifold

Let $\iota:S\to M$ be an immersion into a pseudo-Riemannian manifold $(M,g)$. Its induced bilinear form is

$$h=\iota^*g, \qquad h_p(X,Y)=g_{\iota(p)}(d\iota_pX,d\iota_pY).$$

The pullback is automatically smooth and symmetric. It need not be nondegenerate. Identifying $T_pS$ with its image in $T_{\iota(p)}M$, define the ambient orthogonal complement

$$(T_pS)^\perp =\{V\in T_{\iota(p)}M:g(V,X)=0\ \text{for all }X\in T_pS\}.$$

The radical of the induced form is

$$\operatorname{Rad}(h_p)=T_pS\cap(T_pS)^\perp.$$

Hence $h$ is nondegenerate precisely when this intersection is $\{0\}$. The criterion, rather than the causal label of a chosen coordinate, is the reliable test.

For a hypersurface locally described by $f=0$ with $df\ne0$, the normal covector is $n_\mu=\partial_\mu f$. Raising its index gives $n^\mu=g^{\mu\nu}n_\nu$. Every tangent vector $X$ obeys $n_\mu X^\mu=0$, so $n^\sharp$ spans $(T_pS)^\perp$. If $g^{-1}(n,n)\ne0$, then $n^\sharp$ is transverse to $S$ and can be normalized to squared norm $+1$ or $-1$. If $g^{-1}(n,n)=0$, then $n^\sharp$ is itself tangent and belongs to the radical.

The null-hypersurface calculation

Use Minkowski spacetime in null coordinates,

$$ds^2=-2\,du\,dv+dx^2+dy^2,$$

and take the hypersurface $S=\{u=0\}$. Its tangent bundle is spanned by $\partial_v,\partial_x,\partial_y$. Setting $du=0$ in the pullback gives

$$h=dx^2+dy^2.$$

Thus $h(\partial_v,X)=0$ for every $X\in TS$: the induced form has rank two and kernel $\operatorname{span}\{\partial_v\}$. The normal covector is $du$, while

$$(du)^\sharp=g^{\mu u}\partial_\mu=-\partial_v.$$

It is both normal and tangent, with zero norm. Multiplying it by any nonzero function leaves it null, so no rescaling can produce $g(n,n)=\pm1$. This is the precise reason a null hypersurface has no ordinary unit normal.

The induced-metric construction and its dependence on nondegeneracy are developed in Wendl 2021–2022, §24.3, pp. 206–208, PDF; the explicit null restriction above is an independent Lorentzian check.

Non-null and null intrinsic geometry

For a spacelike or timelike hypersurface, a normalized normal $N$ gives the direct decomposition

$$T_pM=T_pS\oplus\operatorname{span}\{N\},$$

and the inverse induced metric exists. Orthogonal projection and traces intrinsic to $S$ are then unambiguous.

For a null hypersurface the analogous direct sum fails because the normal line is contained in $T_pS$. In an $n$-dimensional Lorentzian manifold, the quotient $T_pS/\operatorname{Rad}(h_p)$ has a canonical positive-definite metric of dimension $n-2$ (two in four-dimensional spacetime): define $\bar h([X],[Y])=h(X,Y)$. Adding a kernel vector to either representative does not change this value, so the quotient construction needs no auxiliary transverse vector.

Alternatively, one may choose a complementary screen subspace inside $T_pS$, often using an auxiliary transverse null vector. The screen is a concrete representative of that quotient, and its choice is extra data. Its metric must not be mistaken for an inverse of $h$ on all of $T_pS$. This separates the canonical quotient geometry from a convenient, noncanonical choice of representatives.

This distinction also prevents a common error: degeneracy of the pullback does not mean the ambient spacetime metric is degenerate. In the example, $\det g=-1$ everywhere while $\det h=0$ on $S$. The failure occurs only after restriction to a tangent subspace containing its own orthogonal direction.

Boundaries of the result

The argument is local and applies to any regular null hypersurface, whether or not it is a horizon. It does not supply a preferred scaling of the null generator, an affine parameter, a foliation by spacelike cross-sections, or global information about generators and endpoints. Those require additional geometry. Extrinsic curvature for null hypersurfaces likewise cannot be obtained by copying the unit-normal construction from a non-null boundary.

The causal type can be read directly from a defining normal covector. For a Lorentzian hypersurface $f=0$, a timelike $df$ has a spacelike kernel, so the induced metric is positive definite. A spacelike $df$ has a Lorentzian kernel, so the hypersurface is timelike. A null $df$ has its raised vector inside that kernel and makes the restriction degenerate. Replacing $f$ by $\widehat f=af$, with $a$ smooth and nonzero on the surface, rescales $df$ there but does not change any of these classifications. In the null case it also rescales the generator, emphasizing that the hypersurface alone fixes a null line field, not a preferred nonzero generator. This three-way test is local; the causal character may change only where the normal becomes null, and a purported smooth unit normal must fail at such a transition.

Exercises

In three-dimensional Minkowski spacetime $ds^2=-dt^2+dx^2+dy^2$, classify the induced metric on each level surface: $t=0$, $x=0$, and $t-x=0$. Identify the kernel when it exists.

Solution

On $t=0$, the pullback is $dx^2+dy^2$, positive definite, so the surface is spacelike. On $x=0$, it is $-dt^2+dy^2$, Lorentzian and nondegenerate, so the surface is timelike. On $t-x=0$, set $dt=dx$ to obtain $h=dy^2$. A tangent basis is $\partial_t+\partial_x,\partial_y$, and $\partial_t+\partial_x$ spans the kernel. The last surface is null and has no unit normal.

References

• Wendl, Chris. Differential Geometry I and II. Humboldt-Universität zu Berlin lecture notes, 2021–2022, §24.3, pp. 206–208. Course PDF.

Further reading

• Tong, David. General Relativity. University of Cambridge lecture notes, 2019, §3.1, pp. 91–99. Official course PDF.