Is there a name for a pseudo-Riemannian manifold that admits no nonzero null vectors? More precisely: For a pseudo-Riemannian manifold $(R,g)$, a null vector is a non-zero vector field $X:M \to TM$ such that $$ g(X,X)(m) = 0, \forall m \in M. $$ As this [question][1] shows - vector like this exist. But can there exist manifolds where they do not exist and do such manifolds have a name?
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$\begingroup$ You are asking about 'nonvanishing null vector fields', not 'nonzero null vectors', and your attempt of '[question][1]' for a link to some other reference question is not working, so we can't see that. $$ $$An explicit example is to consider the torus $T = R/(\pi Z)\times R/(\pi Z)$ and the metric $g = \cos 2x\, (2\,dx\,dy) + \sin 2x\,(dx^2-dy^2)$, which has no nonvanishing null vector field, $X= a\,\partial_x + b\,\partial_y$ because such an $X$ would have $a$ and $b$ be $\pi$-periodic and satisfy either $a\cos x + b\sin x=0$ or $b\cos x-a\sin x = 0$.Either forces $(a,b)$ to vanish somewhere. $\endgroup$– Robert BryantCommented Oct 9, 2021 at 12:03
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2 Answers
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Note first that every pseudo-Riemmanian manifold admits a null vector field which is not identically $0$ (just construct one locally and multiply it by a bump function). So by "non-zero vector field" I assume you mean "nowhere vanishing".
Let $(M,g)$ be a pseudo-Riemannian manifold of signature $(p,q)$. The tangent bundle $TM$ always admits an orthogonal splitting as $E \overset{\perp}{\oplus} F$, where $E$ and $F$ are respectively positive and negative definite (hence of respective rank $p$ and $q$). Moreover this splitting is unique up to homotopy (because, pointwise, the set of such splittings is the symmetric space of the orthogonal group $O(p,q)$, which is contractible).
Proposition: $M$ admits a nowhere vanishing null vector field if and only if $E$ and $F$ both admit nowhere vanishing sections.
Proof: Decompose a nowhere vanishing null vector field $X$ as $X_E + X_F$. Then $g(X_E,X_E) = -g(X_F,X_F)$. If this is $0$ at some point then $X_E$ and $X_F$ vanish at that point (since $g$ is positive definite on $E$ and negative definite on $F$) contradicting the non-vanishing of $X$. Hence $X_E$ and $X_F$ are non-vanishing sections of $E$ and $F$.
Conversely, if $X_E$ and $X_F$ are non-vanishing sections of $E$ and $F$ respectively, then up multiplying $X_F$ them by a positive function, we can assume that $g(X_E,X_E) = -g(X_F,X_F)$. Hence $X_E+X_F$ is a nowhere vanishing null vector field. CQFD
There are thus topological obstructions to the existence of such a vector field (mainly the non-vanishing of the Euler class of $E$ or $F$). For instance, Let $(A,g_A)$ and $(B,g_B)$ be Riemannian manifolds, with $A$ of non-zero Euler characteristic, and consider $(M,g) = (A\times B, g_A \oplus -g_B)$. Then $M$ does not admit a nowhere vanishing null vector field. Indeed, we have the splitting $TM = TA\oplus TB$, and the projection of a null vector field to TA must vanish somewhere since the Euler class of TA is non-zero.
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$\begingroup$ Thank you for the answer! However are you sure that one can always construct a globally defined non-zero vector field. Take the manifold $\mathbb{R}$ and endow $T\mathbb{R}$ with a Riemannian metric that is semi-definite at $0$ and positive definite everywhere else (we can do so by multiplying the metric by a smooth function that vanishes at zero and only at zero). Then the bump fuction argument doesn't seem to work since the "support of the non-positive definitness" is a single point. $\endgroup$ Commented Oct 8, 2021 at 12:21
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$\begingroup$ Your example is not considered a pseudo-Riemannian manifold, since at zero the metric is degenerate. $\endgroup$ Commented Oct 8, 2021 at 18:08
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$\begingroup$ @Pierre: yes of course, thanks for pointing this out. $\endgroup$ Commented Oct 9, 2021 at 7:59
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$\begingroup$ @Pierre: Do the "metrics" considered in my example fit into some more general class? $\endgroup$ Commented Oct 9, 2021 at 9:22
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1$\begingroup$ @NicolasTholozan Although very true, I also like to think of them as degenerate metrics on the cotangent. Depending on what the OP has in mind, maybe their $T\mathbb R$ is actually a $T^*\mathbb R$. That said, even this case would not be considered a sub-Riemannian manifold by many authors. $\endgroup$ Commented Oct 11, 2021 at 15:49
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Such manifolds are precisely the Riemannian ones and their “opposites”.
Suppose $M$ is a pseudo-Riemannian manifold whose signature is not trivial in one direction or the other. We work in a neighbourhood of a point and will define a vector field with arbitrarily small support, so we may as well consider $M=\mathbb R^d$. Suppose also, after a linear change of coordinates, that the metric at zero is $dx_1^2+\cdots+dx_k^2-dx_{k+1}^2-\cdots-dx_d^2$. By hypothesis $0<k<d$.
For all x sufficiently close to 0, define $X(x)$ as the only vector inbetween $\epsilon_1:=(1,0,\ldots,0)$ and $2\epsilon_d:=(0,\ldots,0,2)$ of norm 0. If you accept for a moment that this is a well-defined smooth vector field, then a suitable multiple of $X$ will have very small support, hence be defined everywhere upon extending by zero, but will be precisely $(2/3,\ldots,2/3)$ at zero.
The fact that it is smooth comes from the inverse function theorem. We are looking for $(1-t,0,\ldots,0,2t)$ where $t$ is solution to $$ (1-t)^2g_x(\epsilon_1,\epsilon_1) + 4t(1-t)g_x(\epsilon_1,\epsilon_d) + 4t^2g_x(\epsilon_d,\epsilon_d) = 0. $$ Single roots of a polynomial depend smoothly on the coefficients, so if we show that $t$ is a single root, then it will depend smoothly on the metric. But this is an open condition, and at $x=0$ we have $$ (1-t)^2g_0(\epsilon_1,\epsilon_1) + 4t(1-t)g_0(\epsilon_1,\epsilon_d) + 4t^2g_0(\epsilon_d,\epsilon_d) = (1-t)^2 - 4t^2 = -(3t-1)(t+1) $$ so the root $1/3$ is simple.
