at the border of mathematics and quantum physics

[@MichaelPennMath]

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[@leif1075]

Thanka for sharing this.

[@primenumberbuster404]

please make more mathematical physics content. 😍

[@pacificll8762]

++++

[@hellfire7685]

yes please!! kinda more interesting than actual math

[@davidwright5719]

Physicist here. I didn’t see any mathematical physics in this video.

[@hansolo9892]

@@davidwright5719 Sounds like a skill issue tbh.

[@mastershooter64]

@@davidwright5719 mathematical physics is different from theoretical physics, perhaps you're thinking of the latter

[@synaestheziac]

23:10 “A good place to… look for something in the center”

[@mehrdadassar2542]

It will be very interesting to see the second part of this lecture.

[@dawielindeque4555]

Hey. Love your channel! Here’s a number challenge. With the Rugby World Cup underway, I thought a Rugby related challenge might be nice. Proof: what scores are possible for each team in a rugby game, or, what scores are not possible (clue, I belief all scores are possible > 5). Once that’s done, how many ways are each score possible, i.e.: a team can get to 28 by scoring 5,5,5,5,5,3, or 7,7,7,7. What is the highest score achieved only once way, etc.

[@hankseda]

Congratulations on making math beyond high school more accessible 👏 I subscribe to your channel because you know what you are talking about and focus on content over audio-visual distraction 👍

[@ahsgdf1]

Very interesting. Thank you. Quantum mechanical operators for the conponents of angular momentum can be constructed form SL2 operators via Lx = - i e + i f, Ly = h/2 + f, Lz = e+f-h (1) the Inversion of (1) is e -> (3 I Lx)/4 + Ly/2 + Lz/4, f -> -(I Lx)/4 + Ly/2 + Lz/4, h -> (I Lx)/2 + Ly - Lz/2 The commutation rules are [Lx, Ly] = i Lz and cyclically continued. The casimir operator is the square of the vector of the angular momentum: L^2 = Lx^2+Ly^2+Lz^2

[@ruffifuffler8711]

I once buried 3 different bags of apples along the side of a road, came back 15 years later to find 3 apple bushes, and am still wondering if I had 3 algebra centers, 2 centre's and a center, or 3 centre's?

[@abdulkhadarainur4348]

Quantum Physics and Mathematics (Primarily Integral Calculus) are my two favourite subjects. So it's nice seeing them mingle. And since this is happening like this, guess Lie Algebra overtakes Integral Calculus on my list.

[@nicolasreinaldet732]

See Lie Calculus.

[@abdulkhadarainur4348]

@@nicolasreinaldet732 Well, currently, I'm struggling with Lie Groups.

[@pierrot31511]

All of this was a lecture I had in my 2nd year of Master degree, 11 years ago. Very good memories !

[@elijahberegovsky8957]

Pleeeeeeeease do something on the Virasoro algebra next! I’m reading up on CFT now and a Virasoro themed video would be of immense help for understanding that!

[@georgedoran9299]

You may have already found it but he already made one

[@mickschilder3633]

I absolutely love these more (Lie) algebra focussed videos 👍

[@hadiseigool9889]

more on the last equation please :) thanks for the great content

[@mrl9418]

No it's not a good place to stop. Make it a 100h video [maybe a playlist, ok] where you get into all the literature out there !

[@CM63_France]

Hi, 23:19 : good place to look for something at the center.

[@Mystery_Biscuits]

23:07 good place to look

[@travisporco]

love that you get into some deep stuff

[@jamesfortune243]

I knew a graduate physics student in Boston named Charles M. Blue Jr. He had come up with a lot of combinatorial identities I've never seen before as a hobby. I asked he how he came up with those identities. He claimed that they came out of playing around with noncommutative algebras. He moved to California and I've not heard of him since then.

