Derivative of the induced metric

Question

2-metric $\gamma_{AB}$ induced on the world sheet by the spacetime metric $g_{\mu\nu}$ is $$\gamma_{AB}=g_{\mu\nu}X^{\mu},_A X^{\nu},_B$$

$$\gamma^{AB},_B=-\gamma^{AC}\gamma^{BD}\gamma_{CD},_B$$

How is the derivative of the induced metric explicitly obtained? I guess it is derived from $\gamma^{AB}\gamma_{BD}=\delta^A_D$ $$\gamma^{AB},_B\gamma_{BD}+\gamma^{AB}\gamma_{BD},_B=0 $$ multiplying by $\gamma^{CD}$ $$\gamma^{AB},_B \gamma^{CD}\gamma_{BD}+\gamma^{CD}\gamma^{AB}\gamma_{BD},_B=0 $$ $$\gamma^{AB},_B\delta^C_B+\gamma^{CD}\gamma^{AB}\gamma_{BD},_B=0$$ $C=B$ $$\gamma^{AB},_B+\gamma^{BD}\gamma^{AB}\gamma_{BD},_B=0$$ which is different from the wished result. What I did wrong?

Answer 1

The reason leading to the wrong result in your calculation is that more then two indices are paired, i.e. the B index is used three times in the first line $$\gamma^{AB},_B\gamma^{AB}\gamma_{BD},_B=0.$$ When using Einstein sum convention, you have to be careful to pair only two indices at most, one co- with one contravariant. I.e. every index letter should be at most twice in a term. Thus, instead try $$(\gamma^{DB}\gamma_{CD}),_B = \gamma^{DB},_B \gamma_{CD} + \gamma^{DB} \gamma_{CD},_B = 0.$$ Here, the B and D index are used twice, so we are fine. By raising index with $\gamma^{AC}$ we obtain the desired result: $$\gamma^{AB},_B + \gamma^{AC}\gamma^{BD}\gamma_{CD},_B=0.$$ I hope this answeres your question!