Can we exponentiate d/dx? Vector (fields)? What is exp?

Can we exponentiate d/dx? Vector (fields)? What is exp? | Lie groups, algebras, brackets #4

Рет қаралды 114,639

Күн бұрын

Part 5: • Matrix trace isn't jus...
Can we exponentiate vectors? What does e^(d/dx) mean? Does it make sense to exponentiate a whole bunch of vectors? Well yes! While what these exponentials do seem very different at first, they can be recast into the same framework.
Files for download:
Go to www.mathemaniac.co.uk/download and enter the following password: expderivativeshift
CHAPTERS:
00:00 Introduction
01:03 What is exponentiation?
04:15 Exponentiating vectors
11:23 Exponentiating derivatives
24:04 Exponentiating vector fields
❗Remark❗
1️⃣ I know that many people would be thinking of series expansion of exponentials. I deliberately avoid this because it is not conducive to learning the intuition of the exponential, and more crucially, it does not apply to the exponential of vectors on manifolds. The result is very manifold-dependent, and I will be very impressed if there is a series-like explanation for the exponential map in differential geometry.
2️⃣ However, I want to know: is there a generalisation of the translation operator statement in the video to manifolds? For a flat plane, we have exp(a * nabla) f(x) = f(x + a). And in fact,the exponential map on the flat manifold of R^n gives x + a = exp_x (a). Hence, for flat R^n, we have exp(a * nabla) f(x) = f(exp_x(a)). Can this be generalised to general manifolds? Is it true if I interpret nabla as not a normal gradient, but covariant derivative? Please let me know if you have any ideas for it. I want this to be true because it connects different “exponential” ideas.
📖 Further reading 📖
1️⃣ Exp vectors
Exponential map in Riemannian geometry (if you actually want to know how this is just a generalised exponential map in the usual sense, rather than just having the same “philosophy”, then go to the relationship to Lie theory section - when they say translations, they mean multiply on the left/right by g, a group element of the Lie group): en.wikipedia.org/wiki/Exponen...)
Why exponential map in differential geometry is useful: en.wikipedia.org/wiki/Normal_...
❗You might also need to learn these first before tackling the link above❗
Riemannian metric: www.ime.usp.br/~gorodski/teac...
Connection (the introduction is the most illuminating part): en.wikipedia.org/wiki/Affine_...
math.stackexchange.com/questi...
Levi-Civita connection (a particular kind of connection that makes the metric invariant): en.wikipedia.org/wiki/Levi-Ci...
2️⃣ Exp d/dx
What I have described is kind of solving PDE with the method of characteristics (identifying characteristic curves along which it becomes an ODE): en.wikipedia.org/wiki/Method_...
The partial differential equation is part of the wave equation: en.wikipedia.org/wiki/Wave_eq...
Translation operator in quantum mechanics: en.wikipedia.org/wiki/Transla...)
Time-ordering: solving differential equations of the form ∂f/∂t = X(t) f, where X(t) is a time-dependent differential operator, e.g. t^2*∂^2/∂x^2: en.wikipedia.org/wiki/Ordered...
Time-ordering in example in QM: en.wikipedia.org/wiki/Dyson_s...
3️⃣ Exp vector field
Vector flow: en.wikipedia.org/wiki/Vector_...
This is more related to the video: en.wikipedia.org/wiki/Flow_(m...)
Textbook: www.worldscientific.com/doi/p...
I actually wanted to say the following, but I think the video is long enough and didn’t include it into the script, but vector field is actually related to (and most often described by) differential operators, and in that sense both exponential of (1st-order) differential operators and exponential of vector fields yield very similar things: en.wikipedia.org/wiki/Vector_...
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Пікірлер: 172

@mathemaniac 4 ай бұрын

Sorry that I have been away for so long! I am now a PhD student, and it is very difficult for my brain to switch from my research to the channel. Luckily the Christmas holidays mean that I can focus solely on the video. I do want to continue the series, but I will also have to learn to switch more easily to allow working on videos in the midst of my research. The very rough field of my research is in cosmology, and I am really excited to share it, but the background knowledge is not very accessible, even to this audience. That would mean at least an entire series on GR and QFT, and I am not even sure when this Lie theory video series will end. So that’s really off the table for now.

@Mandragara 4 ай бұрын

There are probably a fair few maths\physics PhDs in your audience. Also laypersons like listening to things they don't understand. I wouldn't write off doing a video on your research completely. Especially if you can use existing material from a conference presentation etc as the basis for a video (or vice versa)

@domenicobianchi8 4 ай бұрын

Good luck for your phd and congrats for this great interesting video. I saw you refer to 3blue1brown video, which hints to a part 2 who was never realized! I wait for the follow up since ages, it would be absolutely great if you could in your video address it, at least in part!

@mathemaniac 4 ай бұрын

@@domenicobianchi8 Which part 2 do you mean? If it is about e^(d/dx), it is already explained in the video, but if you are referring to trace, then it should be the next video on this channel.

@mathemaniac 4 ай бұрын

@@MandragaraI really don't want people who are listening don't understand it - that is what exactly I'm experiencing right now as a PhD student, and that does not feel good, and also something that I don't want to inflict onto others. Of course I don't want to say I will not make a video on my research, but it would have to be of different nature to the usual videos on the channel.

@domenicobianchi8 4 ай бұрын

@@mathemaniac yes, i refer mainly to the Jacobi formula. But also i got the feeling (I could be wrong) that the explanation for the exponentiation of d/dx Grant had in mind is different from the one you presented here.

