@pschwahn@mathstodon.xyz
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pschwahn

@pschwahn@mathstodon.xyz

Sometimes I do differential geometry.

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johncarlosbaez, (edited ) to random
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A while back @pschwahn raised an interesting puzzle here:

You can define a well-behaved cross product of vectors only in 3 and 7 dimensions. The 7d cross product is weird because it's not preserved by all rotations of 7d space. But very smart people have told us there's a way to get the group of 𝟯𝗱 rotations to act on 7d space while preserving the 7d cross product. In fact, you can do it while also getting this group to act 'irreducibly', meaning the only subspaces of 7d space preserved by this action are {0} and the whole space!

The puzzle is: 𝗰𝗮𝗻 𝘆𝗼𝘂 𝘄𝗿𝗶𝘁𝗲 𝗱𝗼𝘄𝗻 𝘁𝗵𝗲 𝗳𝗼𝗿𝗺𝘂𝗹𝗮𝘀 𝗳𝗼𝗿 𝗵𝗼𝘄 𝘁𝗵𝗶𝘀 𝘄𝗼𝗿𝗸𝘀?

I got stuck on this so I asked some of my friends, and now Layra Idarani has outlined a nice way to do it:

https://golem.ph.utexas.edu/category/2024/05/3d_rotations_and_the_cross_pro.html#c063185

Interestingly he doesn't actually give the formulas; he just tells you how to get them. So I will need to do some work to check his answer! If you want to help out, that would be great.

Layra said "The devil of the details is in the eating". I thought the proof was in the pudding. Now I'm hungry for devil's food cake.

pschwahn,
@pschwahn@mathstodon.xyz avatar

Very nice! I've added a few comments there.

I've had another thought: there's nothing that really limits this sort of considerations to the 𝑐𝑜𝑚𝑝𝑎𝑐𝑡 real form of G₂. Say we replace octonions by split-octonions (not a division algebra, but a so-called composition algebra), then the automorphism group G₂* is not compact any more - it does not preserve an Euclidean inner product on ℝ⁷, but instead one of signature (3,4).

Dynkin's classification of maximal subalgebras is undisturbed by this, since G₂ and G₂* have the same complexification. But it is not clear to me what the corresponding subgroups are.

It is known that the maximal compact subgroup of G₂* is SO(4) (which is also maximal as just a subgroup). This suggests that the analogues of SU(3) and SO(3)ᵢᵣᵣ are 𝑛𝑜𝑛𝑐𝑜𝑚𝑝𝑎𝑐𝑡 - but what are they exactly, and how do they act on the imaginary split-octonions?

johncarlosbaez, (edited ) to random
@johncarlosbaez@mathstodon.xyz avatar

There's a dot product and cross product of vectors in 3 dimensions. But there's also a dot product and cross product in 7 dimensions obeying a lot of the same identities! There's nothing really like this in other dimensions.

We can get the dot and cross product in 3 dimensions by taking the imaginary quaternions and defining

v⋅w= -½(vw + wv), v×w = ½(vw - wv)

We can get the dot and cross product in 7 dimensions using the same formulas, but starting with the imaginary octonions.

The following stuff is pretty well-known: the group of linear transformations of ℝ³ preserving the dot and cross product is called the 3d rotation group, SO(3). We say SO(3) has an 'irreducible representation' on ℝ³ because there's no linear subspace of ℝ³ that's mapped to itself by every transformation in SO(3).

Much to my surprise, it seems that SO(3) also has an irreducible representation on ℝ⁷ where every transformation preserves the dot product and cross product in 7 dimensions!

It's not news that SO(3) has an irreducible representation on ℝ⁷. In physics we call ℝ³ the spin-1 representation of SO(3), or at least a real form thereof, while ℝ⁷ is called the spin-3 representation. It's also not news that the spin-3 representation of SO(3) on ℝ⁷ preserves the dot product. But I didn't know it also preserves the cross product on ℝ⁷, which is a much more exotic thing!

In fact I still don't know it for sure. But @pschwahn asked me a question that led me to guess it's true:

https://mathstodon.xyz/@pschwahn/112435119959135052

and I think I almost see a proof, which I outlined after a long conversation on other things.

