David Tong GR Problem Sheet 1 Solutions: Differential Geometry

Source: David Tong, General Relativity, Problem Sheet 1 (manifolds, tensors, Lie and exterior derivatives).

1 Integral Curves of Vector Fields, Q1

Draw the integral curves corresponding to a vector field in $\mathbb{R}^2$ with components, in Cartesian coordinates, given by

$X^\mu = (y,\,-x)$
$X^\mu = (x-y,\,x+y)$

[Hint: you might find life easier in polar coordinates.]

2 Tensors from Linear Maps, Q2

Let $\hat{H}: T_p(M)\to T_p^*(M)$ be a linear map. Define \[H(X,Y) = [\hat{H}(Y)](X).\tag{2.1}\] Show that this map is linear in both arguments (e.g.\ $H(fX+gY,Z) = fH(X,Z)+gH(Y,Z)$ for $f,g\in C^\infty$ and $X,Y,Z\in T_p(M)$) and hence defines a rank $(0,2)$ tensor. Similarly, show that a linear map $T_p(M)\to T_p(M)$ defines a tensor of rank $(1,1)$. What tensor $\delta$ arises from the identity map?

Proof of linearity: \[H(fX+gY,Z) = [\hat{H}(Z)](fX+gY).\tag{2.2}\] Since $\hat{H}(Y)$ is a covariant vector, we have \[H(fX+gY,Z) =f[\hat{H}(Z)](X) + g[\hat{H}(Z)](Y) = fH(X,Z) + gH(Y,Z).\tag{2.3}\] \[H(X,fY+gZ) = [\hat{H}(fY+gZ)](X).\tag{2.4}\] By linearity of the linear map $\hat{H}$, we have \[H(X,fY+gZ) = f[\hat{H}(Y)](X) + g[\hat{H}(Z)](X) = fH(X,Y)+gH(X,Z).\tag{2.5}\] ∎Therefore, $H$ defines a multi-linear map, \[H: T_p(M)\times T_p(M) \to \mathbb{R},\tag{2.6}\] which is a rank $(0,2)$ tensor.

Use he linear map, \[\hat{G}: T_p(M) \to T_p(M),\tag{2.7}\] to define \[G(\omega,X) = \omega(\hat{G}(X)),\tag{2.8}\] with $\omega\in T_p^*(M)$. Now, we shall check the linearity of the $G$ constructed in both entries: \[G(f\omega+g\eta,X) = fG(\omega,X) +gG(\eta,X)\tag{2.9}\] by linearity of covariant vectors, and \[G(\omega,fX+gY) = fG(\omega,X) +gG(\omega,Y)\tag{2.10}\] by linearity of $\hat{G}$. Therefore, $G$ is a rank $(1,1)$ tensor \[G:T_p^*(M)\times T_p(M) \to \mathbb{R}.\tag{2.11}\] The identity map is \[\hat{\delta}(X) = X.\tag{2.12}\] Therefore, it defines the tensor \[\delta(\omega, X) = \omega_\mu X^\mu\tag{2.13}\] with \[\delta(f^\mu, e_\nu) = \delta^\mu_\nu.\tag{2.14}\]

3 The Hessian, Q3

Let $M$ be a manifold and $f:M\to\mathbb{R}$ be a smooth function such that $df=0$ at some point $p\in M$. Let $x^\mu$ be a coordinate chart defined in a neighbourhood of $p$. Define \[F_{\mu\nu} = \frac{\partial^2 f}{\partial x^\mu\partial x^\nu}.\tag{3.1}\] By considering the transformation law for components show that $F_{\mu\nu}$ defines a rank $(0,2)$ tensor. (This is called the Hessian of $f$ at $p$.) Construct also a coordinate-free definition and demonstrate its tensorial properties.