@ryancantpvp 4 ай бұрын

It’s incredible how accessible this series is. Despite requiring knowledge from topics like multivariable calc, linear algebra or group theory, I’m sure even the people who are unfamiliar to those ideas can follow along relatively easily (including me)

@ozachar Ай бұрын

Your initial statement, of understanding the exponent as being the cumulative change from the original state was so illuminating. As a physicist, it connects with understanding why this is, and must be, the mathematical form of all Generator operators (translation, rotation, time evolution, etc...).

@columbus8myhw 4 ай бұрын

This is great! Previously, the main intuition I had for e^(d/dx) being the shift operator was by writing it as I+d/dx+(d/dx)^2/2+(d/dx)^3/6+..., in which case the result is identical to Taylor's theorem.

@mathemaniac 4 ай бұрын

This argument allows more general functions though - the function only needs to be (continuously) differentiable not analytic. [I suspect it only needs to be differentiable, not even continuously differentiable]

@arghyashit6169 3 ай бұрын

@@mathemaniac😅

@TagRLCS 4 ай бұрын

A video regarding the intuition of trace would be absolutely georgous. Despite my best efforts, I am unable to truly find a satisfying geometrical view of the trace. I currently satsify with it's algebraic properties (sum of eigenvalues for example) but I am really dying for something deeper

@mathemaniac 4 ай бұрын

If everything goes well, the next video is about trace.

@05degrees 4 ай бұрын

@@mathemaniac YAY! I’m also down to get some trace intuition. All I know is using actions of a linear operator on exterior algebra to define its det, tr and other coefficients of the characteristic polynomial, but it’s not evidently visual despite being geometric in a way. (Well, algebraic with no coordinates-but like we get instant geometric intuition for det this way okay, but trace needs brain exercise and I just can’t get to it ever.)

@evanlafontaine2536 4 ай бұрын

The exponential of d/dx being a shift reminds me quite a bit of the Laplace Transform where s f(s) is differentiation and e^s f(s) is time delay. In my electrical engineering courses, they were simply given as theorems on a formula sheet without much motivation as to why the latter makes sense. The framework in the video really gave me a satisfying answer to why e^s does this (beyond algebra bashing). Keep up the great work!

@05degrees 4 ай бұрын

Similar for Fourier too, though as we know Fourier is related to Laplace so it’s also no big wonder.

@dkosolobov 4 ай бұрын

This sounds very intriguing. I could not understand how the intuition from the video could be translated to the Laplace transform of e^s f(s) or s f(s) (probably because I'm not well familiar with it). Can you explain this intuition?

@nicolasreinaldet732 4 ай бұрын

It also relates to how in quantum mechanics we see the momentum as the generator of translation and the momentum operator is just a multiple of d/dx.

@binhnguyenquoc3249 11 күн бұрын

@@dkosolobov okay, it's been long since you ask this, but in the video, it mentioned that the exponential of a derivative (of a function) is just that function translate 1 unit (f(x) to f(x+1)). In the case of Laplace Transform, s*f(s) is equivalent to the derivative of f(t) (or f(t)_dot) when you transform it to the time domain, so e^s*f(s) is equivalent to the exponential of the derivative (of the function f(t)) when you transform it to the time domain, which according to the above, equal to f(t+1), or in Laplace transform, you have the general formula L^-1{e^(n*s)*f(s)} is equal to f(t+n).

@_P_a_o_l_o_ 4 ай бұрын

It was extremely deep to visualise why exponentiating a derivative gives a shift operator. I have seen this result being obtained by purely algebraic manipulation, but your geometric visualisation was stunning! Keep up the good work :)

@Rififi50 4 ай бұрын

Nice to see the expontential using the differential definition. I clearly have been too hung up on the series expansion! (Though that does work with derivatives as well, but misses out on the intuitive shift operator connection.)

@mathemaniac 4 ай бұрын

As said in the description, I deliberately avoided the series expansion because of both the intuition and the fact that this does not relate to the exponentiation of vectors.

@goclbert 4 ай бұрын

The series definition is useful for many computations, but the derivative definition offers the most intuition. I feel like so often I was surprised whenever exp would appear but since I've internalized that it's the eigenfunction of the derivative operator, it usually feels very natural.

@snuffybox 4 ай бұрын

@@mathemaniac Out of curiosity what happens if you use the series expansion with a vector and use the geometric product to evaluate it? That set up works out for bivector blades and is equivalent to the complex exponential. I am curious if doing it with a vector is equivalent to the vector exp described in this video. I have to admit I did not fully follow the vector exp described in this video, I am pretty weak in calc, but the description of a vector as complex numbers seems really sus to me as they are not complex numbers, the proper vector interpretation of a complex number is a scalar+bivector.

@mathemaniac 4 ай бұрын

@@snuffybox The vector exponentiation here is very manifold-dependent. It returns another point on the manifold, and if you change the manifold, then you will get a very different point (or in some cases, there might not even be a point). The point of this video is to not use series expansion because it does not apply in this case. However, I will be very happy if there is an even a series-like expansion to exponentiate the vectors.

@pierrebaillargeon9531 4 ай бұрын

Great video. The only thing I'd like to learn in these types of videos is the origin story of the generalizations. I think maths is often taken as mysterious, incomprehensible and magical because we're rarely taught the journey and every result seems to appear out of nowhere. You've given a great explanation for the justification, but not how the current form was first come up with or if there were multiple failed attempts or variations.

