The origin of the commutator

[@xizar0rg]

This is a case where "good place to stop" is homeomorphic to "left as an exercise to the interested reader".

[@goodplacetostop2973]

5:42 Homework 15:27 Good Place To Stop

[@ES-qe1nh]

[@lexinwonderland5741]

YES!! FINALLY MORE LIE GROUP/MANIFOLD VIDEOS! literally it makes my entire week when I see you post an advanced math topic like this on the main channel. Thanks!!

[@aoehler1]

Ditto!

[@music_lyrics-ni7ks]

Wahoo! 🎉

[@ragnarosthefirelord8662]

I'm constantly amazed at your ability to take abstract ideas and present them clearly and succinctly without resorting to sketchy handwaving. Great video as always, Lie Groups are fascinating

[@godfreyw5412]

Thank you Micheal. The videos you made about Lie algebras and groups are super helpful for a physics undergrad like me. I wish if there could be a playlist where you've put all of your videos regarding Lie algebras and groups! 😊

[@emranshafaq1702]

such a great introduction to lie theory. good level of abstraction and high level explanations without getting too in the weeds. thanks for this

[@emranshafaq1702]

as a suggestion, maybe drawing the curves A and B on the torus, showing how they have 0 derivative at the defined identity. Also, drawing out the curve C_s(t), just to show that we've built ourselves a new curve in the domain of the tangent space

[@comrade_marshal]

Glad to see you making videos on topics which I am encountering in Quantum Mechanics classes. The Dirac-δ function and now commutator algebra. Keep it up

[@Reliquancy]

Chalk is getting out of control with the dastardliness.

[@KorawichKavee]

I actually never heard of this concept of Lie algebra until I found out the connection between Quantum mechanics and Lie group. Somehow the Schrodinger's equation got involved.

[@Eta_Carinae__]

No need for quantum. They pop up as a consequence of the Hamiltonian's symplectic nature, in the form of the Poisson bracket.

[@baruchspinoza4979]

No lie, this topic would make a great video series. Thanks for this video!

[@morgengabe1]

I never saw my ODE prof laugh as hard as he did the day i asked in front of the whole class if a Wroskian is a commutator of a function with its derivative.

[@_P_a_o_l_o_]

At 10:45, Michael multiplies A(s) with b. What does it mean to multiply an element from the group G with an element from the Lie algebra g?

[@pwmiles56]

I guess such products arise from the algebra (in the ordinary sense). E.g. if you had a curve in G given by the product A(s)B(s), its corresponding Lie algebra element would be A'(0)B(0) + A(0)B'(0) = a B(0) + A(0) b

[@_P_a_o_l_o_]

@@pwmiles56 Thank you for your reply. However, in your explanation I don't understand what the products "a B(0)" and "A(0) b" would mean. They are products of elements from the Lie group G and its Lie algebra g. How is this operation defined? What sense does it make? Are we embedding the elements of the group in the algebra somehow, or viceversa, or what?

[@pwmiles56]

@@_P_a_o_l_o_ We are working with matrices. A(0) and A'(0)=a are both nxn square matrices so they can be multiplied. As a concrete example, A could be a 3x3 orthogonal matrix (a member of SO(3) which is the Lie group), and a would be a 3x3 skew-symmetric matrix. (Of course A(0)=B(0)=I, the identity matrix, so my example would simplify to a+b)

[@pwmiles56]

@@_P_a_o_l_o_ We are working with matrices. A(0) and A'(0)=a are both nxn square matrices so they can be multiplied. As a concrete example, A could be a 3x3 orthogonal matrix (a member of SO(3) which is the Lie group), and a would be a 3x3 skew-symmetric matrix. (Of course A(0)=B(0)=I, the identity matrix, so my example would simplify to a+b)

[@paolocampodonico5657]

@@pwmiles56 Thanks for the clarification! I see how in the special case of Lie group of matrices, multiplication makes perfect sense. But isn't this an abstract video, that deals with abstract Lie groups and algebras? I missed when Michael said he was focusing on the example of a group of matrices

[@winniedobrokot]

I don’t understand why derivative rules works for abstract group multiplication too. I tried to derive it like analysis textbook, but it requires the vector addition operation distribute with group multiplication to expand sum (A + Δt⋅A’ + o(|Δt|))⊗ (B + Δt⋅B’ + o(|Δt|) where ⊗ is group operation. I’m not sure, but also it seems it requires ⊗ to work on elements outside group which are created with vector operations like Δt⋅A’. Is it implicitly assumed that the Lie group is not some abstract group on manifold but subgroup of matrices?

