# Why can algebraic geometry be applied into theoretical physics?

It is to be said at the outset that I do not have much familiarity with physics beyond what is in a semi-popular book; say, the Feynman Lectures Vol 1 and 2.

As I progressed in math graduate school specializing in number theory and algebraic geometry, it was astounding to discover a certain class of researchers who were doing very serious and nontrivial cutting-edge stuff connecting algebraic geometry and mathematical physics.

At the risk of appearing like a completely naive person, I must confess that I am completely baffled at how can this be possible. From one perspective, algebraic geometry is at its basics about studying solutions of algebraic equations which also have certain geometric aspects. Certainly way of looking at is not enough. How to philosophically explain this connection?

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I know this comment is not very helpful, but so far everything from algebra, geometry and algebraic geometry has been applied to theoretical physics. – Dietrich Burde Oct 15 '13 at 13:53
Your last-but-one sentence looks like it has typos, and it seems to be making an important point, so it might be worth fixing it. – bubba Oct 15 '13 at 13:56
The equations represent geometric things -- curves and surfaces. Many problems in the physical world have a geometric aspect to them. Algebraic geometry is used in my field, too (computer-aided design and manufacturing). We use it in our studies of the shapes of manufactured objects. It's century old algebraic geometry, but it's still algebraic geometry. – bubba Oct 15 '13 at 14:03
Certainly, learning why vector bundles and moduli spaces are useful in physics would be a good start, though I'm afraid I don't know enough about physics to give you good references for that. – Jason Polak Oct 15 '13 at 16:42

To start with, we can look at many different possible spacetimes, and it turns out that looking at certain algebraic varieties (or more generally, symplectic manifolds) is very fruitful. There are many reasons for this, but I'll mention just one: it turns out that the number of algebraic curves in these varieties is something which appears in physics.

Here is an attempt at explaining one piece of this (warning: I only understand the mathematical side, so what I say about physics may be completely wrong. If so, someone who actually knows physics please correct me. Also, note that I do not know how to explain this in a way that is not at least somewhat handwavy, though I will try to not say anything false.)

In classical mechanics, there is the Lagrangian formulation. What this says is that the path of an object must (at least locally) minimize a quantity called the "action" of the path. The calculus of variations lets us prove that this is equivalent to Newton's laws.

Now when you go to quantum mechanics (and in particular quantum field theory, which is where this is really useful), there is the Feynman path integral formulation: A particle may be treated as taking every path it could possibly take. However, most of these paths "cancel out" , and the only ones we actually end up seeing are the ones that are critical points of the action. What this means is that we can evaluate certain things in quantum field theory by integrating over all paths, as most of the paths will cancel out. (This "cancelling out" is something that can happen in quantum mechanics; to give an analogue, think of wave interference.)

Now if you we start working with strings, then as a string moves, we get a 2d-surface instead of a path, called the worldsheet of a string. As with the path integral, we now get integrals over these 2d-surfaces. It now turns out that the surfaces which don't cancel must be pseudoholomorphic curve, and when our spacetime is an algebraic variety, this pseudoholomorphic curve corresponds to an algebraic curve.

So counts of algebraic curves in complex varieties are something which appear in string theory.

Now of course there is no rigorous formulation of infinite-dimensional integration in mathematics. Therefore, many of the results obtainable via the Feynman path integral are not obtainable via traditional techniques, which is why mathematicians are interested in this relation.

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