[@theelk801]

please more of these videos (and more differential forms videos too)

[@Alan-zf2tt]

I sort of agree with sentiments expressed by others about the appeal of physics. Physics seems to have a "wow" factor where an algebra can be part of something "big". But as far as math goes: it is just another algebra. Same with isomorphisms.

[@krisbrandenberger544]

The a and d in the lower left entry of the difference Z*X-X*Z need to be swapped.

[@1.4142]

Each video is more interesting than the last

[@GeoffryGifari]

This one got my brain going for a bit... 1. Can a universal enveloping algebra be built from any lie algebra? 2. Does an algebra always have a center? 3. How many centers can a lie algebra have? 4. Is there a relation between a casimir element of a lie algebra (center of its universal enveloping algebra) and the center of the lie algebra itself? 5. the matrix form of e,f,and h look almost exactly like spin raising/lowering operators and spin projection operator (related to how e and f adds/subtracts "charge"?).... there might even be a deeper connection here 6. Computing Ω in matrix form we'll get a multiple of the identity matrix. Is the identity element connected to the center element (maybe related to its need to commute)? Just some thoughts

[@iabervon]

An algebra always has 0, and it's always in the center. That may be the only element in the center, though, like with (R³, ×), which obviously doesn't have any special elements other than 0.

[@mickschilder3633]

Well the idea of a center is not well defined on a Lie algebra, only on it's enveloping algebra, since a Lie algebra does not in general have a multiplication, so the condition xy-yx=0 is not well defined. Now, if you take the lie bracket as the multiplication you get the condition [x,y]-[y,x]=0, thus 2[x,y]=0, which would probably point to the center to just be {0}.

[@Uoper12]

1. Yes, the construction works for any lie algebra. 2. Yes, if no elements other than the identity are in the center the algebra is said to have a trivial center. 3. The center of a lie algebra is unique and is defined as the set of all elements x of the lie algebra where [x,y]=0 for all y in the lie algebra, and notably if the lie algebra is semisimple the center is trivial. 4. This answer is gonna get a bit in the weeds, but for any finite dimensional Lie algebra you can define a special bilinear form called the Killing Form B(-,-) (if you're interested B(x,y) is the trace of (ad_x)(ad_y)). A casimir element can only exist when B is a non-degenerate bilinear form (this is an equivalent condition to the lie algebra being semisimple). But since the lie algebra is semisimple its center only contains the 0 element. However, there's a pretty well-known linear algebra theorem that if g is a finite dimensional vector space over a field k (lie algebras are vector spaces), then g is isomorphic to its dual g* (which is the vector space of all functions from g to its base field k). Then the way you should think of a casimir element is as an element of g⊗g that corresponds under the isomorphism to the Killing form in g*⊗g*. (This isomorphism isn't unique so B can correspond to any element of the center of U(g) and that's why we don't say casimir element we say casimir element). Tl;dr there isn't a connection to the center of a lie algebra, but a casimir element does come from somewhere. 5. Not a physicist, but I'm pretty sure lie groups and algebras are used extenisively in quantum mechanics so this shouldn't be entirely surprising, though I think they use SO(4) and some others I can't remember off of the top of my head. 6. A fairly common exercise in a linear algebra class is to prove that the only types of matrices that are in the center of a matrix algebra are "scalar matrices" i.e. a multiple of the identity matrix, so it makes sense then that a central element of U(g) can be represented as a scalar matrix.

[@GeoffryGifari]

thanks!@@Uoper12

[@drdca8263]

@@mickschilder3633I would think to define the center of a Lie algebra to be the space {x | for all y, [x,y]=0}

[@rainerzufall42]

Question: If you consider, that ef + fe = h² => ef + fe + 1/2 h² = 3/2 h², isn't it fair to claim, that h² itself is a Casimir element of sl_2(IC) ? You can easily check (with the help of [xy,z] = x [y,z] + [x,z] y), that the whole base commutes with h²: [h²,h] = h [h,h] + [h,h] h = h 0 + 0 h = 0 [h²,e] = h [h,e] + [h,e] h = 2 he + 2 eh = 0 [h²,f] = h [h,f] + [h,f] h = - 2 hf - 2 fh = 0

[@rainerzufall42]

It's easier to visualize, if you consider, that h² = 1 (= [1 0; 0 1]).