[@pseudolullus]

This is important, yes they are matrices. Lie algebras and Lie groups are related through an exponential operation (map, really). The generators of the algebra are indeed matrices up there in the exponent, which generate a vector space themselves.

[@MarceloRobertoJimenez]

This is a good question which is not addressed in the video. It is indeed implied that the group operation plays nice with vectors in the tangent space, but there should be at least a hint to what actualy happens. E.g., why is it possible to multiply the vectors in the Lie algebra by the unity of the Lie group?

[@SplendidKunoichi]

that's because in order to be able to take derivatives, we defined a lie group as a place where you can take derivatives

[@santiagobustamante6192]

@@SplendidKunoichiyeah but those derivaties lie in the tangent space, so we can define a vector A(s)bA^-1(s) as just the tangent vector to the curve C_s(t)=A(s)B(t)A^-1(s) at t=0, however it wasn't discussed if a thing such as A'(s)bA^-1(s) could be "nicely" defined so that usual derivative rules apply. For instance, if you tried to define A'(s)bA^-1(s) as a two-step product, you would have to define either a good gx(gxG) -> g map or a good (gxg)xG -> g map, where G is the Lie group and g its Lie algebra, ending up in a similar situation as the beginning. In the end of the video we write the commutator of a and b as ab - ba, but what is exacty ab and ba? I'm guessing in the end it could just be a tensor product, but the middle steps are a little obscure for me in that sense.

[@Alan-zf2tt]

Outstanding!

[@gavintillman1884]

I wish this had been done in my degree. We did topological groups, and there are aspects of that here such as neighbourhoods of identity, but not tangent spaces.

[@declandougan7243]

Hw at 5:32. I think the solution is to define scalar multiplication i.e. ka where a is a scalar multiple, as the derivative of A(t)^k evaluated at t=0.

[@scottmiller2591]

I assume the sleight of hand of using a matrix identity (d/ds (A^{-1}(s) = ...) for an element which, up to that point, is described without any reference to matrices is based on representation theory.

[@wolphramjonny7751]

I don't understand teh meaning of A(s).b (time 5:01), it multiplies two different objects, unless you assume they are all matrices, but this would only be valid if G is a matrix group

[@isodoubIet]

My man spelling his lower case gs like ფ

[@bernhardriemann5091]

Great lecture!

[@Maths_3.1415]

Always waiting for your videos :)

[@Noam_.Menashe]

Isn't it by definition that a tangent space of a manifold is closed under addition and scalar multiplication?

[@strikeemblem2886]

No, it is not automatic. T1(G) = {tangent vect at the identity 1}, just a set to begin with. You need to fall back to the definition of tangent vectors (via smooth curves 4:30) to show that the vector-space operations that we wish to equip to T1(G) are closed operations.

[@oddlyspecificmath]

Naive question...is a "flat" tangent _plane_ to (e.g.) a sphere a tangent _space_ , i.e., like I spun a vector around a surface normal?

[@Nameless.Individual]

You are indeed correct. No requirements was stated regarding the dimensionality of the tangent space (well, maybe with the exception that it is non-trivial) so planes are indeed a type of tangent space. However, some additional considerations show that it isn't really possible to define any (interesting) Lie structure on the 2-sphere.

[@oddlyspecificmath]

@@Nameless.Individual Thank you; I was hoping for additional information too, but I didn't know what form of it to hope for. This helps.

[@alxjones]

The tangent space is more of an algebraic concept. It's the vector space structure that tangent vectors at a point have. It doesn't really exist *on* the surface geometrically like a tangent plane does, though we often visualize it this way. If we take the "obvious" affine space structure of the tangent plane, we get an associated vector space structure which is isomorphic to the tangent space, so in some sense the distinction in unimportant. However, when we have two such tangent planes, they may intersect, so that points on the line of intersection have affine representation thru either plane. Contrast that with the respective tangent spaces at those points, which are disjoint; two vectors in different tangent spaces cannot generally be compared. There are various additional structures that you can add to a surface (manifold) that allow you to compare vectors from different tangent spaces, one of these being a Lie group structure. A sphere (2-sphere, surface of 3D ball) does not admit any Lie group structure, but even cases that do (like the circle) don't typically end up equating vectors on the line of intersection between planes. The only example I can think of that does is Euclidean space itself.

