Mastering Algebraic Geometry Interview Questions: The Ultimate Guide for Acing Your Next Interview

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Perhaps youre going about this the wrong way. If you want to explain what most algebraic geometers do today, try to explain a problem or set of problems that are pretty clear and easy to understand. I think there are plenty of topics to choose from. Youve already mentioned solving systems of equations. Some algebraic geometers don’t even try to find solutions; they just want to know when there are solutions (Nullstellensatz) and how many parameters are needed. Another important reason in history is the study of elliptic and abelian integrals: you can go from adding laws for integrals to adding laws on curves (Jacobian).

Abhyankars book Algebraic Geometry for Scientists and Engineers doesnt give a short answer, but many long ones, with explicit examples of determining the geometric nature of the solutions of algebraic equations.

In conversations like this, I usually lead with a concrete example of a hard problem. One way to check this is to notice that two surfaces in three-space usually meet in a curve. Then, ask yourself if, given an algebraically defined curve, it is always the intersection of two algebraically defined surfaces. How do you tell the difference between those that are and those that aren’t? This gives you a chance to talk about how important it is to use both geometric intuition and algebraic calculations.

Now generalize to higher dimensions. Now, if they still want more, you can get into more specifics, like the difference between a real complete intersection and a set-theoretic complete intersection. Or give a sequence of increasingly challenging specific cases. Et cetera.

Ive also — though this is sort of cheating — used the example of classifying vector bundles. In the case of topology, this is easy to explain: let’s say you have a circle and you want to connect a line to every point on the circle in a straight line. You can make a cylinder, or you can make a Mobius strip. Do you know when two of these things are “the same”? What else can you make? Keep in mind that the answers to these questions depend on the rules you set for how you’ll build your objects and when you think two of them are “the same.” If you want to study topology, everything has to be continuous. If you want to study algebraic geometry, everything has to be continuous. It’s pretty easy to see that all vector bundles of a certain rank are topologically equivalent if the base space is also a vector space. But it’s not so easy to see this in the algebraic case. Et cetera.

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

An engineer would enjoy this puzzle: what kinds of bent wire can go through a pinhole in a plane without moving at all? These kinds of curves are 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.

1J.ORourke and Emmely Rogers, “Threadable curves,” Proc. 30th Canad. Conf. Comput. Geom., Aug 2018, 328—333. (arXiv abstract).

If you only want to explain what algebraic geometry is and not try to persuade the engineer that it’s important to study, a good place to start is to talk about the different types of conic sections (ellipse, parabola, and hyperbola) and how algebraic geometers try to group the different outcomes that can happen when there are more degrees, equations, or variables.

Another “disconnect” between engineers and mathematicians is that they don’t understand PDEs the same way you do. Engineers often just want to solve PDEs. Mathematicians are interested in solving PDEs too but are also interested in other questions. It might be easier to explain the idea of wanting to understand the qualitative features of a solution in terms of PDEs. Then you can say that the situation in algebraic geometry is similar.

I have had this conversation a few times. I start by explaining the idea of abstract classification of objects. I show them how a mechanical arm that can turn in a circle and has another mechanical arm at the end that can also turn in a circle is, in a way, the same thing as a torus, even though they look different. That being said, the question is how to group things that are defined by rules that look different but are really the same?

I then explain that in A. G. Most of the time, the constraints are polynomials, which lets you use ideas and points of view that totally general constraints don’t allow.

I believe a good explanation should show how algebraic geometry can help us understand what a general point on an irreducible variety really means.

I could use Gerstenhaber’s Theorem as an example. It says that the number of pairs of commuting complex matrices is uncountable. A generic pair is two commuting diagonalizable matrices with different 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 $mathbb{R}^3$ can be diagonalized over $mathbb{C}$. If that’s the case, I would tell them that these matrices aren’t quite general because of the $1$ eigenvalue.

If they want to know more, I’d say that the variety diminishes as the number of matrices increases, which means there is no good reason to think of a generic tuple.

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 $mathbb{R}(sin(t),cos(t))$. To write the point $(sin(t),cos(t))$ in terms of $tan(t/2)$, I then use a unit circle and stereographic projection. This leads to the substitution that solves the antiderivative. And a short explanation of how you discover links between algebraic expressions and geometric objects that help you understand the algebraic object

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. Once they understand the geometry, you can stress the “rationality” part of the construction: “Look, both the slope and the y-intercept are rational, so if one point of intersection is rational, then the other one is too.”

This building has just the right number of logical steps for the engineer to be able to check them if they want to. g. rationality, getting all of them, the necessity of finding a point to start off the process.

Then you could let them play with a quadratic extension (!) to “see what happens.”

This is a good example because both the algebra and geometry are easy enough for your audience to understand.

That’s as far as I’d go with curves.

There are several good answers already so I cannot hope to add much. Still, another way to get engineers’ attention could be to use the well-known subject of linear algebra, especially the idea that solving systems of linear equations is similar to solving systems of polynomial equations.

Start by quickly going over linear systems and why there must be zero, one, or an infinite number of solutions. Show the usual pictures of lines and planes crossing in $mathbb{R}^2$ and $mathbb{R}^3$, such as two planes crossing in a line and a plane and a line crossing in a point. For example, you could use this to start a conversation about the size of solution sets for polynomial equations, the Hilbert Nullstellensatz, and Bezout’s Theorem.

I would say that it’s a subject where you do geometry and think about geometry, but you write about it in a way that sounds like algebra and use algebra, which can be pretty hard to understand. Here is an example from Weibel’s Introduction to Homological Algebra (page 119). Hartshorne first used this idea.

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 $.

Regular sequences make up $P$, $Q$, and $m = P Q = (x_1, x_2, y_1, y_2)R$. This means that the outside terms in the Mayer-Vietoris sequence are also regular sequences.

$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 $H^3_I(R) simeq H^4_m(R) eq 0$. (The cohomology is local cohomology here).

A normal engineer might find this hard to understand, but if you take away the words and jargon, it means that the union of two planes in $mathbb{C}^4$ that meet at a point can’t be described as the solutions of only two equations $f_1 = f_2 =0$. This is a real geometric fact.

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

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