Under the transformation $\tilde{x}^\mu \to x^\mu$, $F_{\mu\nu}$ transforms as \[\tilde{F}_{\mu\nu}\bigg|_p \to F_{\mu\nu}\bigg|_p = \frac{\partial x^\rho}{\partial \tilde{x}^\mu} \frac{\partial}{\partial x^\rho} \left( \frac{\partial x^\sigma}{\partial \tilde{x}^\nu} \frac{\partial f}{\partial x^\sigma} \right)\bigg|_p = \frac{\partial x^\rho}{\partial \tilde{x}^\mu} \frac{\partial x^\sigma}{\partial \tilde{x}^\nu} \frac{\partial^2 f}{\partial x^\rho \partial x^\sigma}\bigg|_p + \frac{\partial^2 x^\sigma}{\partial \tilde{x}^\mu \partial \tilde{x}^\nu} \frac{\partial f}{\partial x^\sigma}\bigg|_p.\tag{3.2}\] Given $df|_p = 0$, the second term vanishes, so \[\tilde{F}_{\mu\nu}\bigg|_p \to \frac{\partial x^\rho}{\partial \tilde{x}^\mu} \frac{\partial x^\sigma}{\partial \tilde{x}^\nu} F_{\rho\sigma}\bigg|_p.\tag{3.3}\] This is exactly the transformation law of a rank $(0,2)$ tensor. Therefore, $f$ defines a rank $(0,2)$ tensor at $p$.

We can also write the tensor in a coordinate-free way as \[F(X_p,Y_p) = X(Y(f))|_p = X^\mu Y^\nu \frac{\partial^2 f}{\partial x^\mu \partial x^\nu}\bigg|_p + X^\mu \frac{\partial Y^\nu}{\partial x^\mu} \frac{\partial f}{\partial x^\nu}\bigg|_p.\tag{3.4}\] Since $df|_p = 0$, this reduces to \[F(X_p,Y_p) = X^\mu Y^\nu \frac{\partial^2 f}{\partial x^\mu \partial x^\nu}\bigg|_p,\tag{3.5}\] which is bilinear and symmetric.

4 Transformation of the Metric Determinant, Q4

Let $g_{\mu\nu}$ be a rank $(0,2)$ tensor. In a basis, one can regard the components $g_{\mu\nu}$ as elements of an $n\times n$ matrix, so that one may define the determinant $g = \det(g_{\mu\nu})$. How does $g$ transform under a change of basis?

5 Lie Derivatives and Cartan's Magic Formula, Q5

Use the Leibniz rule to derive the formula for the Lie derivative of a 1-form $\omega$, valid in any coordinate basis: \[(\mathcal{L}_X\omega)_\mu = X^\nu\partial_\nu\omega_\mu + \omega_\nu\partial_\mu X^\nu.\tag{5.1}\] [Hint: consider $(\mathcal{L}_X\omega)(Y)$ for a vector field $Y$.] Show that the Lie derivative of a $(0,2)$ tensor $g$ is \[(\mathcal{L}_X g)_{\mu\nu} = X^\rho\partial_\rho g_{\mu\nu} + g_{\mu\rho}\partial_\nu X^\rho + g_{\rho\nu}\partial_\mu X^\rho.\tag{5.2}\] For a $p$-form $\eta$, define $\iota_X\eta$ to be the $(p-1)$-form that results from contracting a vector field $X$ with the first index of $\eta$. Show that for a 1-form $\omega$, \[\mathcal{L}_X\omega = \iota_X(d\omega) + d(\iota_X\omega).\tag{5.3}\]