[@oddlyspecificmath]

@@alxjones Thank you. My mental model is improving :)

[@Alan-zf2tt]

Just wondering... link to Naive Lie Theory by Stillwell @ 08:46 ?

[@MichaelPennMath]

I just added it to the description!

[@Alan-zf2tt]

@@MichaelPennMath Thank you Michael - much appreciated. Love the work you are doing for Math

[@franzlyonheart4362]

10:15, this is not obvious. How do we know that differentiation of curves is compatible with the (Lie) group's multiplication? Recall this isn't merely scalar multiplication. Specifically, say for a curve c(t) in G, and for some group element g € G, you claim that (d/dt) (g c(t) g^-1) = g (c'(t)) g^-1, but how can you evidence that?

[@trevorkafka7237]

How is the multiplication in the expression A(t)B(t) defined? (4:50)

[@mbivert]

As far as I understand, the curves (A, B) eat a (real) parameter (t), and spit an element from the manifold, which in this case, is our Lie group G. Hence the multiplication would be the group's operation

[@speedbird7587]

very beautiful explanation! thanks.

[@jms547]

At 11:30 what does Michael meen by "a curve in the tangent space"? I can understand a curve in G (namely A(s)B(0)Ainv(s), which lies in G), but g is a vector space, so there is no notion of "curves".

[@isodoubIet]

A "curve" here is just a (continuous) function of a single scalar parameter taking values in g. He didn't necessarily "prove" continuity here but that should be easy enough given the other properties of a vector space together with some assumptions on A(s) (e.g. we probably require it to be differentiable as well).

[@jms547]

@@isodoubIet thanks. Also I guess that somewhere lurking in the background is a chart c from G to R, so questions of differentiability and continuity (which you can only really talk about for maps from R to R) only make sense after composition with c?

[@douglasstrother6584]

I've always seen simply defined as [a,b] = ab - ba, in the context of Quantum Mechanics. I didn't know that it was connected to anything else.

[@StoicTheGeek]

If there isn’t a Smooth Criminal parody called Smooth Manifold, then there really should be

[@fizikchy]

Leaving aside the teacher's repeated use of the word "ass", it was a great intuitional approach to the subject. Thanks a lot. PS: Sorry for the joke, but I could not curb myself ;)

[@borisborcic]

The title reminds me of a chuckle at the name of "fake monster lie algebra".

[@TheTKPizza]

Could anybody maybe explain to me (as a chemistry student becoming interested in group theory through crystallography) why a Lie algebra can be defined as a tangent space on the identity of the matching Lie group? I really don't get the connection. Something to read up on it would also be very welcome. :)

[@AllAnglesMath]

I recently published a video with a simple example of a Lie group and its Lie algebra. It uses complex numbers. It doesn't go very deep, so it might be a good starting point. You can find it here: kzbin.info/www/bejne/aZjIh3SAq6-AbsU (the relevant part starts around 12:05).

[@aoehler1]

Definitely worth studying as there is a deep connection there

[@Anytus2007]

We know from the state from the definitions of groups, Lie groups, and smooth manifolds that any Lie group, G, must have an identity element and that there exists a tangent space to the manifold at every point, so there must exist a tangent space to the manifold at the point representing the identity. So we're guaranteed that we can find such a tangent space and we can call that tangent space a "Lie algebra" if we want to; that's just giving a name to something that is guaranteed to exist. Proving that the tangent space that we decided to call a "Lie algebra" really is an algebra is what Michael has showed in the video. An algebra over a field is a vector space that also has a bilinear product, which intuitively means that you can add two elements, multiply elements by scalars, and (this is the thing that distinguishes an algebra from a general vector space) combine any 2 elements in a way that is separately linear in each input and get an output that is also in the algebra (we usually call this kind of operation a "product" and note how it differs from addition; addition is not separately linear in each input ie A + c*B =/= c(A+B) for c a scalar and for all A,B in a vector space). The product associated with a Lie Algebra is the Lie bracket that Michael explores in this video. If you're asking why we would want to single out this tangent space and give it a name and study it, the general answer is that Lie algebras are usually easier to study than the associated Lie group, and yet you can still learn something about the group from its Lie algebra. The reason that Lie algebras are often easier is that like any tangent space to a manifold at a point, they are linear spaces. Linear spaces give you access to all sorts of tools that let you write elements in the space as combinations of other elements (via addition, scalar multiplication, and the algebraic product), which lets you understand how the elements are related to each other. By understanding the structure of the Lie algebra, you are learning about the structure of the Lie group in a region around the identity element. That is, there is a natural mapping from the Lie algebra back into the Lie group (at least for finite dimensional algebras), which you can think of as taking a small region of the tangent space near the origin and projecting it back down onto the Lie group. Then you can learn about what is going on near any other arbitrary point in the group by taking the small patch that corresponds to part of your Lie algebra and acting on it with other elements within the group, which has the effect of moving your small patch around the group. Using this strategy, you can determine what the structure of any small neighborhood of the Lie group looks like.

