# What is the geometry in algebraic geometry?

Coming from a physics background, my understanding of geometry (in a very generic sense) is that it involves taking a space and adding some extra structure to it. The extra structure takes some local data about the space as its input and outputs answers to local or global questions about the space + structure. We can use it to probe either the structure itself or the underlying space it lives on. For example, we can take a smooth manifold and add a Riemannian metric and a connection, and then we can ask about distances between points, curvature, geodesics, etc. In symplectic geometry, we take an even-dimensional manifold and add a symplectic form, and then we can ask about... well, honestly, I don't know. But I'm sure there is interesting stuff you can ask.

Knowing very little about algebraic geometry, I am wondering what the "geometry" part is. I am assuming that the spaces in this case are algebraic varieties, but what is the extra structure that gets added? What sorts of questions can we answer with this extra structure that we couldn't answer without it?

I have to guess that this is a little more complicated than just taking a manifold and adding a metric, otherwise I would expect to be able to find this explained in a relatively straightforward way somewhere. If it turns out the answer is "it's hard to explain, and you just need to read an algebraic geometry text," then that's fine. In that case, it would be interesting to try to get a sense of why it's more complicated. (I have a guess, which is that varieties tend to be a lot less tame than manifolds, so you have to jump through more technical hoops to tack on extra stuff to them, but that's pure speculation.)

• Think: conic sections. Commented Oct 26, 2022 at 19:06
• Great question, I hope to see some really well-informed answer. My not-so-well informed answer is that the extra structure is the sheaf of rational functions defined on open sets. Rational functions have a lot more rigidity than smooth ones, so there's more regularity than in the smooth category. Commented Oct 26, 2022 at 19:08
• Your understanding of geometry is waaay to broad. Pretty much everything is a geometry under it. Algebraic geometry in the most generality deals with polynomials and its roots. It is not necessarily some additional structure added to something. This generalizes (in some sense) classical geometry, where polynomial equations over reals play important role as well. Commented Oct 26, 2022 at 19:09
• Algebraic geometry is mainly studying algebraic varieties (and their generalizations) per se. So here, algebraic varieties are analogous to Riemannian manifolds and symplectic manifolds in your examples, rather than underlying smooth manifolds. The "geometry" of smooth manifolds is differential topology, roughly speaking, and the "geometry" of topological spaces is topology. Commented Oct 26, 2022 at 19:13
• Schemes (the object of algebraic geometry) are topological spaces with added structure. The wikipedia page is a bit dense, but you can start there en.wikipedia.org/wiki/Scheme_(mathematics) Commented Oct 26, 2022 at 19:14

This is a big complicated question and many different kinds of answers could be given at many different levels of sophistication. The very short answer is that the geometry in algebraic geometry comes from considering only polynomial functions as the meaningful functions. Here is essentially the simplest nontrivial example I know of:

Consider the intersection of the unit circle $$\{ x^2 + y^2 = 1 \}$$ with a vertical line $$\{ x = c \}$$, for different values of the parameter $$c$$. If $$-1 < c < 1$$ we get two intersection points. If $$c > 1$$ or $$c < -1$$ we get no (real) intersection points. But something special happens at $$c = \pm 1$$: in this case the vertical lines $$x = \pm 1$$ are tangent to the circle. This tangency is invisible if we just consider the "set-theoretic intersection" of the circle and the line, which consists of a single point; for various reasons (e.g. to make Bezout's theorem true) we'd like a way to formalize the intuition that this intersection has "multiplicity two" in some sense, and so is geometrically more interesting than just a single point.

This can be done by taking what is called the scheme-theoretic intersection. This is a complicated name for a simple idea: instead of asking directly what the intersection is, we ask what the ring of polynomial functions on the intersection is. The ring of polynomial functions on the unit circle is the quotient ring $$\mathbb{R}[x, y]/(x^2 + y^2 - 1)$$, while the ring of polynomial functions on the vertical line is the quotient ring $$\mathbb{R}[x, y]/(x - c) \cong \mathbb{R}[y]$$. It turns out that the ring of polynomial functions on the intersection is the quotient by both of the defining polynomials, which gives, say at $$x = 1$$ to be concrete,

$$\mathbb{R}[x, y]/(x^2 - y^2 - 1, x - 1) \cong \mathbb{R}[y]/y^2.$$

This is a funny ring: it has a nontrivial nilpotent! That nilpotent $$y$$ is exactly telling us the sense in which the intersection has "multiplicity two"; it's saying that a function on the scheme-theoretic intersection records not only its value at the intersection point but its derivative with respect to tangent vectors at the intersection point, reflecting the geometric fact that the unit circle and the line are tangent and so share a tangent space. In other words it is saying, roughly speaking, that the intersection is "two points infinitesimally close together, connected by an infinitesimally short vector."

