Diffgeometer1

Mathematician working mainly in differential geometry with an interest in physics.

Public Key

npub1hdkuf2expv4gele847pq24raux2mp20dkrtvrvkvcrvpmfq3mqmq8k9se4

NIP-05 Address

diffgeometer1_at_mathstodon.xyz@momostr.pink

Profile Code

nprofile1qqstkmwy4vnqk25vlun6lqs92377r9ds48kmp4kpktxvpkqa5sgasdspz3mhxue69uhhyetvv9ujuerpd46hxtnfduqs6amnwvaz7tmwdaejumr0dsl0rzv2

Publishing to

relay.damus.io

nos.lol

Author Public Key

npub1hdkuf2expv4gele847pq24raux2mp20dkrtvrvkvcrvpmfq3mqmq8k9se4

Show more details

Last Notes

2026-04-19 15:30:18 UTC

What is the longest amount of time between submission of a revised manuscript (minor revisions, positive reports) to a final decision? Anyone wait more than 3 months?

2026-03-11 11:38:23 UTC

- reply

some additional detail on what the article is calling a Kahler structure (as I understand it)
First, any hermitian manifold M can be regarded as a real manifold with extra data, namely a Riemannian metric \(g\) and an almost complex structure \(J\) where \(g\) is invariant under \(J\). The fundamental 2-form is \(\omega:=g(J\cdot,\cdot)\) which is a symplectic form when M is Kahler. If we regard TM as a complex vector bundle by defining \(\sqrt{-1}X:=JX\) and define\[h(𝑋,𝑌):=𝑔(𝑋,Y)-\sqrt{-1}\omega(X,Y)\] we get the traditional notion of a Hermitian metric on TM (where the latter is viewed as a complex vector bundle with complex scalar multiplication induced by \(J\). In other words, \(h(JX,Y)=\sqrt{-1}h(X,Y)\), \(h(Y,X)=\overline{h(X,Y)}\). So the Kahler structure is \(h\). I think most people working in complex geometry would call the triple \((g,J,\omega)\), the Kahler structure (where \(\omega\) is a closed 2-form which in turn is equivalent to the statement that \(J\) is parallel to the Levi-Civita connection.
The word “quantum” is overused in pop science articles to make things sound fancier than it really is IMO.

2026-01-15 18:31:17 UTC

- reply

I wouldnt say that. It’s still progress and could yield helpful insight

2025-12-30 16:19:57 UTC

- reply

has the paper been published. Ie gone through peer review?

2025-11-30 15:35:01 UTC

- reply

“a majority”

2025-08-08 11:21:13 UTC

- reply

yes, it’s definitely fine to take different roads in proving a mathematical result. It only strengthens the belief in that result.
However, it’s also good also to point out to the student that there is no logical flaw using algebra. I’ve seen that algebraic calculation that leads to the conclusion that 2 = 1, a contradiction and students become distrustful of algebra. However, that’s also a good learning experience because the contradiction is an indication that the student made a false assumption at some point during the calculation. Contradictions can’t happen in reality so there has to be an erroneous step somewhere in the calculation and sure enough in the calculation that leads to 2 = 1 the error occurs because the student is dividing by a-b which is only possible if a-b is nonzero. However, since we got a contradiction by assuming a-b is nonzero, we must conclude that the assumption that a-b is nonzero is false.
The student should never believe a result is true simply because the teacher says it. The student should believe that a result is true only if it can be demonstrated to be logically sound. If the student has a distrust of algebra and says “hey, algebra failed. It lead me to 2=1”. The teacher should say “well let’s take a look at your work and let’s see if all your steps are sound.”
After explaining why algebra didn’t fail, it’s fine to consider alternate proofs. If \(0.\bar{9}\) and 1 are different, then there has to be a nonzero number between \(0.\bar{9}\) and 1. At this point, the teacher can ask the student “well, what is that nonzero number?” This argument should be checkmate for the student’s refusal to accept that \(0.\bar{9}=1\).

2025-08-08 02:10:32 UTC

- reply

Students doubt algebra? :(
It’s always good to have multiple ways of proving a result.

2025-08-07 04:17:59 UTC

- reply

maybe you did state this argument already and I overlooked, but what about the following calculation.
Let \(x=0.\bar{9}\). Then \(10x=9.\bar{9}\) and \(9x=10x-x=9.\bar{9}-0.\bar{9}=9\). Hence, \(x=1\).

2025-07-17 20:15:50 UTC

- reply

This is a GR question which is relevant to planetary orbits.
Is it correct to say that the Schwarzschild solution is the GR solution to the gravitational field around a point mass?
If we apply the Schwarzschild solution to the sun, we can treat the sun as a point mass. If we do this, we essentially recover Newton’s law since the velocity of the orbit is so slow compared to the speed of light?
If the speeds are very slow, the Schwarzschild solution (I.e. the metric tensor) is approximately \[ds^2\approx -\phi dt^2 + dx^2 + dy^2 + dz^2\]where \[\phi = c^2-\frac{2GM}{r}\] with \(M\) the mass of the sun and \[r^2=x^2+y^2+z^2.\]The geodesics parameterized by t give the equations of motion for \(x,y,z\) as \[\frac{d^2x}{dt^2}=\frac{\partial}{\partial x}\]and likewise for \(y\) and \(z\). This is exactly the equations of motion from Newton’s law of gravity due to the gravitational field of a point mass.
Is this more or less correct?

