This was very good and useful and practical. It would be nice to see more things on SO(3) or Lie groups at the nuts and bolts computational level. The formal parts are easy to see, once one does useful examples.
I think he meant the Lie-algebra so(3) with it's vectors being tangent to SO(3). Furthermore the matrix exponential of a 3D skew-symmetric matrix is an element of the special orthogonal group SO(3).
Please! What about an entire playlist about Algebras? Lie Algebras and Hopf Algebras in particular! But also Banach Algebras, why not? Or maybe an entire playlist about Lie Theory, so both Lie Groups and Lie Algebras, and their applications to Physics and Differential Equations! The internet is flooded with calculus, combinatorics, linear algebra; but no one has ever covered these topics. Please, keep ‘em coming!
Would love to see another video on Clifford / Geometric Algebras, where the Cross product is the dual of the Outer product (the Grassman exterior (wedge) product not the tensor outer product) which is defined in more than just R3. I have read that Lie groups are Spinor groups in this formalism, and I would love to see Michael's explanation of that.
He does have one video talking about Clifford Algebras, but it is a very abstract description that describes them in terms of tensor algebras rather than as their own thing.
I find his Clifford vid difficult to grasp. Although the topic lends itself pretty well for a concrete explanation with length, surface and volume (sorry, I am an engineer). I know Clifford algebras are applied for modelling polymer flow in rheology. All very concrete and straightfwd.
As abstract as mathematics can be you could see in relative terms how using some thing that is more concrete and and familiar in our understanding, ie cross products, supports learning something more relatively abstract, Lie algebra. A new area for me though becoming aware it is right under my nose, so to speak. Thank you.
Sometimes the simplest things totally clear up a seemingly very complicated topic. This was very helpful reminding me of something I already knew but swept under the rug, when it was a central link with something simple and totally understood by me. Thank you for this Lie series of Lectures.
10:53 I think it can be shown more easily, take \phi == f since f(a){b} = a * b take f(a * b){c} = ( a * b ) * c = | jacobi id | = - ( b * c ) * a - ( c * a ) * b We want it to be [f(a),f(b)]{c} = f(a){f(b){c}} - f(b){f(a){c}} = a * ( b * c ) - b * ( a * c ) , just some rearrangements = | skew symm 3 times | = - ( b * c ) * a - b * ( a * c ) = f(a * b){c} since c is arbitrary, f(a*b) = [f(a),f(b)] this also holds for any binary operation that has these properties
9:05 - Have you considered putting a prompter in a place where - when you look at it during recording - it would seem as if you are actually looking at the sponsor logo?
Nice video as always! I think you could generalize this to R^n, considering that you have n-1 Linear Independent vectors v1,...,v_(n-1) in R^n, you could have the n-th vector v_n (cross product) as the determinant of a nxn matrix whose 1st row contains the vectors from the canonical basis e1,...,en (as in linear algebra books) such that v_n is orthogonal to v1,...,v_(n-1) and ||v_n|| would be equal to the hypervolume of the hypercube defined by v1,...,v(n-1).
I've come up with a generalisation of cross product that seems obvious to me, but I've never seen it. It is defined for any m n-dimensional vectors where m≤n and is a nxnx...xn (n-m n's) tensor. I didn't know about Lie algebras when I came up with it
"AB - BA" is "the part that does not commute". Are you familiar with the Geometric Product? Simply including directions in space (ie, perpendicular directions: (e1,e2,e3) = (right,up,forward), etc) seem to generate all of this stuff.
Well If you want to calculate a vector v that is perpendicular to x and y, you have to solve the linear system of equation v • x = 0 and v • y = 0 Where • is the euclidean scalar product. When you solve for v you get the Crossproduct Up to scalar multiples. I think that is what He means.
I’m rusty as it’s the 1980s that I studied pure maths, but I still like to dabble. But surely it’s only unit vectors that map to SO(3), presumably R^3 (excluding 0 I guess) maps to O(3)? Put another way, crossing with a unit vector gives a rotation, crossing with and non zero vector gives a rotation composed with a scaling?
well, you need to prove it. An easy way of getting is using a formula for ux(vxw) = (u.w)v - (u.v)w (dot is the inner product, and concatenation is scalar multiple). Add these together and you get the zero.
