# 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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