9 ms·

Why can't you multiply vectors? [video]

115 points by BenoitP 3y ago

BenoitP 3y ago
This post was brought to you by the Geometric Algebra gang. Join us at: https://www.youtube.com/watch?v=60z_hpEAtD8 https://www.youtube.com/watch?v=60z_hpEAtD8 https://bivector.net/ https://bivector.net/ https://enkimute.github.io/ganja.js/examples/coffeeshop.html#pga2d_points_and_lines https://enkimute.github.io/ganja.js/examples/coffeeshop.html...
- contravariant 3y ago
  I never did quite 'get' geometric algebra. I mean the exterior algebra gives a couple of clear generalisations, and for the ones that require a metric you can typically use the Hodge star to find a generalisation. Geometric algebra then blends all of this together, and I'm still not entirely convinced that this improves things. Is it actually ever handy to have to deal with mixed degree multivectors?
  - adrian_b 3y ago
    Even without using mixed-degree multivectors, the fact that the existence and the properties of all the different kinds of multivectors results from a small set of simple axioms is very satisfying in itself. Before all these concepts were unified by geometric algebra, the system of physical quantities as taught in most places was a huge mess of many different kinds of quantities, scalars, polar vectors, axial vectors, pseudoscalars, tensors, pseudotensors, complex numbers, quaternions, spinors and so on. It was not at all obvious why there are so many kinds of quantities, which are the relationships between them, are there any other kinds of quantities besides those already studied, etc. Geometric algebra has brought order in this chaos and it has enabled a much deeper and more complete understanding of physics, by reducing a long list of seemingly arbitrary rules to a much smaller set of axioms, and by deriving all the many kinds of physical quantities from the vectors in the strict sense, i.e. from the translations of the space, which are themselves derived from the points of the affine space (as equivalence classes of point pairs). (Both logically and historically, in physics the vectors are more fundamental than the scalars. The scalars are obtained by dividing collinear vectors, i.e. they are equivalence classes of pairs of collinear vectors. This division operation is a.k.a. measurement and what are now named as "real numbers" were named as "measures" in the past, for more than two millennia. For any Archimedean group it is possible to define a division operation using the Axiom of Archimedes, generating a set of scalars over which the original group is a vector space).
    - contravariant 3y ago
      Yeah but like all of that is true about the exterior algebra as well. Except exterior algebra is a bit more explicit about where you use the metric, which is really convenient when the metric starts to become important.
  - BenoitP 3y ago
    I'm not versed at all into the Hodge star operator but it feels like Maxwell's equations expressed in both systems[1] might locate the answer. With Hodge star: dF = 0 d * F = J Expressed in GA: ∇F = J The eyeball-difference of which (dF = 0) would be your "how GA then blends all of this together", if I understand correctly. I'd guesstimate that dF = 0 to be akin to Gauss' law; and that maybe GA somewhat incorporates that the curl of a gradient is the zero field. [1] https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field https://en.wikipedia.org/wiki/Mathematical_descriptions_of_t...
    - nyssos 3y ago
      These are not quite equivalent, since using the hodge dual explicitly incorporates the geometry of space in a way left implicit by your geometric algebra formulation. The appropriate exterior algebra analogue here is simply `F = dA`. F is the electromagnetic field tensor, d is the exterior derivative, A is the 4-potential.
  - nyssos 3y ago
    Clifford (i.e. "geometric") algebras are of some mathematical interest, but for physics the only real value I've seen from them is in offering a fairly nice presentation of spin groups, which is ultimately where the Dirac matrices come from. The geometric algebra advocacy, so far as I can tell, comes entirely from engineers who were never given a proper account of tensors in the first place: it's certainly an improvement over whatever godawful basis-dependent stuff gets taught there.
    - itishappy 3y ago
      I'd be interested to know your opinion on space-time algebras. They seem like they would provide a nice way of unifying spatial rotations and Lorentz boosts, but that may be blind optimism as your last sentence seems to have been written to describe me specifically...
      - nyssos 3y ago
        Spacetime algebra is just the Clifford algebra Cl_1_3(R), so all of the above applies. > They seem like they would provide a nice way of unifying spatial rotations and Lorentz boosts Yes and no: the right way to unify rotations and boosts is to consider them as the orientation-preserving elements of the Lorentz group (sometimes called the proper Lorentz group). You can construct this from the corresponding Clifford algebra, but it's somewhat technical and not physically well-motivated until you start dealing with spinors. It's also the group of symmetries of spacetime that leave the origin unchanged, which is far, far more natural.
