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Lets allow ourselves to be a bit informal for the sake of exploration. Assume that for any sufficiently nice $f$ in two variables $x$ an $y$, we have $$df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$$ and that $d$ is linear and satisfies the product rule, by which I mean $d(fg) = d(f)\cdot g+f\cdot d(g).$ From a differential forms perspective we have $d(df) = 0$, but lets just ignore that for now; that is, do not assume $d(df) = 0$. Also, lets treat products as symmetrical rather than antisymmetrical.

We compute:$$\begin{align} d(df) &= d\left(\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \right) \\ & = d\left(\frac{\partial f}{\partial x} dx\right) + d\left(\frac{\partial f}{\partial y} dy \right) \\ & = d\left(\frac{\partial f}{\partial x}\right) dx+ \frac{\partial f}{\partial x} d(dx)+ d\left(\frac{\partial f}{\partial y}\right) dy+ \frac{\partial f}{\partial y} d(dy) \\ & = \left(\frac{\partial }{\partial x}\frac{\partial f}{\partial x} dx + \frac{\partial }{\partial y}\frac{\partial f}{\partial x} dy\right) dx + \frac{\partial f}{\partial x} d(dx) + \left(\frac{\partial }{\partial x}\frac{\partial f}{\partial y} dx + \frac{\partial }{\partial y}\frac{\partial f}{\partial y} dy\right) dy + \frac{\partial f}{\partial y} d(dy) \\ & = \frac{\partial }{\partial x}\frac{\partial f}{\partial x} dx^2 + 2\frac{\partial }{\partial y}\frac{\partial f}{\partial x} dx dy + \frac{\partial }{\partial y}\frac{\partial f}{\partial y} dy^2 + \frac{\partial f}{\partial x} d(dx)+\frac{\partial f}{\partial y} d(dy) \end{align}$$

So if we can just find $d(dx)$ and $d(dy)$, then we've found a formula for $d(df)$ that isn't zero. Indeed, it seems to be the quadratic part of the second-order Taylor expansion for $f$. It's therefore tempting to declare that $d(dx)$ and $d(dy)$ are zero. This feels a bit dodgy though. Let's feel less bad about it by declaring that if all second-order partial derivatives of $f$ are zero, then $d(df) = 0$. This yields $d(dx) = 0$ and $d(dy)=0$ as a special case, as desired, but feels less depressing.

With this stipulation, we find that $$d(df) = \frac{\partial }{\partial x}\frac{\partial f}{\partial x} dx^2 + 2\frac{\partial }{\partial y}\frac{\partial f}{\partial x} dx dy + \frac{\partial }{\partial y}\frac{\partial f}{\partial y} dy^2.$$

More generally, I think we find that $d^n f$ tells us the $n$-th order part of the Taylor series of $f$. (I'm not completely sure about this claim actually - thought anyone?). Anyway, assuming this is true, $e^{d} f$ is basically the whole Taylor series, so long as we perform the shady looking computation $$e^d f = \left(\sum_{i = 0}^\infty \frac{d^i}{i!}\right) f = \sum_{i = 0}^\infty \left(\frac{d^i}{i!}f\right)$$

Question. Does this kind of thing go anywhere? This seems relevant.

Remark. I think we should be able to move this kind of reasoning to arbitrary smooth manifolds, by choosing an atlas and computing $d^n f$ locally. We can then either declare that $d(dx_i)$ is zero for each coordinate function $x_i$, or we can do the second-derivative trick; in either case I suspect we'll get the same result, and my guess is that the answer should be atlas-independent.

Further Remark. If we want, we can set all terms above degree $n$ to zero. In this way, we find that $e^d f$ is telling us the $n$-th order Taylor polynomial of $f$ at each point of our space. The case where we kill all terms above degree $1$ to $0$ yields the familiar formula $d(df) = 0$ from differential topology, just by observing that the above formula for $d(df)$ is a linear combination of terms of degree $2$.

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  • $\begingroup$ Right, he is using the symmetric product, and so do you when you write formulas like $ds^2=dx^2+dy^2$. @JesseMadnick $\endgroup$ – Mikhail Katz Apr 17 '18 at 8:56
  • $\begingroup$ @JesseMadnick, I mean the rule $d(fg) = (df)g+f(dg)$. $\endgroup$ – goblin GONE Apr 17 '18 at 9:42
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    $\begingroup$ The "symmetric product" you refer to in this question is usually written $\otimes$ in tensor analysis. It coincides with the wedge product $\wedge$ if one of the factors is a zero-form (a scalar field). The $d$ operator you defined here should be a derivation with respect to $\otimes$, while the usual $d$ is a derivation with respect to $\wedge$, something like that. $\endgroup$ – Giuseppe Negro Apr 17 '18 at 9:52
  • $\begingroup$ Goblin: I'm referring to the fact that you have to clarify what you mean by writing $dx\,dy$, for instance. One could mean the tensor product $dx \otimes dy$ or symmetric product $dx \circ dy$ or wedge product $dx \wedge dy$. All three of these products are sometimes written $dx\,dy$, but it depends on the author. So, when you have something like, say, $d(f\,dx) = df\,dx + f\,d(dx)$, you have to specify which product is being used. $\endgroup$ – Jesse Madnick Apr 17 '18 at 11:51
  • $\begingroup$ @JesseMadnick, "Lets allow ourselves to be a bit informal for the sake of exploration." In short, I don't know what I mean. However, it seems that $dx \circ dy$ fits the bill. Can you tell me what to look up to learn more about this operation? $\endgroup$ – goblin GONE Apr 17 '18 at 13:07

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