Lecture 4: k-Forms (Discrete Differential Geometry)

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

[@williamsamosir4567]

Hi Keenan I just want to say thank you so much for uploading this lecture series, it's such a gift. I came from a 3D art and self-taught programming background, and I can't believe that today I am able to learn of such complex concept just through a youtube video, on the public domain. Even at this point it already transforms the way I look at the possibility of meshes, beyond just sculpting and modelling. Really excited to see where DDG can revolutionize the interactive and creative domain.

[@shiv093]

0:36 k-Vectors and k-Forms - Overview 1:42 Measurement and Duality 2:39 Motivation: Measurement in Curved Spaces 3:49 Vectors and Co-Vectors (Duality) 5:30 Analogy: Row & Column Vectors 6:55 Vectors and Covectors 7:37 Dual Space & Covectors 9:13 Covectors - Example (R3) 11:28 Covectors - Example (Functions) 14:41 Sharp and Flat 16:12 Sharp and Flat w/ Inner Product 18:27 k-Forms 18:41 Covectors, Meet Exterior Algebra 20:06 Measurement of Vectors 20:32 Computing the Projected Length 21:02 1-form 21:47 Review: Determinants & Signed Volume 24:21 Measurement of 2-Vectors 24:59 Computing the Projected Area 26:18 2-form 27:21 Antisymmetry of 2-Forms 30:22 Measurement of 3-Vectors 31:21 Computing the Projected Volume 32:45 3-form 33:48 k-Form 35:47 k-Forms and Determinants 37:47 A Note on Notation 39:31 0-Forms 40:31 k-Forms (Measurement in Coordinates) 41:47 dual Basis 43:59 1-form - Example in Coordinates 46:50 2-form - Example in Coordinates 49:47 Einstein Summation Notation 51:59 Sharp and Flat in Coordinates 54:01 Coming Up: Differential Forms

[@brunomera4965]

Beautiful intuition and remarkable presentation of these abstract concepts! Thanks for sharing! These same kind of ideas also appear in the second quantisation for fermions in many body physics. The exterior algebra is, at least for me, the most natural and beautiful way to think about fermions. However, the subject is not usually presented like this and it makes some of the details obscure.

[@adamhendry945]

The best videos, bar none, for learning discrete differential geometry. You MUST watch these in order to understand it! Thank you Prof. Crane!

[@IntegralMoon]

Thanks so much for uploading this

[@ObsessiveClarity]

I can't believe how enlightening this is. Your explanation of the dual between row and column vectors was so damn good. I watched a video by 3b1b dedicated to the commutativity of the dot product describing this dual geometrically, but this was even more informative. Thank you.

[@familywu3869]

You are such an excellent teacher, Prof. Crane. Thank you!

[@aaronkurtz3119]

These videos are incredibly helpful, thanks so much!

[@ObsessiveClarity]

I wish I saw this 1 year ago. I have been struggling with k-forms and the wedge product and all of that for so long. Why? Because they never talked about k-vectors and this whole duality! Only mentioned the geometric meaning in passing. Thank you :)

[@guangfuwang2515]

best ever for discrete differential geometry tutorial!

[@phmfthacim]

this is wonderful, thank you!

[@444haluk]

Damn, you are the best teacher in the internet.

[@DoobooDomo]

Loving this series! I'm probably overthinking, but on the musical isomorphism slide the notes you used, B and C, only differ by half step. So, Bsharp is an enharmonic spelling to C (they are the same sound). Likewise for Cflat and B. Was this intentional?

[@rudypieplenbosch6752]

This a a great and concise explanation 👏

[@jdsahr]

"Gravitation" (*) by Thorne et al. doesn't use differential forms --- it is tensor based, which is the usual approach in General Relativity. But it does use the terms "vector" and "form" in a way that is similar to the role of "vector" and "form" in these lectures. I have to say that the standard (overloaded) nomenclature and symbols of differential forms and differential geometry is a significant impediment to learning the material. Dr. Crane appropriately and helpfully and apologetically acknowledges this. (*) Affectionately known as "The Black Death."