Consider \[(\mathcal{L}_X\omega)(Y) = \mathcal{L}_X(\omega_\mu dx^\mu)(Y^\nu\partial_\nu) = X^\sigma(\partial_\sigma\omega_\mu)Y^\mu + \omega_\mu\mathcal{L}_X(dx^\mu)(Y^\nu \partial_\nu),\tag{5.4}\] and use the fact that $d$ and $\mathcal{L}_X$ commute, \[\mathcal{L}_X(dx^\mu) =d\mathcal{L}_X(x^\mu) = dX^\mu =\partial_\nu X^\mu dx^\nu.\tag{5.5}\] Therefore, we find \[(\mathcal{L}_X\omega)(Y) = X^\sigma(\partial_\sigma\omega_\mu)Y^\mu + \omega_\mu(\partial_\sigma X^\mu)Y^\sigma \implies (\mathcal{L}_X\omega)_\mu = X^\nu\partial_\nu\omega_\mu + \omega_\nu\partial_\mu X^\nu.\tag{5.6}\] For a rank $(0,2)$ tensor, the Lie derivative is defined as \[\mathcal{L}_Xg = \lim_{t\to0}\frac{((\sigma_{-t}) _*g)_p -g_p}{t},\tag{5.7}\] where the push-forward is given by \[(\varphi_* g)(X,Y)\bigg|_p =g(\varphi^{-1}_*X,\varphi^{-1}_*Y)\bigg|_{\varphi^{-1}(p)},\tag{5.8}\] and $\sigma_{-t}$ is the flow generated by $X\in \mathfrak{X}(M)$. The flow equation, in a local coordinate, is given by \[X^\mu(x) = \frac{dx^\mu}{dt},\tag{5.9}\] which is satisfied by \[x^\mu(t) = x^\mu(0) +tX^\mu(x) + \mathcal{O}(t^2) \tag{5.10}\] to first order in $t$ (note the slight abuse of notation $x^\mu(\sigma_t) = x^\mu (t)$). In a chart, the push-forward is given by \[(\varphi_* g)_{\mu\nu}\bigg|_p = \frac{\partial x^\sigma}{\partial x^\mu(\varphi)}\frac{\partial x^\rho}{\partial x^\nu(\varphi)}g_{\sigma\rho}\bigg|_{\varphi^{-1}(p)},\tag{5.11}\] which gives \[\begin{aligned} ((\sigma_{-t}) _*g)_{\mu\nu}\bigg|_p &= (\delta^\sigma_\mu - t\partial_\mu X^\sigma)^{-1}(\delta^\rho_\nu - t\partial_\nu X^\rho)^{-1}g_{\sigma\rho}\bigg|_{\sigma_t(p)} + \mathcal{O}(t^2) \\ &= (\delta^\sigma_\mu + t\partial_\mu X^\sigma)(\delta^\rho_\nu + t\partial_\nu X^\rho)\left(g_{\mu\nu} + tX^\lambda\partial_\lambda g_{\mu\nu}\right)\bigg|_p + \cdots\\ &=g_{\mu\nu}\bigg|_p + t(X^\rho\partial_\rho g_{\mu\nu} + g_{\mu\rho}\partial_\nu X^\rho + g_{\rho\nu}\partial_\mu X^\rho)\bigg|_p + \mathcal{O}(t^2), \end{aligned}\tag{5.12}\] where we have used (5.10) to Taylor expand $g$ around $p$. Therefore, we find \[(\mathcal{L}_X g)_{\mu\nu} = X^\rho\partial_\rho g_{\mu\nu} + g_{\mu\rho}\partial_\nu X^\rho + g_{\rho\nu}\partial_\mu X^\rho.\tag{5.13}\] First, consider \[\begin{aligned} \iota_X(d\omega) &= \iota_X\frac{1}{2}(\partial_\mu\omega_\nu - \partial_\nu \omega_\mu) dx^\mu \wedge dx^\nu \\&= \frac{1}{2}(\partial_\mu\omega_\nu - \partial_\nu \omega_\mu)X^\mu dx^\nu - \frac{1}{2}(\partial_\mu\omega_\nu - \partial_\nu \omega_\mu)X^\nu dx^\mu\\& = X^\mu\partial_\mu\omega_\nu dx^\nu - X^\mu \partial_\nu\omega_\mu dx^\nu. \end{aligned}\tag{5.14}\] Then, \[d(\iota_X\omega) = d(\omega_\mu X^\mu) = X^\mu\partial_\nu\omega_\mu dx^\nu + \omega_\mu \partial_\nu X^\mu dx^\nu.\tag{5.15}\] Therefore, we find \[\mathcal{L}_X\omega = \iota_X(d\omega) + d(\iota_X\omega).\tag{5.16}\]

