#differential topology
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lipshits-continuous Ā· 1 month ago
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positively-knotted Ā· 1 month ago
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Some Basic Combinatorics Via Not-So-Basic Topology
Proposition. Let P(r; n,k) be the number of partitions of an integer r into n non-negative integers less than or equal to k. Then P(r; n,k) = P(r; k,n).
This is very easy to see with Young diagrams (count columns vs rows). It's also very difficult to see with topology, but that's more fun. This is probably going to be a bit long, but I think the pay-off is worth it.
The standard construction of (real) projective space Pāæ is to call it the space of 1-dimensional subspaces of RāæāŗĀ¹. This generalises pretty naturally to the "Grassmann manifold" Gr(n; n+k) of n-dimensional subspaces of Rāæāŗįµ, so Pāæ = Gr(1; n+1). Hopefully you believe that Gr(n; n+k) is still a compact nk-manifold. If you want, it's also the quotient O(n+k)/O(n) x O(k).
As well as being a manifold, this is naturally a CW complex. Let's think about how. For each point in Gr(n; n+k) (that is, n-dimensional subspace), it has a unique orthonormal basis where the final non-zero coordinate of each basis vector is positive (easy linalg exercise). Taking this basis, we let d_1,...,d_n be the the "dimension" of the basis vectors (i.e. the position of their final non-zero coordinate, so that (-1,2,0) is 2-"dimensional"). We can assume these are strictly increasing by permuting them. Thus, to each point in Gr(n; n+k), we are associating a unique list of n increasing integers between 1 and n+k.
Let e(d_1,...,d_n) be the subset of Gr(n; n+k) which give those integers. This is actually an open disc of dimension Ī£(d_i-i)! Why? To choose the first basis vector, we can pick any unit vector in the upper half-space of dimension d_1. That is, we have a choice in the hemisphere=disc of dimension d_1-1. To choose the ith, we have a choice in a hemisphere=disc of dimension d_i-1, but we have to be orthogonal to the i-1 vectors we have already chosen, giving a choice in a disc of dimension d_i-i. A product of discs is a disc, and we're done!
Checking that this is actually a CW decomposition is kinda tricky, but also unenlightening, it just is true. So let's count how many cells of each dimension it has. An r-cell corresponds to a choice of partition r = Ī£(d_i-i), where 1 ā‰¤ d_1 < ... < d_n ā‰¤ n+k. Equivalently, it is a choice of partition r = Ī£e-i, with 0 ā‰¤ e_1 ā‰¤ ... ā‰¤ e_n ā‰¤ k, by setting e_i = d_i-i. So the number of r-cells is exactly P(r; n,k)!
We're getting close! The final observation is this. Choosing a n-dimensional subspace is equivalent to choosing its orthogonal complement. That is, the map V ā†’ V^āŠ„ is a natural diffeomorphism between Gr(n; n+k) and Gr(k; n+k), and also a CW-isomorphism. (True because everything is naturally defined; slightly tedious to check.) In particular, it has to match up the number of r-cells, so P(r; n,k) = P(r; k,n)! No Young diagrams necessary.
I think this is kinda neat. It's also weird because these Grassmann manifolds completely classify vector bundles on compact manifolds and are how you define characteristic classes, so secretly under all of that differential topology there's really hard partition problems going on.
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superpte Ā· 1 year ago
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Quantum OntologyĀ 
(Among The) Consequences: ā€œOntologicalā€ Existence of God Deconstructed By Quantum And Differential Topologyā€¦ Ontology is the logic ofā€¦ what is. No less. Itā€™s an obscure word that philosophers love to brandish. But the concept has been at the core of the hardest problems in physicsā€¦ for centuriesā€¦ And progress has been madeā€¦ Unbeknownst to most philosophers. My argument against ontology as theā€¦
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thesomebodywho Ā· 15 days ago
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The biggest step of a mathematicians journey to maturity is no longer having to resist giggling at words like homogenous and homomorphic.
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miamaimania Ā· 6 months ago
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Mathematics in bloom āž¤ A stratifold's beauty unfolds
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spectrallysequenced Ā· 8 months ago
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Personally I view this as math showing us that cohomology is a more natural PoV. I mean yes, singular cohomology is quite artificial bit cohomology just ends up being better so often.
In fact sometimes you can't even define homology in a reasonable manner, but you can define cohomology (sheaf cohomology as an example, not enough projectives).
Homology turns out to work in the case of topological spaces, but this is more of a special case than indicative of a larger picture.
Math people, reblog with your fav theorem and why.