    - paholg 3y ago
      I found it a much nicer way to think about physical concepts. Torque and magnetism, for example, make much more sense as bivectors. And never having to do the right-hand rule is a nice plus.
  - smaddox 3y ago
    Yes. In physics. E.g. Maxwell's equation reduces to a single equation in geometric Algebra: https://peeterjoot.com/archives/geometric-algebra/maxwells_ga.pdf https://peeterjoot.com/archives/geometric-algebra/maxwells_g...
- captainclam 3y ago
  Thank you so much for the links! I just happen to be endeavoring to seriously learn this stuff at the moment, driven by interests in physics, differential geometry, and gamedev. I'm seriously jazzed to check these out, thank you.
- uxp8u61q 3y ago
  Somehow, in your links, there isn't a single explanation anywhere (that's not in the form of a video that I don't have time or inclination to watch) about what "geometric algebra" is supposed to mean. I'm trying to gather what this is supposed to be from the catchphrases on the front page, and honestly, this just looks like linear algebra. Can you explain what "geometric algebra" is supposed to be? Or just link to written explanations? I'm a mathematician (algebraic topology), so don't be afraid to get technical. How is this different from linear algebra? The only things I can see are some tensor products and exterior products, and a few couple of division algebra structures. Can you enlighten me? Is there any actually new math behind all this?
  - nyssos 3y ago
    Geometric algebras = Clifford algebras over R.
    - uxp8u61q 3y ago
      And they made dozens of hour long videos about that? That's insane. And the arrogance of calling it "geometric algebra" like they invented a new field of math?!
      - BenoitP 3y ago
        One could argue it is more about physics than math. GA has simpler expressions for a lot of physical topics.
        uxp8u61q 3y ago
        Perhaps, but it's not new math (Clifford algebras are almost 200 years old). Shrouding it in a veil of buzzwords and pizzaz makes you look more like con artists than scientists. A sentence like this one is just absurd: > Clifford's Geometric Algebra enables a unified, intuitive and fresh perspective on vector spaces, giving elements of arbitrary dimensionality a natural home.
  - ngcc_hk 3y ago
- WD40forRust 3y ago
  Based.
  - meindnoch 3y ago
    Based and Grassmann-pilled?
zxexz 3y ago
Oh lovely, can't wait to give this a watch. Freya dives deep, and gives wonderful talks. Her video "The Continuity of Splines"[0] is my favorite watch of the past year. [0] https://www.youtube.com/watch?v=jvPPXbo87ds https://www.youtube.com/watch?v=jvPPXbo87ds
- postmodest 3y ago
  The surprising thing about that video is how it helped me a lot with my CAD.
kubb 3y ago
You can. Thanks for asking.
- mcphage 3y ago
  On one hand, this comment saved me from watching this 50 minute video where I'd also learn that you can multiply vectors. So very efficient usage of time. On the other hand, I'll probably watch the video anyway because I expect it to be more interesting and informative, and go into a lot more depth. So I'm not really sure this comment helped me at all.
- corethree 3y ago
  It's just the title bro. In The end of that video they find a way that's not the cross product or the dot product.
- dventimi 3y ago
  Personally, I found this comment to be helpful. Personally, I don't appreciate it when speakers "get cute" with what I regard as misleading titles like this one has.
manchmalscott 3y ago
Freya is such a talented educator? Storyteller? Her stuff is always a joy to watch.
xigoi 3y ago
I love how she roasts programmers who don't like mathematics.
- sdfghswe 3y ago
  Saying you don't like maths is like saying you don't like how the universe works. It's like... fine, but that's not really actionable.
  - falserum 3y ago
    “I hate math” usually is shorthand for “I hate math lessons, and I don’t feel any need to learn it”.
    - sdfghswe 3y ago
      That's probably true for the vast majority of things, isn't it? I also hate ballet, but I suspect that if I made an effort I would learn to appreciate it. However, I don't make it a point to go around loudly telling everyone that I hate ballet as if that was a badge of honor.