6 Properties of the Exterior Derivative, Q6

Let $\omega$ be a $p$-form and $\eta$ a $q$-form. Show that the exterior derivative satisfies the properties

$d(d\omega) = 0$
$d(\omega\wedge\eta) = (d\omega)\wedge\eta + (-1)^p\omega\wedge d\eta$
$d(\varphi^*\omega) = \varphi^*(d\omega)$ where $\varphi:M\to N$ for some manifolds $M$ and $N$

\[\begin{aligned} d(d\omega) &= d\left(\frac{1}{p!}\frac{\partial\omega_{\mu_1\cdots\mu_p}}{\partial x^\nu} dx^\nu \wedge dx^{\mu_1} \wedge \cdots \wedge dx^{\mu_p}\right)\\ &=\frac{1}{p!}\frac{\partial^2\omega_{\mu_1\cdots\mu_p}}{\partial x^\rho\partial x^\nu}dx^\rho \wedge dx^\nu \wedge dx^{\mu_1} \wedge \cdots \wedge dx^{\mu_p}\\ &=0 \end{aligned}\tag{6.1}\] by antisymmetry of the wedge product.

\[\begin{aligned} d(\omega\wedge\eta) &= d\left(\frac{1}{p!q!}\omega_{[\mu_1\cdots\mu_p}\eta_{\nu_1\cdots\nu_q]} dx^{\mu_1} \wedge \cdots \wedge dx^{\mu_p}\wedge dx^{\nu_1}\wedge \cdots \wedge dx^{\nu_q} \right)\\ &=\frac{1}{p!q!}\left((\partial_{[\mu} \omega_{[\mu_1\cdots\mu_p})\eta_{\nu_1\cdots\nu_q]]} +\omega_{[\mu_1\cdots\mu_p} \partial_{[\mu} \eta_{\nu_1\cdots\nu_q]]}\right)dx^\mu \wedge dx^{\mu_1} \wedge \cdots \wedge dx^{\nu_q}. \end{aligned}\tag{6.2}\] Now consider the wedge product \[\begin{aligned} (d\omega \wedge \eta)_{\mu\mu_1\cdots\nu_q} &= \frac{(p+1)}{(p+1)!q!}(\partial_{[[\mu}\omega_{\mu_1\cdots\mu_p]})\eta_{\nu_1\cdots\nu_q]} \\&= \frac{1}{p!q!}(\partial_{[\mu}\omega_{\mu_1\cdots\mu_p})\eta_{\nu_1\cdots\nu_q]}, \end{aligned}\tag{6.3}\] which is exactly the first term in (6.2). Similarly, the second term of (6.2) can also be written as $\omega \wedge d\eta$ by permuting $\mu$ to the left, which gives a factor of $(-1)^p$. Therefore, we find \[d(\omega\wedge\eta) = (d\omega)\wedge\eta + (-1)^p\omega\wedge d\eta.\tag{6.4}\]

The pull-back can be written as $\partial \tilde{x}^\mu /\partial x^\nu$. By antisymmetry of the exterior derivative, any term involving $d(\partial \tilde{x}^\mu /\partial x^\nu)$ must vanish. Therefore, the exterior derivative must pass through the pull-back and act on the $p$-form, $\omega$. That is \[d(\varphi^*\omega) = \varphi^*d\omega.\tag{6.5}\]

7 Coordinate-Free Definition of the Exterior Derivative, Q7

The exterior derivative of $\omega\in\Lambda^p(M)$ can be defined as \[\begin{aligned} d\omega(X_1,\ldots,X_{p+1}) &= \sum_{j=1}^{p+1}(-1)^{j-1}X_j\!\left(\omega(X_1,\ldots,\hat{X}_j,\ldots,X_{p+1})\right)\\ &\quad +\sum_{j<k}(-1)^{j+k}\omega([X_j,X_k],X_1,\ldots,\hat{X}_j,\ldots,\hat{X}_k,\ldots,X_{p+1}) \end{aligned}\tag{7.1}\] where a hat denotes omission. In a coordinate basis $\omega = \frac{1}{p!}\omega_{\mu_1\cdots\mu_p}\,dx^{\mu_1}\wedge\cdots\wedge dx^{\mu_p}$, use this definition to determine the components of $d\omega$ in the cases $p=1$ and $p=2$.