@mathemaniac 4 ай бұрын

It is linked to Lie theory, see here: hsm.stackexchange.com/questions/3558/how-did-the-exponential-map-of-riemannian-geometry-get-its-name#:~:text=I've%20read%20in%20several,metric%20on%20the%20Lie%20group. There is no definitive answer on the history of this actually :(

@sanewitch8036 4 ай бұрын

This is so cool. So many useful for theoretical physics consepts emerge naturally from just this one definition of exponentiation.

@jdambrine 4 ай бұрын

Very interesting presentation! A few weeks ago I taught my students about the exponentiation of matrices, and they were baffled by the idea that something else than real or complex numbers could be exponentiated. I am considering sending them the link to your video, to show them there is an even more general way of thinking about this. Just a small caveat about the exponential of differential operators (say, D, here) : it is important to keep in mind that exp(tD)f is not always properly defined (in the sense that exp(tD)f has to be in the same space as f). I guess this is one of the problems with working with infinite dimensions! As a counter-example one could consider D=-d^2/dx^2. Then, the "intermediate condition" would be the backwards heat equation which is ill posed (at least for the most classical function spaces). Hence, for D=d^2/dx^2 we end up with a solution generated by a semi-group rather than a group. In the case of reversible PDES (for instance if D is of odd order) there is no problem though. Anyways, thanks a lot for your work!

@mathemaniac 4 ай бұрын

Thanks for bringing up the caveats. We don't need the strong condition of continuous dependence in well-posedness though - just existence and uniqueness. But I don't know enough of PDEs to say definitively that this is not true for backward diffusion, because the usual punchline to the ill-posedness of backwards diffusion is that it violates continuous dependence, not existence / uniqueness. In any case, we can refine it to saying *IF* a solution exists and is unique to the PDE then we can write it in terms of the exponential operator.

@ccc3001 4 ай бұрын

I was struggling and surprised at the exponential map years ago when I was studying Lie group. What a beautiful and generalized concept ! Love to see these stuff.

@danthewalsh 4 ай бұрын

The explanation of the shift operator being generated by the derivative operator also gives an explanation for why the conjugate of position, momentum, has an operator in quantum mechanics that is proportional to a spatial derivative.

@ebog4841 4 ай бұрын

THESE ARE PERFECTLY NORMAL THINGS TO EXPONENTIATE! F the haters!

@HalKworasmi 4 ай бұрын

Exp did nothing wrong! This were perfect exponentiations

@pseudolullus 4 ай бұрын

A pretty nice alternative way to see why momentum is both a derivative operator and the generator of translations :)

@varshneydevansh 4 ай бұрын

Watching this along with my breakfast made my day and yes, this is the first I understood what exponents are.

@ManInTheArena-nl4ti Ай бұрын

Hey, I am doing Honours in Pure Maths and was beginning to feel a bit depresso because my higher algebra and diff geometry courses seemed so abstract as it was being taught. But I stumbled across this series the other day and it has reminded me of why I am doing maths and why I love it! This has reignited my inspiration and motivation to learn the material. Motivation is so important!

@purplenanite 4 ай бұрын

Through geometric algebra, all things are possible.

@jamesyeung3286 4 ай бұрын

geometric algebra supremacy

@apolloandartemis4605 4 ай бұрын

Amen.

@johanngambolputty5351 4 ай бұрын

The pure side gives way to abilities many would consider unnatural Anakin.

@05degrees 4 ай бұрын

This can be defined without GA (via Lie algebra-Lie group correspondence, which the poster is slowly unfolding in this series I presume). I like to think about math in terms of APIs or interfaces (from OOP): actually you don’t always need to know anything about “concrete classes” to do things-you can have code that takes in entities knowing only some of APIs they support, and also returns entities constrained by another interface. Sort of like contracts, too. We can end up talking exclusively about relations of things behaving according to this or that interface, oftentimes being able to factor out unnecessary requirements and simplify our “code” by eventually using all guarantees and construction types we’ve given, and returning a thing that supports no less and no more than these things allow. This corresponds to a clear, invariant proof that can’t be simplified further and is-when you’re okay with understanding APIs that are involved-looke perfectly natural and streamlined and obvious. That’s an ideal state which isn’t often reached, of course, but it’s worth pursuing. In math, it’s of course not irrelevant if we _can actually have_ objects that implement all the interfaces we expect, like it is in code as well-we actually need to get something from somewhere in the end. Though in math also we have luxury of saying “suppose it exists”-and we’ll need to bootstrap endlessly if we don’t postulate anything. We end up relying on some foundations but I don’t think treating GA as an universal foundation is a unilaterally good idea. It can be for computation, but we need to reify and go deeper to understand what we’re operate on more intimately.

@angeldude101 4 ай бұрын

@@05degrees Math has several "interfaces" that various systems can "implement", and you can do many things knowing only the interfaces and automatically know that they work for every instance of said interface. These interfaces include things like magmas, semigroups, groups, monoids, semirings, rings, fields, algebras, lattices, modules, vector spaces, and more. If you don't need division, then you don't need fields and can instead work with rings, whose results extend to fields, but can also apply to structures that aren't fields like integers.

@terdragontra8900 4 ай бұрын

the exponential of derivatice operator being a shift operator is very beautiful to me, wow, wow

@ishtaraletheia9804 4 ай бұрын

A generalized understanding of exponentials as been itching my brain for ages. This made things really click together. Thank you so much.

