Total derivative of a scalar field

Question

I am rereading Spivak's calculus on manifold and on page 89 I notice that he says if $f\colon \mathbb{R}^n\rightarrow \mathbb{R}$, then $Df(p)\in \Lambda^1(\mathbb{R}^n)$. But it became not obvious to me why this is true, if we think $Df(p)$ as a linear transformation from $T_p\mathbb{R}^n$ to $T_p\mathbb{R}$. But the first exterior power of $\mathbb{R}$ is just(naturally isomorphic to) $\mathbb{R}^n$, right? Can anyone helps to clarify what he meant there? Thanks!

Answer 1

I assume $D$ denotes the differential on functions. On $\mathbb{R}^n$ you have an inner product $g$, which gives a canonical isomorphism between $\mathbb{R}^n$ and $(\mathbb{R}^n)^*$ (a vector $v$ is mapped to $v^{\flat} \in (\mathbb{R}^n)^*$ such that $v^{\flat}(u) = g(v,u)$). You know that the differential $Df$ at each point is a linear map from $\mathbb{R}^n$ to $\mathbb{R}$, so $Df(p) \in (\mathbb{R}^n)^*$ for every $p \in \mathbb{R}^n$. But then $$Df(p) \in (\mathbb{R}^n)^* \cong \mathbb{R}^n = \Lambda^1\mathbb{R}^n.$$