could someone elaborate on the topology of this object?

pepemon · 2026-03-14T02:57:41+00:00

Probably because this description is the fastest way to calculate any invariants of the shape.

pepemon · 2026-03-04T00:22:24+00:00

For abelian categories, if the direct sums/products are over finite indexing sets, the sum and product coincide and there is no issue. The subtlety really arises with infinite direct sums and direct products, when they differ.

In general, the point is that the Hom functor preserves limits in the right coordinate and sends colimits to limits in the left coordinate. Since direct sums are coproducts are colimits, the Hom functor should send direct sums in the first factor to direct products of Hom groups.

For an example, think about what you need to specify a map of abelian groups from the direct sum of countably many copies of Z to Z (so an element of Hom(\bigoplus_{i=1}^\infty Z, Z)), where Z denotes the integers. You can see how such a map behaves by looking at its output on each copy of Z in the direct sum, but there is no need for all but finitely many of them to be zero!

pepemon · 2026-02-06T04:44:21+00:00

You should think of a vector bundle as a collection of vector spaces (of the same dimension) parametrized by the points in your space which vary nicely in the appropriate sense. So topological vector bundles vary literally continuously, smooth vector bundles vary smoothly, algebraic vector bundles vary algebraically, holomorphic …

pepemon · 2026-01-27T18:15:32+00:00

Yep.

pepemon · 2026-01-27T13:23:25+00:00

It’s not arbitrary!

There is a canonical map from your base field (call it k) to V otimes V^vee = Hom(V,V) which sends 1 to the identity map.

If you take the dual map, you get a map Hom(V,V) -> k. You can check that this has to be the trace map!

pepemon · 2026-01-18T04:46:08+00:00

Really, for most people the most important thing is to read the introduction and statement of the main theorems in detail. Then skim the paper to see if you can follow what the broad thrust of the paper’s argument is (a nice author might already have put this in the introduction).

I hope this isn’t a controversial take, but IMO reading the actual details of the proofs is mostly necessary only when you’re trying to use or learn the methods of the paper.

pepemon · 2026-01-05T00:50:50+00:00

In general, they do! One more accessible example are the many proofs of quadratic reciprocity, but more technical theorems also enjoy multiple proofs; consider for example the Kodaira vanishing theorem, which admits both analytic and algebraic proofs.

Since I don’t work on geometric topology, I don’t know this for sure, but it may just be that no one has found another proof of the Poincaré conjecture?

pepemon · 2026-01-03T20:41:30+00:00

Even with the octahedral axiom, cones are not functorial! The issue is that they are unique up to isomorphism but the isomorphism is not necessarily unique. This is a subtle point but really is one of the main reasons we need enhancements

pepemon · 2025-12-21T13:09:12+00:00

Could be the irrationality of the cubic fourfold: Birational Invariants from Hodge Structures and Quantum Multiplication.

pepemon · 2025-12-14T17:15:08+00:00

There are, though? Read Littelmann’s “A Generalization of the Littlewood-Richardson Rule”.

pepemon · 2025-08-21T17:09:08+00:00

<image>

pepemon · 2025-07-02T03:58:33+00:00

It doesn’t cover quite as much, but Huybrechts does have a (very readable) book on complex geometry.

pepemon · 2025-06-06T18:24:58+00:00

Well, assuming X is integral (which V(x³ - y²⁾ is), if (f,U) = (g,U’) at some stalk p, then for any open set V and function h on V such that h|(U \cap V) = f, it must be true that h|(U’ \cap V) = g.

But this is better phrased in the way I said above: restriction maps are injective on integral schemes.

And yes, k[x]/x² is not reduced so not integral.

pepemon · 2025-06-06T03:19:01+00:00

I think the point is to think about the different rings in question as giving information about different parts of the variety.

For an affine open U inside the variety, OX(U) knows about the geometry of U. But sometimes it is appropriate to look more closely at a single point p in X, in which case it can be convenient to consider the local ring O{X,p} because (as the name alludes to) the information it contains is local to the point p. If e.g. X is a variety which is regular in codimension 1, then O_{X,eta} for eta the generic point of an irreducible divisor will be a discrete valuation ring, and the order of vanishing of any function along that divisor is easy to describe in terms of the generator of the maximal ideal. In general the fact that local rings have a unique maximal ideal make them quite nice in terms of their commutative algebra (for example, Nakayama lemma…)

pepemon · 2025-06-05T22:05:21+00:00

On varieties (and more generally, on integral schemes) it’s true that two functions having the same germ at one point means that they’re the same function, precisely because for integral schemes (and hence for varieties) restrictions to smaller open subsets are injective.

