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Can one justify Leibniz's formalism in a suitable algebraic or topological context?

We have published some papers recently where we argue that Leibniz's formalism for the calculus wasn't inconsistent as Berkeley claimed. For an insightful review see http://www.ams.org/mathscinet-getitem?mr=3053644

Berkeley claimed that Leibniz wanted to have it both ways: both $dx\not=0$ so as to form the differential ratio, and also $dx=0$ so as to get the right answer (i.e., a "standard" one). Starting about 140 years ago, Berkeley's claim of inconsistency of Leibnizian calculus acquired the status of dogma to such an extent that Robinson himself felt compelled to speak of Berkeley's "brilliant critique" of the calculus, and referred to the hyperreal framework as "a small price to pay for the removal of an inconsistency"--the implied assumption being that such an "inconsistency" was real.

The reason Berkeley was wrong is that Leibniz repeatedly emphasized that he is working with a generalized notion of equality. For example if $y=x^2$, the desired formula $\frac{dy}{dx} = 2x$ does not mean that the residual $dx$ is set equal to zero but rather that it is absorbed into the generalized relation of equality "up to" a negligible term, in an exact sense to be specified. Leibniz called this principle the transcendental law of homogeneity. The principle is mentioned, for example, in the title of his 1710 paper, as reported already in 1974 by Bos. Berkeley did not take this into account and merely misunderstood Leibniz.

Now this is fine for showing that Leibniz was not inconsistent (refuting Berkeley's claim). However, it is not quite enough for showing that Leibniz was actually consistent, or more precisely for formalizing Leibniz's approach. This is because it is not completely clear what the generalized relation is exactly. I will refer to such a generalized notion as "adequality" so as not be have to write "generalized equality up to" every time.

In other words, if we want to be able to work with adequality as we work with the ordinary equality, we need to explain how this is done and why this works and why whenever A=B one can replace A by B in computations. Robinson circumvented the problem by using the standard part function but this isn't completely faithul to Leibniz's formalism.

One attempted solution is to take the adequality 2x+dx=2x to mean that the difference of the two sides is infinitesimal. But if this is our notion of adequality, then this allows us to write down things like dx=0, as well, and if we are allowed to replace dx by 0 in calculations then we end up dividing by zero.

In Lawvere's approach (also Kock, Bell) they replace the ratio formula $f '(x)=dy/dx$ by the multiplication formula $dy=f '(x)dx$. Then they get equality on the nose by working with nilsquare infinitesimals. Thus their adequality is true equality on the nose. In this way they implement (some of) Leibniz's procedures. However, this is not entirely faithful to Leibniz because Leibniz worked with arbitrary order infinitesimals, and also divided by them freely.

Euler worked with what he called a generalized "geometric" equality where A=B means that the ratio of A to B is infinitely close to 1, but this does not automatically allow us to add such relations. Of course if all expressions involved are appreciable, this does work. On the other hand, we can't always assume both sides to be appreciable because this would disallow critical points, certainly a disturbing loss.

How does one address this problem? The idea is to continue working with Euler's "geometric equality" and somehow to make both sides appreciable by working globally rather than at a specific point; or perhaps evaluating the expressions at a generic (or perhaps nonstandard?) point. This would hopefully allow one to manipulate an adequality between expressions as an ordinary equality.

In the specific case of $y=x^2$ the problem is the zero of the derivative where Euler's geometric equality does not work, but at any other point we are OK.

Could one define such a relation in an algebraic (or algebraic-geometric) context? One needs to specify the sort of expressions one is allowed to work with, i.e. introduce a limitation on the objects one is allowed to use. Or perhaps it is enough to declare expressions adequal if they are geometrically equal at a nonstandard point.

Can one redefine the relation "=" in a suitable context, so that for example one could read the chain rule as literally saying $\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}$ ? The relation can be rewritten in a simpler form in terms of differentials: $dz = \frac{dz}{dy}dy$ but this still depends on the "cancellation" of $dy$ in the numerator and denominator. This works with Euler's "geometric equality" but the problem is you can't add such equalities. Already chain rule with 2 variables requires addition.

Aside from the pedagogical value of being able to cancel out the two $dy$'s in $dz/dx =dz/dy\; dy/dx$, an application would be the solution of differential equations by separation of variables, etc. More generally, one would like to be able to take any argument using derivatives, and be authorized to replace $f ′ (x)$ by $dy/dx$ whenever it occurs without changing the argument otherwise. In nested arguments involving multiple derivatives this could be a significant simplification.

