# Variation formula of a metric [closed]

In Terry Tao's notes on the Poincare Conjecture, 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.

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## closed as off-topic by Chris Gerig, Stefan Waldmann, Stefan Kohl, Yemon Choi, Willie WongNov 24 '14 at 13:29

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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
This question has been answered easily in the comments –  Yemon Choi Nov 24 '14 at 12:37

As mentioned in the comments, once you're comfortable with indices, this is just an exercise in calculus. Differentiating both sides of the identity $g^{\alpha\gamma}g_{\gamma\zeta} = \delta^{\alpha}_{\zeta}$ with respect to the parameter $t$, we see that
Replacing the index $\zeta$ by the index $\delta$, you get precisely the expression you were looking for. I refrained from using $\delta$ as an index to avoid confusion with the expressions involving $\delta$ in the first and penultimate lines.