Set Phi=f for short. f(vxw)=[f(v),f(w)] is equivalent to f(vxw)u=[f(v),f(w)]u for any test vector u. The LHS is simply f(vxw)u=(vxw)xu. For the RHS we have [f(v),f(w)]u=f(v)f(w)u-f(w)f(v)u=vx(wxu)-wx(vxu)=vx(wxu)+wx(uxv)+ux(vxw)-ux(vxw)=-ux(vxw)=(vxw)xu, since the cross-product satisfies the Jacobi-identity and is anti-commutative.
It is not good pedagogy to say "The cross product is a lie!" That instantly invalidates every formula with a cross product. That's not of course what you mean. It would be better to say the cross product is a 2-index object, a bivector, which in 3-d space is dual to a vector, so has the same number of independent components.
Pretty interesting. I'd have liked to see how(/if) the matrix form of the cross product relates to the 3D rotation matrix, but looks like I've got a little challenge for the afternoon!
Did you reach somewhere? There's probably a nice geometric interpretation out there, maybe even worth a video. First random thought is that |u x v| = |u| |v| sin(theta) where theta is the angle between u and v. You should be able to see this theta angle appear in the rotation matrix. Second thought is that this relation might be quite indirect. This matrix form of the cross product lives in a Lie algebra, associated to the Lie group of the rotations. So one in probably obtained by differentiating the other. Just random intuitions thrown here. I might just be saying nonsense.
i thought he will say that the idea that a cross product is a perpendicular vector is just a convention and not how vectors really work in physics but it's still an interesting video to watch
what do you mean? In math it's defined as the parpendicular vector making a right-hand basis when added to your input list. (+ condition to normalize it, but whatever)
@@Czeckie depends, in normal vector algebra it's defined as a perpendicular vector as you said, but there are physics who prefer the idea of cross product as the bivector parallel to the two vectors check Clifford Algebra to know what i mean
@@Czeckie yes, the cross product is just a way to "label" each of these 2 forms and their orientation. The labelling is possible with Hodge star operator.
Nice clickbait yo. Tho this was kinda trivial, I guess the commutator is an useful thing to understand how one would generalize the cross product to the external product.
damn. "Lie", not "lie" then. Damn. I don't know math, just wanted to see juicy stuff. And here i am watcing a whole video, wondering where the lie was. But there was no lie, only Lie. I feel cheated, but there was no lie.
This was very good and useful and practical. It would be nice to see more things on SO(3) or Lie groups at the nuts and bolts computational level. The formal parts are easy to see, once one does useful examples.
"The cross product is a Lie"
After learning about the wedge product, I agree!
TOP 10 LIES (algebras) MATHEMATICIANS DON’T WANT YOU TO KNOW ABOUT!!
@@imnimbusy2885 LMAO
it was all LIES (algebras)
Doesn't SO(n) mean the determinant is 1?
I think he meant the Lie-algebra so(3) with it's vectors being tangent to SO(3). Furthermore the matrix exponential of a 3D skew-symmetric matrix is an element of the special orthogonal group SO(3).
Please! What about an entire playlist about Algebras? Lie Algebras and Hopf Algebras in particular! But also Banach Algebras, why not?
Or maybe an entire playlist about Lie Theory, so both Lie Groups and Lie Algebras, and their applications to Physics and Differential Equations!
The internet is flooded with calculus, combinatorics, linear algebra; but no one has ever covered these topics.
Please, keep ‘em coming!
totally agree. Physics is too hard for physicists - Hilbert
Would love to see another video on Clifford / Geometric Algebras, where the Cross product is the dual of the Outer product (the Grassman exterior (wedge) product not the tensor outer product) which is defined in more than just R3. I have read that Lie groups are Spinor groups in this formalism, and I would love to see Michael's explanation of that.
I was just thinking the same thing. :)
He does have one video talking about Clifford Algebras, but it is a very abstract description that describes them in terms of tensor algebras rather than as their own thing.
I find his Clifford vid difficult to grasp. Although the topic lends itself pretty well for a concrete explanation with length, surface and volume (sorry, I am an engineer). I know Clifford algebras are applied for modelling polymer flow in rheology. All very concrete and straightfwd.