@Sanchuniathon384 4 ай бұрын

In the exponentiation of derivatives part, you managed to nicely explain fractional calculus in a purely geometric sense using the function and the surface. This is a really nice way to explain 1D, 2D, and 3D calculus. Exponentials really are magical.

@NathanielAtom Ай бұрын

Isn't fractional calculus a little bit different? There we consider operators like a half derivative or pi'th derivative or i'th derivative; here we're looking at the exp(derivative) operator

@km4168 4 ай бұрын

I have to say, I did not expect d'Alembert's approach to appear so clearly when looking at the exponentiation of the derivative.

@TheJara123 4 ай бұрын

Waiting for a while for your video man!! First thing I do on youtube is to check for your new video!!

@tanchienhao 4 ай бұрын

This is the best intuition I’ve seen for exponentiating a derivative! Awesome video

@MagicGonads 4 ай бұрын

there's a 4th kind of exponentiation: the function set / bicategory the set of functions from A to B (or functors in the bicategory of A and B) is B^A also written [A,B] or A->B but it's not obvious if this has anything to do with those other forms. the only one I can think of it having anything to do with is the power series but we need to define all of the algebra first to even ask the question. It's a genuine question that I can't answer myself. But to get started let's think of it over cartesian closed categories / set-valued types: what's multiplication? the cartesian product what's addition? the tagged/disjoint union what's a number? the number 0 is Void/empty/{}, the number 1 is Unit/()/singleton/{a}, the rest of the naturals can be constructed from what we have so far what's exponentiation? mappings from B to A, if B is a natural representing n then we get A^n as if it was repeated multiplication (well, all of this is only up to isomorphism) well that's all very well defined, let's get into some... murky waters just casually throwing stuff out there as it's midnight for me right now what's division? take the power series for 1/(1-x) which uses just addition and multiplication as we have already defined e.g. List(x) = SUM(0,inf,n) x^n = 1 + x + x^2 + x^3 + x^4 ... AKA a list is nothing or 1 x or 2 xs or 3 xs ... so List(x) = 1/(1-x) ... whatever that means taking this very sus equality seriously for the heck of it: so 1/List(x) = 1-x ... so x = 1-1/List(x) = List(x)/List(x) - 1/List(x) = (List(x) - 1)/List(x) ... so 1/x = List(x)/(List(x)-1) ... great, I guess we interpret List(x)-1 as x + x^2 + x^3 + x^4 + ... and that actually works out? But does that mean List(x)-1 = xList(x)... so List(x) = xList(x) + 1 so we could sub it in for e^x = SUM(0,inf,n) 1/(n!) * x^n well what's n!? it'd be Unit and Bool and ... and n, so it's like a fixed length list but each increasing position can store an extra thing, like making a choice for every number of choices up to n ok so 1/(n!) is gonna be... List(n!)/(List(n!)-1) so the sum is over x^n List(n!)/(List(n!)-1) ... I don't think that helped please let me know if there is some kind of actual connection to be made

@Jaylooker 4 ай бұрын

R^n is a (abelian) Lie group and Lie algebra. The exponential map exp: g - > G is a map between a Lie algebra g and a Lie group G. This exponential map may be constructed as exp(tX) = y(t) where y may be constructed as the integral curve of either a right - or left - invariant vector field. The integral curve solves an ordinary differential equation (ODE) and could for example be a time-invariant differential equation or a vector field as mentioned here. This is inline with exponentiation being parallel transport.

@mishaerementchouk 4 ай бұрын

Funny thing. I do understand this, although thinking about vectors as elements of a Lie algebra goes a bit against of how I usually need to deal with vectors and therefore requires some effort. But I am completely lost in the video.

@05degrees 4 ай бұрын

Much yes! Thank you for doing these!!

@benburdick9834 4 ай бұрын

Great video! I would like to add that I find it helpful to think of the exponential function as a nice way to encode recursion in an equation. For the case of exponentiating derivatives, I think a nice (physics) way of thinking about it is that where you're going depends only on where you are, and translations follow pretty naturally from that.

@valeriibrudanin6753 2 ай бұрын

Man, you are my personal hero. I am a physicyst myself, and I can't help admiring your understanding and skills in visualisation

@sheerun 4 ай бұрын

Holy cow this helps me a lot understanding exponential, thank you!

@gregoired.4660 4 ай бұрын

This video is amazing, exponential maps are absolutly fabulous !

@blinded6502 3 ай бұрын

In study of geometric algebra there are multiple ways to represent positions, one of which acts as vector exponentiation you described.

@racpa5 4 ай бұрын

Another great video. Thanks!

@Hailfire08 4 ай бұрын

This way of defining the exponential links really neatly into the compound interest one if you consider numerically integrating it

@GeoffryGifari 4 ай бұрын

Wow this goes pretty deep. There are some things that I think interesting to explore: 1. For generalized exponentiation, does the power series expansion still work? and does exp(A)exp(B) = exp(A+B) ? 2. Defining exp(x) = g(1) like in the beginning seems like exp(x) is the end product of "tracing a path in the space of functions", the parameter being t. Like we transform a function continuously into other functions until we get to exp(x). 3. While exponentiating a vector, will the end result still be a vector? lets say a vector v is denoted by vec{v}. is exp(vec{v}) still a vector? If that's the case then should exp(vec{v}) also be a velocity (at a point) instead of a point in the manifold? What about the case for vector fields? 4. On the exp(vector field) section we see how the exponent of a matrix works. If we define a matrix mat{A} and a vector vec{v} and knowing that matrices act on vectors, does it make sense to write exp(mat{A})exp(vec{v}) = exp(another vector)? 5. For real numbers, exp(x) has the inverse function ln(x). Can the natural log as an inverse also be generalized in these cases?

