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Consider the following funny argument: Let $d$ be the well-known exterior derivative. We know for any $k$-Form $f \in \Omega^k(M)$ we have $ddf=0$ and $d$ is a linear map. Then $df$ must already have been quite small, as otherwise $ddf=d^2f$ would not vanish identically. Therefore $df$ is infinitesimal.

I agree, that crucial steps in this argument are simply wrong. E.g. what does 'infinitesimally small' mean for standard-$\mathbb{R}^n$? Also any field is a integral domain.

My question here is: 1. Is there any hidden truth in this argument? Especially: Would the intuition of an infinitesimal still work if we for example had $dd=1$ or $dd=d$ or something like that? 2. Is there a nice nonstandard-analytic way of giving reason to this argument? I would assume that in nonstandard-analysis $df$ is defined according to the intuition of Leibniz etc. How does one recover the properties of the exterior derivative from this? So far these questions seems to not have been asked before here. At least I could not find them.

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    $\begingroup$ I don't think the argument conveys the right intuitive idea at all. After all, $\mathrm{d}(3294839) = 0$, but in most circumstances I would not consider $3294839$ to be quite small! So if there is anything reasonable there at all, it's some additional intuitive property that $df$ has that is not explained. $\endgroup$ – Hurkyl Jul 30 '17 at 13:40
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    $\begingroup$ Also, in terms of simply defining the derivative, nonstandard analysis does nothing novel here: see this answer. The novel part is in the surrounding structure (i.e. the halo of a point vs the tangent space). $\endgroup$ – Hurkyl Jul 30 '17 at 13:55
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    $\begingroup$ $d^2 = 0$ is very particular to the exterior derivative, and is not at all about size. It should not even intuitively be thought about in the same sense that $\epsilon^2 = 0$ for an infinitesimal $\epsilon$. You can define differential operators with the same dimensionality as $d$ that do not square to zero, like a covariant derivative. $\endgroup$ – Anthony Carapetis Jul 30 '17 at 14:03
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I think the answer is basically no, since $d^2=0$ is fundamentally a consequence of symmetry and antisymmetry, coming down to partial derivatives commuting, while $d$ is antisymmetric. $\operatorname{tr}(AS)=0$ if $A$ is antisymmetric and $S$ is symmetric, no matter what size $A$ and $S$ are.

Since $d : \Omega^k \to \Omega^{k+1} $, $d^2$ cannot really be equal to either $d$ or the identity, since neither maps $\Omega^k \to \Omega^{k+2}$, but suppose we ignore this. $dd=1$ recalls the split-complex numbers, which can be thought of in terms of shears as complex numbers are to rotations (And of course $i$ is not small, so analogously, neither is $j$).

But if $T^2=1$, $T$ certainly can't make everything small: think of the matrix $\begin{pmatrix} 0 & \epsilon \\ 1/\epsilon & 0 \end{pmatrix}$, which squares to the identity, and for small $\epsilon$, makes it map some vectors to small vectors, some to large. Meanwhile $d^2=d$ is simply a projection, which maps into its eigenspace with eigenvalue $1$. To be made infinitesimal by a projection requires having infinitesimal components to start with, so is also not suitable.

The algebraic nature of most of the manipulations in differential geometry ($d$ being a graded derivation, &c.) means that the differences brought by nonstandard analysis are fairly inconsequential (see the comments). This question has plenty of good answers that help in thinking about the various derivatives in differential geometry.

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