Let $E_1, E_2, E_3$ be an arbitrary frame field on $\mathbb{E}^3$. At each $v\in T_p\mathbb{E}^3$, $\nabla_v E_i\in T_p\mathbb{E}^3$, $i=1,2,3$. So, there exists uniquely 1-forms $\omega_{ij}:T_p\mathbb{E}^3\longrightarrow\mathbb{R}$, $i,j=1,2,3$ such that
\begin{align*}
\nabla_vE_1&=\omega_{11}(v)E_1(p)+\omega_{12}(v)E_2(p)+\omega_{13}(v)E_3(p),\\
\nabla_vE_2&=\omega_{21}(v)E_1(p)+\omega_{22}(v)E_2(p)+\omega_{23}(v)E_3(p),\\
\nabla_vE_3&=\omega_{31}(v)E_1(p)+\omega_{32}(v)E_2(p)+\omega_{33}(v)E_3(p)
\end{align*}
for each $v\in T_p\mathbb{E}^3$. These equations are called the connection equations of the frame field $E_1$, $E_2$, $E_3$. One can clearly see that $\omega_{ij}$ is determined by
$$\omega_{ij}(v)=\nabla_v E_i\cdot E_j(p).$$ The 1-forms $\omega_{ij}$ are called the connection forms of the frame field $E_1,E_2,E_3$. Often the matrix $\omega=(\omega_{ij})$ is called the connection 1-form of the frame field $E_1,E_2,E_3$. The linearity of $\omega_{ij}$ is due to the linearity of the covariant derivative $\nabla E_i$.

Proposition. The matrix $\omega$ is a skew symmetric matrix, i.e. $\omega+{}^t\omega=0$.

Proof. Since $E_i\cdot E_j=0$, the directional derivative $v[E_i\cdot E_j]=0$. On the other hand, by Leibniz rule,
\begin{align*}
v[E_i\cdot E_j]&=\nabla_vE_i\cdot E_j(p)+E_i(p)\cdot \nabla_vE_j\\
&=\omega_{ij}(v)+\omega_{ji}(v).
\end{align*}
Hence,
\begin{equation}\label{eq:skewsymm}\omega_{ij}+\omega_{ji}=0.\end{equation}

If $i=j$ in \eqref{eq:skewsymm}, we get $\omega_{ii}=0$. So, the connection 1-form $\omega$ is written as
$$\omega=\begin{pmatrix}
0 & \omega_{12} & \omega_{13}\\
-\omega_{12} & 0 &\omega_{23}\\
-\omega_{13} & -\omega_{23} & 0
\end{pmatrix}.$$

Remark. The set of all $3\times 3$ skew symmetric matrices is denoted by $\mathfrak{o}(3)$. It is the Lie algebra of the orthogonal group $\mathrm{O}(3)$. The orthogonal group $\mathrm{O}(3)$ is the set of all $3\times 3$ orthogonal matrices and it is a Lie group. Recall that a square matrix $A$ is orthogonal if and only if $A\cdot{}^tA=I$, i.e. $A^{-1}={}^tA$.

The connection equations of the frame field $E_1$, $E_2$, $E_3$
\begin{equation}\label{eq:connecteqns}\nabla_VE_i=\sum_i\omega_{ij}(V)E_j,\ i=1,2,3\end{equation}
where $V$ is a vector field on $\mathbb{E}^3$ become
$$\begin{array}{ccccccc}
\nabla_VE_1&=&&&\omega_{12}(V)E_2&+&\omega_{13}(V)E_3,\\
\nabla_VE_2&=&-\omega_{12}(V)E_1& & &+&\omega_{23}(V)E_3,\\
\nabla_VE_3&=&-\omega_{13}(V)E_1&-&\omega_{23}(V)E_2.
\end{array}
$$
The connections equations are in fact a generalization of the Frenet-Serret formulas.

Let $Y$ be a vector field defined on a region containing a curve $\alpha(t)$. Then $Y_\alpha(t):=Y(\alpha(t))$ defined a vector field on the curve $\alpha(t)$. Then one can easily see that
$$\nabla_{\dot\alpha(t)}Y=\frac{d}{dt}Y_\alpha(t).$$
Let $\alpha(t)$ be a curve with unit speed. Let $E_1=T$, $E_2=N$, $E_3=B$. Then
\begin{align*}
\omega_{12}&=\nabla_{\dot\alpha_(t)}E_1\cdot E_2=\dot T\cdot N=(\kappa N)\cdot N=\kappa,\\
\omega_{13}&=\nabla_{\dot\alpha_(t)}E_1\cdot E_3=\dot T\cdot B=0,\\
\omega_{23}&=\nabla_{\dot\alpha_(t)}E_2\cdot E_3=\dot N\cdot B=(-\kappa T+\tau B)=\tau.
\end{align*}
The connection equations \eqref{eq:connecteqns} are then nothing but the Frenet-Serret formulas
$$\begin{array}{ccccccc}
\dot T&=&&&\kappa N&&\\
\dot N&=&-\kappa T& & &+&\tau B\\
\dot B&=&&-&\tau N.
\end{array}
$$