I suggest you look into geometic algebra.
michael has an unlimited amount of Lie puns
He's Sophusticated.
5:20 “This guy right here…” rather than “This vector right here…”
12:50 “This guy cancels with that guy” vs. “This entry cancels with that entry.”
As abstract as mathematics can be you could see in relative terms how using some thing that is more concrete and and familiar in our understanding, ie cross products, supports learning something more relatively abstract, Lie algebra. A new area for me though becoming aware it is right under my nose, so to speak. Thank you.
Superb presentations! Thank you! I look forward to a video on the exponential map and left-invariant vector fields.
Sometimes the simplest things totally clear up a seemingly very complicated topic. This was very helpful reminding me of something I already knew but swept under the rug, when it was a central link with something simple and totally understood by me. Thank you for this Lie series of Lectures.
One of your better videos. Nice!
This vid is a really good one! Basic, concrete and tangible, thank you Sir!
10:53 I think it can be shown more easily, take \phi == f
since f(a){b} = a * b
take f(a * b){c} = ( a * b ) * c = | jacobi id | = - ( b * c ) * a - ( c * a ) * b
We want it to be [f(a),f(b)]{c} = f(a){f(b){c}} - f(b){f(a){c}} = a * ( b * c ) - b * ( a * c ) , just some rearrangements
= | skew symm 3 times | = - ( b * c ) * a - b * ( a * c ) = f(a * b){c}
since c is arbitrary, f(a*b) = [f(a),f(b)]
this also holds for any binary operation that has these properties
This video doesn't load at all... Is youtube faking it's position as platform again?
9:05 - Have you considered putting a prompter in a place where - when you look at it during recording - it would seem as if you are actually looking at the sponsor logo?
Nice video as always!
I think you could generalize this to R^n, considering that you have n-1 Linear Independent vectors v1,...,v_(n-1) in R^n, you could have the n-th vector v_n (cross product) as the determinant of a nxn matrix whose 1st row contains the vectors from the canonical basis e1,...,en (as in linear algebra books) such that v_n is orthogonal to v1,...,v_(n-1) and ||v_n|| would be equal to the hypervolume of the hypercube defined by v1,...,v(n-1).
en.m.wikipedia.org/wiki/Cross_product#External_product
I think this is the Levi-Civita symbol.
I've come up with a generalisation of cross product that seems obvious to me, but I've never seen it. It is defined for any m n-dimensional vectors where m≤n and is a nxnx...xn (n-m n's) tensor. I didn't know about Lie algebras when I came up with it
"AB - BA" is "the part that does not commute". Are you familiar with the Geometric Product? Simply including directions in space (ie, perpendicular directions: (e1,e2,e3) = (right,up,forward), etc) seem to generate all of this stuff.
TIL my chalkboard is broken.
That capitalized Lie is highly sus
The headline is excellent!
is there a way to represent both vectors as matrices so that their matrix product is the same matrix you would expect from the cross product vector?
12:29 third column third row should be -bf-ae.
I also noticed.
What happened at 11:06? Please don't say its trivial.
Never mind.
Lmao the pun
The volume seems low...
Somebody pls post the thing he skipped at 1:16 where the fact that the output is perpendicular implies the vector of the anti symmetric differences
Well If you want to calculate a vector v that is perpendicular to x and y, you have to solve the linear system of equation
v • x = 0 and v • y = 0
Where • is the euclidean scalar product.
When you solve for v you get the Crossproduct Up to scalar multiples. I think that is what He means.
Beautiful, thank you Michael.
I’m rusty as it’s the 1980s that I studied pure maths, but I still like to dabble. But surely it’s only unit vectors that map to SO(3), presumably R^3 (excluding 0 I guess) maps to O(3)? Put another way, crossing with a unit vector gives a rotation, crossing with and non zero vector gives a rotation composed with a scaling?
Wait how is
ux(vxw)+wx(uxv)+vx(wxu)=0 ???
well, you need to prove it. An easy way of getting is using a formula for ux(vxw) = (u.w)v - (u.v)w (dot is the inner product, and concatenation is scalar multiple). Add these together and you get the zero.
