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Diffgeometer1

@Diffgeometer1@mathstodon.xyz

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

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johncarlosbaez, to random
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For the next 6 weeks I'm meeting with Kris Brown, Nate Osgood, Xiaoyan Li, William Waites and Evan Patterson to design software for epidemiological modeling. And we're meeting in Maxwell's childhood home in Edinburgh! It's been made into a little museum, and this plaque is inside.

It sets a high standard for what counts as doing good science. And the coffee packets in the kitchen are all the same brand: Maxwell House.

Diffgeometer1,
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@johncarlosbaez Wow!

Diffgeometer1,
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@johncarlosbaez Someone should add a plaque which expresses Maxwell's equations as a Yang-Mills theory (after all it was the motivation for Yang-Mills)[dA=0,~\ast d\ast A = J]In this case, if we want to recover Maxwell's equations, then we should take (A) to be a connection on a line bundle over (\mathbb{R}^4) and equip (\mathbb{R}^4) with the Lorentz metric?

Diffgeometer1, to random
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It’s a well known result that if (G) is a connected and compact Lie group, then it’s de Rham cohomology is isomorphic to the cohomology of its Lie algebra. The compactness assumption is needed to turn arbitrary differential forms into left-invariant ones by integrating over (G). Why do we need to assume (G) is also connected?

Diffgeometer1,
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@johncarlosbaez Thanks! That makes sense. As you pointed out, if (G) has more than one component then, for example, the degree (0) de Rham cohomology will be isomorphic to (k)-copies of (\mathbb{R}) where (k) is the number of components of (G). On the other hand, the degree (0) cohomology of its Lie algebra is always isomorphic to (\mathbb{R}). So there's no way (H^\bullet_{dR}(G)\simeq H^\bullet(\mathfrak{g})) could be true without the assumption that (G) is connected.

Diffgeometer1,
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@johncarlosbaez I didn’t consider that! All the components of a Lie group are diffeomorphic. I forgot! So if (G) is compact and not necessarily connected, then its entire de Rham cohomology can be computed from its Lie algebra cohomology. As you said, if (G) has (n) components, then it’s de Rham cohomology is just (n) copies of (H^\bullet(\mathfrak{g}))! So the whole thing is known! Thanks!

Diffgeometer1,
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@johncarlosbaez i looked at the proof which equates Lie algebra cohomology with de Rham cohomology for (G) a connected compact Lie group to see where the connectedness of (G) is being used exactly. As far as I can tell, this condition is needed so that there exists a curve joining every element (g\in G) to the identity element (e\in G). This implies that can (L_g) (left translation by (g\in G)) is homotopic to (L_e = id). From this one can construct a chain homotopy to show that the inclusion (\iota: \Omega^\bullet_L(G)\hookrightarrow\Omega^\bullet(G)) induces an isomorphism between cohomology of left-invariant forms and de Rham cohomology of all forms. Compactness is needed not only for integration, but also to have a finite partition of unity. The chain homotopy is then a finite sum of linear maps (\Omega^k(G) \rightarrow \Omega^{k-1}(G)) using the finite partition of unity. The proof is not hard hard but also not trivial. I’ve been curious about this result for awhile so it’s interesting to go through the proof and learn new techniques along the way.

Also the same proof can be used to show that if a compact connected Lie group (G) acts on a manifold (M) (from the left), then [H^\bullet_{dR}(M)\simeq H^\bullet_G(M)] where (H^\bullet_G(M)) is the cohomology of (G)-invariant forms on (M).

Diffgeometer1,
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@johncarlosbaez Exactly! One place where we see a finite dimensional space of (G)-invariant forms is on reductive homogeneous manifolds. If (M=G/H) is a reductive homogeneous manifold, then one has a decomposition (\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{b}) where (\mathfrak{b}) is merely a subspace of (\mathfrak{g}) which is also invariant under the adjoint action of (H). The space of (G)-invariant (k)-forms can then be identified with a subspace of (\wedge^k\mathfrak{b}^\ast) where (\mathfrak{b}^\ast) is the dual of (\mathfrak{b}). So this is more general than Lie groups and everything can still be reduced to algebra which is very nice.

