How to explain to an engineer what algebraic geometry is?

Question

This question is similar to this one in that I'm asking about how to introduce a mathematical research topic or activity to a non-mathematician: in this case algebraic geometry, intended as the most classical complex algebraic geometry for simplicity.

Of course some of the difficulties of the present question are a subset of those of the linked question. But I think I want to be more precise here, about what's the pedagogical/heuristic obstacle I want to bypass/remove/etc, which is, after all, a detail.

So, say the engineer is happy with the start

"Algebraic geometry is the study of solutions of systems of polynomial equations in several variables..."

An engineer certainly understands this.

"...with complex coefficients..."

Here she's starting to feel a bit perplexed: why complex numbers and not just reals? But she can feel comfortable again once you tell her it's because you want to have available all the geometry there is, without hiding anything - she can think of roots of one-variable polynomials: in $(x-1)(x^2+1)$ the real solutions are not all there is etcetera.

Happy? Not happy. Because the engineer will inevitably be lead to think that what algebraic geometry consists of is fiddling around with huge systems of polynomial equations trying to actually find its solutions by hand (or by a computer), using maybe tricks that are essentially a sophisticated version of high school concepts like Ruffini's theorem, polynomial division, and various other tricks to explicitly solve systems that are explicitly solvable as you were taught in high school.

Question. How to properly convey that algebraic geometry mostly (yeah, I know, there are also computational aspects but I would contend that the bulk of the area is not about them) doesn't care at all of actually finding the solutions, and that algebraic geometers rarely find themselves busy with manipulating huge polynomial systems, let alone solving them? In other words, how to explain that AG is the study of intrinsic properties of objects described by polynomial systems, without seeming too abstract and far away?

Also, how would you convey that AG's objects are only locally described (or rather, in the light of the previous point, describable) by those polynomial systems in several variables?

Answer 1

Perhaps you're going about this the wrong way. Instead of trying to describe what most algebraic geometers do today, try to describe a problem, or set of problems, that is reasonably concrete and accessible, and go from there. I think there are plenty of topics to choose from. You've already mentioned solving systems of equations. Even if most algebraic geometers don't try to find solutions, they certainly want to know when solutions exist (Nullstellensatz) and how many parameters are required (dimension). Another historically important motivation is the study of elliptic/abelian integrals: you can go from addition laws for integrals to addition laws on the curve/Jacobian...

Answer 2

Abhyankar's book Algebraic Geometry for Scientists and Engineers doesn't give a short answer, but many long ones, with explicit examples of determining the geometric nature of the solutions of algebraic equations.

Answer 3

In conversations like this, I usually lead with a concrete example of a hard problem. Complete intersections seem to work well: Observe that in general, two surfaces in three-space meet in a curve, and then ask whether, given an (algebraically defined) curve, it's always the intersection of two (algebraically defined) surfaces. How do you recognize those that are and those that aren't? This gives you a chance to talk about the value of bringing both geometric intuition and algebraic computations to the table.

Now generalize to higher dimensions. Now (if they seem to want more) you can talk about subtleties like the distinction between a true complete intersection and a set-theoretic complete intersection. Or give a sequence of increasingly challenging specific cases. Et cetera.

I've also --- though this is sort of cheating --- used the example of classifying vector bundles. This is easy to explain in the topological case: You've got, say, a circle and you want to attach a line at every point in a continuous way. You can make a cylinder, or you can make a Mobius strip. What else can you make? When do you want to consider two of these things "the same"? Now observe that the answers to these questions depend partly on the rules for how you're going to build your objects in the first place and the rules for when you consider two to be the same. If the rules are that everything has to be continuous, you're doing topology; if the rules are that everything has to be algebraic, you're doing algebraic geometry. Mention Quillen-Suslin: If the base space is itself a vector space, it's pretty easy to see that all vector bundles of a given rank are topologically equivalent, but quite hard to see the same thing in the algebraic case. Et cetera.

Answer 4

This is along the lines suggested by @DonuArapura: "describe a problem [...] that is reasonably concrete and accessible, and go from there."

Here is a problem an engineer would appreciate: Which bent pieces of wire can pass through a pinhole in a plane via rigid motions? Such curves have been called threadable curves.¹

Deciding whether a given planar algebraic curve $C$ is threadable depends on the number of bitangents. For a curve of degree $d$, this number is $O(d^4)$, a result of Schubert. See the MO question, Number of bitangents to connected algebraic curve.

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¹J.O'Rourke and Emmely Rogers, "Threadable curves," Proc. 30th Canad. Conf. Comput. Geom., Aug 2018, 328—333. (arXiv abstract).

Answer 5

If you're just trying to communicate what algebraic geometry is, without trying to convince the engineer that it's worth studying, then one simple starting point is to recall the classification of conic sections (ellipse, parabola, hyperbola) and say that one thing algebraic geometers try to do is classify the different possibilities that can occur with larger degree/number of equations/number of variables.