The octonions keep surprising me.

https://en.wikipedia.org/wiki/Seven-dimensional_cross_product

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez Here's an idea: Let's encode the cross product as a 3-form, and understand the irreducible representation of SO(3) on ℝ⁷ as the space of harmonic cubic homogeneous polynomials on ℝ³, i.e. ℝ⁷≅Sym³₀ℝ³. (Here Sym means the symmetrized tensor power, and the subscript 0 means we take only those elements which are trace-free, corresponding to harmonic polynomials).

So what we are looking for is an invariant 3-form on this space, i.e. an SO(3)-invariant element of Λ³(Sym³₀ℝ³)*.

My hunch is that this may be written as the restriction of an SL(3,ℝ)-invariant element of Λ³(Sym³ℝ³)* (dropping the 0 because taking a trace is not an SL(3,ℝ)-invariant operation). Indeed, a calculation in the software LiE confirms that both representations have a one-dimensional trivial subspace.

Since Sym³ℝ³ is spanned by the cubes X³ for X∈ℝ³, and the alternating powers are spanned by wedge products, my naive guess for an invariant 3-form on this space would be to map
X³∧Y³∧Z³ ↦ det(X,Y,Z)³.

One would of course have to check that this extends to a well-defined 3-form on the whole of Sym³ℝ³, which I am currently not sure how to do.

But supposing we have found this 3-form, we can then restrict it to Sym³₀ℝ³ and then re-interpret it as a cross product (raising/lowering indices is still an SO(3)-invariant operation).

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez Yes, but it this the right representation? Perhaps you missed my earlier reply, but I think this representation is not irreducible. Aren't span{i,j,ℓ}, span{k,iℓ,jℓ} and span{kℓ} invariant subspaces?

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez It seems this maximal SO(3) is more mysterious than anticipated!

Looking at Dynkin's paper (Table 16), there are four conjugacy classes of 3-dimensional subalgebras of 𝔤₂. Two of those are the 𝔰𝔲(2) factors of the 𝔰𝔬(4) subalgebra, one belongs to the SO(3)⊂SO(4) acting reducibly on Im 𝕆, and one belongs to the maximal subgroup SO(3)ᵢᵣᵣ.

Each 𝔰𝔬(3) subalgebra contains what Dynkin calls a "defining vector", that is, an element in a fixed maximal torus of 𝔤₂. Not just any element of the torus can be a defining vector: Dynkin gives the possible coordinates in the basis of simple roots.

From this point of view all 𝔰𝔬(3) subalgebras are on equal footing, so I'm not sure whether one can speak of "generic" objects here. At a glance it looks like they all have the same degrees of freedom, namely a choice of maximal torus in 𝔤₂ plus a choice of simple roots.

pschwahn, to random German
@pschwahn@mathstodon.xyz avatar

The compact Lie group 𝐺₂, usually defined as automorphism group of the octonion algebra 𝕆 has (up to conjugacy) three maximal connected subgroups:

  • the subgroup preserving the algebra of quaternions ℍ⊂𝕆, which is isomorphic to SO(4),
  • the subgroup preserving some imaginary element like i, which is isomorphic to SU(3),
  • the subgroup SO(3)ᵢᵣᵣ given by the image of the irreducible, faithful 7-dimensional real representation of SO(3). This representation may be realized as the space of harmonic cubic homogeneous polynomials on ℝ³, or if you are a chemist, the space of f-orbital wavefunctions.

Now I wonder whether SO(3)ᵢᵣᵣ also has some interpretation in terms of the octonions. What irreducible action of SO(3) on the imaginary octonions is there?

@johncarlosbaez , do you perhaps have an idea?

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez I took that from nLab, Prop. 3.3 of the page on 𝐺₂: https://ncatlab.org/nlab/show/G2

Unfortunately I cannot find the original reference, but here is a sketch of my understanding:

Each automorphism φ of 𝕆 that preserves ℍ gives rise to an automorphism of ℍ, simply by restriction. But there is more gonig on: φ also preserves the orthogonal complement of ℍ (let's identify it with ℝ⁴), and its restriction to this 4-dimensional subspace will be some element of SO(4). Now it is not hard to show that φ is already completely determined by its restriction to the complement of ℍ, and perhaps slightly more difficult to see that every element of SO(4) gives a well-defined automorphism of 𝕆 in this way.

On the level of Lie algebras and their representations, here's what happens: Split

𝕆=ℝ⊕(Im ℍ)⊕ℝ⁴.