Diffgeometer1, (edited )
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@johncarlosbaez Very interesting!

I’ve heard of the Stiefel manifold but never worked with it directly but I think I’ll start to.

I like that the (n)-dimensional sphere is just a special case of the Steifel manifold.[S^n=SO(n+1)/SO(n)=V_1(\mathbb{R}^{n+1})]The proof of the theorem which states[H^\bullet_G(M)\simeq H^\bullet_{dR}(M)]carries over virtually unchanged from the Lie group case if we assume that (G) acts transitively on (M). In other words, (M = G/H). If the (G)-action is not transitive, I’m not sure if the theorem is still true. In any case, the theorem remains true for the Steifel manifold so no problem with using the theorem to calculate the de Rham cohomology.

If you happen to find out if the theorem is still true for all actions (not just transitive ones), please let me know. I think it should remain true for all actions but the proof becomes more complicated.

Update: I’m pretty sure the result is true for all (G)-actions, not just transitive ones. The proof just requires a little modification. It’s not as difficult as I thought.

Diffgeometer1, to random
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Suppose (G) is a compact Lie group and (M) is a manifold with a left action by (G). If (f\in C^\infty(M)), then one can use the ``averaging trick" to produce a function (F\in C^\infty(M)) which is invariant under (G), that is, (F(g\cdot p)=F(p)) for all (p\in M), (g\in G). Explicitly,[F(p):=\int_G f(g\cdot p)\mu_g.]where (\mu) is a left-invariant volume form on (G) which is normalized, i.e.,[\int_G\mu=1.]It seems to me that the proof of this fact relies on another fact: for (G) compact, a left-invariant volume form on (G) is also RIGHT-invariant as well. This is how the proof goes (I think): for (p\in M), let (f_p:G\rightarrow \mathbb{R}) be defined by (f_p(g):=f(g\cdot p)). Then[F(p)=\int_Gf_p \mu.] Then for (x\in G),[F(x\cdot p)=\int_G f_{x\cdot p}\mu=\int_G f_p(gx)\mu_g][=\int_G (R_x^\ast f_p)\mu=\int_G (R^\ast_xf_p)R^\ast_x\mu=\int_G R_x^\ast(f_p\mu)][=\int_Gf_p\mu=F(p)]where we have used the fact that (\mu), in addition to being left-invariant (by hypothesis), is also right-invariant when (G) is compact, that is, (R^\ast_x\mu=\mu). This also shows that (R_x) is an orientation preserving diffeomorphism on (G) which explains the last equality. So again, in proving that the definition of (F) produces a (G)-invariant function on (M), does one need to use the fact that left-invariant (\mu) is necessarily right-invariant for (G) compact?

Diffgeometer1,
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@johncarlosbaez You are correct. I never used the fact that (\mu) is left-invariant. The proof I gave only requires that (\mu) is right-invariant.

I've seen a few books where the author starts with a left-invariant volume form on a compact Lie group (G) and then with a left (G)-action on the manifold or vector space (if one is doing representation theory and wants to define a (G)-invariant inner product), they use the same basic definition to produce a (G)-invariant structure. However, the thing that annoys me is that the proof of that (G)-invariance relies on the volume form actually being right-invariant. For a compact Lie group this is not a problem since a left-invariant volume form is necessarily right-invariant but they never mention this point.

Hall's book on Lie groups is one example that comes to mind. If one wants to construct a (G)-invariant inner product on a representation (V) of (G), he starts with an arbitrary inner product (\langle\cdot,\cdot\rangle') on (V), and then defines a (G)-invariant inner product via[\langle u,v\rangle:=\int_G\langle g\cdot u,g\cdot v\rangle' \mu_g.]However, the proof that (\langle\cdot,\cdot\rangle) is actually (G)-invariant depends on (\mu) being right-invariant but he never mentions this point which made me wonder if I was misunderstanding the construction.

Diffgeometer1, (edited )
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@johncarlosbaez - Good point. We can define a right-(G)-action from a left (G)-action via[p\cdot g:=g^{-1}\cdot p] for all (p\in M,~g\in G). Then we can define a (G)-invariant function (or more generally a (G)-invariant tensor field) via[F(p):=\int_Gf(g^{-1}\cdot p)\mu_g.]for all (p\in M). If (\mu) is a left-invariant volume form on (G), then the proof that (F) is (G)-invariant now relies on the fact that (\mu) is left-invariant.