@mathemaniac 4 ай бұрын

1. No and no. I deliberately didn't use series expansion because it does not apply to the vector case on a manifold - I would be very impressed if there is even a series-like approach to that because it is very manifold dependent. exp(sA) exp(tA) = exp((s+t)A) is true, because it is going along the journey by time s and then by time t. But in general, exp(A) exp(B) does not equal exp (A+B) - see matrices. 3. Well in the first section of the video, exp_p(v) is a point on the manifold, not a vector. That's how it is defined. That might also answer your Q4. 5. I would imagine yes, because you can definitely define logarithm on matrices, and perhaps vector flows as well (but I don't imagine this to be unique). But not sure about the other cases.

@GeoffryGifari 4 ай бұрын

@@mathemaniac _Well in the first section of the video, exp_p(v) is a point on the manifold, not a vector. That's how it is defined. That might also answer your Q4._ From what I understand, the exponent of a square matrix will also result in a matrix (and can have power series expansion). So exp(mat{A}).vec{v} will result in a vector, but it seems like exp(vec{v}) is another thing entirely. Maybe those "exponents" are distinct after all...

@MissPiggyM976 4 ай бұрын

What a wonderful video, one of the best on you tube!

@ramkitty 4 ай бұрын

Fantastic explanation and animation ch3 translated me further into Needham visual differential geometry and forms. Thanks

@michaelmann8800 4 ай бұрын

This was fantastic!

@eigenchris 4 ай бұрын

Interesting approach. I'm also studying Lie Algebras, and I was satisfied by interpreting e^(d/dx) using the Taylor series (#2 method at the start) and using a bit of algebra to show that e^(d/dx) f(x) = f(x+1). But the approach here seems much more general. I also tried learning some quantum field theory last year, and have trauma from the phrase "time-ordering". I don't think I really understood what was going on there. Do you have any math sources handy for learning about time-ordered exponentials? I feel like most physics sources I've looked at just leave me confused.

@mathemaniac 4 ай бұрын

Hi eigenchris!!! The time-ordering bit is something that I couldn't imagine to be put in as simple as the video illustrated for the intuition, but in terms of the Dyson series (the time-ordering bit), basically the time-ordering is necessary because H(t') and H(t) might not commute. It's difficult to explain it in a comment, but hopefully if you somehow haven't come across David Tong's notes on QFT, they will be helpful (p.51 onwards in www.damtp.cam.ac.uk/user/tong/qft/qft.pdf).

@stenzenneznets 4 ай бұрын

@@mathemaniachey, the link is not working

@ghkthILAY 4 ай бұрын

anxiously waiting for your next video on spinors just saying :D

@jaborl 4 ай бұрын

remove the closing parentheses at the end and it does@@stenzenneznets

@stevenfallinge7149 4 ай бұрын

Try the book "Quantum Fields and Strings: a Course for Mathematicians" volumes I and II. One of the first concepts it talks about are graded vector spaces, a topic of abstract algebra, so abstract algebra among other things is a prerequisite. Chapter II then pivots to manifolds and sheaves, topics of differential geometry and algebraic topology.

@drdca8263 4 ай бұрын

I had not expected the time ordered exponential to show up here, so that was a fun surprise

@raxirex6443 4 ай бұрын

good luck on your phd endeavours!

@eanerickson8915 4 ай бұрын

I remember when I first saw this stuff. Math got harder for me. Good explanation.

@Tucan_-wj5qo 4 ай бұрын

I am trying to visualize Nil geometry but am bad at math, thanks for this series on basics of Lie theory it nice and easy to understand for such a complex topic

@anisingh5437 4 ай бұрын

Great video .

@dkosolobov 4 ай бұрын

Thank you very much for this wonderful video! Somehow, knowing all the ingredients, I perceived the matrix exponent only as the Taylor series or as the solution for y' = Ay, not as a flow operator as, it now seems for me, it should be understood.

@ImaGonnar 4 ай бұрын

That was great! Now let’s see tetration of vector-fields.

@whoeverofhowevermany 4 ай бұрын

Exp stands for experience points, and it is exactly the real world situation that math is primarily used for.

@Ithirahad 4 ай бұрын

"These are not hallucinated..." But I thought you said the theory is a Lie?

@asdfasdfasdf1218 4 ай бұрын

One important note is that the first and second thing are often viewed as the same thing in differential geometry, where the tangent space is represented as the vector space of differential operators.

@mathemaniac 4 ай бұрын

Not entirely true. Well, true if your choice of coordinates happen to be geodesic normal coordinates. But for other coordinate choice, the exponential maps of these two do NOT yield the same thing.

@howwitty 4 ай бұрын

Thank you

@LamontGranquist 2 ай бұрын

Is there a further connection to QM by taking Green's functions solutions of the PDE equation in the exponential map for various differential operators?

@Czeckie 4 ай бұрын

can't put my finger on it, but I feel like all of these examples are exponentiation of vector fields? The vector is a field parallel to a connection; the operator d/dx is a simple vector field in a coordinate patch (but second derivative isn't). Is the common generalization somehow along these c̶u̶r̶v̶e̶s̶ lines? Good video, even though I'm somewhat familiar with these concepts, I lack the deep intuitive understanding.