Adding nilpotents to geometry takes some getting used to but it turns out to be very useful; among other things it is possible to define tangent spaces in algebraic geometry this way (Zariski tangent spaces), hence to define Lie algebras of algebraic groups in a purely algebraic way.

So, this is one story you can tell about what kind of geometry algebraic geometry captures, and there are many others, for example the rich story of arithmetic geometry and its applications to number theory. It's difficult to say anything remotely complete here because algebraic geometry is absurdly general and the sorts of geometry it is capable of capturing veer off in wildly different directions depending on what you're interested in.

• @The_Sympathizer: it's just the thing I wrote down; there's a system of equations defining the intersection and it's given by $x^2 + y^2 = 1$ and $x = c$. From these you can get the single equation $y^2 = 1 - c^2$. Now the crucial point is that when $c = \pm 1$ so that the equation reads $y^2 = 0$, you do not conclude from here that $y = 0$, because in algebraic geometry $y$ could be a nontrivial nilpotent. In other words we really take seriously that $y^2 = 0$ is a different equation from $y = 0$ even though they have the same solutions over, say, the real numbers. Commented Oct 27, 2022 at 7:22
• You can consider the jump to taking nilpotents seriously as being analogous to the jump to complex numbers, where we take seriously that $y^2 = -1$ can have solutions; here we take seriously that $y^2 = 0$ can have nonzero solutions! Commented Oct 27, 2022 at 7:22
• @The_Sympathizerwell, the scheme has a structure added to the set of points (as well as a few "extra" points than those you can draw). This structure is (simplyfying) the ring(s) of polynomial functions. So you can have a point with just trivial functions or a point with more functions and they are very different schemes even when they're just one point. Naturally this also holds for higher-dimensional varieties. Commented Oct 27, 2022 at 19:40
• @The_Sympathizer: so I glossed over this but we do this by changing the meaning of "the intersection." There are a few different ways to say this depending on what level of sophistication you want to work at; above I did this in the simplest way, by saying "impose both equations $x^2 + y^2 = 1$ and $x = c$, but don't simplify $y^2 = 0$ to $y = 0$." Formally you take the quotient by the ideal generated by these two polynomials, or take the tensor product of their coordinate rings. Commented Oct 27, 2022 at 21:30
• @The_Sympathizer You can think of a scheme as more than just “a solution set to a system of polynomial equations”. This gives the underlying set of the scheme, but the scheme has the additional structure of a choice of system of defining equations. That’s where the extra structure comes in. Commented Oct 28, 2022 at 15:03

A classical manifold is a space that locally looks like $$\mathbb{R}^n$$; or, via results like the Whitney embedding theorem, a suitably nice subspace of some $$\mathbb{R}^N$$. If "looks like" involves some notion of smoothness, for example, then we can expand into differential geometry and talk about constructions like tangent spaces and differential forms. If we stick to just continuity, then we can still work with some constructions like homology and cohomology (just not, say, de Rham cohomology), and we can deal with more pathological spaces and functions between them.

A natural question to ask, then, is what's so special about $$\mathbb{R}^n$$? We can consider spaces that locally look like an arbitrary Banach space, for example. (I don't think this is a particularly popular approach, at least, at the undergrad/early grad school level, but Abraham, Marsden, and Ratiu works in this category.) The starting point of algebraic geometry is wanting to deal with spaces over an arbitrary commutative ring. It's not clear how continuity or smoothness should map over to this case, but at the very least polynomials make sense over an arbitrary ring, and we can look at the space like $$V(f) = \{x\in k^n:\, f(x) = 0\}$$ for a polynomial $$f\in k[X_1, \dots, X_n]$$. But that's not exactly what we want either; for the important case of $$k$$ finite, for example, $$V(f)$$ is just a finite collection of points.

The analogy that turns out to work is going in the opposite direction, and trying to generalize the idea of functions on a manifold. To that end, algebraic geometry works with locally ringed spaces, which are pairs $$(X, \mathcal{O}_X)$$ with $$X$$ a topological space and $$\mathcal{O}_X$$ a sheaf of rings on $$X$$ satisfying properties roughly analogous to what you'd expect for, say, smooth functions on a manifold. In rough terms, what you wind up with is a space that locally looks like the spectrum of a commutative ring--- but unlike the case of real manifolds, that ring can vary along the space. That's admittedly a vague analogy, and it takes a lot of technical results to even talk about the resulting object. But if you're familiar with vector bundles, for example, then consider Swan's Theorem: For a smooth, connected, closed manifold $$X$$, the sections functor $$\Gamma(\cdot)$$ gives an equivalence between vector bundles over $$X$$ and f.g., projective modules over the ring $$C^\infty(X)$$.