Using some Lie theory one can show that the subalgebra 𝔰𝔬(4)≅𝔰𝔬(3)⊕𝔰𝔬(3) of 𝔤₂ acts as the adjoint representation of the left 𝔰𝔬(3) factor on Im ℍ, and as the standard vector representation on the complement ℝ⁴. That means, it preserves the quaternion subalgebra, and the projection to the left factor should correspond on the level of groups to the surjection SO(4)→Aut(ℍ)≅SO(3).

pschwahn, (edited )
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez Exactly! And this gives rise to a short exact sequence SU(2)→SO(4)→SO(3), as noted on the nLab page.

I take it that by "in a nice way" you mean something like in terms of quaternions? That is, we represent elements of SO(4) as pairs (q₁,q₂) of unit quaternions modulo a common sign.

Here's an idea: Pick some (unit length) e in the complement of ℍ once and for all. Just looking at the type of representation of 𝔰𝔬(4) on Im 𝕆, my guess is that SO(4) has to act on the complement of ℍ by xe↦(q₁xq₂⁻¹)e, where x∈ℍ. Stipulating that this is the restriction of an action of SO(4) on 𝕆 that preserves the octonion product, and using the available identities of the octonions, I think I can show that the action on x=-(xe)e∈ℍ is given by x↦q₁xq₁⁻¹. Note that this depends only on q₁, just as we expect from the representation of 𝔰𝔬(4) on Im ℍ!

So if we equivalently write an octonion as a pair of quaternions x,y, then the action of [q₁,q₂]∈SO(4) is given by

(x,y) ↦ (q₁xq₁⁻¹, q₁yq₂⁻¹).

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez Thanks! Right now I'm still stuck at showing that my SO(4)-action above really preserves octonion multiplication... but searching the internet I found another argument in a set of lecture notes of Mark Reeder:

Any φ∈G gives (by restriction) rise to an orthogonal transformation of the complement ℝ⁴, as discussed earlier. This representation is faithful: if φ acted trivially on ℝ⁴, it would fix both e and xe for any x∈ℍ, and then we can conclude that it also fixes x, so it is the identity on 𝕆. Hence G embeds into O(4).

On the other hand, G surjects onto Aut(ℍ): any f∈Aut(ℍ) can be extended to an automorphism of 𝕆 by setting f(x+ye):=f(x)+f(y)e for x,y∈ℍ.

Together with the other end of the short exact sequence this should be enough to see that G≅SO(4).

Now, time to try and wrap my head around SO(3)ᵢᵣᵣ...

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez Oh, thanks - the Cayley-Dickson rule is really much easier to work with... and using this I see now that I made a mistake, the correct action is
(x,y) ↦ (q₂xq₂⁻¹, q₁yq₂⁻¹),
and this does in fact preserve multiplication!

Anyway, let's try your guess: Indeed, when I take a basic triple 𝑒₁,𝑒₂,𝑒₃ and apply some 𝐴∈SO(3), then the triple 𝐴𝑒₁,𝐴𝑒₂,𝐴𝑒₃ is also basic: 𝐴 preserves inner products, so the transformed triple is still orthonormal, and multiplying out the product (𝐴𝑒₁)(𝐴𝑒₂) I get a linear combination of 𝑒₁𝑒₂, 𝑒₂𝑒₃, 𝑒₃𝑒₁, which by assumption are orthogonal to 𝑒₁,𝑒₂,𝑒₃, which 𝐴𝑒₃ is a linear combination of.

So each 𝐴∈SO(3) indeed defines a unique automorphism of 𝕆 in this way, and the correspondence is injective by construction. But there is a problem. Denoting also with 𝐴 the extension to 𝕆, we have by definition 𝐴(𝑒₁𝑒₂)=(𝐴𝑒₁)(𝐴𝑒₂) which, as I just mentioned, lies in the span of 𝑒₁𝑒₂, 𝑒₂𝑒₃, 𝑒₃𝑒₁. Same goes for 𝐴(𝑒₂𝑒₃) and 𝐴(𝑒₃𝑒₁). So this representation has an invariant subspace, unfortunately.

I believe that 1, 𝑒₁𝑒₂, 𝑒₂𝑒₃, 𝑒₃𝑒₁ span a quaternion subalgebra of 𝕆, which is thus preserved by this representation of SO(3). So this SO(3) must be conjugate to a subalgebra of SO(4)⊂𝐺₂.

pschwahn,
@pschwahn@mathstodon.xyz avatar

I thought me mentioning the f-orbital was just a crackpot idea.