I'll start, the Wedderburn-Artin theorem is a beautiful structure theorem on semisimple rings which says they decompose uniquely as a product of matrix rings over division rings. This is a beautiful result but it also underlies a lot of very cool theory like Brauer Theory, Galois Cohomology and the theory of Galois and Ɖtale Algebras.
What's yours?
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isomorphismes Ā· 1 month ago
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Test functions and [tempered] distributions require the notion of topological vector space ā€¦ distributions can be traced back toĀ Green's functionsĀ in the 1830ā€™s to solve ordinary differential equations ā€¦ the 1936 work ofĀ Sergei SobolevĀ onĀ hyperbolic PDEā€™s.
Laurent Schwartz introduced the term "distribution" by analogy with a distribution of electrical charge, possibly including not only point charges but also dipoles and so on.
tempered distribution
topological vector space
(^ if there is a dipole, there must be a notion of subtraction, hence the need for a vector, and to speak of this very conceptually, use a TVS)
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nablavii Ā· 1 month ago
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So today was the second course day into algebraic geometry. We've studied how projective geometry involves non-measurable stuff. I mean, once you step into real analysis or some algebra, you find things like metric, norm, and so on, but sometimes you gotta step back a little, where things lose measure and you only have a ruler and something to make circumferences. Then, you can ask: Are there any relations between geometrical symmetries and polynomial roots? Well... You've gotta find out.
ESFM IPN - Dr. CĆ©sar Lozano Huerta.
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vsnotresponding Ā· 2 months ago
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happy storyteller saturday! have you been dragged down any research rabbitholes for your projects? what are some interesting things you've learned? :3
hewo and thanks for the ask :]
not exactly a research rabbithole but you'd be surprised how many stuff that i've learnt in my math degree has had an impact on my understanding on how the world in untitled project works
in up time is circular, and two of the main players are pseudo eternal beings that are linked to each other. on one of my classes last year we learnt about stable and unstable invariant manifolds (dynamical systems stuff), where you basically want to study where your fixed point is. on the stable manifold you reach this point by going to infinity (and get infinitely close to it but never reach it when you go back), while on the unstable manifold you reach it by going to minus infinity (and get infinitely closer to it but never to it when times runs forwards).
this point, where the two manifolds meet, is a source of chaos
for some reason (i know the reason. it's that im a nerd) it made me think about these two players. they are both linked by a death (the fixed point). ely (the stable manifold) keeps going back to it without fully reaching it, doomed to encounter it in their future, while the fls keeps forcing herself to keep going when the only way for her to find peace is to confront what happened and go back
here's a series of posts of me losing my mind as it all happened
i never do much research for my stories tbh (i just make stuff up informed by whatever im surrounded by atm/what im learning). at most it'd be (also for up) to read a bunch of philosophy on time and watching kurzgersagt videos about black holes and dark matter which i'd read and watch anyway lol
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positively-knotted Ā· 5 months ago
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Gauss-Bonnet is p sick, and my Masters Dissertation was on (two specific) higher-dimensional analogues, i.e. results linking the sign of scalar curvature to homotopy type.
Interestingly, nothing too similar to GB can be true in higher dimensions: every closed manifold in 3 or more dimensions admits a metric with negative scalar curvature, unlike SĀ² and TĀ². So we can only place restrictions on the existence of positive or zero scalar curvature.
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<3
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lipshits-continuous Ā· 2 months ago
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A half of my time this academic year will be dedicated to developing tools to study holes
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positively-knotted Ā· 5 months ago
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Dissertationposting 2: Curvature & Hypersurfaces
(Note: I'm not gonna say anything about what I mean by "curvature", as if you haven't already seen it in some context I don't think you're gonna get much out of these posts :/. But for clarity, I'll only ever talk about scalar curvature and call it R. If you're used to Gaussian curvature written K, this is the same except R = 2K. Oh also all my manifolds are oriented.)
It may not be immediately obvious why we would want to consider hypersurfaces when trying to understand curvature - after all, RĀ³ is flat, but contains surfaces with all kinds of curvature. So a motivating example may be in order. This is in fact the example that spurred all of the developments we'll talk about, and is of particular importance in theoretical cosmology! Maybe I'll say more about that later.
Consider a manifold homeomorphic to the n-torus Tāæ - or more properly, Tāæ with a choice of geometry (Riemannian metric). Intuitively, this shouldn't be able to have R > 0 everywhere, in the same way that TĀ² can't. Proving this is surprisingly hard, but one sensible approach would be to try induction. After all, TĀ³ is just TĀ² Ɨ SĀ¹, and SĀ¹ is easy to understand. Thought of the other way around, we want to find nice hypersurfaces and do inductive descent until we reach dimension 2, then apply Gauss-Bonnet.
