I find myself waiting for these videos everyday. Watching 3 minutes after posting. Prof Borcherds, your videos are teaching me a lot of mathematics which I have been meaning to learn. Thanks a lot!
I find it easier to think of Lie groups as exponentials of Lie algebras. e.g A = exp(B). If B is anti-symmetric than A is orthogonal etc. Then you get exp(A)exp(B) = exp(A+B+[A,B]/2+...) which gives the Lie bracket.
This leads to some problems though. The exponential map of sl_n into SL_n or PSL_n is a different map. Furthermore the exponential map is not surjective in general and even taking the group it generates only gives you the connected component of the identity.
I would say this isn't supposed to be a full formal course, but rather as an addition to a more serious text(I am not sure what to recommend, tho, but there are plenty of good books on Lie theory). As for prereq, you definitely need to know some Algebra, Linear algebra, and some topology/differential geometry.
@@annaclarafenyo8185 Gotta disagree. If you don't know what a manifold is, or what a ring is, or what a vector space is, it would be very hard to follow this lecture. These aren't extremely fancy or niche concepts, but they're certainly prerequisites!
If you get lost here, it means you don't have a solid idea of how Lie Algebras are infinitesimal matrices. Try a physics source, like Lie Algebras by Howard Georgi, this has concrete presentation. Once you understand the matrix examples, it's completely trivial to turn it abstract. Going the other way, unfortunately, from abstract to concrete, is impossible. This might be enough: Consider the group of rotations in n dimensions, i.e. matrices O such that O-transpose times O is the identity. O* O = e (star means transpose, e is the identity matrix) An infinitesimal group element is, by definition, mostly the identity e, plus an additional infinitesimal matrix A. So an infinitesimal rotation is written as e+ A. In order to be in the group of rotation, there is a condition A inherits from the rotation: (e* + A*) (e + A) = e i.e. A + A* = 0 That means A is antisymmetric. So the antisymmetric matrices are the Lie Algebra of the Orthogonal group of rotations. That's the construction being abstracted. By taking matrix exponentials of antisymmetric matrices, you reconstruct the rotations. That means, taking (e+A)^N where A is infinitesimal and N is infinite. This works because rotations form a group. When abstracting out the essence of this construction, you want to know what data you specify just on the A's to recover the group multiplication law. To reconstruct the exponentiation here, you needed not just the A's as an abstract vector space, but also how they are embedded in a larger matrix group, to know what A^2, A^3, etc mean in relation to A. If you remove the embedding into a matrix group, you want to know what you need to know about the A's, just as a vector space, so that HOWEVER you embed them into a matrix form, you recover the same group structure at least locally, near the identity. The way to do this is by multiplying elements of the group together to second order, so that the first order parts cancel. Take U = (e+A + A^2/2 ) V= (e+B +B^2/2) and construct the inverses to second order U^(-1) = (e- A + A^2/2), V^(-1)=(e-B +B^2/2). The simplest product that cancels to first order but not to second order is the commutator, and it is equal to: U V U^(-1)V^(-1) = e + (AB - BA) Since it cancels to first order, AB-BA must be in the Lie Algebra again, because you can now consider the (AB-BA) as the leading order infinitesimal. That means, given A and B in the Lie Algebra, AB-BA is also in the Lie Algebra. So if you choose a basis for the Lie Algebra, you learn that you can find (AB-BA), which is the definition for matrices of the bracket [A,B], for any two basis elements and re-express this in the basis. Calling the basis elements u^i, [u^i , u^j ] is a linear combination of the u's. That's written as C^i_{jk} u^k, where the constants C^i_{jk} are called the structure-constants of the Lie Algebra. They express the commutation structure in terms of a basis. The structure constants are enough to reconstruct the group multiplication order by order in a power series. up to choice of basis for the Lie Algebra. That means that any two groups with the same Lie Algebra (at least, for finite dimensions) will coincide in a neighborhood of the identity. The presentation in this lecture simply does this construction abstractly, without a matrix embedding, without an explicit basis, just using the manifold structure and the group structure we are given. The result is equivalent to defining structure constants. The problem with the mathematics presentations is allergy to matrix presentations and allergy to infinitesimals. The part you are getting lost on is a trivial change of notation from talking about infinitesimals to talking about vectors as differential operators, which is really trivial, but anti-pedagogical.
Lie groups and lie algebras are optional subject in many universities offering pure math programs in india. Probably these arose in connection with physics. So some motivation for pure math sense is good
29:16,The result of (1+e A) (1+fB) (1-eA)(1-fB) isn't 1+ ef [A,B], it's 1 + ef [A,B] - e^2 A^2 - f^2 B^2 . To get the right answer, you need to work consistently to second order: (1 + e A + e^2 A^2 /2 ) and (1+ f B + f^2 B^2/2), and now the no good parts cancel. Of course, this had to work out to the right answer by the group condition, but the calculation you posed doesn't get the cancellation to happen, and this will confuse students who work it out.
The calculation in the lecture is ignoring all terms involving e^2 and f^2 (in other words working modulo the ideal (e^2,f^2).This is not quite the same as working to second order. I agree that it may confuse people and should probably have been explained better.
I think for students, it's best to write the Jacobi identity as: [ [A , B] , C] = [ A , [B, C] ] - [ B , [ A, C] ] Now it's so self-evident, it doesn't even require verification-- [A,B] adjointly acting on C is B followed by A acting on C, minus A followed by B acting on C.
I find myself waiting for these videos everyday. Watching 3 minutes after posting. Prof Borcherds, your videos are teaching me a lot of mathematics which I have been meaning to learn. Thanks a lot!
This is a top-rate explanation: clear, concise, rigorous. Thank you Professor Borcherds!
These lectures are immensely enjoyable. Many thanks!