Notice, by the way, that there is a similar "disconnect" between engineers and mathematicians when it comes to PDEs. Engineers often just want to solve PDEs. Mathematicians are interested in solving PDEs too but are also interested in other questions. The concept of wanting to understand the qualitative features of a solution may be easier to explain in the context of PDEs, and then you can say that the situation in algebraic geometry is analogous.

Answer 6

I have had this conversation a few times. I warm them up by introducing the concept of abstract classification of objects: I explain how a mechanical arm which can rotate in a circle, and which has another mechanical arm at the end which can also rotate in a circle, is in some sense the same thing as a torus, despite their apparent differences. The question is, then, how we can classify things that are defined by constraints that appear to be different but in some underlying way are the same?

I then explain that in A. G. the constraints are typically polynomials and this allows the use of ideas and points of view that completely general constraints don't.

Answer 7

I think a good explanation should give an idea of how algebraic geometry can make precise the idea of a generic point on an irreducible variety.

One example I might try is Gerstenhaber's Theorem, that the variety of pairs of commuting complex matrices is irreducible; a 'generic pair' is two commuting diagonalizable matrices with distinct eigenvalues. I think one could give a good idea of this without (explicitly) using the group action or topology. The engineer might already know that rotations of $\mathbb{R}^3$ are diagonalizable over $\mathbb{C}$: if so I'd explain that these matrices are still not quite 'generic' because of the $1$ eigenvalue.

If they want more, I'd go on to say that for larger number of matrices, the variety is usually reducible, so there is no reasonable idea of a 'generic tuple'.

Answer 8

One of the things engineers are very familiar with is integration. "What kind of substitution should I make to explicitly find antiderivative?" is a very natural question. And, you know, Algebraic Geometry sometimes helps with that.

My favorite answer for this kind of question is to start with an integral of something in $\mathbb{R}(\sin(t),\cos(t))$. Then I draw a unit circle and stereographic projection to express the point $(\sin(t),\cos(t))$ in terms of $\tan(t/2)$, leading to the substitution that solves the antiderivative. And a brief explanation that you make connections between the algebraic expressions and the geometric objects that give you an insight into the algebraic object.

Answer 9

I would start by showing them to how find rational points on a conic. If you have a rational point then you can draw lines and find more. They will be comfortable with the geometric aspect and then you could stress the “rationality” part of the construction ie “look, the slope and the y-intercept are rational, so if one point of intersection is rational then the other one is too”

This construction has enough but not overwhelmingly many logical steps which the engineer will be able to verify should they want to E.g. rationality, getting all of them, the necessity of finding a point to start off the process.

Then you could go up to a quadratic extension (!) to “see what happens” and let them play around

I like this example because the algebra and the geometry are both at the level your audience should be comfortable with.

I wouldn’t even go to elliptic curves & the group law, in my experience it takes more mathematical exposure to appreciate those phenomena

Answer 10

There is an answer by D. Mumford to biologists, valid also for engineers: Can one explain schemes to biologists, blog post (2014) (link).

Answer 11

There are several good answers already so I cannot hope to add much. That said, another approach to reach engineers could be by the familiar subject of linear algebra, in particular, solving systems of linear equations as a special case of solving systems of polynomial equations.

Perhaps start with a quick review of linear systems and why we must have either zero, one or an infinite number of solutions. Show the usual pictures in $\mathbb{R}^2$ and $\mathbb{R}^3$ of lines and planes intersecting, including two planes intersecting in a line and a plane and a line intersecting in a point. Use this to lead into discussing solution sets of polynomial equations (with pictures), their dimension, the Hilbert Nullstellensatz and Bezout's Theorem, for instance.

Answer 12

I would just say that very roughly speaking it's a subject where you are doing geometry and thinking about geometry, but you write about it formally like it is algebra and you use algebra (which can be fairly esoteric). I have an example from page 119 of 'Introduction to Homological Algebra' by Weibel (the application is originally due to Hartshorne).

Let $R = \mathbb{C}[x_1, x_2, y_1, y_2]$, $P=(x_1,x_2)R$, $Q=(y_1,y_2)R$, and $I = P \: \cap \: Q $.

Since $P$, $Q$, and $m = P+Q = (x_1, x_2, y_1, y_2)R$ are generated by regular sequences, it can be shown that the outside terms in the Mayer-Vietoris sequence

$H^3_P(R) \bigoplus H^3_Q (R) \rightarrow H_I^3(R) \rightarrow H^4_m(R) \rightarrow H^4_P(R) \bigoplus H^4_Q (R) $

vanish, which tells us that $H^3_I(R) \simeq H^4_m(R) \neq 0$. (The cohomology is local cohomology here).

Esoteric and abstract to the typical engineer? Perhaps, but leaving aside the words and jargon, this implies that the union of two planes in $\mathbb{C}^4$ which meet at a point cannot be described as the solutions of only two equations $f_1 = f_2 =0$, which is a concrete geometric fact.