But in the AMS volume "Selected Papers of E. B. Dynkin with Commentary" (which also contains Dynkin's original discovery of SO(3)ᵢᵣᵣ) one finds a short review by Yuval Ne'eman, titled "Dynkin Diagrams in the Physics of Particles, Fields and Strings". The whole thing is a delight to read, but he writes something particularly interesting about an idea of physicist Giulio Racah:

"Racah found ways of applying various simple algebras in classifying higher spectra. His methods, later developed and extended by such as L. Biedenharn and M. Moshinsky, exploited higher rank Lie algebras applied to the representation spaces of SO(3). I recall Racah enjoying (anecdotically) the fact that he had found an application for Cartan's exceptional G(2), in studying the f-shell in atomic spectra. One defines an SO(7) algebra acting on some constructs involving the 7-dimensional f-shell representation of SO(3) - and the inclusion G(2)⊂SO(7) does it.. In these very complicated atomic spectra of the lanthanides, it provides some physical insights."

I really wonder that these physical insights are...
@johncarlosbaez

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez Now I have this stupid image in my head of someone in a lab coat looking through an optical spectrometer and noting to themselves "ah yes, that's 𝐺₂ there..."

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez This is fascinating, and so is the note written by Borthwick and Garibaldi linked in this thread - a fine addition to my reading list.

https://arxiv.org/abs/1012.5407

johncarlosbaez, (edited ) to random
@johncarlosbaez@mathstodon.xyz avatar

@TruthSandwich got me interested in the vibrational modes of bells. They're not harmonics with frequencies 1, 2, 3, 4, ... times the lowest frequency: they're much more complicated! That's why bells sound clangy. This chart shows how they sometimes work.

The lowest frequency vibrations are called:

• the 'hum' (the lowest frequency)

• the 'prime' (with frequency roughly 2 times that of the hum)

• the 'tierce' (roughly 2.4 times the hum, so a minor third above the prime)

• the 'quint' (roughly 3 times the hum, so a major fifth above the prime)

• the 'nominal' (roughly 4 times the hum, so an octave above the prime)

and so on. If you think these names are illogical, join the club! One reason it's tricky is that the loudest vibration is not the lowest one: it's the 'prime'.

The numbers I just gave you should be taken with a big grain of salt. They really depend on the shape of the bell, and you'd have to be great at designing bells to make them come out as shown here. It's not like a violin string or flute, where the math is on your side.

This quote helps explain the chart:

"Modern theory separates the modes of vibration into those produced by the "soundbow" and those produced by the remaining bell "shell". The bell vibrates both radially and axially and the principal vibrational modes are shown in the diagram together with their classification using the scheme proposed by Perrin et al. This scheme consists of the mode of vibration (RIR - Ring Inextensional Radial, RA - Ring Axial, R=n - Shell driven), the number of meridians (where “m” is half the number of meridians) and the number of nodal circles (n)."

Starting to sound like orbitals in quantum mechanics!

(1/3)

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez As someone who likes to dabble in spectral geometry, I now wonder whether it's possible to hear the shape of a bell.

Assuming the linear approximation is not far from the truth, what are the correct boundary conditions? Can't be just Dirichlet at the center, right?

Thanks to your post I just realized bells are nice examples of why one would want to study the spectrum of a Riemannian manifold of cohomogeneity one - that might come in handy for the one or other research proposal ;)

pschwahn,
@pschwahn@mathstodon.xyz avatar

@tsrono @johncarlosbaez Did some digging and had to show to someone: In one of the experimental papers cited here (Grützmacher et al.: Akustische Untersuchungen an Kirchenglocken/Acoustic study of church bells, Acustica 16, 1965) one finds this cool image of the sound field of a model bell, displaying both phase and amplitude of the sound waves simultaneously.

The picture is itself taken from a work of Schroeder (Ein optisches Verfahren zur amplituden- und phasengetreuen Darstellung stationärer Schallfelder, Acustica 13, 1963).

Really gives me some 60's horror movie vibes!

pschwahn, to random German
@pschwahn@mathstodon.xyz avatar

Non-semisimple Lie groups are so weird. Weyl's unitarian trick does not work for them. So I need to constantly remind myself that:

  1. representations of GL(n,ℂ) are not determined by their character,
  2. not every finite-dimensional representation of GL(n,ℂ) is completely reducible,
  3. Finite-dimensional GL(n,ℂ)-representations are not in 1:1-correspondence with finite-dimensional U(n)-representations.