@mathemaniac 4 ай бұрын

Yes, they are! I kind of make a mention of there needing to be more advanced concepts at play to see the connection (pun intended) and also in the description about that vector fields are related to and often described by 1st-order differential operators.

@robertadamovski3822 Ай бұрын

@mathemaniac 14:17 can you tell me why the rate of evolution along t axis is f’(x) at t=0?

@brainstink 4 ай бұрын

What are the actual chances of that! I have just started learning about taking the derivative of vector's, and you drop a masterpiece of a video! Do you have an email that I could ask you a question pertaining to computing a tricky gradient (derivative) of particular function? It would mean a lot to get a second opinion to see if I am doing it correctly!

@ohault 4 ай бұрын

We can ask ourselves, on the one hand, whether the other two definitions will provide the same generalized response, and on the other hand, whether these three definitions are three different points of view of the same concept and whether the latter can be linked to the definition of an exponential object in category theory ?

@Tomyb15 4 ай бұрын

I'm gonna have to rewatch this one with headphones to be able to pay better attention. But from the gist I've gotten so far, then can the schrodinger equation be re-written as an application of the exponential operator with the hamiltonian to the initial state of the wave function?

@mathemaniac 4 ай бұрын

Yes, in fact this is how typically physicists write down their solutions to Schrodinger equation (most usually for Dirac bra-ket states notation).

@KeniAlquist 4 ай бұрын

14:20 Slope in white is derivative with respect to 't', after that in the "exponentiation mould" the 'initial velocity' is df/dx. I don't understand why it is not denoted df/dt there.

@KeniAlquist 4 ай бұрын

or is it just that, x = x(t) (somehow), or is it that f(x) = g(0,x) and the 'initial velocity' should be: par g(t,x)/par t at t=0. ...then because of the PDE par g/par x = par g/par t at t=0 it is then df/dx = par g/par x|t=0 = par g/par t |t=0. To you, lovely human who reads math comments and is willing to dive into what I am trying to understand, thank you.

@mathemaniac 4 ай бұрын

It should be df/dx, because f is itself just a function of x, and as you said, f(x) = g(0,x). And the explanation was already given by you!

@jaeimp Ай бұрын

Anyone knows the term or name for the modification of the exponential @1:35?

@imrebalint6809 4 ай бұрын

It is impressively visualised.......

@BentoEmanuel 3 ай бұрын

So e to the squiggly line is possible???

@TheAzuratis 4 ай бұрын

If possible, can you set the subtitle Settings of your Videos to Viewer Default? Your videos are ignoring my subtitle settings (off)

@christoffel840 4 ай бұрын

Can anyone explain why the equation dg/dx = dg/dt should apply for the exponential of a derivative? I was confused on that point.

@DensityMatrix1 4 ай бұрын

I don’t understand the motivation for the initial explanation of exponentiation. Is there a text where this is treated formally?

@davidyang102 4 ай бұрын

how do you determine what the natural xg(x) is for each type of entity? if we choose a different definition for the velocity we would get a different exponential, why is the chosen velocity the correct one? clearly it has nice properties, but what makes this velocity more exponential than any other velocity

@mathemaniac 4 ай бұрын

I agree it is somewhat subjective - and you are free to define different velocities to be different from these "standard" definitions of exponentiation. What is happening here is simply that these are standard definitions of exponentials, but they can be lumped together in the same framework.

@Miguel_Noether 4 ай бұрын

These are just operators and function of operators for our physicist fellows

@jaeimp Ай бұрын

Isn't the equation of partials wrt to t and x @15:34 totally unmotivated?

@joonasmakinen4807 4 ай бұрын

Is there some integral way to take log of d/dx or sqrt of d/dx operators? At least the latter appears in QM as well.

@noahbarton8020 4 ай бұрын

If you're familiar with the Fourier series, then F{D^n g}(s) = (is)^n F{g}(s), where D=d/dx and F{.} is the Fourier transform. Thus, assuming convergence, you can take any of function of D by creating a Taylor series like so: F{h(D) g}(s) = h(is) F{g}(s) ---> h(D) g(x) = F^(-1){h(is) F{g}}(x) by taking the inverse Fourier transform of both sides. This integral certainly would not be easy to compute, but it is one possible way of defining a function composed with a differential operator (I am sure there are far more precise ways of doing this, but I have yet to take functional analysis). Hope that helps!

@joonasmakinen4807 4 ай бұрын

@@noahbarton8020 Wow, thanks tons! I didn’t think of Fourier route but it makes sense a lot! This seems to make sqrt-case very simple indeed. Gotta try that for log of d/dx. (Perhaps Pade approximation would yield better accuracy than Taylor.) Since log of d/dx or T-log(d/dx) is just an inverse of T-exp(d/dx), then I wonder now how the derivation on the video could be reversed to derive its equations with boundary conditions.

@noahbarton8020 4 ай бұрын

@@joonasmakinen4807 No problem! Yeah, I had never seen the derivation as presented in the video before, so there's definitely a lot there to explore. I attached a link to the wikipedia page for "quantum gravity," which if you scroll down to the section titled "Quantum gravity as an effective field theory," it whos a method of defining the log of the d'Almbertian, which is essentially a 4d Laplacian. Perhaps just replacing the expression with d/dx would work. Good luck! Link: en.wikipedia.org/wiki/Quantum_gravity

@ianthehunter3532 4 ай бұрын

I liked previous thumbnail, looking forward to the video nonetheless.

