Spinors for Beginners 13: Ideals and Projectors (Idempotents)
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- Опубликовано: 9 июл 2024
- Full spinors playlist: • Spinors for Beginners
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Powerpoint slide files + Exercise answers: github.com/eigenchris/MathNot...
0:00 - Matrix Projectors
7:23 - Clifford Algebra Projectors
11:12 - Ideals
18:19 - Projectors create Ideals
The plot thickens...
i continue to think it should be spinors for beginnors
Chris, thank you for making these videos. Hate from the internet is inevitable, your video series was not. I am grateful to you for your sacrifice in making them.
Complexly intelligent, sophisticated, clear, keenly enthusiastic !!!
I pretty much knew all of the individual aspects of the things you covered, but I had never had the connection between them spelled out for me. Ideals were something we studied in abstract algebra, and projectors were something we studied in linear algebra. While I did have a course on the way to my masters that did a little bit of of both of these two types of algebras, I don't think we actually connected them in this way - it was just that the class was supposed to cover certain topics because it was a required course to go further, and there were certain things that we had to learn. Seeing this kind of stuff actually connect together is really interesting, and is part of the reason I generally like mathematics so much. It's too bad that grad school really sucked all the fun out of the subject.
Yeah, I hear you on school sucking the fun out of math. I remember when I was trying to piece all this together about 2 years ago, there were surprisingly few sources that "spelled everything out" like this. They would often leave things out, or say it in such complicated language that I had no idea what was going on. I hope this video fills in the gaps for anyone trying to learn.
@@eigenchris It does definitely!
i finally understood ideals!!! :)
as always, you manage to explain something like no other
upd: and the projector explanation omg!!!!!!!!!!!!!!!! ❤️❤️❤️❤️❤️❤️❤️
It's the first time I really understand Ideals. Wow.
Thanks for your amazing videos ❤
As someone who didnt have theory in college, it really helps
Petition for a full blown course on quantum mechanics!
Amazing content regardless of a quantum mechanics course
Yeah, I said it before, this man can give wondrous teachings. Making convoluted topics graspable. A QM or QFT course from him, that would be amazing. I'd sign right away!
For the general ring case, it needs to be stated that a [left/right] ideal must also contain the additive inverse of each of its elements so that it is a subgroup of the ring, as in the general case the ring might not have a multiplicative identity and hence won't have a -1 element that would automatically imply this based on the other conditions.
All the definitions I've seen have always specified that a ring must have a multiplicative identity. I think the more general structure which doesn't have to have one is called a "rng" (pronounced like "rung").
@@seneca983 The general definition does not specify a multiplicative identity, a ring with a specified multiplicative identity is called a ring with unit or unitary ring - regardless of what rubbish people write on Wikipedia that then gets cloned all over the internet.
@@M.athematech I have also seen that definition in math textbooks and university lectures, not just on Wikipedia. I also suspect the definition of ring used in this series includes the multiplicative identity.
every ideal has {0} and itself as a subideal. minimal ideals have *only* {0} and themselves as a subideal, but themselves are not {0} (a bit like primes, although primes are closer to maximal ideals)
24:10 To nitpick a bit, I guess you should say that there are no *non-trivial* sub-ideals so as to exclude {0}.
Yeah, that's a good point. I might mention that correction in the next video.
Best explanation on this topic, I ever have seen. Many thanks for it😊
The identification of idempotents and projector operators is such a key concept, and the null space plus image decomposition too. Really glad that you start the video giving the intuitive motivation, and encouraging the viewer to prove it. For anyone reading this before watching the video I really recommend stopping and trying for yourselves when he prompts you.
Clearly a projector is idempotent, but is an idempotent always some sort of projector? Why?
@@-minushyphen1two379 If T is an idempotent in V, than V is the direct sum of Im(T) and ker(T) and T projects into Im(T). The proof is left for the reader 😃
@@-minushyphen1two379 it's a linear operator, so any element being acted on can be broken down into a sum of two parts: a part that gets kept unchanged and a part that gets annihilated, with the operator acting separately on each (distributing into the sum). You can show these parts are orthogonal, which means it's projecting onto a subspace: if any nonzero elements get annihilated, the dimension of the space is being reduced (the identity operator also counts as a trivial projector, it's the only one that doesn't do this), and any elements left can be added together while remaining in the subspace. So elements are being projected onto a subspace, or with the identity operator being projected onto the whole space itself.
