Introduction to Riemannian manifolds problem $4-14$ related to connection $1$-forms

Question

Let $M$ be a smooth $n$-manifold and $\nabla$ a connection in $TM$, let $(E_i)$ be a local frame on some open subset $U \subset M$, and let $(\varepsilon^i)$ be the dual coframe. Show that there is a uniquely determined $n \times n$ matrix of smooth $1$-forms $(\omega_i^j)$ on $U$, called the connection $1$-forms for this frame, such that $$\nabla_X E_i = \omega_i^j(X)E_j.$$

I'm working on this and what I've managed to get at is that if we write $X = X^jE_j$, then by the properties of the connection we can deduce that $$\nabla_XE_i=\nabla_{X^jE_j}E_i=X^j \Gamma_{ji}^kE_k$$ but I do not know where these $1$-forms should pop up? $X^j$ is a real number and $\Gamma$ maps from $T_pM \times T_pM \times T_pM \to \mathbb{R}$.

Answer 1

Defining $\omega^j_i(X) = \varepsilon^j(\nabla_XE_i)$, one has $\nabla_XE_i = \sum_{j=1}^n \omega^j_i(X)E_j$, for the very simple reason that $\{\varepsilon^1,\ldots,\varepsilon^n\}$ is the dual coframe of $\{E_1,\ldots,E_n\}$. Uniqueness comes from the fact that $\{E_1,\ldots,E_n\}$ is a frame: the coefficients of the vector field $\nabla_XE_i$ are unique in this frame. Finally, by $\mathcal{C}^{\infty}$-linearity of $\varepsilon^j$ and of the covariant derivative in the lower entry, it follows that $\omega^j_i$ is a $1$-form on $U$.

Answer 2

This boils down to basic linear algebra, and really has nothing to do with manifolds. Let $V$ be an $n$-dimensional vector space over any field $\Bbb{F}$, let $\beta=\{e_1,\dots, e_n\}$ be a basis for $V$, and let $\beta^*=\{\epsilon^1,\dots, \epsilon^n\}$ be the dual basis. Now by definition of $\beta$ being a basis for $V$, what does that tell us? Well, for any vector $v\in V$, there must exist unique scalars $c^1,\dots, c^n\in\Bbb{F}$ (depending on $v$ of course) such that \begin{align} v&=\sum_{i=1}^nc^ie_i. \end{align} In fact we have that for each $1\leq i\leq n$, the coefficient $c^i$ is equal to $\epsilon^i(v)$. The proof is simple; apply $\epsilon^j$ to the equation above, then you get $\epsilon^j(v)=\sum_{i=1}^nc^i\epsilon^j(e_i)=\sum_{i=1}^nc^i\delta^j_i=c^j$. In other words, the role of the dual basis is to pick out the coefficients in the expansion of a vector in terms of a given basis. Once again, summarizing, given the basis $\beta$ and dual basis $\beta^*$, we have for each $v\in V$, that \begin{align} v&=\sum_{i=1}^n\epsilon^i(v)e_i.\tag{$*$} \end{align} By the way, just as a side remark, if you unwind the definition of the isomorphism $\text{End}(V)\cong V^*\otimes V$, then the above equation is saying that the identity map on $V$ corresponds to $\sum_{i=1}^n\epsilon^i\otimes e_i$.

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By applying the equation ($*$) at each tangent space, you’ll see that (being careful to not overload indices) \begin{align} \nabla_XE_i&=\sum_{j=1}^n\epsilon^j(\nabla_XE_i)\,E_j. \end{align} So, the definition of the connection 1-forms is obvious: simply set $\omega^j_i(X):=\epsilon^j(\nabla_XE_i)$. In words, it says $\omega^j_i(X)$ is the $j^{th}$ coefficient function in the expansion of the local vector field $\nabla_XE_i$ relative to the local frame $\{E_1,\dots, E_n\}$.

Finally, as for why $\omega^j_i$’s are 1-forms (on $U$), this is obvious once you note the dependence on $X$ is $C^{\infty}(U)$-linear since $\nabla_XY$ behaves (by definition) in a $C^{\infty}(U)$ linear fashion with respect to $X$, and so is each $\epsilon^j$.