[@robertwilsoniii2048]

Wow. This is powerful. The pay off of learning all that hard multilinear algebra is that these dual vector things simplify down to the kronecker delta making calculations a breeze now. This must be the insight Einstein had about this notation.

[@strft]

beautiful stuff thanks brotha

[@theodorostsilikis4025]

It's a good thing that the notation (a^b)*(u,v) is used and not (a^b)*(u^v). turns out that those two are opposite in geometric algebra. (a^b)*(u^v) gets an extra minus sign from the (e1^e2) squared for being in the same space.

[@qinzhengyang4000]

thank you！

[@98danielray]

my guess is the notation is of applying a k-form to "k-vectors" is because the objects are different. instead of wedging vectors spaces and then dualizing k-vector spaces, the standard way is to dualize vector space and then wedge those duals. the constructions are isomorphic either way. the only difference is instead of having the alternating, distributive, associative properties come from the wedge product, they now come from the space of alternating, multilinear maps given by the construction through the wedge of dual spaces.

[@columbus8myhw]

What is that opening jingle?

[@columbus8myhw]

Random thought. There's a community of people out there who like to create languages; they call their creations "conlangs" (for "constructed languages"), and the activity is "conlanging" (and the people are "conlangers"). Mathematical descriptive tools such as the exterior calculus are often called "languages". Does that mean that whoever invented the exterior calculus was doing a form of conlanging?

[@444haluk]

4:40 a minor correction, in the philosophy of the east, yin is female & disorder & dark, so yin measures the yang which is male & order & light. If yang has no order, it is discarded :D

[@benwifi]

I'm taking differential geometry at Pitt, was weird to see the Pittsburgh tag on the video!

[@adamhendry945]

At 39:00, I believe the reason arguments to k-forms are written as lists is because k-forms are generally operators and can thus be thought of as functions operating on vectors. Quantities that transform covariantly (co-vectors and k-forms) are typically quantities like change per unit length, like gradient, whereas quantities that change contravariantly (vectors) are typically quantities like differential length. Hence, at least to me, it makes sense to write k-forms in a kind of functional operator notation.

[@johnwaczak8028]

Be careful with how you use the terms "transform covariantly" and "transform contravariantly". This applies to the components of the form or vector in a particular basis. This is one of the reasons why working with coordinates is often confusing. I think one of the reasons for the arguments-as-lists format comes from people who learn to use tensors before forms. From that direction, one can define the tensor product of two 1-forms and this will act on a list of of vectors (i.e. there are two "slots" you can put vectors into). Then the wedge product is just the anti-symmetric combination of the tensor product (i.e. it looks like the cross product). See this: en.wikipedia.org/wiki/Exterior_algebra#Alternating_tensor_algebra

[@adamhendry945]

@@johnwaczak8028 Be careful with how you read the words in my comments. I said covariant-LY and contravariant-LY, not covariant and contravariant. I agree with you that these are antiquated terms because they only describe how components transform (as well pointed out in "A First Course in General Relativity, by Bernard Schutz", Section 3.3, pg. 60), but I'm using the adverbs in the sense described in "A Student's Guide to Vectors and Tensors, by Daniel Fleisch". It is important to understand both sets of terminology, as Prof. Crane also mentions, if for nothing else because different papers/books may use both types of terms. Daniel Fleisch properly explains in that book that the above mentioned quantities transform in that kind of manner. Alternatively, I could have said "direct transformation" and "inverse transformation". It is important to understand all ways to describe how these quantities are chaning. At the time of this post, no one says these quantities transform "covectorly" or "k-formedly". There is an analogue to the terms in statistics as well, so it would be familiar to the stats student as well (e.g. covariant = in direct proportion, contravariant = in inverse proportion). I'm trying to explain how they change, not what they are.