@@BlazeOrangeDeer Thanks! I noticed that this also applies to abelian groups, so if f is an idempotent homomorphism from an abelian group A to itself, then A = ker f + im f, so that f can be thought of as projecting f onto the image. And it seems like a deformation retract from topology.
can't wait for the next vid!
I'm so happy I found your channel 😊
Wonderful, I never saw such a clear explanation on this topic! It took me a pretty long time to see this connection. I think this is the didactical best way to approach the concept of ideals.
Wonderful!
Very good 👏👏👏
Nice job! 👍
Great work! Thanks …
Sir you are awesome..l did your all the courses except from spinors.. I will start it very soon..you are a excellent teacher.. I am a student of physics from Bangladesh..we have not enough teachers and resources on the field of theoritical physics... when I found you it was a blessing for me.. always take my love dear eigenchris..❤
I want to be a theoritical physicist and I will always remember you in my rest of the life ❤ 1:43 1:45 1:47
Thank you.
Some thoughts for discussions.....
1. If we have an operator P acting on a vector space and it turns out that PP = P,
does this operator always "project" to something? (in another words, Is P a projector *if and only if* PP = P ?)
2. Are the number of projectors we can find acting on a vector space finite/infinite? minimal ideals/minimal operators seems like they can be enumerated...
3. So when comes from a projector, ideals work like picking a coordinate?
4. From the logic in the video, can the matrices
| 0 1 | | 0 0 |
| 0 0 | | 1 0 |
act like projectors, multiplied to the right? you might remember them as the ladder operators for 2 level systems
5. Is there a difference in writing the pauli spinor as
| A |
| B |
instead of
| A 0 |
| B 0 |
? Reminds of of the concept of "direct sum"
What do you mean by the third question?
@@-minushyphen1two379 in the illustration it seems like a 2x2 matrix can be written in terms of left/right ideal or up/down ideal, which reminded me of position being written in cartesian or polar coordinates
i wondered if this is an apt analogy
Still think it's better to say that spinor spaces are minimal left ideals. That makes it clear that all your spinors have to come from the *same* ideal, not just *any* ideal
Why they can't come from any ideal?
Hi chris, over time doing theoretical physics, do you feel like gravitating more and more to the physics side/math side?
As always, such a great video! Never understood this stuff and this was a super clear exposition.
One quick question, regarding the ideals are axes metaphor. I know you told us to not take it too literally, but bear with me. The confusion arises because I already think of basis vectors of a vector space as being the "axes" of the vector space. So is there some sense in which ideals are basis vectors, or at least are related to them?
I haven't looked too deeply into it. But you can write a wring as a "direct sum" of various ideals, the same way you can write a vector space as the "direct sum" of the axes. The direct sum can be done with vector spaces of any dimension, not just 1-dimensional "axes". I guess all algebras are already vector spaces by definition, so with ideals you're just break up an algebra into lower-dimensional subspaces, which you can then combine back together with the direct sum to get the full algebra back.
Does this projection have anything to do with the resultant of two polynomials?
Awesome video as always! I had a question about a property of projection matrices.
If seems that if P and Q are projection matrices and PQ = 0, does that imply QP = 0? Or is that an extra requirement we must impose?
Yeah that's a fair point. I probably should have defined orthogonal as PQ=QP=0
@@eigenchris Though it's not necessary. QP=0 is implied. I'm too lazy to give a rigorous proof, but it's not difficult to see. One possible argument is that tr(QP)=tr(PQ)=0. Indempotent matrices are always diagonalizable and can only have eigenvalues of 1 and 0 so tr(QP)=0 implies that all of its eigenvalues are 0 and thus QP=0.
I think there is a problem with the formula in 8:11
Teorically is correcto because has the properties of a projection, it doesn't project to the basis vector you are using. Just to give an example, the operator (1+e1) acting on e1 should give e1 by your explanarion, but the resulta is e1+1, a multivector.
