dg.differential geometry – Metrics $g_1leqslant g_2$ implies the Ricci current $g_1(t)leqslant g_2(t)$? Answer

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dg.differential geometry – Metrics $g_1leqslant g_2$ implies the Ricci current $g_1(t)leqslant g_2(t)$?

Let M breathe an entire,n dimensional Riemannian manifold with out border. Suppose $g_1,g_2$ are two metrics on M and $g_1leqslant g_2$. Suppose that there exists $T>0$ such that for $i=1,2$, the Ricci current $frac{partial g}{partial t}=-2 Ric$ with preliminary situation $g(0)=g_i$ live for $tin [0,T]$. Do we now have $g_1(t)leqslant g_2(t)$ for $tin (0,T]$?

When n=2, the metric can breathe written within the conformal coordinates
$$
g_i(t)=e^{2u_i(t)}(dr^2+r^2 dtheta^2).
$$
$u_i(t)$ satisfies the parabolic equation
$$
frac{partial u_i}{partial t}=e^{-2u_i}Delta u_i=-K_i
$$
the place $K_i$ are the Gauss curvature. We have $u_1(0)leqslant u_2(0)$, we necessity to show $u_1(t)leqslant u_2(t)$.

Let $v(t)=min_{xin mathbb{R}^2} u_2(t,x)-u_1(t,x)$, suppose the minimal are attained at $x_t$, then $Delta(u_2-u_1)|_{x_t} leqslant 0$ and thus
$$
frac{partial v}{partial t} leqslant (e^{-2u_2}-e^{-2u_1})Delta u_1|_{x_t}.
$$
It appears that we won’t anticipate that $frac{partial v}{partial t}leqslant 0$. So how can we show $u_1(t)leqslant u_2(t)$ ?

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