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We define the tangent space $T_p\mathbb{R}^n$ by $T_p\mathbb{R}^n=\text{span}\{\frac{\partial}{\partial{x^1}}|_p,\dots,\frac{\partial}{\partial{x^n}}|_p\}$ and the cotangent space $(T_p\mathbb{R}^n)^*$ by $(T_p\mathbb{R}^n)^*=\text{span}\{d_px^1,\dots,d_px^n\}$. We also have the following definition:

Definition: Let $f:\text{dom}(f)\subset\mathbb{R^n}\to\mathbb{R}$ be a smooth function and let $p\in\text{dom(f)}$. We define the differential of $f$ at $p$ by $d_pf\in(T_p\mathbb{R}^n)^*$, given by $d_pf(V)=V(f)$ (where $V\in{T_p}\mathbb{R}^n$), and the assignment of $p\mapsto{d_pf}$ gives a differential 1-form $d_pf$ on $\mathbb{R}^n$.

To demonstrate the example given in the lectures I am watching, let $f=x^2y+z^3$. It follows that if, say $V=V^1\frac{\partial}{\partial{x}}|_p+V^2\frac{\partial}{\partial{y}}|_p+V^3\frac{\partial}{\partial{z}}|_p$, then $$\begin{align}d_pf(V)&=V(x^2y+z^3)\\&=(V^1\frac{\partial}{\partial{x}}|_p+V^2\frac{\partial}{\partial{y}}|_p+V^3\frac{\partial}{\partial{z}}|_p)(x^2y+z^3)\\&=V^1(2xy)|_p+V^2(x^2)|_p+V^3(3z^2)|_p.\end{align}$$ My issue is that if $d_pf\in(T_p\mathbb{R}^n)^*$ and $V=V^1\frac{\partial}{\partial{x}}|_p+V^2\frac{\partial}{\partial{y}}|_p+V^3\frac{\partial}{\partial{z}}|_p$, then in the above it appears as if $f=x^2y+z^3$ is the differential 1-form. This obviously doesn't seem right (maybe I am misreading the definition), but this is the only thing that acts on $V\in{T_p}\mathbb{R}^n$ when we expand out the expression.

My question then is what is the differential 1-form? Is it $$V^1(2xy)|_p+V^2(x^2)|_p+V^3(3z^2)|_p?$$ Does the differential 1-form then instead act on the point $p$? Can someone just point to the differential 1-form when everything has been expanded out?

I usually think of a differential 1-form as an expression involving $dx^i$ (which makes sense when we consider $(T_p\mathbb{R}^n)^*=\text{span}\{d_px^1,\dots,d_px^n\}$); where does that fit into all of this? I don't see any expression involving $dx^i$.

I hope I haven't made a complete mess of this but any help would be greatly appreciated.

Seth
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A differential 1-form takes a point $p \in M$ and a tangent vector at $p$ and returns a real number. For every $p$, you get a (different, generally speaking) 1-form on $T_pM$. Let $(U,\varphi)$ be a coordinate chart on $M$, i.e. $\varphi\colon\ U \to {\mathbb R}^n$. Call $q=\varphi(p) \in {\mathbb R}^n$ the coordinate representation of $p$. In your case, $$ (2xy)_q \cdot V^1 + (x^2)_q\cdot V^2 + (3z^2)_q\cdotp V^3 \in {\mathbb R} $$ is the numerical value produced by the 1-form $$ (2xy)_q \cdot d_qx + (x^2)_q\cdot d_qy + (3z^2)_q\cdotp d_qz \in T^*_q{\mathbb R}^n \cong T^*_pM $$ on the tangent vector $$ V^1\left(\frac{\partial}{\partial x}\right)_q + V^2\left(\frac{\partial}{\partial y}\right)_q + V^3\left(\frac{\partial}{\partial z}\right)_q \in T_q{\mathbb R}^n \cong T_pM. $$ Keep in mind that $d_qx^i\left(\frac{\partial}{\partial x^j}\right)_q = \delta_{ij}$.

User1610788
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  • Quick follow up, but it seems as though the terms $2xy,x^2$ and $3z^2$ are the result of the 'partial' operators (from the tangent vector) acting on the function even though it hasn't acted on the tangent vector. Is there something I'm implicitly missing here? – Seth Sep 23 '21 at 00:17
  • $(2xy)q = \omega_q\left(\frac{\partial}{\partial x}\right)_q$, where $\omega$ is your form. Ditto $x^2$ and $3z^2$. Tangent vectors acting on functions is irrelevant here. You understand linear algebra, so you'll understand this too because it's basically the same thing. In linear algebra, you'd write ${e_1,\ldots,e_n}$, ${e^1,\ldots,e^n}$, and you'd have $e^i(e_j) = \delta{ij}$. Same here. – User1610788 Sep 23 '21 at 03:45