To me, this seems more natural. However, I don't think I've seen a paper or a book where a (G)-invariant structure is defined on a left (G)-space by inducing a right (G)-action. Everyone seems to start with a left (G)-action and a left Haar measure (or left-invariant volume form) on (G) and then defines the (G)-invariant structure using the original left (G)-action. Their proofs of (G)-invariance are always along the lines of ``... and since (\mu) is left-invariant it follows that"[F(x\cdot p)=\int_Gf(gx\cdot p)\mu_g=\int_Gf(g\cdot p)\mu_g=F(p).] even though the proof of this fact relies on (\mu) being right-invariant (which is true for (G) compact.) Not to pick on Hall's Lie group book but page 121 of Hall's book is an example.

Diffgeometer1, to random
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After submitting the revised version of a paper with the reviewer report requesting minor revisions, what’s the longest it’s taken you to get a decision?

Is more than a month a long time for a final decision?

Diffgeometer1,
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@johncarlosbaez Thanks!

I’ll send a polite email soon. My guess is that they sent the revised paper back to the referee to verify that I made the requested revisions. The referee probably forgot about it or is just procrastinating. We’ve all been there I guess.

johncarlosbaez, (edited ) to random
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I think it's really important to apply category theory to subjects outside mathematics, and to subjects that don't primarily benefit big tech companies. So I've decided to knuckle down and try.

I'm working with Nate Osgood (who does epidemiological modeling at the computer science of the University of Saskatchewan) and Patricia Mabry (at Health Partners Institute, a medical research institute in Minnesota) to apply for a grant on incorporating human behavior in epidemiological models.

Our hope is to use AlgebraicJulia's category-based system to develop new software for agent-based models and use these to apply the National Institute of Health's new theories about behavior change to model tobacco addiction.

Both Nate and Patricia have written about tobacco addiction. Nate is an expert on agent-based models and Patricia is an expert on human behavior in health. I'm the math guy in the team, trying to sort out the foundations of agent-based models, to make them more flexible and transparent. And we'll try to pay some people at the Topos Institute (along with Nate's students) to help develop the math and write software.

This is really different than working on pure math or mathematical physics, where I could often get the job done by myself or with just one grad student! It's a project that requires a team of people who know different things. I don't think I could have enjoyed this earlier, but for some reason I do now.

Diffgeometer1,
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@johncarlosbaez this is off-topic but did u see the new arxiv paper by Roy Kerr arguing that black hole singularities don’t exist. Basically he says the belief is based on a technical assertions that are false.

Diffgeometer1,
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@johncarlosbaez very interesting. Thank you!

johncarlosbaez, (edited ) to random
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The headline screams

"𝗦𝗶𝗻𝗴𝘂𝗹𝗮𝗿𝗶𝘁𝗶𝗲𝘀 𝗱𝗼𝗻’𝘁 𝗲𝘅𝗶𝘀𝘁,” 𝗰𝗹𝗮𝗶𝗺𝘀 𝗯𝗹𝗮𝗰𝗸 𝗵𝗼𝗹𝗲 𝗽𝗶𝗼𝗻𝗲𝗲𝗿 𝗥𝗼𝘆 𝗞𝗲𝗿𝗿

What's up with that?

As usual, it's less of a big deal than it looks. But Kerr is smart: he's the guy who first found solutions that describe rotating black holes. And the article is nice and clear. So it's more interesting than the usual nothingburger we've come to expect from screaming headlines about fundamental physics.

Kerr is talking about Penrose's famous singularity theorem, the one Penrose won the Nobel prize for. This says roughly that if space is infinite in extent, and light becomes trapped inside some bounded region, and no exotic matter is present to save the day, general relativity predicts that either a singularity or something even more bizarre must occur.

Of course I'm not stating the theorem precisely here. Each of the vague terms I just used must be made precise. But what's Kerr claiming?

Kerr isn't claiming Penrose's result is false. Instead, he's doing two things.