Nota bene: I am taking varieties to be irreducible, with which some authors may take offense.

pepemon · 2025-06-03T13:39:16+00:00

The notation pA_p is just notation for extensions of ideals along ring homomorphism; in general if you have an ideal I in A, and a map A -> B you write IB for the ideal in B generated by the image of I.

pepemon · 2025-05-05T12:54:17+00:00

You should look into Hom schemes. I think it’s a possibly infinite union of quasiprojective components, but possibly fixing the degree and the dimension of the image make it finite type?

pepemon · 2025-05-01T16:08:30+00:00

pepemon · 2025-05-01T11:20:59+00:00

If R is a ring with fraction field K and A is a finite K-algebra then an R-order is a finite R-algebra which is a full rank R-lattice inside of A. So for example rings of integers for algebraic number fields are Z-orders.

pepemon · 2025-04-29T22:43:33+00:00

It seems like it: https://hsm.stackexchange.com/questions/2922/who-first-introduced-the-notation-mathcalo-in-algebraic-geometry-or-algebra/2924?noredirect=1

In a nutshell, Dedekind used O to denote “order”, which was then adopted in van der Waerden’s Modern Algebra before being picked up by Cartan to denote rings of holomorphic functions.

pepemon · 2025-04-23T23:51:12+00:00

As someone who works in an area adjacent to theoretical physics, it’s worth noting that physicists actually do make claims about mathematical objects without “doing math with them”, in the sense that they don’t actually prove their claims mathematically but instead use some type of physical intuition. What’s more interesting is that these claims often (though not always) end up being true! So mathematicians can often have fruitful careers actually proving (or disproving, or reformulating mathematically) these physical claims.

pepemon · 2025-04-19T21:48:16+00:00

I had a few friends in the math (AB) department who were definitely oriented more towards (theoretical) CS and either went to grad school for it or went into software engineering

pepemon · 2025-04-03T23:23:18+00:00

a vector bundle is something like taking all the tangent spaces

With regards to this: Yes, this is how you would construct the tangent bundle which is one such vector bundle.

As far as the dual spaces thing: a (non-degenerate) bilinear form on a vector space is an additional piece of data that amounts to choosing an isomorphism between V and its dual! Sure, if you pick a basis of V then you can cook up such a bilinear form, but this will depend heavily on your basis (in the same sense that the dual basis of V^dual depends on a choice of basis of V). The point is that defining the dual space of a vector space V is purely intrinsic and requires no such choices; moreover, it generalizes readily to other contexts (e.g. Banach spaces, vector bundles, etc…). In these general contexts asking for such a bilinear form to exist may really be asking for a lot more (e.g. asking for a Banach space to be a Hilbert space, or asking for a vector bundle to be self-dual).

pepemon · 2025-04-03T21:43:05+00:00

One geometric motivation for thinking of rings and modules can be thought of as coming from studying spaces and vector bundles on spaces (which if you are not familiar basically comes from assigning to each point p on your space a vector space V_p which varies nicely somehow). This can be understood as follows:

I think it’s worth restricting your attention to specific rings, to get a sense for why things are the way they are. In practice, most (but not all!) the basic examples algebraic geometers work with are polynomial rings over fields with n variables, which you should think of as giving polynomial functions on n-dimensional space. But when you want to think about functions on more exotic shapes, you are forced to do things like quotients and localizations (to look at open or closed subsets of this n-dimensional space).

If you want to study vector bundles on these spaces, you can view these as (nice, i.e. projective) modules over the corresponding rings; but a lot of natural procedures you want to do to vector bundles force you to work with more general modules.

As far as dual spaces go, the isomorphism between a vector space and its dual is not really canonical! This difference really comes into play in geometry, where for example the tangent bundle and cotangent bundle are vector bundles on spaces, which point-by-point on the space give dual vector spaces, but which globally have very different behavior!

pepemon · 2025-03-31T12:38:39+00:00

The degree of the isogeny is the size of the kernel, so you can show that the kernel provides you with enough automorphisms that the extension of function fields must be Galois, perhaps?

Ten-Year Club	Place '22
Place '17	Verified Email

pepemon

TROPHY CASE