[@TheTKPizza]

@@Anytus2007 Thank you so much, that was incredibly helpful! It makes a lot of sense, why you would study the tangent spaces instead of the groups themselves, as those obviously have a lot more "structure" (for lack of a better term) from their linearity, than any of the often even practically rather high dimensional and even mostly non-commutative Lie-groups. I did not know, that general definition of an algebra, you just explained, since I didn't dive into "pure" math much more than basic group and field structures, since stuff like that is not part of my curriculum at all. So that was one important connection, because I do know some linear algebra and differential geometry. So basically mapping from a Lie algebra to the associated group is kinda like mapping from a tangent line on the complex unit circle to the circle itself through the exponential function, does that make sense? If I have a a suitable mapping, that should kinda work for any group, I will probably run into issues, once I get near the boundaries of my manifold, won't I? Does that mean, Lie groups cannot have boundaries? I gotta read up on some of that. Yeah, those are the kind of connections that can sometimes be difficult to learn just from literature and youtube, so thank you very much! :)

[@Anytus2007]

@@TheTKPizza Excellent example. The circle (S^1 in manifold-speak) is probably the simplest example of a Lie group that is not itself also a vector space. Sometimes it may also be referred to as U(1), the two are isomorphic. The intuition about the tangent line and the exponential function is exactly the right one. In fact this exact idea generalizes to all finite-dimensional Lie groups. We can always find a basis for the Lie algebra (these are called generators) and then reconstruct a neighborhood of the Lie group via exponentiating elements of the Lie algebra. But we should indeed be worried about what may happen far away from the identity. In the case of spheres you can map all of the sphere minus a single point to Euclidian space via stereographic projection. Since you are missing only a single point, any sequence of points that diverges "to infinity" in the algebra will converge to the missing antipodal point in the group, so you haven't actually lost anything (and this generalizes to the fact that there is a one-to-one correspondence between finite dimensional Lie algebras and compact, simply connected Lie groups). But in general there may be points in the group that cannot be reached via exponentiation of elements from the Lie algebra or any sequence thereof. It is, however, true that we do NOT have to worry about boundaries. Suppose x is in G, a Lie group, and x is on the boundary of G. Because it is a group x must have an inverse x^-1 with x x^-1 = 1, the identity. Because it is a lie group, we also know that x, x^-1, and 1 must also all have smooth (and therefore invertible) maps to Euclidian space (or half-space in the case of a boundary). Now consider this, what happens if I apply the smooth map, f, that takes a neighborhood of x to Euclidian space, and then apply the *inverse* of the smooth map, ginv, for a nbhd of x^-1? So I start at x in G, map to Euclidian space via f, then map from Euclidian space back to G via ginv. This is a map from G to itself (start and end somewhere in G) and where does this map send x? Since x x^-1 = 1, we BETTER end up at the identity otherwise our maps have failed to represent the group structure (a contradiction with the definition). But that means that ginv composed with f is a smooth map from x to the identity. Therefore, a neighborhood of x is diffeomorphic to a neighborhood of the identity, so if x is on the boundary, then so is 1 and every other point in G! But having every point be on the boundary is a contradiction because it implies that there is no point in G that is diffeomorphic to an open ball in Euclidian space (they'd all have to be half-balls), which contradicts the definition of a smooth manifold. So in fact Lie groups never have boundary and the way that you should think about points in a Lie group is that they are all identical, up to relabeling. We may pick out a point and call it the identity but we could just as easily label that point x and then move all the points by multiplying by x. That sends a -> x a and x^-1 -> x x^-1 = 1 is now the identity. Hopefully this makes it clear why Lie groups are tied to symmetries; they're extremely symmetric objects. Every element has the same local structure and can be transformed into any other element via group multiplication (just apply it's inverse and then the element for where you want to map it to). This is what makes them so interesting; every element of the group is simultaneously representable as a point in Euclidian space AND as a map that maps the group into itself, preserving the group structure. Lie groups can then be in some sense considered spaces that are the same "shape" as their own symmetries.