I was quite shocked couse I read "Lie" as "lie". ahahahah
Hi,
Could you do something about bi-vector theory? I find it a killing theory, or formalism, that simplify bunches of things.
Title is worth the upvote alone
If this cross prpduct is a Lie then what is true
wedge product
My whole life is a lie.
14:01
Way better than my physics professors lecture on this. Thanks
which cross product? the 7-dimensional one is not a lie algebra
Set Phi=f for short. f(vxw)=[f(v),f(w)] is equivalent to f(vxw)u=[f(v),f(w)]u for any test vector u. The LHS is simply f(vxw)u=(vxw)xu. For the RHS we have [f(v),f(w)]u=f(v)f(w)u-f(w)f(v)u=vx(wxu)-wx(vxu)=vx(wxu)+wx(uxv)+ux(vxw)-ux(vxw)=-ux(vxw)=(vxw)xu, since the cross-product satisfies the Jacobi-identity and is anti-commutative.
BeLIEve that!
Gotta give this a like just for the wonderful pun in the title
It is not good pedagogy to say "The cross product is a lie!" That instantly invalidates every formula with a cross product. That's not of course what you mean. It would be better to say the cross product is a 2-index object, a bivector, which in 3-d space is dual to a vector, so has the same number of independent components.
Goddammit hahahahah the title got me good 😂😂😂😂
Pretty interesting. I'd have liked to see how(/if) the matrix form of the cross product relates to the 3D rotation matrix, but looks like I've got a little challenge for the afternoon!
Did you reach somewhere? There's probably a nice geometric interpretation out there, maybe even worth a video.
First random thought is that |u x v| = |u| |v| sin(theta) where theta is the angle between u and v. You should be able to see this theta angle appear in the rotation matrix.
Second thought is that this relation might be quite indirect. This matrix form of the cross product lives in a Lie algebra, associated to the Lie group of the rotations. So one in probably obtained by differentiating the other.
Just random intuitions thrown here. I might just be saying nonsense.
I don't think of it as a lie. It's just a very *convenient* truth. So good it hurts.
Also, I am ashamed I didn't see the entendre there. Very nice.
The cross product is really just a specific type of triple dot product :)
Cross product is not a lie, whatever you mean with lie. Using a, b, c instead of u1. u2, u3 is less helpful
i thought he will say that the idea that a cross product is a perpendicular vector is just a convention and not how vectors really work in physics
but it's still an interesting video to watch
what do you mean? In math it's defined as the parpendicular vector making a right-hand basis when added to your input list. (+ condition to normalize it, but whatever)
@@Czeckie depends, in normal vector algebra it's defined as a perpendicular vector as you said, but there are physics who prefer the idea of cross product as the bivector parallel to the two vectors
check Clifford Algebra to know what i mean
@@Czeckie I think what is being referred to is polar vs. pseudo vectors.
@@vonakakkola alright, but that would be the exterior product, right? Written as a wedge.
@@Czeckie yes, the cross product is just a way to "label" each of these 2 forms and their orientation. The labelling is possible with Hodge star operator.
Minute 13:03 you lodt a minus in front of bf
Cool shirt tho, where did you get it from? Just asking(*)
oh, thank you for the pronounciation. I'm about to sue my former college's mathematics department for teaching me a lie.
Nice clickbait yo. Tho this was kinda trivial, I guess the commutator is an useful thing to understand how one would generalize the cross product to the external product.
damn. "Lie", not "lie" then. Damn. I don't know math, just wanted to see juicy stuff. And here i am watcing a whole video, wondering where the lie was. But there was no lie, only Lie. I feel cheated, but there was no lie.
Clifford algebra for the win in $\mathbb{R}^n$
Therefore, cross product is a piece of cake. This completes the proof.
I have been duped
Make a full course on Linear Algebra sir pls... it'll be very very benifitial for us...love from India ❤️✨
Refer to Gilbert Strang's excellent book and MIT course video series which you can find here on UA-cam.
@@JivanPal okay brother, thanks ❤️
He does it on his other channel, MathMajor
Well
Your wish has been granted! Check out mathmajor
WOWWWWWWWWWWWWw
You got me there
The derivative is also a lie algebra.
:)
pr໐๓໐Ş๓ 😎