But if any old hypersurface doesn't say much, what about particularly nice ones? The obvious candidates are so-called stable minimal hypersurfaces. These are minimal points for area, in the sense that if you perturb them very slightly, their area must increase (e.g. the meridian of a standard torus in the motivating example). The big result about these is as follows:
Lemma 1.
If Ī£ is a stable minimal hypersurface in a manifold (M, g) and f is any function on Ī£, then [1]
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Here the left side is the gradient of f (in the normal calculus sense), and \II is the second fundamental form, a number that tells us something about how Ī£ sits inside M. You can think of it as a kind of error term, but the important bit is that it's squared, so non-negative. R is the scalar curvature of the ambient manifold M, and R_\Sigma is the induced scalar curvature of Ī£.
You've probably never seen this before, even if you've done differential geometry. I don't know why this formula isn't better known, it's super useful, as with clever choices of f it gives big results. Just taking f=1 shows that for a stable minimal hypersurface, the intrinsic curvature has to be greater on average than the external curvature. This is a pretty big step forward!
Ok, so we have our choice of useful hypersurfaces. Now we just need to show that they always exist. And they do! In fact, they exist for each homology class:
Lemma 2.
Let (M, g) be a closed n-manifold, 3 ā‰¤ n ā‰¤ 7. Then for any class Ī± āˆˆ H_{nāˆ’1}(M), we can find a closed stable minimal hypersurface Ī£ such that [Ī£] = Ī± in homology.
Let's unpack this a bit. We can think of hypersurfaces as living in a particular homology class in the usual geometric way. It's a standard result from differential topology [2] that we can in find a hypersurface living in any homology class, and that slightly perturbing it won't change that class. The interesting bit here is that we can always take a stable minimal representative, in sufficiently small dimensions. [3] Annoyingly, this means everything from here on out only applies in dimensions less than 8, even though there's no good topological reason why it isn't true more generally. (It definitely is true more generally, we just need a stronger version of this lemma and that makes it an analysis question.)
Next time, we'll use this to show that Tāæ doesn't admit positive scalar curvature! Notes below the cut.
[1] Proof of Lemma 1. Start with the usual Gauss equation. Take the trace to pass to Ricci curvature, then again to pass to scalar curvature. It turns out that the irritating terms from varying dimensions cancel out! More detail in the screenshots below:
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[2] Remember that there is a bijection between HĀ¹(M) and homotopy classes of maps M -> SĀ¹, pairing [f] with f^*(Ļ‰), where Ļ‰ generates HĀ¹(SĀ¹). We can assume f is smooth, choose a regular value t, and set Ī£ = f^{-1}(t). Then [Ī£] is PoincarĆ© dual to f^*(Ļ‰).
[3] The actual argument here is pretty standard for the analysis I'll be omitting. In general, if we want to show that an object in space X with property P exists, we'll pass to a bigger space Y where standard results give us an object with that property; then somehow argue that P implies that it's actually of type X. Here, we pass from submanifolds to currents, and minimise area for currents in a given homology class. By measure theoretic magic, this minimisation forces the singular set of the current to have codimension at most 7, so in dimensions less than 8 it is in fact a submanifold.
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didsomeonesaysoup Ā· 7 months ago
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me after being 100% honest with my therapist
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klavierpanda Ā· 2 years ago
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3 exams down, 5 to go
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shamelessnightlight Ā· 8 months ago
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How could they not be? Look at all the differential structure! The smoothness! The metric! So curve, much geodesic, Morse functions. I think I had better go lie down and perhaps fan myself.
fun fact: Boston Museum of Science calls their evening lecture series ā€œSubSpaceā€, which would be a totally innocuous math term except for the fact that, to make sure you know these lectures are higher-level and not aimed at their usual audience (kids), they chose to subtitle it ā€œSubSpace: Adult Experiencesā€
šŸ˜¶
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incorrect-moriarty-sherliam Ā· 1 month ago
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William sighs as he begins counting sheep. Instead of simply counting the fuzzy creatures, he starts to calculate permutations and combinations of numbers in his head, forming complex sequences.
William: "Five... Euler's number, pi. Six... Riemannian topology. Seven... Fourier series. Eight... differential equations. Nine... Pythagorean theorem..."
Sherlock: "Liam, aren't you supposed to be asleep?"
William, at the end, creates a new theorem... because he can XD
+++++++
@missshello contribution! Yeah!
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