For $p=1$, we have \[\begin{aligned} d\omega(X,Y) &= X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y])\\ &= X^\mu\partial_\mu(\omega_\nu Y^\nu) - Y^\nu\partial_\nu(\omega_\mu X^\mu) - \omega_\rho(X^\mu\partial_\mu Y^\rho - Y^\mu\partial_\mu X^\rho)\\ &= (\partial_\mu\omega_\nu - \partial_\nu\omega_\mu)X^\mu Y^\nu, \end{aligned}\tag{7.2}\] so the components read $(d\omega)_{\mu\nu} = \partial_\mu\omega_\nu - \partial_\nu\omega_\mu = 2\,\partial_{[\mu}\omega_{\nu]}$, and \[d\omega = \partial_\mu\omega_\nu\,dx^\mu\wedge dx^\nu.\tag{7.3}\] For $p=2$, with $\omega = \frac{1}{2}\omega_{\mu\nu}dx^\mu\wedge dx^\nu$, we have $\omega(Y,Z) = \omega_{\mu\nu}Y^\mu Z^\nu$, and \[\begin{aligned} d\omega(X,Y,Z) &= X(\omega(Y,Z)) - Y(\omega(X,Z)) + Z(\omega(X,Y))\\ &\quad - \omega([X,Y],Z) + \omega([X,Z],Y) - \omega([Y,Z],X). \end{aligned}\tag{7.4}\] Expanding the first piece, \[X(\omega(Y,Z)) = X^\rho(\partial_\rho\omega_{\mu\nu})Y^\mu Z^\nu + \omega_{\mu\nu}X^\rho(\partial_\rho Y^\mu)Z^\nu + \omega_{\mu\nu}X^\rho Y^\mu(\partial_\rho Z^\nu),\tag{7.5}\] and similarly for the other two, while each commutator contributes terms such as \[\omega([X,Y],Z) = \omega_{\rho\nu}(X^\sigma\partial_\sigma Y^\rho - Y^\sigma\partial_\sigma X^\rho)Z^\nu.\tag{7.6}\] All terms involving derivatives of the vector fields cancel pairwise after relabeling dummy indices and using antisymmetry of $\omega_{\mu\nu}$. What remains is \[d\omega(X,Y,Z) = \left(\partial_\rho\omega_{\mu\nu} + \partial_\mu\omega_{\nu\rho} + \partial_\nu\omega_{\rho\mu}\right)X^\rho Y^\mu Z^\nu,\tag{7.7}\] so the components read \[(d\omega)_{\rho\mu\nu} = \partial_\rho\omega_{\mu\nu} + \partial_\mu\omega_{\nu\rho} + \partial_\nu\omega_{\rho\mu} = 3\,\partial_{[\rho}\omega_{\mu\nu]},\tag{7.8}\] which gives \[d\omega = \frac{1}{2}\,\partial_\rho\omega_{\mu\nu}\,dx^\rho\wedge dx^\mu\wedge dx^\nu.\tag{7.9}\] Both cases fit the general pattern $(d\omega)_{\mu_0\mu_1\cdots\mu_p} = (p+1)\,\partial_{[\mu_0}\omega_{\mu_1\cdots\mu_p]}$.

8 Maurer-Cartan Structure Equations on $S^3$, Q8

A three-sphere can be parameterized by Euler angles $(\theta,\phi,\psi)$ where $0<\theta<\pi$, $0<\phi<2\pi$, $0<\psi<4\pi$. Define the following 1-forms \[\begin{aligned}\sigma^1 &= -\sin\psi\,d\theta + \cos\psi\sin\theta\,d\phi,\\ \sigma^2 &= \cos\psi\,d\theta + \sin\psi\sin\theta\,d\phi,\\ \sigma^3 &= d\psi + \cos\theta\,d\phi.\end{aligned}\] Show that $d\sigma^1 = \sigma^2\wedge\sigma^3$ with analogous results for $d\sigma^2$ and $d\sigma^3$.

