Q:

Connection for the induced metric

2-metric $\gamma_{AB}$ induced on the world sheet by the spacetime metric $g_{\mu\nu}$ is $$\gamma_{AB}=g_{\mu\nu}t^{\mu}_A t^{\nu}_B$$
where $t^{\mu}_A=\frac{\partial X^{\mu}}{\partial \xi^A}$.
The connection generated $\gamma_{AB}$ is known to be equal to:
$$\Gamma^{C}_{AB}=\frac{1}{2}\gamma^{CD}(\gamma_{DA,B}+\gamma_{DB,A}-\gamma_{AB,D})
$$
How to prove the identity?
$$
\Gamma^C_{AB}=\gamma^{CD}(\Gamma_{\mu\nu\lambda} t^{\mu}_A t^{\nu}_B t^{\lambda}_D+t_{A,B}\cdot t_D)
$$

A:

Let me first write down my interpretation of the notation you've used, then I'll proceed to write down my solution. I guess $t_A=t^\mu_A {\partial \over \partial x^\mu}={\partial x^\mu \over \partial \xi^A}{\partial \over \partial x^\mu}={\partial \over \partial \xi^A}$. So, $t_{A,B}={\partial^2 \over {\partial \xi^A \partial \xi^B}}$ and $t_{A,B}\cdot t_D=g_{\mu\nu}t^\mu_{A,B}t^\nu_D$. And $t_{A,B}=t_{B,A}$, since partial derivatives commute. Now we can proceed to solve the problem. Let's calculate $\gamma_{AB,D}$. $$\gamma_{AB,D}=(g_{\mu\nu}t^\mu_A t^\nu_B)_{,D}={\partial g_{\mu\nu}\over \partial \xi^D}t^\mu_At^\nu_B+g_{\mu\nu}t^\mu_{A,D} t^\nu_B+g_{\mu\nu}t^\mu_A t^\nu_{B,D}={\partial g_{\mu\nu}\over \partial x^\lambda}{\partial x^\lambda \over \partial \xi^D}t^\mu_At^\nu_B+g_{\mu\nu}t^\mu_{A,D} t^\nu_B+g_{\mu\nu}t^\mu_A t^\nu_{B,D}=g_{\mu\nu,\lambda}t^\lambda_Dt^\mu_At^\nu_B+g_{\mu\nu}t^\mu_{A,D} t^\nu_B+g_{\mu\nu}t^\mu_A t^\nu_{B,D}$$ Calculating $\gamma_{DA,B}$ and $\gamma_{DB,A}$ in a similar manner, and substituting it in the equation of $\Gamma^C_{AB}$, and rearranging, we get $$\Gamma^C_{AB}={1\over 2}\gamma^{CD}[(g_{\mu\nu,\lambda}t^\lambda_Bt^\mu_Dt^\nu_A+g_{\mu\nu,\lambda}t^\lambda_At^\mu_Dt^\nu_B-g_{\mu\nu,\lambda}t^\lambda_Dt^\mu_At^\nu_B)+g_{\mu\nu}(t^\mu_{D,B}t^\nu_A+t^\mu_Dt^\nu_{A,B}+t^\mu_{D,A}t^\nu_B+t^\mu_Dt^\nu_{A,B}-t^\mu_{A,D}t^\nu_B-t^\mu_At^\nu_{B,D})]$$ Changing the dummy variables in first parenthesis ($\mu \rightarrow \lambda, \nu \rightarrow \mu, \lambda \rightarrow \nu$ for the first term, and $\mu \rightarrow \lambda, \lambda \rightarrow \mu$ for the second to take $t^\lambda_Dt^\mu_At^\nu_B$ out of the parenthesis) and applying $t_{A,B}=t_{B,A}$ on the expression in the second parenthesis, we get $$\Gamma^C_{AB}={1\over 2}\gamma^{CD}[(g_{\lambda\mu,\nu}+g_{\lambda\nu,\mu}-g_{\mu\nu,\lambda})t^\lambda_Dt^\mu_At^\nu_B+2g_{\mu\nu}t^\mu_Dt^\nu_{A,B}]=\gamma^{CD}[\Gamma_{\mu\nu\lambda} t^\lambda_Dt^\mu_At^\nu_B+t_{A,B}\cdot t_D]$$ which is the expression you wanted to arrive.