[@Sinjakin]

Awesome video, thanks.

[@VaradMahashabde]

Damn, books shouldn't be jumping to Lie algebras!

[@wagsman9999]

this channel has a lot of range

[@qqqrrrsss]

Dear Michael Penn! I would like to ask you: does the commutativity of matrices have the property of transitivity? Namely, if A commutes with B, and B commutes with C, does it follow that A commutes with C?

[@schweinmachtbree1013]

Commutativity is not transitive - if you try and come up with a proof that A and C commute you will get stuck. Since B and C commute, so do B^(-1) and C, as B^(-1)C = CB^(-1) is equivalent to CB = BC. Now let's try and use the information that AB = BA and BC = CB (and B^(-1)C = CB^(-1)) to get to AC = CA: we have AC = ABB^(-1)C = BACB^(-1), but this isn't going to help us move the A past the C. So at this point we suspect that the conjecture is false and start looking for a counterexample. You could try some randomly chosen 2 by 2 matrices and you would probably find a counterexample straight away, or you can make the following clever little argument: the identity matrix commutes with everything, so taking B = I we have that A commutes with B and B commutes with C, but then clearly A does not commute with C in general because that would mean _all_ matrices commute.

[@Anonymous-zp4hb]

Didn't quite get it. Are the curves supposed to uniquely define the element in the tangent space? If so, then what dictates the magnitude of the tangent vectors? Or put another way, how would the curves that describe two different points in that space (colinear with the identity) differ from one another?

[@ramsey121kamar]

Perhaps this is a stupid question….but in proving closure under addition, how do we know that A(t)B(t) is a curve within the manifold?

[@pwmiles56]

The manifold i.e. the Lie group isn't closed under addition (necessarily). E.g. orthogonal 3x3 matrices don't (usually) add to another orthogonal matrix. A(t) and B(t) are assumed to be manifold members from the way they are constructed. Given that, the group closure property implies that A(t)B(t) is also a member of the manifold.

[@leetingfung]

I don’t understand why d/dt A(t)B(t) at t=0 has to be in the tangent space. We only know A(0)B(0) would pass through the identity but its tangent could be anything

[@otterlyso]

I'd say: the curves A(t) and B(t) are formed from points in the Lie group and their product is calculated pointwise. We know - as stated early in the video - that the product of any two points in the Lie group (x & y) is another point in the group (xy). So all the points of the product curve A(t)B(t) are also in the group i.e. the curve A(t)B(t) is also in the space - and as you said we know the curve passes through the identity. So the curve A(t)B(t) has a tangent at the identity that is also in the tangent space.

[@__hannibaal__]

I stop Lie algebras many years and i more oriented to complex analysis, analytic number theory, YOUR Course push (‘my be is English word to what i feel’) me to return back to Geometry, analytic geometry, multidimensional calculus, Geodesic, variational calculus, Hamiltonian Mechanics (& Lagrangian ) , Manifold, Differential Manifolds, Calculus on manifolds, Hodge Operator, differential form, ………….. Let stop little bit C/C++ programming and dev, and return to abstract mathematics again.

[@klausolekristiansen2960]

Do you have anything interesting to say about Analytic and Algebraic Topology of Locally Euclidean Metrization of Infinitely Differentiable Riemannian Manifold?

[@Wabbelpaddel]

Get a good mathematician you can just plagiarize from. Just change some words here and there.

[@zandere1]

Does someone know how the "homework" with scalar multiplication works? I would guess, you use something like B(t)=A(t)^n, for integer n, as a curve and then B(0)=1, and B'(0)=n A'(0) or b = n a. But then to get from integer n to real (or complex) numbers is not quite that trivial, or is it?

[@mbivert]

There's a proof/construction of the vector-space structure for a tangent space in kzbin.info/www/bejne/ppbThpJtg6-Gfa8 which relies on the chain rule. Here's what I think it would give: let a ∈ T_1(G): ∃ a curve A(t) such that (i) A(0) = 1 and (ii) A'(0) = a. Let k be a real, and C(t) a curve such that C(t) = A(kt). Let f(t) = kt. Hence, C(t) = A(f(t)). By the chain rule, C'(t) = f'(t)A'(f(t)) = kA'(kt). Observe that C'(0) = kA'(0) = ka, and C(0) = A(0) = 1, i.e. ka ∈ T_1(G)

[@zandere1]

@@mbivert Thank you!