To get the correcto form, you can expand the formula for linear algebra: Proj_u (v) = (v*u)u, where u has unitary length. Then you expand that multiplication, like 1/2 (vu+uv)u = 1/2 (v+uvu), taking the v out gives Proj_u () = 1/2 ( 1+u • u) , where the • symbol indicates where to put the vector proyected.
Also, that formula came more naturally using the identity of the Clifford product: vu = v*u + v^u, then multiply on the left by u then you get u = (v*u)u + (v^u)u where the first term corresponde to the projection, and the second is the rejection, and if you take the formula for the rejection, is 1/2(1-u•u), which is the orthogonal proyection, as expected.
The clifford algebra projectors don't behave in a straightforward way like standard 3D space projectors do. That's why I said they were more abstract. Cl(3,0) is an 8-dimensional space (1 scalar + 3 vectors + 3 bivectors + 1trivector), and the projector (1 + e1)/2 projects this 8-dimensional space down to a 4-dimensional space. This ends up being the 4 real components of a 2x1 complex Pauli spinor. I'll talk about all of this in the next video.
@@eigenchris I'd wait for ypur next video then, but the formula is incorrect if you want to project vector to some basis tho. And given that the pauli matrices already span Cl(3,0), for now, I dont see the point of truncate the matrices.
How do you prove that a projector/idempotent like the projector onto x is a minimal projector? Perhaps in a vector space any projection onto a 1D subspace is minimal?
I'll consider that in the next video. I'm not quite sure yet.
3:15 the same way, two measures don't always commute in QM. Second, I presume RM Ideals are dual spinors (Am I correct?)
Not sure about your first comment, but I believe minimal right ideals are related to dual spinors. I'll talk about this in my next video. I'm still working it out.
multiplying the entire ring (from the left/right) always yields a (left/right) ideal. What makes projectors (like any zero divisors i think?) special is that it's a nontrivial ideal, it's neither zero nor the entire ring
I should have specified "non-trivial projector". 1 and 0 are both projectors that give the entire ring or zero element. I'm ignoring those.
The projectors P and 1 - P feel a lot like the relation between the projection and rejection operators usually used in Geometric algebra, except written as single multivectors rather than and formulae: proj_x(y) = (y • x)/x and rej_x(y) (y ∧ x)/x. I know that the formulae work with any pair of blades and I think also work when projecting a versor onto a blade, but i don't know what the limitations of the Projectors P and 1 - P.
Hold on... If there no actual difference between an Algebra and a Ring? The only difference i heard was that Algebras are defined on vector spaces, but any additive group, like would be found in any Ring, could be considered a 1-dimensional vector space, automatically making every Ring also an Algebra. Meanwhile, Ring didn't seem to have anything specifically preventing its corresponding set from being a multidimensional vector space.
Rings and algebras are not the same. There are two requirements for rings that algebras might not satisfy:
1) Multiplication in a ring must be associative whereas in an algebra it doesn't have to be associative (though it can be).
2) A ring must have a multiplicative identity (i.e. 1) whereas an algebra might not have one (though it can have one).
However, the tensor algebras, Grassmann/exterior algebras, and Clifford algebras also happen to fulfill both of these requirements so they are rings.
"any additive group, like would be found in any Ring, could be considered a 1-dimensional vector space, automatically making every Ring also an Algebra"
No, I don't think that (always) works because the definition of a vector space requires a *field* of scalar coefficients. An arbitrary ring won't do (at least not obviously) but if the ring happens to be a field then it does indeed work as a 1-dimensional vector space and algebra.
@@seneca983 Huh. Checking Wikipedia, it seems that specifically _unital associative_ algebras are rings, as you said. However it mentions that sometimes those traits can be implied. It also mentions a generalization of "algebra over a field" to "algebra over a ring," and I remember seeing that Clifford Algebras can be defined over rings rather than just fields (though fields are the most common). This does mean that you can still call any ring a "1D associative unital algebra over a ring," with the scalars just being themselves.
