# Proving that this metric tensor is Riemannian

Let $(M,g)$ be a Riemannian $n$-manifold, and $\varphi: M \to \Bbb R$ be a smooth map. Define another metric tensor by: $$\widetilde{g} = g - {\rm d}\varphi \otimes {\rm d}\varphi$$I know that $\widetilde{g}$ is Riemannian if $\|{\rm grad}_M\varphi\| < 1$ and Lorentzian if $\|{\rm grad}_M\varphi\| > 1$ $(\|\cdot\|$ induced by $g$, of course$)$.

To see the second case it is easy, it suffices to evaluate $\widetilde{g}$ in ${\rm grad}_M\varphi$ itself - it will be a timelike vector.

Is there an easy way to check that $\widetilde{g}$ will be positive definite? I don't quite know if there's a result about this, but I feel that computing the metric on ${\rm grad}_M\varphi$ should be enough. I'm not sure how to exactly compute $\det \widetilde{g}$ here... Thanks.

Take a vector field $v$, and compute $\bar g(v,v)$. You need to show this is strictly positive when $v \ne 0$. But $\bar g(v,v) = g(v,v) - (d\varphi(v))^2$. Isn't this enough to make it work?
• Oh yes, it is. Thanks, I don't know what I was thinking. $${\rm d}\varphi(v)^2 = g({\rm grad}_M\varphi, v)^2 \leq g(v,v).$$ – Ivo Terek Jul 15 '15 at 0:41