Derivative of identity map and moving forms So it's possible this question is simply "why is the derivative of the identity map on a manifold the identity on the tangent spaces?", but the context comes from reading Spivak's book about moving frames. I'll give enough here that you don't need that reference, I think.
So I understand we have a moving frame $(X_1(p),X_2(p),...,X_n(p))$ at a point $p$ on manifold $M_n$, and we can easily define "a" one-form,
$$\tilde{\omega}=\tilde{\theta}^iX_i$$
If $\tilde{\theta}^i$ are any particular one-forms on $M_n$ (or $T^*M_n$ I guess). And I also get that if we pick the dual one-forms such that $\theta^i(X_j)=\delta^i_j$, then that one form becomes an identity map, such that acting on a vector $V_p=V^jX_j(p)$
$$\omega(V_p)=\theta^i(V^jX_j)X_i(p)=V^j\delta^i_jX_i(p)=V_p.$$
(ok, I guess I'm slightly confused by this, since one-forms should act on vectors and produce scalars, but I think that's for a scalar-valued one-form...here we have a vector valued one-form $\omega$, so acting on a vector produces a vector...I think!)
The next thing is confusing to me. Apparently, if we take the identity map $P:M_n\to M_n$, we can actually identify $\omega$ with $dP$. This I don't understand - if I was going to take the exterior derivative of $P$, I would do something like
$$dP=\frac{\partial P}{\partial X_i}X_i+...$$
That's in the frame, not a coordinate system, so maybe I would do
$$\frac{\partial P}{\partial X_i}=\frac{\partial P}{\partial x^j}\frac{\partial x^j}{\partial X_i}$$
So even if I started thinking about Veilbien $x^i=e^{ij}X_j$ so that maybe $\frac{\partial x^j}{\partial X_i}=e^j_i$, so that $\theta^i=\frac{\partial P}{\partial x^j}e^{ij}$? But $\partial P / \partial x^j=1$ for each $j$....there's no index there and I'm just getting more confused.
Can anyone straighten this story out?
 A: This discussion is confused. Tildes appear and then disappear. A $1$-form is scalar-valued, unless you specifically speak of a vector-valued $1$-form or, more generally, a $1$-form with values in a vector bundle.
Remember that a linear map $T\colon V\to V$ can be viewed as a $(1,1)$-tensor, i.e., an element of $V^*\otimes V$. So, if $X_i$ is a basis for $V$ and $\theta^i$ is the dual basis for $V^*$, then the identity map is represented as the $(1,1)$-tensor $\sum \theta_i\otimes X_i$. That's all Spivak is doing with tensor fields (locally) on $M$. (If you consider the case that $M\subset\Bbb R^N$ for some $N$, then you can actually make sense of $\sum \theta^iX_i$ as an $\Bbb R^N$-valued $1$-form whose values happen to land in the tangent spaces of $M$.)
The discussion of Élie Cartan's $dP$ is meant to align with terminology from a century ago. But that's all that's going on. If you take the identity map $M\to M$, its derivative at a point is the identity map $T_pM\to T_pM$, and then you apply this discussion.
