# How exactly does it make sense to differentiate a function whose input is a point on a manifold?

In this lecture, prof Frederic Scheuller introduces the concept of the tangent vector space defined the following way by introducing the concept of velocity at a point:

Let $$(M,\theta,A)$$ be a smooth manifold, and a curve $$\gamma: R \to M$$ at least $$C^1$$(One differentiable and derivative is continous). Suppose $$\gamma(\lambda_o)=p$$, then the velocity at $$p$$ is defined as: $$v_{\gamma,p} : C^{\infty} (M) \to R$$

Where $$C^{\infty}(M) := \{ f: M \to R | f \text{ is a smooth function} \}$$ equipped with an addition of the form $$(f+g)(p)= f(p) + g(p)$$

and s multiplication of the form $$( \lambda \cdot g)(p) = \lambda g(p)$$ where the addition and scaling is in the same sense as operations we define on the real numbers

From 1:23 of the video , time stamped link

The above is fine and more or less intuitive, but what bothers me is what happens at 52:51, where he takes the derivative of the following object:

$$[ \partial_i( f \circ x^{-1})] x (p)$$

I am not sure what the object of $$\partial$$ is precisely meant to mean but I can say that (this is one of my doubts) :

$$p$$ is the geometric point on the manifold

$$x$$ is the function which takes us from the manifold to the point on the chart corresponding to it

$$x^{-1}$$ is the function which inverts the point on the chart to the point on the manifold

$$f$$ is a function from a point on the manifold to real number line

This is where I get confused, if $$x^{-1}$$ maps from points on the chart to geometric points on the manifolds and f takes that as an input, so what would it mean to take derivative of such a thing? The professor even points this out at around 55:40 but doesn't explain what exactly he has written.

Thanks in advance.

• Honestly, in my answer, I haven't said anything that wasn't already mentioned in the lecture, and it's not really clear to me where your confusion is because the lecturer points out all these issues in that part of the lecture. Commented May 1, 2021 at 20:36

## 2 Answers

Very clearly $$(\partial f\circ x^{-1})x(p)$$ is not what was written on the board; I assume you just made a typo. What was written is $$[\partial_i(f\circ x^{-1})](x(p))$$. This is simply the $$i^{th}$$ partial derivative of the function $$f\circ x^{-1}:x[U]\subset\Bbb{R}^d\to\Bbb{R}$$ evaluated at the point $$x(p)$$. Now this is really just basic multivariable calculus at this point. If you want everything explicitly in terms of limits, we have \begin{align} \frac{\partial f}{\partial x^i}(p)&:= [\partial_i(f\circ x^{-1})](x(p))\\ &:=\frac{d}{ds}\bigg|_{s=0}(f\circ x^{-1})(x(p)+se_i)\\ &:=\lim_{s\to 0}\frac{(f\circ x^{-1})(x(p)+se_i) - (f\circ x^{-1})(x(p))}{s}\\ &=\lim_{s\to 0}\frac{(f\circ x^{-1})(x(p)+se_i)-f(p)}{s}, \end{align} where $$e_i=(0,\dots, \underbrace{1}_{\text{i^{th} spot}},\dots, 0)\in\Bbb{R}^d$$. The first equal sign is a definition for the symbol on the LHS in terms of usual partial derivatives. THe next equal sign is simply the definition of the $$i^{th}$$ partial derivative. The next equal sign is just the definition of a (single-variable) derivative in terms of a difference quotient. The last equal sign is just by definition of composition.

Here everything makes perfect sense. $$f\circ x^{-1}$$ is a function from an open subset $$x[U]$$ of $$\Bbb{R}^d$$ into $$\Bbb{R}$$. $$x(p)$$ is an element of $$x[U]$$. So, by openness, if $$s\in\Bbb{R}$$ is small enough, then the points $$x(p)+se_i$$ lie in $$x[U]$$. Thus, it makes sense to plug these into $$f\circ x^{-1}$$.

• I don't think this answer was what the professor was intending, at 55:57 the professor states that this is NOT supposed to be the partial derivative Commented May 1, 2021 at 20:39
• he says "that (referring to $\left(\frac{\partial f}{\partial x^i}\right)_p$) is our notation for this (referring to $[\partial_i(f\circ x^{-1})](p)$)". He says $\left(\frac{\partial f}{\partial x^i}\right)_p$ is NOT a partial derivative in the naive sense that it is not defined as $\lim_{h\to 0}\frac{f(p+he_i)-f(p)}{h}$, because such an expression is clearly meaningless because it makes no sense to add points on the manifold. He means it is a partial derivative only when properly interpreted in terms of the chart's partial derivative. Commented May 1, 2021 at 20:43
• Once we involve the chart, the quantity $[\partial_i(f\circ x^{-1})](x(p))$ is truly a 100% correct notation for a partial derivative. (which he does say around the 50 minute mark). My suggestion is you rewatch the entire thing from the 50:00 and listen carefully to everything he writes and says. Commented May 1, 2021 at 20:44
• Oh. Now, it is all clear. I had misunderstood what he had said at that point. Commented May 1, 2021 at 20:46

Let $$M$$ be an $$n$$-dimensional manifold, $$p \in M$$ a point, $$x: U \subset M \to x(U) \subset\mathbb{R}^n$$ a chart around $$p$$ and $$f: V \subset M \to \mathbb{R}$$ a function such that $$p \in V$$ (with $$V$$ open in $$M$$). A priori it doesn't make any sense to say $$f$$ is smooth at $$p$$. But we can differentiate functions with Euclidean (open) domains and co-domains, and the coordinate representation of $$f$$ with respect to the chart $$x$$ fits that exact description! So you ask, what is the coordinate representation of $$f$$? Well, since $$x$$ is a chart, we can map points of $$x(U)$$ to $$M$$ via $$x^{-1}$$ and look at which of those points belong to $$V$$ so we can apply $$f$$ to those points. Doing that, we get a function $$\hat{f}: x(U \cap V) \subset \mathbb{R}^n \to \mathbb{R}$$, and we do know how to differentiate such a function. So we shall say that $$f$$ is smooth at $$p$$ iff its coordinate representation $$\hat{f} \doteq f \circ x^{-1}$$ is smooth at $$x(p)$$. Since $$x(p)$$ is a point in $$\mathbb{R}^n$$, all of this makes perfect sense. In particular, when $$f$$ is smooth, we can talk about the partial derivative of $$f$$ at $$p$$ by defining them to be the partial derivatives of its coordinate representation relative to a chart (again, we do know very well how to take derivatives in the Euclidean case) evaluated at... well, at the only place where it makes sense to evaluate it: $$x(p)$$. Since $$\hat{f}$$ is a function from $$\mathbb{R}^n$$ to $$\mathbb{R}$$, there are $$n$$ such partial derivatives. Naturally, they are given by:

$$\frac{\partial f}{\partial x^i}(p) \doteq \frac{\partial (f \circ x^{-1})}{\partial r^i}(x(p)), \text{for each 1 \leq i \leq n }$$

where on the left side the partial derivative is with respect to the chart $$x$$ and on the right side, it's the ordinary Euclidean partial derivative (which is again a partial derivative with respect to a chart, just the most trivial one of them all). And of course, all of this generalizes very naturally to functions between general manifolds.

• Thank you for the answer, it is very clear what you have written. I had accepted the other one as @peek-a-boo had pointed out the misunderstanding I hade made of what the lecturer was saying. Commented May 1, 2021 at 20:48