The frame $E_1,E_2,E_3$ can be written in terms of the natural frame $U_1,U_2,U_3$ as
\begin{align*}
E_1&=a_{11}U_1+a_{12}U_2+a_{13}U_3,\\
E_2&=a_{21}U_1+a_{22}U_2+a_{23}U_3,\\
E_3&=a_{31}U_1+a_{32}U_2+a_{33}U_3.
\end{align*}
Each real-valued function $a_{ij}:\mathbb{E}^3\longrightarrow\mathbb{R}$ is uniquely determined by $a_{ij}=E_i\cdot U_j$. The matrix $A=(a_{ij})$ is called the attitude matrix (also called rotation matrix or orientation matrix) of the frame field $E_1,E_2,E_3$. One can clearly see that the attitude matrix $A$ is an orthogonal matrix. In the above remark, I mentioned that the set of all $3\times $ skew symmetric matrices is the Lie algebra $\mathfrak{o}(3)$. The Lie algebra $\mathfrak{g}$ of a Lie group $G$ is defined to be the tangent space $T_e G$ to $G$ at the identity element $e$. (A Lie group is a differentiable manifold, so it make sense to talk about tangent spaces to $G$.)

Let us define a curve $\gamma: \mathbb{R}\longrightarrow\mathrm{O}(3)$ by
$$\gamma(t)=A(t)\cdot{}^tA(0).$$
Then $\gamma(0)=I$.
Hence $\dot{\gamma}(0)=\frac{dA(t)}{dt}|_{t=0}\cdot{}^tA(0)$ is a tangent vector to $\mathrm{O}(3)$ at the identity matrix $I$. That is, $\dot{\gamma}(0)\in\mathfrak{o}(3)$. Hence one can easily expect that the following theorem holds.

Theorem. If $A=(a_{ij})$ is the attitude matrix and $\omega=(\omega_{ij})$ the connection 1-form of a frame field $E_1, E_2, E_3$, then
$$\omega=dA\cdot{}^tA$$
or equivalently
$$\omega_{ij}=\sum_k da_{ik} \cdot a_{jk}\ \mbox{for}\ i,j=1,2,3.$$

Proof. For each $v\in T_p\mathbb{E}^3$,
$$\omega_{ij}(v)=\nabla_vE_i\cdot E_j(p).$$
In terms of the natural field $U_i$, $i=1,2,3$,
$$E_i=\sum_ka_{ik}U_k,\ i=1,2,3.$$
So,
\begin{align*}
\nabla_vE_i&=\sum_k v[a_{ik}]U_k(p)\\
&=\sum_k da_{ik} U_k(p).
\end{align*}
Hence,
$$\omega_{ij}=\sum_k da_{ik}a_{jk},$$
i.e.
$$\omega=dA\cdot{}^tA.$$

Remark. In general, if $G$ is a Lie group then its Lie algebra $\mathfrak{g}$ is given by the set of differential $1$-forms
$$\mathfrak{g}=\{g^{-1}dg:\ g\in G\}=\{(dg^{-1})g:\ g\in G\}.$$

Example. Let us compute the connection forms of the cylindrical frame field. The attitude matrix is
$$A=\begin{pmatrix}
\cos\theta & \sin\theta & 0\\
-\sin\theta & \cos\theta & 0\\
0 & 0 & 1
\end{pmatrix}.$$ Thus
$$dA=\begin{pmatrix}
-\sin\theta d\theta & \cos\theta d\theta & 0\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
0 & 0 & 0
\end{pmatrix}.$$
Hence,
\begin{align*}
\omega&=dA\cdot{}^tA\\
&=\begin{pmatrix}
-\sin\theta d\theta & \cos\theta d\theta & 0\\
-\cos\theta d\theta & -\sin\theta d\theta & 0\\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix}
\cos\theta & -\sin\theta & 0\\
\sin\theta & \cos\theta & 0\\
0 & 0 & 1\end{pmatrix}\\
&=\begin{pmatrix}
0 & d\theta & 0\\
-d\theta & 0 & 0\\
0 & 0 & 0
\end{pmatrix}.
\end{align*}
The connection equations of the cylindrical frame field are then
\begin{align*}
\nabla_VE_1&=d\theta(V)E_2=V[\theta]E_2,\\
\nabla_VE_2&=-d\theta(V)E_1=-V[\theta]E_1,\\
\nabla_VE_3&=0
\end{align*}
for all vector fields $V$. As expected the vector field $E_3$ is parallel.