[@johnwaczak8028]

​@@adamhendry945 Apologies if my comment came across as rude, that was not my intention. I think my point still stands. Prior to 39:00 in the video, we had used a coordinate free approach. Without specifying a coordinate system, the phrases "transform covariantly" and "transform contravariantly" don't make sense as k-vectors and k-forms are geometric objects independent of any choice of coordinates. The application of a two form to a two vector by plugging the vectors into the slots of the 2-form has nothing to do with coordinates. You will get the same value no matter what coordinates you use. This was a big point of confusion for me when I learned about forms, tensors, etc in my physics classes because physicists tend to define tensors solely in terms of their components.

[@adamhendry945]

@@johnwaczak8028 No, not at all, the fault is mine. I'm sorry for misreading your comment! I see now what you are saying: that's very interesting. I do need to read up and learn more about it. Thank you for your comments!

[@98danielray]

my guess is the notation of applying a k-form to "k-vectors" is because the objects are different. the notation is actually describing maps acting on k vectors. instead of wedging vector spaces and then dualizing k-vector spaces, the standard way to construct k-forms is to dualize the vector space and then wedge those duals. the constructions are isomorphic either way. the only difference is instead of having the alternating, distributive, associative properties come from the wedge product, they now come from the space of alternating, multilinear maps given by the construction through the wedge of dual spaces.

[@jonharper1630]

Can someone help me see the light at the end of the tunnel? I'm getting a bit of theory fatigue. I’m a programmer (Julia is currently my favorite language), and I'm trying to find an itch that I can scratch using these elegant tools. I need some inspiration! Are you guys using this to ultimately gain an enlightened perspective in physics? Are you using this to implement some graphical art and/or 3D simulations? Anyone have recommendations how to apply to machine learning? I want to learn more, but I'm really wanting to code something so I'll have an excuse to learn more :) Any inspiration is appreciated! Keenan, this entire lecture series is beautiful and incredibly articulated, thank you!

[@keenancrane]

Thanks! You can see all the algorithms we'll implement in the class on last year's webpage, here (in reverse order): brickisland.net/DDGSpring2020/category/assignments/?order=desc These applications are largely in mesh processing, along the lines discussed here: www.cs.cmu.edu/~kmcrane/Projects/Other/SwissArmyLaplacian.pdf The same concepts are useful in developing solvers for physically-based animation. For some cool examples, see, e.g., the fluids papers here: cseweb.ucsd.edu/~alchern/projects/ This is just a random sampling of examples… in general there are a *lot* of places these concepts & tools come up in algorithms!

[@jonharper1630]

@@keenancrane wonderful. Plenty to chew on there. Thanks!

[@jonharper1630]

Followup: Keenan's "Computer Graphics" series is precisely what I was looking for as conceptual scaffolding (maybe prerequisite) for DDG. I'm enjoying it very much: kzbin.info/aero/PL9_jI1bdZmz2emSh0UQ5iOdT2xRHFHL7E

[@adityaprakash256]

you said for 2-form alpha and beta need not be othonormal, but is it necessary for them to be orthogonal?

[@fragileomniscience7647]

Don't they just need to be linearly independent? Since by the definition of multilinear alternating maps, any linear dependencies evaluate to zero.

[@adamhendry945]

At 30:55, when you say "all 3-vectors have same 'direction'", are you assuming the 3-vector and 3-form are embedded in 3-space and not some higher-dimensional space?

[@johnwaczak8028]

This applies to R^3 as there is only one basis 3-vector or basis 3-form in R^3. This would be different in R^4 for example.

[@adamhendry945]

@@johnwaczak8028 My question is more are we speaking "extrinsicly" or "intrinsicly"? Are we an ant on the surface as in "Flat Land" (in which case I only truly have a sense of 2 directions, as hinted at by Prof. Crane's discussion of "manifold"), or are we able to step back and look at the surface in 3-space?

[@98danielray]

@@adamhendry945 everything seems to be extrinsic for now. so we are embedded in R^3

[@adamhendry945]

@@98danielray and @John Waczak Thank you both for the clarification!