9 Coordinate-Induced Bases and the Poincaré Lemma, Q9

Let $\{e_\mu\}$ be a basis of vector fields, with \[[e_\mu, e_\nu] = \gamma^\rho{}_{\mu\nu}\,e_\rho.\tag{9.1}\] The functions $\gamma^\rho{}_{\mu\nu}$ are known as commutator components. For a choice of coordinates $\{x^\mu\}$, we often work with the coordinate induced basis $e_\mu = \{\partial_\mu\}$. Show that, in this case, $[e_\mu, e_\nu] = 0$.

The purpose of this question is to show the converse: that $[e_\mu,e_\nu]=0$ only for a coordinate induced basis. Consider a general basis $\{e_\mu\}$ and the dual basis $\{f^\mu\}$ of one-forms. In general, these can be expanded as \[e_\mu = e_\mu{}^\rho\frac{\partial}{\partial x^\rho} \qquad \text{and} \qquad f^\mu = f^\mu{}_\rho\,dx^\rho\tag{9.2}\] where $e_\mu{}^\rho f^\nu{}_\rho = \delta^\mu_\nu$. Show that \[e_\mu{}^\sigma\frac{\partial e_\nu{}^\lambda}{\partial x^\sigma} - e_\nu{}^\sigma\frac{\partial e_\mu{}^\lambda}{\partial x^\sigma} = \gamma^\rho{}_{\mu\nu}\,e_\rho{}^\lambda.\tag{9.3}\] Hence deduce that \[e_\mu{}^\sigma e_\nu{}^\lambda\frac{\partial f^\rho{}_\lambda}{\partial x^\sigma} - e_\nu{}^\sigma e_\mu{}^\lambda\frac{\partial f^\rho{}_\lambda}{\partial x^\sigma} = -\gamma^\rho{}_{\mu\nu},\tag{9.4}\] and finally that \[\frac{\partial f^\rho{}_\sigma}{\partial x^\lambda} - \frac{\partial f^\rho{}_\lambda}{\partial x^\sigma} = -\gamma^\rho{}_{\mu\nu}f^\mu{}_\lambda f^\nu{}_\sigma.\tag{9.5}\] Use this result, together with the Poincaré lemma, to show that if $[e_\mu,e_\nu]=0$ $\forall\,\mu,\nu$ then the basis is coordinate induced.

For coordinate bases, we clearly have \[[\partial_\mu, \partial_\nu] = 0.\tag{9.6}\]

