Lambda index for Christoffel symbols So I watched this video where it states that the Riemann tensor can be written in terms of Christoffel symbols by
${\displaystyle R^{\alpha}{}_{\beta\mu \nu }=\partial _{\mu }\Gamma ^{\alpha}{}_{\beta \nu}
+\Gamma ^{\lambda}{}_{\beta \nu}\Gamma ^{\alpha}{}_{\lambda \mu}
-\partial _{\nu }\Gamma ^{\alpha}{}_{\beta \mu}
-\Gamma ^{\lambda}{}_{\beta \mu }\Gamma ^{\alpha}{}_{\lambda \nu}}$
But I am really not sure what the lambda index for the Christoffel symbols represent or how to obtain them.
For example. if I have $R^0 _{101}$ , I know that alpha is $0$, beta is $1$, mu is $0$, and nu is $1$. However, in this case, how would I know what value the lambda index is?
 A: The formula you quote is using Einstein notation: this means that you should take a sum over repetead indexes in a formula, such as $\lambda$ is in this case.
To be more precise, if $M$ is $n$-dimensional, and we use coordinates indexed from $1$ to $n$, $\vec{x}=(x^1,\dots,x^n)$, to obtain the curvature tensor you should compute these sums:
\begin{equation}
R^\alpha_{\beta\mu\nu}=\partial_\mu\Gamma^\alpha_{\beta\nu}-\partial_\nu\Gamma^\alpha_{\beta\mu}+\sum_{\lambda=1}^n\left(\Gamma^\alpha_{\lambda\mu}\Gamma^\lambda_{\beta\nu}-\Gamma^\alpha_{\lambda\nu}\Gamma^\lambda_{\beta\mu}\right).
\end{equation}
For example, say that $M$ is $3$-dimensional with coordinates $(x^0,x^1,x^2)$. So $\lambda$ can be $0$, $1$ or $2$. Using the previous formula we get
\begin{equation}
\begin{split}
R^0_{101}=\partial_0\Gamma^0_{11}-\partial_1\Gamma^0_{10}+\left(\Gamma^0_{\lambda 0}\Gamma^{\lambda}_{11}-\Gamma^0_{\lambda 1}\Gamma^{\lambda}_{10}\right)_{\lambda=0}+\left(\Gamma^0_{\lambda 0}\Gamma^{\lambda}_{11}-\Gamma^0_{\lambda1}\Gamma^{\lambda}_{10}\right)_{\lambda=1}+\left(\Gamma^0_{\lambda 0}\Gamma^{\lambda}_{11}-\Gamma^0_{\lambda1}\Gamma^{\lambda}_{10}\right)_{\lambda=2}=\\
=\partial_0\Gamma^0_{11}-\partial_1\Gamma^0_{10}+\left(\Gamma^0_{0 0}\Gamma^0_{11}-\Gamma^0_{0 1}\Gamma^0_{10}\right)+\left(\Gamma^0_{1 0}\Gamma^{1}_{11}-\Gamma^0_{11}\Gamma^1_{10}\right)+\left(\Gamma^0_{2 0}\Gamma^2_{11}-\Gamma^0_{21}\Gamma^2_{10}\right).
\end{split}
\end{equation}
The paretheses are added just to make it clearer where we substituted $0$, $1$ or $2$ for $\lambda$.
If instead $M$ was a $2$-dimensional manifold, $\lambda$ can only get two possible values, $0$ and $1$. So you would have a slightly simpler expression:
\begin{equation}
R^0_{101}=\partial_0\Gamma^0_{11}-\partial_1\Gamma^0_{10}+\left(\Gamma^0_{0 0}\Gamma^0_{11}-\Gamma^0_{0 1}\Gamma^0_{10}\right)+\left(\Gamma^0_{1 0}\Gamma^{1}_{11}-\Gamma^0_{11}\Gamma^1_{10}\right).
\end{equation}