[@markd.shattuck4139]

At 14:19 you mention that the covectors of the integrable functions f(x) are not functions, but it seems like all functions g(x) in \int_0^1g(x)f(x)dx are covectors. g(x)=1 is the first covector you mention and g(x)=\delta(x) (dirac delta) is the second. Or do you just mean that f(x) can not also be a delta function, since the integral of the delta function squared is not integrable?

[@Aetheraev]

Well, with an appropriate definition of inner product you can identify any vector space with its dual. In this case a natural choice is (g,f) := \int_0^1g(x)f(x)dx and so the dual space can be also be thought of as integrable functions over the same range (indeed google L^p spaces, this is L^1). Note however that this is a choice and we could make a different choice and get a different identification or refrain from making the choice at all and they remain different spaces.

[@markd.shattuck4139]

@@Aetheraev Yes. I agree. What I was questioning was Prof Crane's idea that sometimes the covectors are not from the same space as the vectors. In his example ,a vector was any integrable function on the interval [0,1] (I assume he meant all square integrable functions) but a co-vector was the average function or the delta function which are somehow different than regular functions. It seems to me that covector and vector provide a symmetric description. For example, from a general inner product with mass matrix we have = C' M' V, with M positive definite. We could think of the covectors as M C and vectors as V. Then the covectors would be "different" but we could equally decompose M=Q sqrt(Lm) I sqrt(Lm) Q', where Q is the orthogonal eigenvector matrix and Lm is the diagonal eigenvalue matrix of M. Then we could define a new vector V*=sqrt(Lm) Q V and new covector C*=sqrt(Lm) Q C. Now C* and V* are symmetric. Maybe I'm missing something and there is a way to come up with covectors that are not int he same space as the vectors.

[@Aetheraev]

@@markd.shattuck4139 I'm not really sure what you mean by "symmetric" here. But we can explore this straightforwardly in the language of vector spaces. If a vector space is finite dimensional then it has the same dimension as it's dual so they are isomorphic (although there is no canonical choice of isomorphism here). An inner product defines such an isomorphism by the Riesz representation theorem. However if the vector space is infinite dimensional we get some weirder stuff going on. Then the dual space (defined as the set of linear functional on V) is actually "larger" than the original space. This is the situation we are looking at. L^1 is infinite dimensional indeed uncountably infinite. So even with the inner product we only get some of the dual space. There are thus other elements not described as integrable maps even allowing for the choice of inner product

[@walkquietlyby]

@@Aetheraev I think it's also important here to make the distinction between the dual space (set of linear form on V) and the *topological* dual space (set of continuous linear form). If V = L1 as in his example, then (if I am not mistaken) the map delta he defines is not continuous and not a member of the topological dual. However the topological dual is isometric to L∞ which is another function space. (hope I didn't say anything wrong...)

[@98danielray]

@@markd.shattuck4139 you already stated your answer. dirac delta is not in the vector space of integrable functions.

[@simpaticode]

At 11:00 you assert that all linear maps from R3->R take the form ax+by+cz. But doesn't any polynomial work, too? E.g. ax^2+bx + cy^3 + dy + ez.

[@dantong5623]

Is it still linear if that's the case?

[@epicmorphism2240]

do u know what linear means

[@simpaticode]

@@epicmorphism2240 no

[@epicmorphism2240]

@@simpaticode f is linear if: 1) it’s additive: f(x+y)=f(x)+f(y) 2) and homogeneous cf(x)=f(cx). Now obviously this doesn‘t apply to polynomials. Also, not to be rude: If you don‘t know what linear maps are you should revisit linear algebra before lesrning differential geometry.

[@alivecoding4995]

The 2-form result of -40 doesn't make any intuitive sense for me, since the two areas look quite similar w.r.t. their areas.

[@user-wt8vl3bi6y]

43:10

[@randyhelzerman]

After watching this, I feel like I've been beaten by sticks.....

[@yizhang7027]

Math is giving different names to the same thing.

[@AmentasOnIce]

Duality is giving different names to the same thing!