So, what makes this algebraic thing we've constructed look geometric? Smoothness doesn't make sense outside of $$\mathbb{R}^n$$, but if we're working with polynomials, they have a formal derivative that allows us to do roughly the same thing. More generally, a local ring $$(R, \mathfrak{m})$$ has a cotangent space $$\mathfrak{m}/\mathfrak{m}^2$$ that's roughly analogous to the cotangent space of a manifold; and with a bit of work, we can get something that at least has some of the formal properties one wants for a tangent or cotangent space. Even though the topology we're working with turns out to be much more complicated than the case of manifolds (the Zariski topology, for example, is generally non-Hausdorff), we still have a notion of cohomology (the simplest being Cech cohomology with a sheaf). There's a massive jump in abstraction and technical requirements compared with the more geometric case, but algebraic geometry turns out to be the right extension of more familiar geometry when dealing with things such as, say, number fields.

• Thanks for this very nice answer. It pains me to have to choose which of the answers to accept, but since I do have to choose, I've decided to give the checkmark to Qiaochu Yuan. I found both answers very helpful, and if I could give out two checkmarks, I would.
– d_b
Commented Oct 28, 2022 at 18:01
• @d_b: It's no problem whatsoever (and I upvoted his too). Commented Oct 28, 2022 at 18:39
• What do you mean by "but unlike the case of real manifolds, that ring can vary along the space"? For manifolds, the structure sheaf is not constant, either. Maybe you want to say that a (paracompact) manifold is somehow completely determined by its ring of global sections? It is a nontrivial theorem, and maps between mamifolds don't seem to be the same as ring maps. Commented Nov 2, 2022 at 11:42
• @Yai0Phah: I just mean that a (say, closed and smooth) manifold is modeled on $\mathbb{R}^n$ for some fixed $n$ by, e.g., invariance of domain, whereas a scheme is modeled on affine schemes that can vary. But "modeled" is in fact a pretty loose term here, and one of the leaps of abstraction in going from classical geometry to algebraic geometry is focusing on the structure sheaf. Commented Nov 2, 2022 at 12:34
• OK, I see. What you said is similar to saying that the stalks are isomorphic (but not canonically). One can avoid structure sheaves in the definition — the functor of points is another framework to say "gluing from affine schemes", which also incorporates classical objects such as manifolds and simplicial sets, say. Commented Nov 2, 2022 at 12:52

there are some great answers already here and I'm just adding a few (very informal) thoughts myself:

Take an affine algebraic scheme $$X=\mathbf{Spec}(A)$$ for some $$\mathbf{C}$$-algebra $$A$$ (i.e. polynomials modulo some relations). This could be an affine line, plane or maybe the zeros of a polynomial equation inside of an affine space (i.e. a nice curve in the plane, a surface embedded in three-dimensional space etc.).

Then algebraic geometry tells you that the ring $$\mathcal{O}_X$$ of "holomorphic functions" on $$X$$ is just $$A$$. In differential/complex geometry, you'd see $$X$$ as a manifold and $$\mathcal{O}_X$$ is this ring of $$C^\infty$$ functions on $$X$$.
Now the cool part of algebraic geometry is that starting with any abstract unitary commutative ring $$A$$, you get a topological space $$X=\mathbf{Spec}(A)$$ whose ring of regular functions is the $$A$$ you started with. I.e., elements of an abstract ring $$A$$ can be seen as regular functions on a topological space $$\mathbf{Spec}(A)$$ (manifold whatever...).

This is even an equivalence of categories $$\{ \text{affine schemes }\}^{op} \leftrightarrow \{\text{commutative }\text{rings with 1}\}$$.

When this was introduced 80(?) years ago, this created a big echo - in this approach, a function $$f$$ (element of a ring $$A$$) is not uniquely defined by all of its values $$f(x), x \in X$$. For example, take $$A=\mathbf{C}[t]/(t^2)$$ then the function $$t$$ is zero at the only point $$0 \in \mathbf{Spec}(A)$$ but $$t\neq 0$$ is not the zero function.

However, if you add nice properties to $$X$$ (reduced, proper, separated etc) you get at least in the complex case a space that is categorically equivalent to a complex variety (see Serres GAGA theorems).

since you were asking about geometry: This funny space $$X=\mathbf{Spec}(\mathbf{C}[t]/(t^2)$$ has an interesting application. Let me just use a blackbox that this space consists of one point (because $$A$$ has only one prime ideal), but one can show that functions $$f$$ on $$X$$ look like $$f(t)=a+bt$$ for $$a,b\in \mathbf{C}$$ so can be seen as $$1^{st}$$ order Taylor approximations to polynomial functions.
Now If you want to understand the tangent space at a point $$y \in Y$$ of some other space $$Y$$, you can look at maps $$\phi: X \rightarrow Y$$ mapping the unique point of $$X$$ to $$y$$. The fact that $$X$$ keeps track of tangent directions allows you to see $$\phi$$ as $$1^{st}$$ order deformations of $$y \in Y$$, i.e. is a choice of tangent vector of $$\mathbf{T}_yY$$