However these work when you look only at irreducible representations, or when you replace GL by SL (and U by SU). The archetypical counterexample is given by the (reducible but indecomposable) representation
[\rho: \mathrm{GL}(1,\mathbb{C})=\mathbb{C}^\times\to\mathrm{GL}(2,\mathbb{C}):\quad z\mapsto\begin{pmatrix}1&\log |z|\0&1\end{pmatrix}.]
(Example shamelessly stolen from: https://math.stackexchange.com/questions/2392313/irreducible-finite-dimensional-complex-representation-of-gl-2-bbb-c)

Turns out that entire StackExchange threads can be wrong about this (for example https://math.stackexchange.com/questions/221543/why-is-every-representation-of-textrmgl-n-bbbc-completely-determined-by), so be wary!

pschwahn,
@pschwahn@mathstodon.xyz avatar

Huh? Am I tripping, or is Wikipedia also wrong on this?

https://en.wikipedia.org/wiki/Unitarian_trick

"An important example is that in which G is the complex general linear group, and K the unitary group acting on vectors of the same size. From the fact that the representations of K are completely reducible, the same is concluded for those of G, at least in finite dimensions."

So it claims that all finite-dimensional representations of GL(n,ℂ) are completely reducible?
Surely I must have misunderstood something??

pschwahn,
@pschwahn@mathstodon.xyz avatar

@AxelBoldt Thanks for chiming in! I'm going to reply here since I'm not familiar with the Wikipedia rules/etiquette.

GL(n,ℂ) is certainly reductive (see Fulton-Harris, Ex. 9.25). Perhaps there is a version of the Weyl trick which does apply to reductive groups, but I am not aware of its formulation.

I think your suspicion in the Wikipedia talk about algebraic groups is correct: The (finite-dimensional) RATIONAL representations ρ of GL(n,ℂ), i.e. those where the entries of ρ(g) are rational functions of the entries of g∈GL(n,ℂ), are indeed determined by their character, completely reducible, and in 1:1-correspondence with finreps of U(n). And the counterexample I wrote down is not a rational representation!

See e.g. these lecture notes: https://pages.uoregon.edu/belias/WARTHOG/infcomm/LectureNotes/Lecture7Notes.pdf

Unfortunately I don't have a reference for the converse (that every completely reducible representation of GL(n,ℂ) is rational), but I strongly suspect this is the case.

pschwahn,
@pschwahn@mathstodon.xyz avatar

@AxelBoldt Fulton-Harris describes Weyl's trick in Chapter 9 and it quite adamant about that it only works for semisimple Lie groups. (I don't think analyticity of representations is required however, just smoothness.)

Generally, the trick works by establishing a 1:1-correspondence between the (finite-dimensional) representations of a complex Lie group G with that of a compact real form G₀. And I think this can be done if the group G is reductive (hence a compact real form exists) provided we consider only rational representations.

The critical step, I believe, is that a complex representation of (\mathfrak{g}=\mathfrak{g}_0^\mathbb{C}) is determined by its restriction to (\mathfrak{g}_0). This fails for our counterexample, but seems plausibly true for rational (or even analytic?) representations. We just need the infinitesimal representation (d\rho: \mathfrak{g}\to\mathfrak{gl}(n,\mathbb{C})) to be complex-linear, right?

pschwahn,
@pschwahn@mathstodon.xyz avatar

@AxelBoldt If I find anything about the rational case, I'll get back to you!

Right, the operations of G are holomorphic - but representations of Lie groups are usually taken as being just smooth (the representation doesn't necessarily "see" the complex structure of G). The problem (existence of non-rational representation) probably also applies to other (noncompact) real forms of G.

pschwahn,
@pschwahn@mathstodon.xyz avatar

@AxelBoldt you're right, that does make sense. I just realized that (z\mapsto z^k|z|^w) is a representation of (\mathbb{C}^\times) for all fixed (k\in\mathbb{Z}, w\in\mathbb{C}) but only analytic when (w=0).

So we should really be talking about complex analytic representations.

However for a group such as GL(𝑛,ℝ) it looks like the issue persists - a representation such as 𝑔↦|det(𝑔)|ʷ for non-integer w will be real-analytic, but it will not correspond to a representation of U(n).