[@MDNQ-ud1ty]

Is this the actual origin? E.g., can one confirm that the very first usage of the commutator was in Lie theory? The commutator shows up in Group theory and is very useful for Abelianization. Which one really came first?

[@Tehom1]

As colorful as the video description is, it's upstaged by the "Naive Lie Theory" link sitting quietly below it, especially if you didn't know how to pronounce "Lie".

[@lookmath4582]

You my friend seem so interested in this stuff in particular . The jacobi identity , the commutator , differential algabras etc

[@pseudolullus]

It's his research topic as a mathematician, he says so in other vids

[@lookmath4582]

@@pseudolullus ahah , I see

[@gbessinpenieleliezerhoumba3337]

Hi dear Professor, I have a topic that might interest you and I would like to see a video about it: PROVE THAT FOR EVERY REAL NUMBER NOT A MULTIPLE OF 2 CONTAINS IN ITS DECOMPOSITION AT LEAST ONE ODD NUMBER

[@CM63_France]

Hi, I can't say if it is more and more clear or less and less confusing for me 😁

[@tomholroyd7519]

... the algebra is the tangent space at the identity ... 🤯

[@Necrozene]

You lost me when you said eat 9:40 that A(s) was A of zero, but s could be any member of the Lie group. Doesn't affect the result though.

[@Ron_DeForest]

This is going to be nothing for you but I thought I’d ask. Given any point in 3 space is noted as ( X, Y, Z ). How do you find a point that’s approx equa distant to three other points in 3 space?

[@angeldude101]

The standard arithmetic mean should work. (P1 + P2 + P3)/3. More generally, normalize(∑{i=1→n} P_i) If you want an actual equal distance point rather than just the average/centre-of-mass, that would probably be best done by connecting the points with a circle and then finding the circle's center. This should be possible by taking the perpendicular bisectors between each point and seeing where they meet. Then again, this is only unique in 2D. In 3D, you'd need 4 points forming a sphere for a unique solution. With only 3 points in 3D, the desired solution would probably be the one in the same plane as them rather than the entire orthogonal line.

[@cngrinder9423]

I don't understand, lie bracket=commutator doesnt come from enveloping algebra?

[@sonarbangla8711]

Atiya was the master of abstract algebra and at the fag end of his life he admitted that he understood nothing. That is when I stopped trying to understand. I fail to understand why this is Penn's favorite subject? In this video, he not only explained why this subject is his favorite, nor why he resorts to trickery?

[@wjrasmussen666]

What chalk does he use?

[@Milan_Openfeint]

A sentient, malevolent chalk. He's using it so we don't have to.

[@mrgadget1485]

Damn! Watching this video, cost me the price of the book :)

[@MrRyanroberson1]

if A(0) = 1 then shouldn't A^-1(1) = 0?

[@oliverherskovits7927]

This inverse doesn't mean the functional inverse, it means the inverse within the group

[@alegal695]

A and B are curves, so they are column vectors, however Michael consider A as a square matrix. Am I missing something?

[@pwmiles56]

The curves are A(s) and B(t) with s and t real parameters. A and B are indeed matrices of the Lie group.

[@alegal695]

@@pwmiles56thanks, but it is not clear in the video how the square matrices pop up.

[@pwmiles56]

@@alegal695 It's because we are looking at continuous groups of matrices with matrix multiplication as the group operation. Michael's demonstration isn't specific to any particular group but examples are SU(2), unitary 2x2 matrices and SO(3), orthogonal 3x3 matrices, both with determinant 1.

[@alegal695]

@@pwmiles56 thanks, but it doesn't answer my question.

[@pwmiles56]

@@alegal695 Well, they aren't column vectors -- that's how you would store them as data. A and B are square matrices with each matrix element a smooth, differentiable function of the parameter. It's implied they fulfil the group membership criterion e.g. with 2D rotation matrices they could be [cos(theta) -sin(theta)] [sin(theta) cos(theta)] with theta the parameter.

[@utkarshanand9706]

Since the scalar product involves derivatives, this video might prove to be useful. kzbin.info/www/bejne/aKHXd6F_n5xkia8

[@MichaelPennMath]

This is one of my favorite math videos on KZbin!