Now, consider the commutation relation \[[e_\mu{}^\rho\partial_\rho,e_\nu{}^\sigma\partial_\sigma] = \gamma^\lambda{}_{\mu\nu}e_\lambda{}^\delta\partial_\delta.\tag{9.7}\] The left-hand side is \[[e_\mu{}^\rho\partial_\rho,e_\nu{}^\sigma\partial_\sigma] = e_\mu{}^\rho(\partial_\rho e_\nu{}^\sigma)\partial_\sigma - e_\nu{}^\sigma(\partial_\sigma e_\mu{}^\rho)\partial_\rho.\tag{9.8}\] After relabeling, we find \[e_\mu{}^\sigma\frac{\partial e_\nu{}^\lambda}{\partial x^\sigma} - e_\nu{}^\sigma\frac{\partial e_\mu{}^\lambda}{\partial x^\sigma} = \gamma^\rho{}_{\mu\nu}\,e_\rho{}^\lambda.\tag{9.9}\] Then, consider the expression \[e_\mu{}^\sigma\frac{\partial (e_\nu{}^\lambda f^\rho{}_\lambda)}{\partial x^\sigma} - e_\nu{}^\sigma\frac{\partial (e_\mu{}^\lambda f^\rho{}_\lambda)}{\partial x^\sigma}.\tag{9.10}\] Since $e_\mu{}^\rho f^\nu{}_\rho = \delta^\mu_\nu$, the above expression must vanish. Therefore, we must have \[\begin{aligned} 0=\;&e_\mu{}^\sigma\frac{\partial (e_\nu{}^\lambda f^\rho{}_\lambda)}{\partial x^\sigma} - e_\nu{}^\sigma\frac{\partial (e_\mu{}^\lambda f^\rho{}_\lambda)}{\partial x^\sigma} \\ =\;& e_\mu{}^\sigma f^\rho{}_\lambda \frac{\partial e_\nu{}^\lambda}{\partial x^\sigma} - e_\nu{}^\sigma f^\rho{}_\lambda\frac{\partial e_\mu{}^\lambda }{\partial x^\sigma} + e_\mu{}^\sigma e_\nu{}^\lambda\frac{\partial f^\rho{}_\lambda}{\partial x^\sigma} - e_\nu{}^\sigma e_\mu{}^\lambda\frac{\partial f^\rho{}_\lambda}{\partial x^\sigma}\\ =\; & \gamma^\rho{}_{\mu\nu} + e_\mu{}^\sigma e_\nu{}^\lambda\frac{\partial f^\rho{}_\lambda}{\partial x^\sigma} - e_\nu{}^\sigma e_\mu{}^\lambda\frac{\partial f^\rho{}_\lambda}{\partial x^\sigma}. \end{aligned}\tag{9.11}\] Therefore, we find \[e_\mu{}^\sigma e_\nu{}^\lambda\frac{\partial f^\rho{}_\lambda}{\partial x^\sigma} - e_\nu{}^\sigma e_\mu{}^\lambda\frac{\partial f^\rho{}_\lambda}{\partial x^\sigma} = -\gamma^\rho{}_{\mu\nu}.\tag{9.12}\] Then, contracting with $f$, we find \[\frac{\partial f^\rho{}_\sigma}{\partial x^\lambda} - \frac{\partial f^\rho{}_\lambda}{\partial x^\sigma} = -\gamma^\rho{}_{\mu\nu}f^\mu{}_\lambda f^\nu{}_\sigma.\tag{9.13}\] The Poincaré lemma states that, on $M = \mathbb{R}^n$, closed implies exact. Multiplying the previous expression by $dx^\lambda \wedge dx^\sigma$ and using antisymmetry of the wedge, \[d\theta^\rho = \frac{1}{2}\!\left(\frac{\partial f^\rho{}_\sigma}{\partial x^\lambda} - \frac{\partial f^\rho{}_\lambda}{\partial x^\sigma}\right)dx^\lambda\wedge dx^\sigma = -\frac{1}{2}\gamma^\rho{}_{\mu\nu}\,\theta^\mu\wedge\theta^\nu,\tag{9.14}\] where $\theta^\rho = f^\rho{}_\sigma\,dx^\sigma$ is the dual basis to $e_\mu = e_\mu{}^\rho\partial_\rho$. Now suppose $\gamma^\rho{}_{\mu\nu}$ vanish. Then $d\theta^\rho = 0$, so $\theta^\rho$ is a closed 1-form. By the Poincaré lemma, on $M = \mathbb{R}^n$ (or any contractible neighbourhood) closed forms are exact, so there exist functions $y^\rho$ with \[\theta^\rho = dy^\rho.\tag{9.15}\] The $y^\rho$ are functionally independent because $\theta^\rho$ are linearly independent at every point, so they form a coordinate system. In these coordinates, $\theta^\rho(e_\mu) = \delta^\rho_\mu$ becomes $\partial y^\rho/\partial y^\mu_{(e)} = \delta^\rho_\mu$, i.e.\ $e_\mu = \partial/\partial y^\mu$, a coordinate basis.

Conversely, if $e_\mu = \partial/\partial y^\mu$ for some coordinates $y^\mu$, then $[e_\mu, e_\nu] = [\partial_{y^\mu}, \partial_{y^\nu}] = 0$, so $\gamma^\rho{}_{\mu\nu} = 0$. Therefore $\{e_\mu\}$ is (locally) a coordinate basis if and only if $\gamma^\rho{}_{\mu\nu} = 0$.