@symmetrysystem1642 4 ай бұрын

since you did bring up the exponentiation of matrices, couldn't you interpret a vector as a rectangular matrix? or would the math just work out the same or does the method with matrices only works with square ones? (i could also just be severely misunderstanding bc i was just listening to the video in the background while at work)

@jinjunliu2401 4 ай бұрын

How you would normally define the exponential for matrices is via the series expansion, which would only work for square matrices

@symmetrysystem1642 4 ай бұрын

@@jinjunliu2401 ah, thank you

@mathemaniac 4 ай бұрын

Actually in the video you can also kind of see why the matrices have to be square - in n dimensional space R^n, we want to flow to different places to R^n, then the linear vector field would need to have vectors in R^n.

@3ff3xgaming68 4 ай бұрын

Something that always bugged me, can you Derive/Integrate a Matrix? Would it have any meaning?

@KrasBadan 4 ай бұрын

2:51 I lost my understanding from this moment. What does it mean to have speed of x? g(t) is a function that spits out another function exp(tx) for every t. So it's a function of functions. If you follow the rules of derivatives, sure, the derivative is x exp(tx), which is x for t=0. But what does this derivative mean? Like, the function exp(0x) is just 1. The derivative of a constant is 0. If you make a small step ∆t then the function would be exp(∆t x). The difference between this function and previous one is neither x nor is it even proportional to x. Where to look to find the derivative of x?

@mathemaniac 4 ай бұрын

x is a variable, or if you prefer, an unknown. If you really want to, then set x = x_0 for a fixed x_0. g(t) here would be a number for each t, a function of t (and not a function of x, because x is fixed).

@KrasBadan 4 ай бұрын

@@mathemaniacif x is a variable and t is a variable, then it's a 2d function f(t, x). By fixing the x, you slice the 2d function into many 1d functions. Which of these 1d functions do you differentiate to get a derivative equal to x? What do you mean by "have a velocity" of x or xg(t)?

@KrasBadan 4 ай бұрын

@@mathemaniac I figured it out. Each of the slices has a derivative at t=0 equal to the x coordinate of the slice, and then it adjusts according to the height of the function at this point. I think the way you could've shown it better (the way I would've understood) is as follows: Let's say we wanna calculate e^a. Let a=2 for this example. Show the coordinate plane and starting point (0, 1). We want it to grow such that at the start it grows with the speed of a, and after that it grows with the speed of a*(the height of the point). Then you show the ax+1 line to show the growth speed at x=0 and after that you show its trajectory after it starts to adjust its growing speed. Then you bring out that this trajectory is actually just the e^ax function, and you can show it by differentiating (you get ae^ax). And of course at x=1 this function just equals e^a. So at this point you have on screen the a=(some value) text, the coordinate plane, e^ax graph, x=1 line, ax+1 line and maybe also y=e^a line to show that the intersection point has y coordinate that we want. Move the value of a a little just to show that it always works. And after that you can bring the 3d-ness of the e^ax graph (input space is 2d so the graph is 3d). You can show the 3d graph and different slices of a. Also you could zoom to the x=0, y=1 line to show that for every a, the slope of a tangent is equal to a.

@pdp11 4 ай бұрын

@@KrasBadan Yeah, I was completely lost from the beginning for this exact same reason. It was way too quickly glanced over, and I still don't think I understood it. Not intuitively at least. As a computer scientist what was really missing was types. What types are these functions, and what are the types of the operations (transformations) that we do with the terms. Concepts like "velocity" are being asked of the viewers to imagine, but these concepts have not been defined at all! It is also not clear which constructions are arbitrary (like the choice of 0 and 1 in g), and which are important (why do we keep the projection of the vector constant on the manifold?). I think these explanations make sense only for someone already familiar with the material. I am familiar with exponentiation operators from quantum mechanics and I still could not follow what is being presented here. The whole video is filled with a mix of utterly trivial concepts that the author takes a frustratingly long time to explain tied together by incomprehensible concepts that the author glances over in just seconds because they are supposed to be intuitive. I am sure they are if you know the material! Sadly, for me, they are anything but. Granted, this is video #4 of a series, and I have not watched any previous videos. When you are generalizing something, you need to show that a restriction of the generalization induces the original object you are generalizing. And this needs to be done only by using the algebraic properties of the object in question, not by appeal to a model that has not yet been defined nor proven sound or adequate. Only after the generalization has been established a model (hopefully intuitive) of the generalization can be attempted. But even that, the model has to be well defined mathematically before you can ask of people to imagine it geometrically. It cannot be done the other way around. You cannot ask people to imagine a model of something that has not yet been shown to have any interesting properties.