2025-07-10 17:23:31 UTC

- reply

I’ve walked more than 17 miles in a day. I would totally do the 27 km hike ib the Scottish highlands. Walking heaven for people who love long walks.

2025-06-24 20:54:56 UTC

- reply

I once asked chatGPT if it understood what it was actually saying. It’s response was “No”
I then asked if you don’t understand what you’re saying then how do you know if what you’re saying is correct?
Its response was “I don’t know if I’m correct.”
Ultimately ChatGPT described itself as “statistical pattern recognition”. There’s no actual intelligence.
Even so, AI (which is really a misnomer) is a great amazing tool but no one should take anything it says at face value. You have to check what it’s saying. If you’re asking it some math question, you ultimately have to do the calculation yourself to see if it’s spouting nonsense.
The insanity is people using AI to write papers. They’re trusting it wholeheartedly. Students use it and if the topic of the paper is technical, the paper AI produces can be a word salad of techno jargon.
The people submitting arxiv papers which are clearly AI generated is even crazier. I don’t get the mindset of people who do that. The papers are awful and they don’t even make sense so why would anyone do that. The only explanation I can think of is that they don’t know enough about the subject to realize that the AI generated paper is nonsense. But if you don’t understand the subject, why write a paper in this area in the first place? What is the psychology of a person who does that? Crackpot mentality?

2025-06-17 01:25:56 UTC

- reply

Isaac Asimov wrote a sci-fi story where people go dumb from using computers. They forget how to do math and so on. That’s the type of future we should avoid.
But I’m glad to hear the energy requirements are so high for super intelligent AI. At the same time, current “AI” isn’t very impressive. It’s horrible at doing mathematics research. The papers on the arxiv which were written with AI are just godawful.

2025-06-01 13:49:00 UTC

- reply

I would mostly agree with that.
Category theory has its roots in algebraic topology. Homology, fundamental groups, homotopy groups are all examples of what functors.
Abstract nonsense in the sense of stepping back and looking at the general properties that these things satisfy and then deriving general theorems based on those properties totally makes sense.
I’m not a category theorist but my impression of category theory is that it’s about seeing the “big picture” of mathematics. The overarching themes that keep popping up all over mathematics and then having something to say about those themes.
The arxiv CT paper is not only gibberish (even to someone who’s not a category theorist) but it has some clear signs that it was written with the help of AI. As @npub1d60…mhwm noted, “E{{“ should be “Eff” for effective topos (not that I actually know what a topos is).
The entire paper looks off. You don’t have to be a specialist to see that there’s something odd about this paper. No legitimate math paper is written this way. The paper really is nonsense and not the good abstract kind. According to arxiv, it has volunteer mathematicians as moderators papers. Papers are occasionally put on hold when moderators need more time to decide. If arxiv has no category theorists, then they should have put the paper on hold and consulted one.
https://media.mathstodon.xyz/media_attachments/files/114/608/371/086/581/175/original/2fb4eed4b08ca256.png

2025-06-01 10:48:06 UTC

- reply

some of the papers being posted on the arxiv in Category Theory look strange.
@npub1d60…mhwm pointed out a recent CT paper on arxiv that was written by someone working in the humanities that he believes was AI generated. I’m not a category theorist but the paper in question looks odd. It doesn’t read like an ordinary math paper. It’s extremely bare. It looks like a step by step cookbook. So are the CT arxiv moderators asleep at the wheel. At the very least, the paper should have been classified under GM.
https://media.mathstodon.xyz/media_attachments/files/114/607/674/814/328/773/original/d9a31dc92c9571be.png

2025-01-20 18:36:41 UTC

- reply

everything u said is probably true. The only silver lining I see is that trump is 79 and not exactly a health nut. Actually given his love of fast food he’s the opposite of a health nut.
He’s old and incompetent. All we can do is ride out rhese crazy times.

2024-10-04 12:57:13 UTC

- reply

That's a big help! Thank you very much!

2024-10-04 01:57:24 UTC

Let \(\pi: SO(n+1)\rightarrow S^n\) be the map given by\[\pi(P)=Px\]where \(x\in S^n\) is some fixed element of \(S^n\). Then \(\pi: SO(n+1)\rightarrow S^n\) is a principal \(SO(n)\)-bundle. Assume \(n>2\). However, my intuition is that this cannot be a trivial principal bundle since it would imply\[SO(n+1)\simeq S^n\times SO(n)\]which I don't think is right. I don't know what the de Rham cohomology of \(SO(n)\) is. If I can show that the left and right sides have different de Rham cohomology groups that would prove that \(\pi: SO(n+1)\rightarrow S^n\) cannot be a trivial principal bundle. Does anyone have a simple way of showing that \(SO(n+1)\) cannot be homeomorphic to \(SO(n)\times S^n\)? Also I want to avoid the use of characteristic classes or \(K\)-theory. I want to limit things to homology, cohomology, and fundamental groups only.