Differential geometry is a rich source of examples where salvaging Leibniz's formalism could simplify many proofs. Beyond being able to say that the center of curvature is the point of intersection of two infinitely close normals, numerous arguments in Gauss and Riemann are simpler in their original infinitesimal form (including formula for curvature) than their modern reformulations; see discussion in Spivak, Differential geometry, volume 2, and related MSE thread.

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    $\begingroup$ Could you write a list of explicit examples of Leibniz's formalism at work, comprehensive enough that we get a good idea of its flavor? That way, it won't be necessary for specialists in the modern mathematics that may form a potential answer to also be experts in the works of Leibniz. $\endgroup$
    – S. Carnahan
    Commented Aug 11, 2014 at 14:31
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    $\begingroup$ Scott, I appreciate your interest. First, I am looking for a restricted class of functions and a generalized relation "=" such that one can literally write $f'(x)=\frac{dy}{dx}$ where $dy$ is the $y$-increment corresponding to the infinitesimal $x$-increment $dx$. In such a framework one should be able to interpret chain rule $\frac{dz}{dx}=\frac{dz}{dy}\frac{dy}{dx}$ literally. Furthermore one should be able to add relations $A=B$ and $C=D$ to obtain $A+C=B+D$. The framework should be powerful enough to allow solving a differential equation $\frac{dy}{dx}=g(x)h(y)$ literally by separation... $\endgroup$ Commented Aug 12, 2014 at 9:26
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    $\begingroup$ ... of variables. Also the center of curvature of a plane curve at a point should literally "=" the intersection of a pair of infinitely close normals (a definition right out of Cauchy, by the way). Infinitesimal calculations in Gauss and Riemann should be literally true, e.g., formula for curvature in terms of second order differentials. As you may know, Lie's original approach to Lie algebras was by means of infinitesimal displacements in the Lie group. This can be done by means of elementary nilpotent infinitesimals (this does not require Lawvere's framework), but it should fit into such... $\endgroup$ Commented Aug 12, 2014 at 9:30
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    $\begingroup$ @MikhailKatz: I am posting this little historical morsel here since (0) I just stumbled upon this fact, (1) think you might both be interested in it and perhaps even you might perhaps never heard of it (though I only searched some of your papers for the keyword "Gordan", and (2) this thread is the most fitting 'venue' I found within the five minutes I allowed myself to search for such a 'venue': Paul Gordan, which via the Hilbert-basis-theorem-story seems to have come to be seen as a proponent of constructive, calculational precision, in his doctoral disputatio [...] $\endgroup$ Commented Sep 13, 2017 at 19:59
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    $\begingroup$ [...] on March 1, 1862 at University of Berlin, had to debate, in Latin, a questio disputata which was essentially saying that the use of infinitesimals is no less precise than the use of limits. Max Noether gives the question on p. 3 of this obituary for Gordan in Mathematische Annalen 75. Literally, what Gordan had to defend says, translated to English: "The method of the infinitely small is, I claim, no less precise than the method of limits." $\endgroup$ Commented Sep 13, 2017 at 20:04

2 Answers 2

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You seem to think that synthetic differential geometry only handles squarenil infinitesimals, but this is not so. The generalized Kock-Lawvere axiom allows us to work with infinitesimals of any order, and actually much more than that. So as far as I can tell synthetic differential geometry captures all the essential features of Leibniz's infinitesimals, and more, while working with a completely standard notion of equality. This is important because for algebraic manipulations we want a notion of equality that allows us to replace equals for equals.

It might be helpful to find out what sources you are using to find out about synthetic differential geometry. Already Bell's A Primer of Infinitesimal Analysis, the most basic text on SDG, talks about infinitesimals of order $n$.

Supplemental: since the discussion below talks about the chain rule, here is its derivation in SDG, for reference, where $dx$ is a square nilpotent infinitesimal: $$g(f(x + dx)) = g(f(x) + f'(x) \cdot dx) = g(f(x)) + g'(f(x)) \cdot f'(x) \cdot dx,$$ We just used twice the principle of Microaffinity. We can now directly read off $$(g \circ f)'(x) = g'(f(x)) \cdot f'(x).$$ I do not see how this derivation would be more straightforward if we were allowed to magically divide by infinitesimals.