Thank you for doing this, sir. I started learning about Lie groups just recently and these lectures have helped enormously. This is great!!!
I was lost for a long time at the product at 6:02 until I realized it's the product rule.
I find it easier to think of Lie groups as exponentials of Lie algebras. e.g A = exp(B). If B is anti-symmetric than A is orthogonal etc. Then you get exp(A)exp(B) = exp(A+B+[A,B]/2+...) which gives the Lie bracket.
This leads to some problems though. The exponential map of sl_n into SL_n or PSL_n is a different map. Furthermore the exponential map is not surjective in general and even taking the group it generates only gives you the connected component of the identity.
my mind is blown, king
What's the prerequisite for this course? I feel lost
True Algebras.
I would say this isn't supposed to be a full formal course, but rather as an addition to a more serious text(I am not sure what to recommend, tho, but there are plenty of good books on Lie theory). As for prereq, you definitely need to know some Algebra, Linear algebra, and some topology/differential geometry.
@@markfjord5148 Not really. There are no pre-requisites.
@@annaclarafenyo8185 Gotta disagree. If you don't know what a manifold is, or what a ring is, or what a vector space is, it would be very hard to follow this lecture. These aren't extremely fancy or niche concepts, but they're certainly prerequisites!
If you get lost here, it means you don't have a solid idea of how Lie Algebras are infinitesimal matrices. Try a physics source, like Lie Algebras by Howard Georgi, this has concrete presentation. Once you understand the matrix examples, it's completely trivial to turn it abstract. Going the other way, unfortunately, from abstract to concrete, is impossible.
This might be enough: Consider the group of rotations in n dimensions, i.e. matrices O such that O-transpose times O is the identity.
O* O = e (star means transpose, e is the identity matrix)
An infinitesimal group element is, by definition, mostly the identity e, plus an additional infinitesimal matrix A. So an infinitesimal rotation is written as e+ A. In order to be in the group of rotation, there is a condition A inherits from the rotation:
(e* + A*) (e + A) = e
i.e. A + A* = 0
That means A is antisymmetric. So the antisymmetric matrices are the Lie Algebra of the Orthogonal group of rotations. That's the construction being abstracted. By taking matrix exponentials of antisymmetric matrices, you reconstruct the rotations. That means, taking (e+A)^N where A is infinitesimal and N is infinite. This works because rotations form a group.
When abstracting out the essence of this construction, you want to know what data you specify just on the A's to recover the group multiplication law. To reconstruct the exponentiation here, you needed not just the A's as an abstract vector space, but also how they are embedded in a larger matrix group, to know what A^2, A^3, etc mean in relation to A. If you remove the embedding into a matrix group, you want to know what you need to know about the A's, just as a vector space, so that HOWEVER you embed them into a matrix form, you recover the same group structure at least locally, near the identity.
The way to do this is by multiplying elements of the group together to second order, so that the first order parts cancel. Take U = (e+A + A^2/2 ) V= (e+B +B^2/2) and construct the inverses to second order U^(-1) = (e- A + A^2/2), V^(-1)=(e-B +B^2/2).
The simplest product that cancels to first order but not to second order is the commutator, and it is equal to:
U V U^(-1)V^(-1) = e + (AB - BA)
Since it cancels to first order, AB-BA must be in the Lie Algebra again, because you can now consider the (AB-BA) as the leading order infinitesimal. That means, given A and B in the Lie Algebra, AB-BA is also in the Lie Algebra. So if you choose a basis for the Lie Algebra, you learn that you can find (AB-BA), which is the definition for matrices of the bracket [A,B], for any two basis elements and re-express this in the basis.
Calling the basis elements u^i, [u^i , u^j ] is a linear combination of the u's. That's written as C^i_{jk} u^k, where the constants C^i_{jk} are called the structure-constants of the Lie Algebra. They express the commutation structure in terms of a basis.
The structure constants are enough to reconstruct the group multiplication order by order in a power series. up to choice of basis for the Lie Algebra. That means that any two groups with the same Lie Algebra (at least, for finite dimensions) will coincide in a neighborhood of the identity. The presentation in this lecture simply does this construction abstractly, without a matrix embedding, without an explicit basis, just using the manifold structure and the group structure we are given. The result is equivalent to defining structure constants.
The problem with the mathematics presentations is allergy to matrix presentations and allergy to infinitesimals. The part you are getting lost on is a trivial change of notation from talking about infinitesimals to talking about vectors as differential operators, which is really trivial, but anti-pedagogical.
Beautiful. Thank you dear Sir.
Lie groups and lie algebras are optional subject in many universities offering pure math programs in india. Probably these arose in connection with physics. So some motivation for pure math sense is good
love these videos
29:16,The result of (1+e A) (1+fB) (1-eA)(1-fB) isn't 1+ ef [A,B], it's 1 + ef [A,B] - e^2 A^2 - f^2 B^2 . To get the right answer, you need to work consistently to second order: (1 + e A + e^2 A^2 /2 ) and (1+ f B + f^2 B^2/2), and now the no good parts cancel. Of course, this had to work out to the right answer by the group condition, but the calculation you posed doesn't get the cancellation to happen, and this will confuse students who work it out.
The calculation in the lecture is ignoring all terms involving e^2 and f^2 (in other words working modulo the ideal (e^2,f^2).This is not quite the same as working to second order. I agree that it may confuse people and should probably have been explained better.
Lie algebras: Lie groups
yeeeeeeeee
I think for students, it's best to write the Jacobi identity as:
[ [A , B] , C] = [ A , [B, C] ] - [ B , [ A, C] ]
Now it's so self-evident, it doesn't even require verification-- [A,B] adjointly acting on C is B followed by A acting on C, minus A followed by B acting on C.
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