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[EDIT: I know what the notation means, and I can easily show that $x=\epsilon \otimes 1 + 1\otimes \epsilon$ satisfies $x^3=0$ but $x^2 \neq 0$. That's not what this question is about. It might be better to restrict the question to Synthetic Differential Geometry where it makes more sense. There are infinitesimals $\epsilon$ and $\delta$ in SDG such that $\epsilon^2=\delta^2=0$ but $\epsilon\delta \neq 0$. What are these objects really? How can I understand this phenomenon intuitively?]

One of the flaws with the Dual Numbers is that they don't contain numbers that cube to zero. So while it's possible to take linear approximations, you can't take quadratic approximations.

One suggested fix is to use $\mathbb R[\epsilon] \otimes \mathbb R[\epsilon]$. Or continue the sequence ad-infinitum $\mathbb R[\epsilon] \otimes \mathbb R[\epsilon]\otimes \mathbb R[\epsilon]\otimes \mathbb R[\epsilon]\otimes\dotsc$

But this results in there being a pair of "infinitesimals" $\epsilon$ and $\delta$ such that $\epsilon^2 = \delta^2=0$, but $\epsilon\delta\neq0$. Why is this weird? Because if $x$ and $y$ are two real numbers, the number $xy$ must lie somewhere in the interval between $x^2$ and $y^2$; if we now imagine that $x^2$ and $y^2$ are "so small" that their square is zero, then $xy$ has to also be zero.

In spite of this pathology, this approach gets used in Synthetic Differential Geometry. Dan Piponi shows you can use it to compute Lie Brackets.

More than that, in some SDG books they carry out geometric arguments using such generalised dual numbers. For instance, simple length and area formulas are derived using them. But they do this while one of the main laws of geometry, the ordering of points on the number line, is violated. How can they do that?

I can sort of makes sense of it in the following way. You've got a function $f(x,y)$. You can derive a linear approximation $f(x + dx,y) = f(x,y) + dx\frac{\partial f}{\partial x}(x,y)$ using dual numbers. That tells you one of $f$'s partial derivatives. It then makes sense to make the partial derivative bit non-infinitesimal, so that the expression $f(x,y) + dx\frac{\partial f}{\partial x}(x,y)$ becomes the vector $[f(x,y), \frac{\partial f}{\partial x}(x,y)]$ and then increase $y$ to $y+dy$. This gives you the expression $[f(x,y) + dy\frac{\partial f}{\partial y}(x,y), \frac{\partial f}{\partial x}(x,y) + dy\frac{\partial^2 f}{\partial x\partial y}(x,y)]$, which you can then make into the vector $[f(x,y), \frac{\partial f}{\partial y}(x,y), \frac{\partial f}{\partial x}(x,y), \frac{\partial^2 f}{\partial x \partial y}(x,y)]$. This sequence of steps would give you the same result if you partially differentiated $x$ first or $y$ first. But this is never described anywhere, and it doesn't seem very intuitive.

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  • $\begingroup$ Assuming I'm understanding your question (the titular question) clearly isn't this just because $(1\otimes\varepsilon+\varepsilon\otimes 1)^3=1\otimes \varepsilon^3+3(\varepsilon^2)\otimes \varepsilon+3(\varepsilon\otimes\varepsilon^2)+1\otimes\varepsilon^3$ and all those terms are zero? $\endgroup$ – Alex Youcis Apr 8 '19 at 11:12
  • $\begingroup$ @AlexYoucis That's clear, but it's not really what I asked. I'm aware you can prove that $1\otimes\varepsilon+\varepsilon\otimes 1$ is nilcube using the binomial theorem $\endgroup$ – ogogmad Apr 8 '19 at 11:43

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