# How does the differential $df$ act on an element of $T_pM$?

Let $f$ be a smooth real valued function on a smooth manifold $M$. The differential of $f$ is the covector field $df$ defined by

$$df_p(v) = v(f)$$

where $v \in T_pM$ and where we are now thinking of $v$ as an element of $\operatorname{Der}(M)$. Now I am trying to see if I can understand this definition in the following concrete case. Let $f : \Bbb{S}^2 \to \Bbb{R}$ be the smooth real valued function that just picks out the $z$ - coordinate of a point $p \in S^2$. Now if $p \in S^2 - N$ where $N$ is the north pole then choose the coordinate chart $\{S^2 - N, \sigma_N\}$ on $S^2$ where $\sigma_N$ is stereographic projection from the north pole. With this chart, I have computed $df_p$ in coordinates to be

$$df_p = \left( \frac{4x}{\left(x^2 + y^2 + 1\right)^2} dx + \frac{4y}{\left (x^2 + y^2 + 1\right)^2} dy \right)\Bigg|_{\sigma_N(p)}.$$

My question is: Having computed $df_p(v)$, how can I see what it does to an arbitrary vector $T_pS^2$? For example if $v$ is the vector $(1,1,0)$ based at the point $p = (0,0,-1)$, what is $df_{(0,0,-1)}( v)$?

Thanks.

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$df_p$ is a linear form, that is, a linear map from $T_pM$ to $\mathbb{R}$. To the vector $X = X^1\frac{\partial}{\partial x^1} + \dots + X^n\frac{\partial}{\partial x^n}$, the differential $df_p = f_1 d x^1 + \dots + f_n d x^n$ acts just by $$df_p (X) = X^1 f_1 + \dots + X^n f_n.$$

Was that your question or am I understanding it wrong?

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Dear Kofi, your $X^i$ are real numbers yes? –  user38268 May 11 '13 at 8:40
@Bejna: Yes, because $X$ is a tangent vector. If $X$ was a vector field, then the $X^i$ would be scalar fields. –  Hurkyl May 11 '13 at 8:51
@Hurkyl Thanks. –  user38268 May 11 '13 at 10:35
@Hurkyl Is it true that the space of all derivations at a point $p$ on a manifold $M$ has basis $\frac{\partial}{\partial x^i}|_p$, $1\leq i \leq n$? –  user38268 May 11 '13 at 11:30
Yes, of course, as is written in any introductionary book on differential geometry! –  Kofi May 11 '13 at 16:24

If you view vectors $v$ as derivations, then the action is very simple: $$df(v)(h)=v(h\circ f)$$ for all functions $h$ defined on the codomain of $f$.

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Thanks for your answer. I forgot that the differential can also be thought of a kind of map between tangent spaces. However if now I think of $v$ as a vector on $S^2$, how do I compute this? For example, if $p$ is the south pole, then I can consider the vector $(1,1,0)$ based at $p$. How can I know what $df(v)$ is? –  user38268 May 11 '13 at 8:31
Well, I was responding to the question you asked originally: now that you edited... this has become rather disconnected with it :-/ –  Mariano Suárez-Alvarez May 11 '13 at 8:34
Sorry Mariano, I realised that what was more meaningful to ask is the edited version of my question. I'm sorry; could you help me with this edited version? Thanks. –  user38268 May 11 '13 at 8:36

You can use plain linear algebra; you can first evaluate the effect of the pushforward on basis tangent vectors, and then write (1,1,0) as a combination of these basis vectors, say, $\partial/\partial x$ and $\partial/\partial y+\partial\partial z$ , using the isomorphism that identifies a vector v with the directional derivative in the direction of v. Let { ${\partial/\partial x' ,\partial/\partial y'}$} be a basis for $T_p \mathbb R^2$. We then first calculate the pushforward :

$df_p(\partial / \partial x)=4x/(x^2+y^2+1) (\partial/\partial x')$

$df_p(\partial /\partial y+\partial/\partial z)=4y/(x^2+y^2+1)(\partial/\partial y')$

Now, using the above isomorphism, write $(1,1,0)$ as a linear combination of the basis vectors $(1,0,0)$ and $(0,1,1)$ , and use linearity of $df_p$ to find its value at (1,1,0)and evaluate your expression numerically at the point p=$(0,0,-1)$

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