# Variation formula of a metric

In Terry Tao's notes on the Poincare Conjecture http://terrytao.wordpress.com/2008/03/28/285g-lecture-1-ricci-flow/#more-293 He makes a jump I can't understand! From differentiating the identity $g^{\alpha \beta}g_{\beta \gamma} = \delta^\alpha_\gamma$ we obtain the variation formula $\frac{d}{dt}g^{\alpha \beta} =-g^{\alpha \gamma}g^{\beta \delta}\dot{g}_{\gamma \delta}$ But I don't understand this leap. Can anyone show me what I'm supposed to be doing, I'm familiar with tensors and the variational calculus but putting the two together is hard for me.

-
Ever heard that if a matrix-valued function $t\mapsto A(t)$ takes non-singular values, then $\frac{d}{dt}A(t)^{-1}=-A^{-1}\frac{dA}{dt}A^{-1}$ ? This is a calculus exercise. –  Denis Serre Mar 1 '12 at 21:22
And this is done most easily using implicit differentiation of $A A^{-1} = I$. –  Deane Yang Mar 1 '12 at 21:40
It's also worth noting that the two comments above are equivalent to just differentiating $g^{\alpha\beta}g_{\beta\gamma} = \delta^\alpha_\gamma$ with respect to the variation parameter and then solving for $\dot{g}^{\alpha\beta}$ by multiplying the differentiated equation by $g^{\eta\alpha}$. –  Deane Yang Mar 2 '12 at 17:35
Ok so it was a lot simpler than I thought. –  Peadar Coyle Mar 3 '12 at 11:03