Lecture 3: Exterior Algebra (Discrete Differential Geometry)

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Keenan Crane

Keenan Crane

Күн бұрын

Full playlist: • Discrete Differential ...
For more information see geometry.cs.cmu.edu/ddg

Пікірлер: 82
@shiv093
@shiv093 3 жыл бұрын
0:30 Why Learn Exterior Calculus? 5:18 Where Are We Going Next? 6:45 Why Are We Going There? 8:11 Vector Space (Warm-up Multiplication) 11:26 Review: Vector Spaces 14:39 Vector Spaces - Geometric Reasoning 16:51 Review: Inner Product 19:12 Review: Span 21:32 Wedge Product (^) 24:08 Degeneracy 25:28 Associativity 27:17 Distributivity 28:14 k-Vectors 28:59 Visualization of k-Vectors 30:48 0-vectors as Scalars 32:07 Hodge Star (Review: Orthogonal Complement) 34:11 Orthogonal Complement 36:00 Hodge Star (*) 38:28 Hodge Star - 2D 40:26 Exterior Algebra - Recap 41:46 Coordinate Representation 42:43 Basis - Visualized 43:32 Basis & Dimension 44:34 Basis k-Vectors - Visualized 45:57 Basis l-Vectors - How Many? 51:46 Hodge Star - Basis k-Vectors 53:54 Exterior Algebra - Formal Definition 54:25 Sanity Check 55:07 Exterior Algebra - Example 1:00:53 Summary
@NewDeal1917
@NewDeal1917 2 жыл бұрын
this has to be one of the best courses not only on differential geometry, but on more or less 'abstract math' in general. for some unknown reason the overwhelming majority of authors assume the only way to teach that kind of stuff is by using the most general language possible right from the beginning, with little or no motivation and no numerical examples for illustrations. sadly, that applies both for textbooks, articles and youtube videos. your work is a perfect proof that even complicated things can be tought in an approachable manner without the need to humiliate the student and submit him\her to the realm of vague symbol manipulation. thanks for that gem! don't think you'll notice the comment, but still, are there any places where I could find your material on tensors? your illuminating style could really build the intuition, still lacking after all the usual learning materials. thanks again and please keep on delivering!
@keenancrane
@keenancrane 2 жыл бұрын
Thank you for your very kind comments. No material on tensors, I'm afraid, but certainly a worthy subject to cover! (And one that is indeed poorly explained/motivated in many cases…)
@NewDeal1917
@NewDeal1917 2 жыл бұрын
@@keenancrane wow, grateful for your response! If you allow, I'd like to share some of my ideas on how to avoid the widespread tensor confusion (speaking from personal experience). Perhaphs you consider some of them useful... or not. I've checked literally dozens of books, articles and lectures on tensors and have to say they all suffer greatly from the same problems - all the explanations assume the vast prior knowledge and experience in undergrad math which most random students don't have. Except for "ML - approach", where you just stumble into "it's an n-dimensional matrix, load your Python, gentlemen!") The only examples of a somewhat clear explanation would be the article 'Tensor product demistified' by Tai-Danae Bradley and some chapters in Nathaniel Johnston's 'Advanced Linear and Matrix Algebra". Other than that it's a dead end of manifolds and covariant transformations appearing out of thin air. The least painful way (as I see it) is describing tensors as multilinear maps, which requires no reference to any basis or coordinates, but only a very basic understanding of linear maps and rudimentary matrix algebra (you could actually motivate the idea by showing how the well known cross-product and n x n determinants are nothing more than tensors in disguise). That way the generalization of all the linear stuff actually makes sense! You can easily classify all the p-forms, metric tensors, quadratic forms as well as their decompositions with almost no mental pain. It greatly coincides with your idea of learning a language to simplify things and understanding, which you've referenced while introducing the concepts in exterior algebra, rather than enduring the abuse of unmotivated symbolic manipulations. I apologise if I've wasted too much of your time, thanks for the awesome content and well structured courses, rather than separated videos about 'cool stuff', which tend occupy the majority of math-related KZbin nowadays. Have a great day!
@alinayan7767
@alinayan7767 Жыл бұрын
Thanks so much for explaining k-forms (and so on)! Unfortunately, in my university's differential geometry course, we were just given definitions without any explanation. Lectures 3 to 7 of this course opened my eyes to the motivation behind all these notions =)
@LowestofheDead
@LowestofheDead 7 ай бұрын
I've watched a few videos on Geometric/Clifford Algebra and similar topics, and this one has by FAR the best presentation and teaching. Everything is made intuitive with a visual example, it's wonderful. Thank you so much for making this series.
