16

I've recently started studying differential geometry and was really hoping that in doing so I'd finally have an answer to something that's been bugging me since I first learnt calculus - what is $dx$?!

As far as I understand, in differential geometry $dx^{i}$ is a linear functional that maps vectors in a tangent space $T_{p}M$ at a point $p\in M$ on a manifold $M$ to the set of real numbers $\mathbb{R}$, i.e. $$dx^{i} :T_{p}M\rightarrow\mathbb{R}$$ In this sense the differential form $dx^{i}$ maps a vector $v\in T_{p}M$ to its $i^{th}$ coordinate with respect to the coordinate basis $\frac{\partial}{\partial x^{i}}$, i.e. $dx^{i}(v)=v^{i}$.

In elementary calculus I was always told when I asked the question "what is $dx$?", that it is an infinitesimal change in the x-coordinate. This has never rested easy with me as e.g. if we have the formula $$ df=\lim_{\Delta x\rightarrow 0}\Delta f = \lim_{\Delta x\rightarrow 0}f'(x)\Delta x $$ then due to the properties of limits this can be expressed as $$\lim_{\Delta x\rightarrow 0}f'(x)\lim_{\Delta x\rightarrow 0}\Delta x$$ and clearly $\lim_{\Delta x\rightarrow 0}\Delta x =0$ which seems inconsistent.

So my main question is: what actually is $dx$ and is there any intuitive (perhaps geometric) explanation as to how it relates to an infinitesimal line element?

Perpetual learner
  • 599
  • In the elementary calculus case, don't think of $df$ and $dx$ as infinitesimals -- just think of them as finite changes. See my answer to this question for my interpretation of the elementary calculus case. Your differential forms explanation is spot on though. –  Mar 17 '15 at 15:29
  • I'm struggling a bit to get away from the idea that $df$ is an infinitesimal change in $f$. Is the idea that $df$ is a new function (parametrised by $\Delta x$) that gives an incremental change along the tangent line (tangent to the function $f$ at the point $x$) such that $df=df(x,\Delta x)=f'(x)\Delta x$?! Also is there any intuitive link to differential forms (Wikipedia claims so, but I can't follow what's written there very well)?! – Perpetual learner Mar 17 '15 at 17:58
  • That is the idea, yes. I'm not entirely sure about the intuitive link to differential forms, though. –  Mar 17 '15 at 18:16
  • Do you know of a good reference (textbook or otherwise) that gives a good description of differentials in the way I put in my previous post? – Perpetual learner Mar 17 '15 at 22:43
  • My familiarity with differential forms is through the lens of geometric calculus. I haven't really studied the classical methodologies. Though, as your post Is there an intuitive... asks about a geometric interpretation, this might be worth studying for you. If so, you could look at David Hestenes & Garret Sobczyk's Clifford Algebra to Geometric Calculus. –  Mar 17 '15 at 22:51

3 Answers3

4

In dimension one:

$$df = f'(x)dx$$

or better:

$$dy = f'(x)dx$$

Here $${T_x}R$$ and it's dual $$({T_x}{R) }^ *$$

are 1-dimensional vector-spaces which are identified with $R$ itself as reals. In this case you are allowed to do divisions. There is no mystic behind:

$$\frac{{dy}}{{dx}} = f'(x)$$

$dx$ can be viewed as 1-dimensional volume-element, and function $f$ as change of variables. Volume-element $dy$ with respect to $x$ has shape like $dy = f'(x)dx$ $$f'(x)$$ is the 1-dimensional Jacobian!, if you want. By the way if the limit $$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}}$$ at a point exists, then we have $$\frac{{dy}}{{dx}}({x_0}) = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}} = f'({x_0})$$ In general $\frac{{\Delta y}}{{\Delta x}}$ and $\frac{{dy}}{{dx}}$ are fractions, but different ones. Namely slope for secant-line and second slope for tangent-line.

And because they are reals, you can draw them in 1-dimensional calculus as slope for tangent-line.

In higher dimensions the coorinates for $$df = \sum\limits_i {\frac{{\partial f}}{{\partial {x^i}}}(x)d{x^i}}$$ are the well known components for gradient, which lives in $({T_x}{R^n) }^ *$ in case of n-dim. vectorspace or manifold.

Frieder
  • 1,655
0

Your understanding of the calculus version of $dx$ is wrong. While there are ways to do calculus using infinitesimals (non-standard analysis), $dx$ is certainly not the limit of $\Delta x$ as $\Delta x \to 0$: that would be simply $0$.

One way to understand $dx$ and $dy$ in calculus is as new variables expressing changes in $x$ and $y$ on the tangent line to the graph of your function. See e.g. Differential.

hardmath
  • 37,015
Robert Israel
  • 448,999
  • 1
    The argument I gave for the "$\Delta x$" part was taken from the introduction of a differential geometry book (unfortunately I can't remember which one off hand). Can $dx^{i}$ be considered as an infinitesimal change in the coordinate function $x^{i}:\mathbb{R}^{n}\rightarrow\mathbb{R}$, i.e. as an element in the cotangent space? – Perpetual learner Mar 17 '15 at 15:26
0

Intuitively they are same and can be taken as simple. I state about how it appears to me. We consider small scalars and worry more about what is to be done among them in group operations rather than be bogged down what it deeply and definitively is.I say this as you are comfortable at advanced level and not so at the fundamental level.

The Pythagoras theorem $ s^2 = x^2 + y^2 $ in the plane has become in calculus in minuscule $ ds^2 = dx^2 + dy^2 $. In differential geometry of two dimensions/ surface theory we have non-linear simple versions of the same curved coordinate lines $ ds^2 = E dx^2 + 2 F dx dy + G dy^2 $ which are now called metric, linked together in tangent space of a simple surface manifold. Further generalized into multidimensional Riemannian manifolds.

Narasimham
  • 40,495