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Igor Khavkine
Igor Khavkine
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This is the original Nomizu-Ozeki article:

Nomizu, Katsumi; Ozeki, Hideki, The existence of complete Riemannian metrics, Proc. Am. Math. Soc. 12, 889-891 (1961). ZBL0102.16401.

Applying their proof to a family of metrics $h_t$, the conformal factors $\omega_t(x)$ need only satisfy the property that $\omega_t(x) > 1/r_t(x)$, where $r_t(x)$ is the sup of all radii of precompact metric balls at $x$. The existence of a smooth $\omega_t(x)$ only relies on the continuity of $r_t(x)$, which they prove (and second countability of the underlying manifold $\Sigma$, which goes hand in hand with Riemannian metrizability).

A priori, we do not know that $\omega_t(x)$ can be chosen to be smooth in $(t,x)$, nor that $r_t(x)$ itself is continuous in $(t,x)$. But the logic of the proof still applies if we replace $\Sigma$ by $M = \mathbb{R}\times \Sigma$ and instead choose a smooth $\Omega(t,x)$ so that $\omega_t(x) = \Omega(t,x) > 1/R(t,x) > 1/r_t(x)$$\omega_t(x) = \Omega(t,x) > 1/R(t,x) \ge 1/r_t(x)$ for some $R(t,x)$ that is continuous on $M$. It's simple to define by analogy the function $R(t,x)$ to be the sup of all radii of precompact balls at $(t,x)$ in the metric space $(M,\Delta)$ where $\Delta((t,x), (t',x'))$ is the Riemannian metric distance on $(M, dt^2 + h_t)$. Exercise: repeat the arguments from Nomizu-Ozeki to show that one can still reduce to the case where $R(x,t) < \infty$ everywhere and that $R(x,t)$ is continuous.

This is the original Nomizu-Ozeki article:

Nomizu, Katsumi; Ozeki, Hideki, The existence of complete Riemannian metrics, Proc. Am. Math. Soc. 12, 889-891 (1961). ZBL0102.16401.

Applying their proof to a family of metrics $h_t$, the conformal factors $\omega_t(x)$ need only satisfy the property that $\omega_t(x) > 1/r_t(x)$, where $r_t(x)$ is the sup of all radii of precompact metric balls at $x$. The existence of a smooth $\omega_t(x)$ only relies on the continuity of $r_t(x)$, which they prove (and second countability of the underlying manifold $\Sigma$, which goes hand in hand with Riemannian metrizability).

A priori, we do not know that $\omega_t(x)$ can be chosen to be smooth in $(t,x)$, nor that $r_t(x)$ itself is continuous in $(t,x)$. But the logic of the proof still applies if we replace $\Sigma$ by $M = \mathbb{R}\times \Sigma$ and instead choose a smooth $\Omega(t,x)$ so that $\omega_t(x) = \Omega(t,x) > 1/R(t,x) > 1/r_t(x)$ for some $R(t,x)$ that is continuous on $M$. It's simple to define by analogy the function $R(t,x)$ to be the sup of all radii of precompact balls at $(t,x)$ in the metric space $(M,\Delta)$ where $\Delta((t,x), (t',x'))$ is the Riemannian metric distance on $(M, dt^2 + h_t)$. Exercise: repeat the arguments from Nomizu-Ozeki to show that one can still reduce to the case where $R(x,t) < \infty$ everywhere and that $R(x,t)$ is continuous.

This is the original Nomizu-Ozeki article:

Nomizu, Katsumi; Ozeki, Hideki, The existence of complete Riemannian metrics, Proc. Am. Math. Soc. 12, 889-891 (1961). ZBL0102.16401.

Applying their proof to a family of metrics $h_t$, the conformal factors $\omega_t(x)$ need only satisfy the property that $\omega_t(x) > 1/r_t(x)$, where $r_t(x)$ is the sup of all radii of precompact metric balls at $x$. The existence of a smooth $\omega_t(x)$ only relies on the continuity of $r_t(x)$, which they prove (and second countability of the underlying manifold $\Sigma$, which goes hand in hand with Riemannian metrizability).

A priori, we do not know that $\omega_t(x)$ can be chosen to be smooth in $(t,x)$, nor that $r_t(x)$ itself is continuous in $(t,x)$. But the logic of the proof still applies if we replace $\Sigma$ by $M = \mathbb{R}\times \Sigma$ and instead choose a smooth $\Omega(t,x)$ so that $\omega_t(x) = \Omega(t,x) > 1/R(t,x) \ge 1/r_t(x)$ for some $R(t,x)$ that is continuous on $M$. It's simple to define by analogy the function $R(t,x)$ to be the sup of all radii of precompact balls at $(t,x)$ in the metric space $(M,\Delta)$ where $\Delta((t,x), (t',x'))$ is the Riemannian metric distance on $(M, dt^2 + h_t)$. Exercise: repeat the arguments from Nomizu-Ozeki to show that one can still reduce to the case where $R(x,t) < \infty$ everywhere and that $R(x,t)$ is continuous.

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Igor Khavkine
Igor Khavkine
  • 21.5k
  • 2
  • 60
  • 113

This is the original Nomizu-Ozeki article:

Nomizu, Katsumi; Ozeki, Hideki, The existence of complete Riemannian metrics, Proc. Am. Math. Soc. 12, 889-891 (1961). ZBL0102.16401.

Applying their proof to a family of metrics $h_t$, the conformal factors $\omega_t(x)$ need only satisfy the property that $\omega_t(x) > 1/r_t(x)$, where $r_t(x)$ is the sup of all radii of precompact metric balls at $x$. The existence of a smooth $\omega_t(x)$ only relies on the continuity of $r_t(x)$, which they prove (and second countability of the underlying manifold $\Sigma$, which goes hand in hand with Riemannian metrizability).

A priori, we do not know that $\omega_t(x)$ can be chosen to be smooth in $(t,x)$, nor that $r_t(x)$ itself is continuous in $(t,x)$. But the logic of the proof still applies if we replace $\Sigma$ by $M = \mathbb{R}\times \Sigma$ and instead choose a smooth $\Omega(t,x)$ so that $\omega_t(x) = \Omega(t,x) > 1/R(t,x) > 1/r_t(x)$ for some $R(t,x)$ that is continuous on $M$. It's simple to define by analogy the function $R(t,x)$ to be the sup of all radii of precompact balls at $(t,x)$ in the metric space $(M,\Delta)$ where $\Delta((t,x), (t',x'))$ is the Riemannian metric distance on $(M, dt^2 + h_t)$. Exercise: repeat the arguments from Nomizu-Ozeki to show that one can still reduce to the case where $R(x,t) < \infty$ everywhere and that $R(x,t)$ is continuous.