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!
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…)
@@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 RUclips nowadays. Have a great day!
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 =)
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.
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❤️🇮🇩
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*
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.
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...)?
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.
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!
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?
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.
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!
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.
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.
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.
@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?
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!
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!
@@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
@@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).
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.
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)?
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
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 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.
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.
@@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").
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?
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.
@@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 :)
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.
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!
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?
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?
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.
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.
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
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!
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…)
@@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 RUclips nowadays.
Have a great day!
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 =)
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.
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❤️🇮🇩
Many thanks for your kind words. Not enough hours in the day! But I will continue to add videos and explanations when I can.
I almost feel like I stole something getting to watch such awesome, informative, and well-made lectures for free...thanks much!
This lecturer is unbelievable
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*
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.
With this lecture one can grasp good intuition.
Thank you so much for uploading this amazing introduction to exterior algebra!
This is one of the most insightful videos I have seen.
This is pure gold.
Wow this course is moving fast! 5 hours of lectures already 😅
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...)?
@49:22 there's a minor typo: in the last three vector, you've switched to upper indexes from lower.
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.
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!
www.cs.cmu.edu/~kmcrane/faq.html#figures
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?
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.
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!
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?
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.
Yes, they will. It's the beginning of the semester here and we haven't handed out any assignments yet. :-)
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.
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.
@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?
"Gin" is silver. "Kin" is gold. It should be read "kin"
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.
1:00:00 Can you show how (2e1 ^ e2) ^ 3e3 = 6e1 ^ e2 ^ e3 ?
Shouldn't the answer at 1:00:42 be 1/3 ?
Since the determinant should be equal to 1 ?
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!
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!
@@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
@@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).
Why do you depict the orientation of a bivector as a normal vector instead of as a counterclockwise vs clockwise direction?
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.
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)?
see the next lecture on k-forms, but basically a k-form measures a k-vector
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.
Thanks a lot. Things get more clear after lecture 4 :)
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
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?
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!
@@keenancrane I should have said that this is a great video, one of the best I have seen for quite a while.
@@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.
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.
@@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").
I was wondering if professor can speak German when professor said "steinar formula" with a very proper /ˈʃt/... Now Japanese! 1:11 😲..wow!
The hodge star explains the formula for the determinant in R^n.
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?
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.
@@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 :)
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.
I thought so too, except that |v1+v2|
@@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.
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.
at 57:00 you also used the antisymmetry property.
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!
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?
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.
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?
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.
So basically wedge products are like plane segments with finite length sides.
I'm here after some linear algebra and some geometric algebra
N choose K possible bases for K-vectors. I think I now understand why they are called k-vectors in R^n.
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?
Great lecture but I won’t forgive that assault on Bavarian food. 😂
I am coming for that gold.
1:24 i pointed my phone's camera at the screen and used google lens. The translation only replaced "gold" by "money"
Came for the maths, stayed for the Bavaria bashing.
41:48 5234
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.