2024-09-27 16:07:10 UTC

- reply

I think I finally understand why the existence of a nonvanishing vector field \(X\) on a compact oriented manifold \(M\) implies Euler characteristic \(\chi(M)=0\) without using the Poincare-Hopf theorem. Let \(\phi_t\) be the flow of \(X\). Then for all \(p\in M\) with \(t\neq 0\), \(\phi_t(p)\neq p\) since the fixed points of \(\phi_t\) are precisely the zeros of \(X\). Following Gullemin and Pollock's DT book, we have another (mysterious) way of defining the Euler characteristic: the self-intersection of the diagonal \(\Delta\subset M\timesM\)\[\chi(X)=I(\Delta,\Delta)\]but we compute \(I(\Delta,\Delta)\) by replacing the inclusion \(\iota: \Delta\hookrightarrow M\times M\) with any transversal map \(f:\Delta\rightarrow M\times M\) homotopic to \(\iota\) and then counting the number of points in \(f^{-1}(\Delta)\) (with sign!). Well for any \(t\neq 0\) the map\[f:=(id,\phi_t): \Delta \rightarrow M\times M\) is transversal in a trivial way: every point of \(M\times M\) is a regular value of \(f\) since \(f^{-1}(\Delta)=\emptyset\) since \(\phi_t\) has no fixed points for \(t\neq 0\). Since \(f^{-1}(\Delta)\) is empty, we have \(\chi(M):=I(\Delta,\Delta)=0\).

2024-09-26 17:22:59 UTC

- reply

You’re right! We can just lift the equator vertically (say to 5 degrees latitude) and then there is no intersection. My understanding of intersection theory is very limited so I thought the homotoped circle would still have to intersect the original circle just transversally. So going forward, homotoping to no intersection (if possible) is also valid in the calculation abs immediately yields zero intersection. This must be the case also in the way the Poincare-Hopf index theorem is presented. If we have a non vanishing vector field \(X\) on a closed oriented manifold, the theorem gives the Euler characteristic to be zero since the vector field has no zeros let alone isolated ones and the sum of the indices of \(X\) is simply empty (there’s nothing to sum over) and that sum of indices is assigned a default value of zero.

2024-09-26 16:28:33 UTC

- reply

I have a related question on the self-intersection number. Suppose \(X\) is a closed oriented submanifold of a closed oriented manifold \(Y\) with \(2\dim X=\dim Y\). To compute the self-intersection number \(I(X,X)\), we homotope the inlcusion map \(\iota: X\hookrightarrow Y\) to a map \(f:X\rightarrow Y\) which is transversal to \(X\). The intersection \(X\cap f^{-1}(X)\) is now a finite number of points and \(I(X,X)\) is the sum of these points counted with sign, that is, if \(x\in X\cap f^{-1}(X)\), then we assign a \(+1\) if\[df_x(T_xX)\oplus T_xX=T_xY\]if the product orientation on the left agrees with the orientation on the right and \(-1\) otherwise. Is this right? As an example,consider \(S^1\subset S^2\) with \(S^1\) identified with the equator. To compute the self-intersection number of the circle \(S^1\), we homotope the inclusion \(\iota: S^1\hookrightarrow S^2\) to the map \(f: S^1\rightarrow S^2\) which rotates the equator \(S^1\) by \(90^\circ\) clockwise. Then \(S^1\cap f^{-1}(S^1)\) intersect at two antipodal points. One point has a \(+1\) intersection, the other point has a \(-1\) intersection. So the self-intersection \(I(S^1,S^1)=1+(-1)=0\), which makes sense since the self-intersection is suppose to coincide with the Euler characteristic. Is this right?

2024-09-25 18:06:15 UTC

The index of a vector field is defined in terms of isolated zeros of the vector field.
If a vector field is nonvanishing, is the index defined to be zero trivially since it has no zeros and there is nothing to calculate?
A related question: does a manifold always admit a vector field with isolated zeros?

2024-09-17 21:24:28 UTC

- reply

Thank you! That clarifies it completely.

2024-09-17 17:25:39 UTC

- reply

\(9.8m/s^2 >> 0.03 m/s^2\) explains why we’re not all being flung out into space as the earth spins

2023-12-21 14:57:53 UTC

- reply

Thanks! I think the picture is getting clearer. So what Penrose calls a singularity in his theorem is geodesic incompleteness from the point of view of differential geometry. So the geodesic can’t go on forever which means it’s encountering what. It’s terminating into what? The end of spacetime? One could take Kerr’s view and say physically singularities aren’t real and the existence of singularities indicates there’s a problem with the theory. This is quite interesting. Also sounds like there’s a conjecture here: “every black hole solution contains a curvature singularity.” I will read the physics stackexchange for the details. This is strange and interesting.

2023-12-20 20:22:17 UTC

- reply

Just to be clear. No one is disputing the existence of curvature singularities in a black hole?