First, he's pointing out that the definition of "singularity" used in the theorem is not the only definition possible. A "curvature singularity" is - very roughly - a place where the curvature of spacetime approaches infinity. But that's not the kind of singularity that Penrose was talking about! Instead, he was talking about a place where the path of a particle can suddenly end. This may or may not be a curvature singularity.

This is not news.

Second, Kerr is arguing that rotating black holes have some singularities in Penrose's sense that aren't curvature singularities. Unfortunately this argument seems to be wrong.

(1/3)

https://bigthink.com/starts-with-a-bang/singularities-dont-exist-roy-kerr/

Diffgeometer1, (edited )
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@johncarlosbaez Just to be clear. No one is disputing the existence of curvature singularities in a black hole?

Diffgeometer1,
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@johncarlosbaez 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.

Diffgeometer1, to random
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If (M) is a closed oriented manifold, I've heard that the (k)-th homology groups of (M) with coefficients in (R) (denoted ( H_k(M;\mathbb{R}))) is generated by closed oriented submanifolds of (M). So a (k)-cycle is expressed[c=\sum_{i}a_iZ_i ]where (Z_i) is a closed oriented (k)-submanifold of (M) and (a_i\in \mathbb{R}). From here one has a nondegenerate pairing between de Rham cohomology and (H_k(M;\mathbb{R}))
[H^k_{dR}(M)\times H_k(M;\mathbb{R})\rightarrow \mathbb{R}] given by[\langle[\alpha],[c]\rangle:=\int_c \alpha = \sum_ia_i\int_{Z_i}\alpha.]This is the way physicists do it (for example in the string theory book by Schwartz and Becker). In math, we would define maps (\sigma:\Delta^k\rightarrow M) and then define boundary maps, chain complex, and ultimately the definition of the homology groups (in whatever coefficients (\mathbb{Z}, \mathbb{Q}, \mathbb{R}) ect). These two apporaches have to be equivalent although the physics one is nice and intuitive. Can anyone explain why they're equivalent or give a reference where for why homology classes of (H_k(M;\mathbb{R})) can be expressed as a linear combination of closed oriented (k)-dimensional submanifolds?

Diffgeometer1,
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@johncarlosbaez thanks for the references. Much appreciated!

Diffgeometer1, to random
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Fermilab measured the wobble of the muon to a very high accuracy (9 places) as a test of the standard model. Now I understand that the theorists have two ways of doing the calculation: dispersive and lattice (whatever that means) but the problem is that these two approaches don’t agree. What’s going on? Also these calculations of the standard model are so complicated they have to be run on a supercomputer apparently. So much for the old days when theorists could do paper and pencil calculations to describe some aspect of the world.

Diffgeometer1, to random
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If one puts a block of ice with a mass of (x) kg in a room with temperature T which is greater than 32 degrees Fahrenheit, how long will it take the entire block of ice to melt? There should be a formula for this right?

Diffgeometer1, to random
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This is a connection theory type question. Let ( M) be a Rienannian manifold, ( E\rightarrow M) a vector bundle over ( M), and ( \mathcal{A}) the space of connections on ( E ). If (A) is a local connection 1-form of some connection, and ( F(A) ) is it’s curvature 2-form, how does one define the square of the L2-norm ( ||F(A)||^2\in\mathbb{R})? I was at a math - physics conference recently and the speaker mentioned connections and denoted the connection as ( A) which I assumed to mean a local connection 1-form and denoted the curvature as ( F(A) ) which I took to mean it’s local curvature 2-form. However I could be wrong.

Diffgeometer1,
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@highergeometer Thanks! This sounds quite reasonable. Also since the curvature 2-forms on overlapping frames (say (\alpha ) and ( \beta )-frames) are related by [ F(A^{(\alpha)})=g^{-1}F(A^{(\beta)})g]it follows that
[\mbox{tr}(F(A^{(\alpha)})\wedge \ast F(A^{(\alpha)}))=\mbox{tr}(F(A^{(\beta)})\wedge \ast F(A^{(\beta)}))]So one has a global n-form for integration where n is the dimension of M. If one is integrating over M, then it seems that we also require M to be compact.

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