@Seleno_phile
@Seleno_phile 2 жыл бұрын
Prof Keenan, greetings from Yemen and thanks a lot for the time and efforts you put in these lectures and slides. You have a radio voice and great way to introduce concepts. In the future, when you have time, I wish you can publish general mathematical concepts for the public. You have everything to be a great science (math) communicator. This channel will become very popular , I am sure. Thanks again❤️🇮🇩
@keenancrane
@keenancrane 2 жыл бұрын
Many thanks for your kind words. Not enough hours in the day! But I will continue to add videos and explanations when I can.
@jhnflory
@jhnflory 2 жыл бұрын
I almost feel like I stole something getting to watch such awesome, informative, and well-made lectures for free...thanks much!
@undisclosedmusic4969
@undisclosedmusic4969 2 жыл бұрын
This lecturer is unbelievable
@exploringcs7484
@exploringcs7484 Жыл бұрын
Professor, I hope they are awarding you for your class. This in one of the best lectures on abstract math, I have ever had. I can't downplay the fact that I knew most of the concepts. But despite knowing them, I thoroughly enjoyed the lecture. *knock* *knock* *knock*
@matthewgrasinger4788
@matthewgrasinger4788 3 жыл бұрын
This entire lecture series is fantastic, but I especially enjoyed the motivation for learning multiple mathematical languages and the geometric connection to why 3x4 = 4x3. These perspectives and levels of understanding are powerful. Thank you, professor.
@GermanTutorials
@GermanTutorials 2 жыл бұрын
Thank you so much for uploading this amazing introduction to exterior algebra!
@tensorfeld295
@tensorfeld295 3 жыл бұрын
With this lecture one can grasp good intuition.
@gnikkings
@gnikkings Жыл бұрын
This is one of the most insightful videos I have seen.
@robertwilsoniii2048
@robertwilsoniii2048 Жыл бұрын
This is pure gold.
@bryanbischof4351
@bryanbischof4351 3 жыл бұрын
Wow this course is moving fast! 5 hours of lectures already 😅
@PeeterJoot
@PeeterJoot 3 жыл бұрын
@49:22 there's a minor typo: in the last three vector, you've switched to upper indexes from lower.
@unspeakablevorn
@unspeakablevorn 3 жыл бұрын
I'm kind of surprised that the Hodge star *keeps* the magnitude instead of *inverting* it: I'd have expected *(3 e1 ^ e2 ^ e3) to be 1/3, so that x ^ *x is just exactly the highest dimension k-vector in the space. Of course if that were the case we wouldn't get the generalization of cross product as *(a ^ b), which is a reasonably useful property.
@keenancrane
@keenancrane 3 жыл бұрын
There's an even more basic reason: if the Hodge star inverted the magnitude, it wouldn't be a linear map! And wouldn't be well-defined on k-vectors of zero magnitude. Linearity is important for later developing ideas like cohomology, and for translating the Hodge star into computation based on solving sparse linear equations.
@christianaustin782
@christianaustin782 3 ай бұрын
​@keenancrane I'm cool with it being a linear map because linear maps are great, but then how does it satisfy that determinant property as OP mentioned? If you scale a k vector, wouldn't it's Hodge star need to scale (roughly) inversely so that the determinant is 1?
@adamhendry945
@adamhendry945 3 жыл бұрын
Thank you for these wonderful videos! Around 37:25, it would be helpful if you could add that "n is the dimension of the vector space". It took me a minute to realize we have to know the dimension of the space our k-vector is embedded in to compute the Hodge star.
@keenancrane
@keenancrane 2 жыл бұрын
Yes, good suggestion. The fact that the Hodge star depends on the dimension of the space is of course extremely important. I had hoped the concept of orthogonal complement would make this clear-but good reminder to be explicit rather than implicit!
@gurki110
@gurki110 3 жыл бұрын
What an excellent lecture! Appreciate all the work that went into this. Does anyone know how the visualizations were created, maybe LaTeX/Asymptote, or VRay? Those are gorgeous!
@keenancrane
@keenancrane 3 жыл бұрын
www.cs.cmu.edu/~kmcrane/faq.html#figures
@hermogene706
@hermogene706 3 жыл бұрын
Thanks for such a clear getting into business. When you add, (e.g.) two 2-vectors as in example for distributivity in previous lecture, how do you combine orientations (aligned , opposite, at an angle...)?