@KrasBadan 4 ай бұрын

@@pdp11 If you watched previous videos of the series, I don't think it would've helped here. I can try to explain it, as I think that now I get it. So the main idea is that exponentiation is the result of the process that lasts 1 unit of time, where you choose the starting point, you choose the starting velocity and constantly adjust it according to some rule. Output is the coordinate of point at time 1. For numbers it looks like this: We need to find e^x. Imagine t-y plane. The starting point is (0; 1). The starting velocity x means that the literal derivative dy/dt is equal to x at that point. You can think of the line y=tx+1 to imagine it. X is starting speed and it affects the endpoint, which is output. The rule of adjusting is constantly multiply the speed by the height of the point at that time. For example, at t=0 we don't need to do anything because currently height y is 1 and multiplying by 1 doesn't do anything. This is why we chose (0; 1) as the starting point. If some small time ∆t passes, then the point will be a little higher than 1 (assuming x>1), it will be about 1+x∆t because ∆t is small and the growth at that scale is pretty much linear. So now we need to multiply the current speed of x by 1+x∆t, getting x+x²∆t. That's a derivative of our function at that point. We do this constantly until 1 unit of time passes, and the value of y at t=1 is the output, e^x. This whole function is actually equal to e^ax, where x is a constant. You can check it by differentiating: (e^ax)'=xe^ax (starting speed multiplied by current height). And of course the function of e^ax at a=1 is just equal to e^x. For vectors it's all the same except you choose any point at the manifold and the adjustment rule is now to keep the point within manifold. The thing I really disliked about this video is how it says that there're 3 levels of understanding connection between vector and number exponentiation, and just says that explaining 3rd level is too hard so just believe it bro, they are actually the same thing.

@ArthurRainbow 17 күн бұрын

I must admit I'm very confused at 2'50''. I didn't get how do we have a $x$ in $g'(t)=xg(t)$. My first intuition is that $g$'s type was $\mathbb R\to(\mathbb R\to \mathbb R)$. Which would explain how $g'(t)$ could introduce a new variable. The derivative of the function $g(t)$ would be $x\mapstop xg(t)$. And so $g(0)=1$ meant that $g(0)$ is the constant function 1. After some more time, I guess $x$ a fixed constant, in which case there are an infinite number of function $g_x(t)$. And not indicating the paramater explicitly still seems really counter-intuitive to me.

@tombouie 4 ай бұрын

Thks buts; Occam's razor would suggest that e^x is just a power series. If x is a matrix, it must be square. Plus this way its conerent wih/to the rest of math & science.

@mathemaniac 4 ай бұрын

The series expansion does not apply to exponentiation of vectors this way though, and they are how mathematicians really exponentiate vectors.

@Cyrusislikeawsome 3 ай бұрын

So, quick question: isn't this possible for basically any function? Like, didn't all this depend depend more on the dialing up process than the fact that f(x)=e^x? Couldn't we do the same with any other f=f(x) by condtructing g=f(tx)?

@Sanchuniathon384 4 ай бұрын

Also, a question: If the intermediate point for vector fields is the vector attached to g(t), what is the intermediate point for tensor fields? Is it also the tensor attached to g(t)?

@mathemaniac 4 ай бұрын

You need to have the tensor being a velocity... which I can't imagine how unless the tensor is a vector itself.

@Lolwutdesu9000 4 ай бұрын

At 1:00, you made a mistake. What you referred to as exponentiation is simply the standard or natural exponential function. In English, the process of exponentiation is simply raising anything (number, function, etc) to some power.

@user-bf3ko7ts5e 4 ай бұрын

We can calculate the "exponential of a square matrix".

@xicad1533 3 ай бұрын

What about exponentiating a toquos?

@marc-andredesrosiers523 4 ай бұрын

For my part, I think of exponentiatiin starting from multi0lication.

@MrJdcirbo 4 ай бұрын

Shouldn't the definition be f'(x)=x'f(x) since x in this case isn't a single variable?

@abada00zhanghongbing 4 ай бұрын

be equal to 1+Kε，here ε is infinitesimal, K is a matrix, vector, or operater.

@DeathSugar 4 ай бұрын

I don't really like how you use word "normal" so much within different context (normal numbers, normal translation etc), considering you have to actually normalize stuff and have normals/tangents to the the plane/manifolds.

@BleachWizz 4 ай бұрын

ok but why having a concrete application means it wasn't hallucinated by someone, this stuff seems very crazy 👀

@Fire_Axus 4 ай бұрын

where is the exponentiated derivative in practice

@jewelly4270 4 ай бұрын

I am only a high school graduate but I found this explanation clear and satisfying, thank you for making these interesting parts of math more accessible!

@ianweckhorst3200 4 ай бұрын

My question is: can you use a number other than e?

@karolakkolo123 4 ай бұрын

Using a number other than e is just equivalent to changing the t to be something other than 1. For example 2^x = e^(x*ln2), so the endpoint would be at g(ln(2)) instead of g(1)

@andrewsantopietro3526 4 ай бұрын

is it just me or does the exponentiation thing just look very similar to a laplace transform shift

@forheuristiclifeksh7836 2 ай бұрын

1:40

@fabibi_ha 4 ай бұрын

For the first 10 minutes of your video i was super irritated about your definition of expontiating (Starting p., intermediate p., end p.). Suddenly it clicked and i remembered the oral presentation about Eulers Number (e=2.71828...) i had to give as a high school student ten years ago. School had only given us the number and the magic property e^x has derivative e^x. I had wanted to know what was so special about 2.71828... in particular. Watching exponential processes in finance or biology and their rates of growth, there is this linear relationship between function and its derivative e.g. 2^x -> ~0.7 * 2^x; 3^x -> ~1.1 * 3^x. Constructing e as the number a for which a^x derives to 1 * a^x, yielded a starting point of 1 and an endpoint e after 1 unit of 'time' has passed. More precisely: 1+1 is a very crude estimate to Eulers Number; (1+1/2)*(1+1/2) is a better estimate; with lim n->inf: (1+1/n)^n being the exact number. It's like 100% interest rate being accounted for infinitely many times in a period of one year. Man do i love math