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    $\begingroup$ Andrej, thanks for your answer. Leibniz, though, divided by his infinitesimals. The simplest test case is the chain rule: it seems to me it would be awkward to write it down if one couldn't use ratios. I realize there are drawbacks to tampering with equality, and the price to pay may require limiting the class of objects one works with. But Leibnizian functions were all definable :-) $\endgroup$ Commented Aug 11, 2014 at 10:33
  • $\begingroup$ Have you read SDG treatments of the chain rule? I don't find them particularly impractical. It's not clear to me what you'd like. SDG recovers all the essential ideas at a price (intuitionistic logic, can't divide by an infinitesimal). Are you looking for a better price, or are you hoping there is free lunch? $\endgroup$ Commented Aug 11, 2014 at 11:01
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    $\begingroup$ Bell's book does not really treat chain rule but gives it as an exercise on page 69. The derivatives are treated as single symbols rather than ratios. This is of course fine mathematically but it does not salvage Leibnizian formalism. $\endgroup$ Commented Aug 11, 2014 at 11:13
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    $\begingroup$ It is not at all clear to me what you mean by "salvage Leibnizian formalism". SDG keeps all the essentials of it, so it salvages it. If you worry about little details such as "we do not divide but rather cancel on both sides" then yes, SDG does not "salvage" Leibniz's formalism. $\endgroup$ Commented Aug 11, 2014 at 11:17
  • $\begingroup$ With regards to the chain rule of derivatives, in Bell's primer it is called "composition rule" and is treated in detail on page 28, end of section 2.1. Or are we talking about two different things? I cannot find any exercises on page 69. I am looking at ISBN 0 521 62401 0 (hardback), printed in 1998. $\endgroup$ Commented Aug 11, 2014 at 11:19
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This is a little long for a comment, but not really an answer. One way I think you could argue that Leibniz "got it wrong" is his treatment of higher order differentials. As far as I know he would write

$$ d^2f = f''(x)dx^2 $$

When in fact we should be writing something like

$$ d^2f = f''(x)(dx)^2+f'(x)d^2x $$

This new formula satisfies the chain rule, and so gives an invariant definition on manifolds. It is related to Andrej Bauer's answer as well: This is related to the Weil Algebra $W = \mathbb{R}[x,y]/(x^2,y^2)$. $M^{Spec(W)}$ is the tangent bundle to the tangent bundle of $M$, and the above formula (basically) gives the second derivative as a map from that second order tangent bundle (in the case $M = \mathbb{R}$). For a reference developing this kind of stuff without reference to intuitionistic logic, you might want to check out Kolar, Michor, and Slovak's "Natural Operations in Differential Geometry".

Also, for first order differentials, what is wrong in your opinion with differential forms?

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  • $\begingroup$ "as far as you know" is not correct. Leibniz knew perfectly well how to work with second differentials. See for example link.springer.com/article/10.1007/BF00327456 $\endgroup$ Commented Aug 11, 2014 at 13:17
  • $\begingroup$ Differential forms are fine, but if you consult Spivak's book on differential geometry you will find him struggling to translate infinitesimal arguments in Gauss and Riemann into modern terminology and usually comes up with a more cumbersome result. $\endgroup$ Commented Aug 11, 2014 at 13:26
  • $\begingroup$ I finally got a chance to look up Kolar et al. This seems to focus on naturality in differential geometry and it is hard to see how this relates to developing a framework for infinitesimals. Perhaps @Peter Michor could comment? $\endgroup$ Commented Aug 12, 2014 at 12:11
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    $\begingroup$ A Weil algebra is an $\mathbb{R}$ algebra of the form $\mathbb{R} \oplus N$ with $N$ a nilpotent ideal. This ideal gives the "form" for the "infinitesimals" you want to consider. A major point in the book is that to each such algebra, there is a product preserving functor from the category of manifolds to the category of bundles over manifolds. For example, the tangent bundle corresponds to $\mathbb{R}[x]/(x^2)$, the tangent bundle to the tangent bundle to $\mathbb{R}[x,y]/(x^2,y^2)$, the space of $k$-jets to $\mathbb{R}[x]/(x^k)$. Essentially the algebra is keeping track of $\endgroup$ Commented Aug 12, 2014 at 13:40
  • $\begingroup$ the algebra of infinitesimals in your manifold. $\endgroup$ Commented Aug 12, 2014 at 13:40

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