@alimamdouh1764
@alimamdouh1764 Жыл бұрын
first thanks for your great content form Egypt. could you explain what have you meant by "union of two vector" in 28:00. does that mean the sum of the two areas? because i dont think that would be the case, however if we add the two vectors resulting from the wedge with v1 and the wedge from v2 , its magnitude will equal to the area of v1+v2 widging with u, and it would be perpendicular to that plan.
@KyussWhitewater
@KyussWhitewater 3 жыл бұрын
Thank you for creating this lecture series and for making it open to the public. I have a couple of questions: a) reg. the Hodge star and the $k \mapsto n-k$ comment. The hodge star takes a k-form, which can be reasonably thought of as a signed scalar (dim=1 ?) and outputs an n-k vector (?), what is n in the general case? b) also at about 38:16 the convention for positive orientation is introduced: $z \wedge \ast z$. But is that operation well defined? Up until this point the wedge product was an operation between vectors, the convention introduces a wedge product between a k-form and a vector. How should one intuitively reason about this?
@AkamiChannel
@AkamiChannel Жыл бұрын
"Gin" is silver. "Kin" is gold. It should be read "kin"
@xxh7595
@xxh7595 3 жыл бұрын
I was wondering if professor can speak German when professor said "steinar formula" with a very proper /ˈʃt/... Now Japanese! 1:11 😲..wow!
@michaelhartmann9585
@michaelhartmann9585 3 жыл бұрын
At 28:01 you are arguing that distibutivity can be explained by adding the area of parallelograms. However, the sum of the areas of the two smaller parallelograms could vary quite a lot without changing v_1 + v_2, hence the sum of the areas of the smaller parallelograms will in general not be the area of the large one. Moreover, orientation isn't taken into account. Do we just have to accept this rule as not very illustrative or is there maybe another more accurate explanation, e.g. involving projections?
@keenancrane
@keenancrane 3 жыл бұрын
I would love to have a better picture here. I am sure there is one. I just haven't spent the time to figure it out!
@michaelhartmann9585
@michaelhartmann9585 3 жыл бұрын
@@keenancrane I should have said that this is a great video, one of the best I have seen for quite a while.
@j_razavi
@j_razavi 2 жыл бұрын
@@keenancrane I think when you say "union" at that point it's just a slip of the tongue: you mean the sum of the wedges equals the wedge of the sums. It's a bit confusing because sum of 2-vectors and "distributivity" over sum are introduced together. One of the many beautiful things I'm learning from this series is that the picture is the justification for both of them: it justifies the sum because we expect that adding up the faces of the (closed) prism should give zero (and the triangles cancel out), and distributive because the side of the sum is visibly the sum of the sides. I think you just tried to say both these things at the same time.
@anuman99ful
@anuman99ful 2 жыл бұрын
What do you mean you can change the area of the smaller parallelograms without changing v_1+v_2 and this resulting in not summing the area of the large parallelogram? I can't see how varying v_1 and v_2 in a way that the area (at least not signed area) of the sum of these two isn't equal to the area of the one which is made from the sum of v_1 and v_2.
@j_razavi
@j_razavi 2 жыл бұрын
@@anuman99ful You can see the issue by just thinking in 2D: the length of the base of a triangle is not the sum of the lengths of the other sides. You could have a tall, narrow triangle and a squat, flat one with the same base length. Similarly, if you think of the figure we're discussing as a roof, you could make the roof tall and pointy or flat, which would change the area of the parallelograms being added, but not the area of the sum (which is the area of the "room under the roof").
@rasclausen
@rasclausen 3 жыл бұрын
Will the assignments be available online on your website? I have checked and it currently says nothing found. I am a computer science student from University of Copenhagen currently planning to do a project and bachelors project in geometry processing, and the course notes and lectures are very useful.
@keenancrane
@keenancrane 3 жыл бұрын
Yes, they will. It's the beginning of the semester here and we haven't handed out any assignments yet. :-)
@florankacaku64
@florankacaku64 Жыл бұрын
Im a little confused with the last example, when you take the *(3 e1^e2^e3) shouldnt that be 1/3 in order to make the determinant =1?
@marcushendriksen8415
@marcushendriksen8415 3 жыл бұрын
Is it correct to say that u wedge u equals zero, because of the anti symmetry rule? That is, u wedge u = -(u wedge u), which automatically makes it zero
@jdsahr
@jdsahr Жыл бұрын
at 57:00 you also used the antisymmetry property.
