Let $M,N$ be smooth manifolds, $\psi: M \rightarrow N$ a $C^{\infty}$ and $M_m, N_{\psi(m)}$ the tangent spaces at $m \in M$ and $\psi(m) \in N$. The differential of $\psi$ at $m$ is the linear map $$d\psi: M_m\rightarrow N_{\psi(m)}$$ defined by setting $$d\psi(v)(g)=v(g\circ\psi),$$ where $v \in M_m$ is a tangent vector and $g$ is $c^{\infty}$ in a neighborhood of $\psi(m)$.

The dual map $$\delta\psi:N^*_{\psi(m)} \rightarrow M^*_m$$ is defined by requiring that $$\delta \psi(\omega)(v)=\omega(d\psi(v))$$ whenever $\omega \in N^*_{\psi(m)}$ and $v\in M_m$


I am having some trouble understanding the following:

In the special case of a $C^{\infty}$ function $f:M\rightarrow \mathbb{R},$ if $v\in M_m$ and $f(m)=r_0,$ then $$df(v)=v(f)\frac{d}{dr}\bigg |_{r_0}.$$ In this case, we usually take $df$ to mean the element of $M^*_m$ defined by $$df(v)=v(f).$$ That is, we identify $df$ with $\delta f(\omega)$, where $\omega$ is the basis of the $1$ dimensional space $\mathbb{R}^*_{r_0}$ dual to $(d/dr)|_{r_0}$

My Interpretation

Since $df(v)$ is a tangent vector at $f(m)=r_0$ and since $\frac{d}{dr}\bigg|_{r_0}$ is a basis for the tangent space at $r_0$, we should have $$df(v)=a\frac{d}{dr}\bigg|_{r_0}.$$ Now, evaluating both sides at $r$ we obtain $$df(v)(r)=a.$$ Now $df(v)(r)=v(r \circ f)=v(f)$ so we have the desired result.

Next, I am not sure what the text means when it states that $df(v)=v(f)$. We just showed that $df(v)=v(f)\frac{d}{dr}\bigg |_{r_0}$. In fact, I do not understand how $df$ is being associated with $\delta f(\omega)$ (or why this association even matters). Is the text just trying to say that since $df(v)$ maps into the reals, it is a linear functional on $M_m$?


The text I am referring to is Foundations of Differentiable Manifolds and Lie Groups by Frank W. Warner


1 Answer 1


Elements of $M^*_m$ are functionals $\phi:M_m\to\mathbb{R}.$ From definition, you have written, we see that $$df:M_m\to\mathbb{R}_{r_o}.$$ Manifold $\mathbb{R}$ thanks to cannonical parameterisation $id=r:\mathbb{R}\to\mathbb{R}$ has cannonical one element base of tangent space $\mathbb{R}_{r_0}$ at each point $r_0.$ Namely $\frac{d}{dr}|_{r_0}.$

Hence we can indentify spaces $\mathbb{R}$ and $\mathbb{R}_{r_0}$ just by indentifying basis $e\mapsto\frac{d}{dr}|_{r_0}.$ After this indentification we can say that $$df:M_m\to\mathbb{R}.$$ But from definition of $M^*_m$ we see that in above interpretation $df\in M^*_m.$

To see the second part, indentify $\mathbb{R}$ with $\mathbb{R}^*_{r_0}$ again by basis and then indentify linear maps from $\mathbb{R}$ to $M_m^*$ with elements of $M_m^*.$


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