[@GeoffryGifari]

do mathematicians really call those fancy letters "fancy"? lol

[@honor9lite1337]

Only fancy mathematicians do.

[@jean-francoistremblay7744]

I'm actually a math student at a French speaking University in Canada and when these fancy letters come around, the teacher refers to them as "calligraphique" which evidently means "calligraphic", "in a calligraphy manner"...and yes, it's specified all the time =D

[@goodplacetostop2973]

29:22

[@yuseifudo6075]

You're a Legend

[@goodplacetostop2973]

@@yuseifudo6075 I know 😎

[@peasant12345]

lie bracket is a 'derivation'. You can use this property to simplify a little bit.

[@RoboticusMusic]

I need visuals to understand any of this. Team up with some talented programmers, there are many people who make art with math and code.

[@trueriver1950]

Whether you are doing maths or physics is a question of motive, not of the actual thoughts you write down. If you like playing with definitions and symbols and questions like "what can I logically derive from this assumption?" then you're doing maths. If you are turning the handle on the process in order to make predictions that you hope will more closely match real experimental results, then you're doing theoretical physics Thirdly: if you're doing the same process again but in order to create a new machine or a new technical process, then you're doing engineering.

[@archismanrudra9336]

The definition of alternating property that the video is using assumes that the characteristic of the underlying field is not 2. I assume it is better to actually assume [x,y] = -[y,x] ?

[@PsyKosh]

Am a bit unclear. If you have a polynomial with terms of different charge, how's the charge of the whole defined? Is it simply undefined in that case? like, does e + h have a defined charge?

[@Browl22LPs]

What I find quite intriguing is that if ch[x,z] = ch(x)+ch(z) and we want [x,z] = 0 for all x \in A, then for the casimir element with ch(z) = 0 we would have ch[x,z] = ch(0) = ch(x) where ch(x) could be any value as x can be any 'polynomial'. That makes the "vacuum" element '0' have every charge at the same time. Would that be a bug or a feature?

[@mrl9418]

I was wondering the same thing. From what it looks like, maybe the element 0 will formally have - infinite or +infinite charge like in degrees of polynomials. But I don't know

[@TheEternalVortex42]

20:00 isn't charge(-he -eh) = 2 ? why does he say it's 1

[@Nikolas_Davis]

we're only counting each term's charge. In other words, multiplication contributes to charge, but not addition. Addition merely puts together many terms that may have the same or different charges. In terms of physics, each term would signify a state of fixed charge, or spin, or some other conserved quantity; and the fact that all terms have the same "charge" means that "charge" is a well-defined "quantum number" of the total state.

[@chemicalbrother5743]

29:22 good place to what?

[@TymexComputing]

Yes - thank you! 777th thumb up!

[@annaairahala9462]

A lot of people don't realize that math and physics aren't any different. Math is just the language used while Physics takes that language and applies it to our observations of the real world

[@talastra]

Math and physics are different precisely in the fact that math is idealized and physics is not. Working at a scale where the idealizations of math fit into the thresholds of physics are where math works; otherwise, it doesn't.

[@mrgadget1485]

Please address the representations of the Poincare group.

[@wolliwolfsen291]

Interesting video but somehow clickbaiting. The concrete link to physics is totally missing. Please fulfil what you promise with the title of this video. But anyway you neither will read this comment nor you give an answer

[@alanthayer8797]

PLEASE MK PHYSICS MATH VIDEOS LIKE DA OTHA COMMENT B4 said

[@kkanden]

how does the sl2(C) group laugh? [e,f]e [e,f]e [e,f]e [e,f]e

[@mrl9418]

Omg

[@baronvonbeandip]

brilliant lol