@English1108
@English1108 Жыл бұрын
whoa, thank you for noticing that! I visually missed the fact that the term was "e2^e1" and "e1^e2" so I had no idea how that negative sign got dropped. Thanks for the commenting!
@adelelopez1246
@adelelopez1246 3 жыл бұрын
Why do you depict the orientation of a bivector as a normal vector instead of as a counterclockwise vs clockwise direction?
@keenancrane
@keenancrane 3 жыл бұрын
Good question. There are many possible ways to diagrammatically indicate orientation, and probably the right thing is to consider using all of them, depending on the context. Here, I am trying to draw a connection between the orthogonal complement (which students may be familiar with from linear algebra) and the Hodge star. So, drawing a vector (which points along a line or ray) feels appropriate. I use the orientation arrows for oriented simplices, so it may indeed be worth making this connection visually when I draw k-forms. There are also nice visualizations that I won't use anywhere in the lectures-such as the "onion" visualization for differential forms: citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.126.1099&rep=rep1&type=pdf I think the short answer for anything about mathematical visualization is: (unfortunately) nobody takes mathematical visualization very seriously as an academic discipline. So, the way we draw is usually very ad hoc/learned informally; more effort could definitely be put into finding the "right" pictures/approaches to visual communication.
@baptiste-genest
@baptiste-genest 3 жыл бұрын
Hi Mr.Crane, I absolutely love your videos, I find them incredibly insightful and I'm so much grateful! But at 59:50 isn't there a sign error? because beta ^gamma = -gamma ^ beta and -gamma = e1^e2 so -gamma^beta = (e1 ^ e2) ^ 3e3 = 3e1^e2^e3 so the final result must be 9 e1^e2^e3 and not 3e1^e2^e3, or am I missing something in the computation? Thank you again!
@keenancrane
@keenancrane 3 жыл бұрын
Thank you! Hmm… e1 ^ e2 ^ e3 is an odd permutation of e3 ^ e2 ^ e1: you can get from one to the other by exchanging just one term. So, the sign flips. Hopefully I have that right!
@baptiste-genest
@baptiste-genest 3 жыл бұрын
@@keenancrane Indeed you must be right but I can't see the error in my original comment ;-( don't we have : e3^e2^e1 = e3^(e2^e1) = e3^(-e1^e2) = -e3^(e1^e2) = (e1^e2)^e3 =e1^e2^e3 ? I'm sorry it must be a really dumb mistake xD
@keenancrane
@keenancrane 3 жыл бұрын
@@baptiste-genest It's actually a very subtle mistake: you swapped a 1-form and a 2-form, which in this case is equivalent to swapping two 1-forms twice (even permutation). So, you have a negation where you shouldn't: going from -e3^(e1^e2) to (e1^e2)^e3. To convince yourself which one is right, I'd lean less on the algebra and go back to the geometry (e.g., think of u^v as a cross product, and then in any given wedge product u^v^w or w^u^v or whatever, ask whether w points in the same or opposite direction as u^v).
@nehalkalita1
@nehalkalita1 2 жыл бұрын
1:00:00 Can you show how (2e1 ^ e2) ^ 3e3 = 6e1 ^ e2 ^ e3 ?
@PierreGleize
@PierreGleize 2 жыл бұрын
Shouldn't the answer at 1:00:42 be 1/3 ? Since the determinant should be equal to 1 ?
@amiltonwong
@amiltonwong 3 жыл бұрын
At 36:44, the term two-form (2-form) is mentioned in the first time. What's the relationship between 2-form and 2-vector (as previously mentioned as the wedge product of two input vectors)?
@michaeldeakin9492
@michaeldeakin9492 3 жыл бұрын
see the next lecture on k-forms, but basically a k-form measures a k-vector
@keenancrane
@keenancrane 3 жыл бұрын
Sorry, this is a flub. I should have said "2-vector" here. As Michael Deaken mentions below, k-forms will be defined in the next lecture.
@amiltonwong
@amiltonwong 3 жыл бұрын
Thanks a lot. Things get more clear after lecture 4 :)
@bainsonic
@bainsonic 3 жыл бұрын
the wedge product mentioned here looks very similar to the cross product. Are they equivalent or is one a special case of the other or are they just two different things?