@@angeldude101 Maybe it can be but I think usually "algebra" without qualifications refers to an algebra over a field; at least in this series it probably is.
Also, googling this I came across a site called ProofWiki and according to it an algebra over a ring still requires that the ring be commutative (because otherwise defining bilinearity becomes difficult) so an arbitrary ring might not do.
One more note on terminology. I think even if you're talking about an algebra over a ring you probably can't call it a vector space. The generalization of vector spaces where the scalars can be from any ring, not just a field, is called a "module". (I'm not sure whether commutativity is required here or not.)
I like that the ideal is called RP1 like the popular rocket fuel xD
I have a brain inch that it should always be possible to construct a set of minimal, (mutually) orthogonal projectors that sum to the identity projector, but I'm struggling with the proof... (Admittedly I haven't made a very serious effort yet lol)
My initial thoughts are that basically, if we consider some minimal projector P, then we do know that P + (1-P) = 1, and that P and (1-P) are orthogonal. If 1-P is itself minimal, then we are done...but otherwise 1-P is not minimal, and the proof continues. But then, I'm struggling to move forward. One lemma could be that any set of (not necessarily minimal) projectors that together sum to some projector Q which is orthogonal to P, all those projectors are themselves orthogonal to P. (In the same sense that project-x is orthogonal to project-yz, as well as (and perhaps because that it is orthogonal to) project-y and project-z.) Another lemma could be that any projector can be written as the sum of (orthogonal?) minimal projectors. ...this is as far as I've gotten.
Of course, this proof is probably way easier for the specific case of matrix projectors (sum of all unique matrixes whose only non-zero element is a 1 somewhere on the primary diagonal), but I want the extra challenge of a general proof.
I'm actually struggling with how to prove a given projector is minimal. I need to figure it out soon, as it's the topic of the next video. For matrices you're right that it's easy, and you're exactly right that it's a single 1 on the diagonal.
After giving your response more thought, perhaps you could reach backwards from ideals and prove minimality of a projector by the minimality of the ideal it projects onto? For matrices we have the concept of rank that allows us to claim something to the effect of "the resulting ideal contains matrices of only rank 1", I don't know how readily that extends to ideals, or how transparent that is to find for any given ideal...
I think you've skipped the following property: if projectors P and Q are orthogonal, so are Q and P. You don't state it, but use it
a
mong us
i-DEM-po-tents, not I-dem-PO-tents
I've heard it pronounced both ways. Google and most online dictionaries pronounce it the way I do, though.
@@eigenchris It just doesn't flow correctly in English, particularly when used as an adjective ("an idempotent function"). Compare with "incompetent" or "incontinent". It's the long vowel in the "po". "Independent" can have the stress on the "pen" because it's a short vowel. I've already leased a grave on this hill, so come at me, bro!
Great video series though. I've been waiting for it since I saw those long ass videos you posted years and years ago when you were doing research and going through Road to Reality (I think). 👍👍
@@fixed-point I think the main problem here is that English is kind of awful and makes no sense. :) I'm not too worried if someone pronounces it differently than me, though.
Yeah, those videos are about 2 years old now, and I've learned more since then. It's been very frustrating putting this all together, but hopefully it saves the frustration for other people.
@@fixed-point At least Merriam-Webster says the primary stress is on "i" and secondary stress is on "po", not on "dem". I don't know what you mean by "doesn't flow correctly". There isn't any kind of objective definition for a "correct" flow of spoken language. Maybe you like one pronunciation better than another but that's just a subjective experience.
@@seneca983 I'm not saying my way is correct because that's what any recognized authority claims. I'm simply observing that it is the correct way to pronounce it.
You say it's subjective, but is it? Is it subjective whether the Earth revolves around the Sun?
Really mathematicians, please come up with better names...
idempotents? what drug was being taken when that name was invented...
That's simply latin.
Wikiepdia says the term was invented in 1870. "idem" = same and "potent" = potency = power. I prefer "projector", although not all idempotents have an obvious geometric "projection" interpretation. I think a lot of abstract concepts in math suffer from naming problems.
This channel lost traction along time ago but continues, sort of a train reck.
I never did learn how to drive a train.