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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 Wong Nov 24 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
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And this is done most easily using implicit differentiation of $A A^{-1} = I$. –  Deane Yang Mar 1 '12 at 21:40
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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 at 12:37

1 Answer 1

up vote 6 down vote accepted

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

\begin{align*} \frac{d}{dt}\left(g^{\alpha\gamma}g_{\gamma\zeta}\right) &= \frac{d}{dt}\delta^{\alpha}_{\zeta}\\ \dot{g}^{\alpha\gamma}g_{\gamma\zeta} + g^{\alpha\gamma}\dot{g}_{\gamma\zeta} &= 0\\ \dot{g}^{\alpha\gamma}g_{\gamma\zeta} &= -g^{\alpha\gamma}\dot{g}_{\gamma\zeta}\\ \dot{g}^{\alpha\gamma}g_{\gamma\zeta}g^{\beta\zeta} &= -g^{\alpha\gamma}\dot{g}_{\gamma\zeta}g^{\beta\zeta}\\ \dot{g}^{\alpha\gamma}\delta_{\gamma}^{\beta} &= -g^{\alpha\gamma}g^{\beta\zeta}\dot{g}_{\gamma\zeta}\\ \dot{g}^{\alpha\beta} &= -g^{\alpha\gamma}g^{\beta\zeta}\dot{g}_{\gamma\zeta}.\\ \end{align*}

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.

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Thanks Michael, I have actually stopped studying Mathematics - I moved into the world of software engineering (like we all do). But thanks nevertheless for the clear answer and the cleaning up of my question in accordance with the standards of MathOverflow –  Peadar Coyle Nov 25 at 13:22

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