@keenancrane
@keenancrane 3 жыл бұрын
The cross product captures some of the spirit of the wedge product, but is much more specialized. (In some countries, by the way, students learn to write the cross product as u ^ v!) For one thing, the cross product operates only on a pair of vectors u, v. For another thing, these vectors must be from R^3. Also, formally, the cross product of two vectors yields a vector, where as the wedge product of two vectors yields a 2-vector. However, in R^3 one can "implement" the cross product using the wedge product and the Hodge star: u x v = *(u^v). In R^3, the star turns the 2-vector u ^ v into a 1-vector, equal to the cross product.
@bainsonic
@bainsonic 3 жыл бұрын
@@keenancrane Oh thank you. I thought these parallelograms are just how we imagine 2-vectors with them actually being normal vectors where we interpret the length as the area of said parallelogram. But it is much clearer now :)
@robertwilsoniii2048
@robertwilsoniii2048 Жыл бұрын
The hodge star explains the formula for the determinant in R^n.
@reinerwilhelms-tricarico344
@reinerwilhelms-tricarico344 Жыл бұрын
Would have been nice to show an example how to actually use this determinant formula to determine *a from a given a k-form a.
@yizhang7027
@yizhang7027 2 жыл бұрын
28:04 how do the two smaller areas add up to the large one? |v1+v2| < |v1| + |v2| right? I think they are only equal in 2d.
@MrTeeFilter
@MrTeeFilter 2 жыл бұрын
I thought so too, except that |v1+v2|
@diribigal
@diribigal 2 жыл бұрын
@@MrTeeFilter well, an inner product gives rise to a norm. And also when this video talked about lengths and areas, the standard meanings were used.
@diribigal
@diribigal 2 жыл бұрын
That was indeed an error in the narration. What you could say is that for 2-vectors in 3D that share a vector in the wedge, we *define* addition using the picture on top. Then you can get distributivity by noticing the triangle formed by v1, v2, and their sum.
@codatheseus5060
@codatheseus5060 3 ай бұрын
I'm here after some linear algebra and some geometric algebra
@christianaustin782
@christianaustin782 3 ай бұрын
Ik this video is a few years old now, but I'm a bit confused about Hodge star. It's supposed to be a linear map, but whenever you wedge a k vector with its Hodge star you get determinant 1? How is this consistent? Wouldn't the hodge star need to invert any scalar multiples on a k vector instead of preserving them as a linear map would do?
@TheRsmits
@TheRsmits 2 күн бұрын
It's only when you wedge a k vector built up of basis vectors with its Hodge star that you get determinant 1, otherwise it is linear.
@English1108
@English1108 Жыл бұрын
either there is a mistake at 57:34 or something wasn't introduced properly. that should be -e1^e2+4e1^e2=3e1^e2 why did the negative sign on that term get dropped?
@English1108
@English1108 Жыл бұрын
I found the answer thanks to another commenter. the first term is e2^e1, not e1^e2 -- its reversed. so the asymmetry property tells us we can flip the order if we flip the sign - aka drop the negative.
@robertwilsoniii2048
@robertwilsoniii2048 Жыл бұрын
So basically wedge products are like plane segments with finite length sides.
@robertwilsoniii2048
@robertwilsoniii2048 Жыл бұрын
N choose K possible bases for K-vectors. I think I now understand why they are called k-vectors in R^n.
@user-sv5vb1mj1q
@user-sv5vb1mj1q Жыл бұрын
e3 ^ e2 ^ e1 = (e3 ^ e2) ^ e1 = -e1 ^ (e3 ^ e2) = e1 ^ e2 ^ e3. e3 ^ e2 ^ e1 = -e3 ^ e1 ^ e2 = e1 ^ e3 ^ e2 = -e1 ^ e2 ^ e3 What I am doing wrong?
@webgpu
@webgpu 2 жыл бұрын
1:24 i pointed my phone's camera at the screen and used google lens. The translation only replaced "gold" by "money"
@shoopinc
@shoopinc 2 жыл бұрын
I am coming for that gold.
@reinerwilhelms-tricarico344
@reinerwilhelms-tricarico344 Жыл бұрын
Great lecture but I won’t forgive that assault on Bavarian food. 😂
@user-wt8vl3bi6y
@user-wt8vl3bi6y Жыл бұрын
41:48 5234
@aoeuable
@aoeuable Жыл бұрын
Came for the maths, stayed for the Bavaria bashing.
@dsm5d723
@dsm5d723 3 жыл бұрын
Are you mapping the skin effect in solids that are not a toroidal slice, or wire?? I like dihedron algebra from Wildberger. And his calculus. I gave God's math. No $$$$ for perfection now. Only toys. And efforting.
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