This is turning out more subtle than I had imagined...

pschwahn, (edited )
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez @AxelBoldt oh yeah, that does make sense! I was trying to compare representations of GL(n,ℝ) and U(n) because they have the same complexification, but of course, the maximal compact of GL(n,ℝ) is O(n).

This description of irreducible smooth representations does sound plausible! However I don't think every finite-dimensional smooth representation of GL(n,ℝ) is completely reducible - take the example from the beginning,
[\mathrm{GL}(1,\mathbb{R})\to\mathrm{GL}(2,\mathbb{R}),\quad t\mapsto\begin{pmatrix}1&\log|t|\0&1\end{pmatrix}.]
But this counterexample of course goes away if we only care about rational representations.

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez @AxelBoldt Yep, I think rational representations of GL(n,ℝ) can without problems be complexified to rational representations of GL(n,ℂ), which are then complex analytic (so in particular they correspond to the smooth finreps of U(n)). And complex analytic representations of GL(n,ℂ) are discussed in Fulton-Harris Section 15.5 - they are exactly as you said!

johncarlosbaez, (edited ) to random
@johncarlosbaez@mathstodon.xyz avatar

So you wake up one day wanting to invent a 2-dimensional number system. This requires a new number 𝑖 that's at right angles to 1. So you figure multiplying by 𝑖 must rotate numbers by 90°. So multiplying by 𝑖² rotates by 180°, so

𝑖² = -1

Cool!

Then you notice something else. The derivative of a function in the 𝑦 direction must be 𝑖 times its derivative in the 𝑥 direction, because the derivative is linear and you get the 𝑦 direction by rotating the 𝑥 direction by 90°: that is, multiplying it by 𝑖. So you get this equation:

[ \frac{\partial f}{\partial y} = i \frac{\partial f}{\partial x} ]

Cool!

Then you notice something else. If you use this equation twice you get

[ \frac{\partial^2 f}{\partial y^2} = i \frac{\partial f}{\partial x\partial y} = i^2 \frac{\partial^2 f}{\partial x^2} = - \frac{\partial^2 f}{\partial x^2} ]

so

[ \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = 0 ]

Wow! Every function with a second derivative obeys the Laplace equation!

You decide this one is a keeper.

https://en.wikipedia.org/wiki/Cauchy%E2%80%93Riemann_equations

pschwahn,
@pschwahn@mathstodon.xyz avatar

@vacuumbubbles @johncarlosbaez I've never heard of the geometric derivative! I'm guessing it is something like (d+d^\ast) in the language of differential forms?

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez @vacuumbubbles I think a notational accident has occurred: * may be read as the Hodge star operator or the L² adjoint.

I meant (d^\ast) to denote the codifferential, i.e. adjoint of (d), which is indeed (\pm\ast d \ast) depending on degree.

johncarlosbaez, (edited ) to random
@johncarlosbaez@mathstodon.xyz avatar

This curve is not an elliptic curve - because even though you can write it in as

y² = P(x)

with P a cubic polynomial, elliptic curves need to be smooth! We say this curve is 'singular', not smooth everywhere, because it crosses itself at one point, making a kind of X shape. Mathematicians call this point a 'node'. So this curve, which I'd rather write as

y² = x³ - x²

is called a 'nodal cubic'.

It's still fun to count the solutions of this equation in a finite field. Let's do it!

(1/n)

pschwahn,
@pschwahn@mathstodon.xyz avatar

@johncarlosbaez @dimpase I feel you there completely! Especially the constant digging through the internet to find a guide for doing basic thing on whatever OS/software/programming language always gets on my nerves.

I might add that the WSL setup works, but is suboptimal - in particular computations are much slower if done through a virtual machine.

Recently I bit the bullet and installed Ubuntu on my laptop on a new partition. I also use it for other stuff but I must say that for Sage alone it was worth it - there's just so much stuff one can do with Sage!

The whole procedure just took two major steps. First, installing Ubuntu from a bootable USB stick: ubuntu.com/tutorials/create-a-usb-stick-on-ubuntu

It takes care of creating the partition and everything automatically! Of course, Ubuntu is not strictly required here, but I found it to be the simplest option.

Second, installing Sage with Jupyter notebook - and this is done just by typing "sudo apt install sagemath-jupyter" into the console.

And then you're good to go! To launch it type "sage -n jupyter".

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