      - Kranar 3y ago
        Do you really "hate" ballet though? Like let's be honest here.
        sdfghswe 3y ago
        Totally honest, I find it revolting. Why does this surprise you? It doesn't surprise you that some people hate math.
  - Kranar 3y ago
    No it's not. Plenty of people don't like math and it has nothing to do with how the universe works. The way that math is taught to most people is absolutely atrocious, involving rote memorization, following rules for the sake of following rules, very little intuition involved. If anything, most people who express a hatred of math do so because it's taught to them in a way that completely divorces it from the universe or anything whatsoever. Most people I know who appreciate math did not learn it at school, but learned it either at home from their parents, or learned it independently. I have also made it my own responsibility to make sure my daughter learns math from me and applies math to every day situations and can develop a basic mathematical intuition.
mo_42 3y ago
Was expecting something about vector spaces and multiplication operations or the lack of them. Watched a quick ride towards quaternions. Also nice.
- chpatrick 3y ago
  I don't think it's really about explaining quaternions, but more how quaternions are just a special case that arise from this general framework.
whacked_new 3y ago
clicked out of curiosity and was hooked to the end. fantastic, masterful talk, thanks.
ewgoforth 3y ago
I guess I'm unclear on what physical concept she's getting at by multiplying vectors. Depending on which concept you're trying to calculate with your multiplication, you have dot products like work and cross product like torque.
- itishappy 3y ago
  Or rotations like phasors or whatever the heck spinors do or... (Spoiler: it's all of them)
- gpm 3y ago
  Defining "multiplication" to be "any function that takes two arguments and outputs one" isn't a very interesting definition of multiplication. The word is a lot more useful if you put constraints on it like saying for something to be a multiplicative operator it has to respect (a + b) * c = a * c + b * c (the distributive property). Once you put a "reasonable" set of constraints on it... you discover that you can't actually multiply vectors (no function exists that satisfies the properties you want). Though the talk isn't about proving that (or justifying the set of constraints that mean you can't multiply vectors) and instead goes off in another direction of extending your vector space to a bigger space (like how the complex numbers are a bigger space than the reals) where you can define a reasonable multiplication operator.
  - dventimi 3y ago
    Doesn't the scalar product have the distributive property?
    - gpm 3y ago
      Yes, that was an example of one property you probably want, not a set sufficient to make it such that no such operator exists. Another property you want (and the talk uses) is that the operator is that the operator is from V x V to something. I.e. we are multiplying two vectors (because that's what we asked for in the title) not a scalar and a vector. That excludes your counter example, but still isn't nearly enough to make it so that no multiplication operator exists. I'll be honest and say I'm not listing properties here because I don't remember what properties are needed to make it so you can't define the operator... hopefully someone who has studied this a bit more recently or thoroughly than me can chime in.
m3kw9 3y ago
Why not if the vector is 0,2 and then multiply it by 0,2 you get double the magnitude 0,4?
- raphlinus 3y ago
  This is known as the Hadamard product and is covered in the video. The tl;dr is that, while it certainly has uses, it doesn't represent multiplication of vector spaces in any reasonable way (in particular, it gives different results when there's a change of basis, while other notions of "product", including dot product, are invariant). There's a deeper explanation here: https://math.stackexchange.com/questions/185888/why-dont-we-define-vector-multiplication-component-wise https://math.stackexchange.com/questions/185888/why-dont-we-...
  - joshlemer 3y ago
    Maybe that's okay? If you look at regular multiplication, it seems vulnerable to choice of "basis" as well. For example: 6 x 6 = 36 But if we choose 2 as basis, rather than 1, then we have 3 x 3 = 9 (aka 3(2) x 3(2) = 9(2)) But 9(2) != 36 So even regular multiplication isn't invariant under chosen "basis"
    - gizmo686 3y ago
      With regular multiplication, the numbers are the "real" thing you are interested in. In a vector space, the fundemantal object is the vector. Writing it as a sequence of numbers with an assumed basis is a notational convience. If you want to define a function for vectors, then you need that function to give the same result regardless of the basis you use to represent the vectors. When you here mathaticians talk about proving that a function is "well defined" this is what they are talking about. Of course, you might be working with some spefic structure where component wise multiplication is meaningfull and well define. That structure might happen to also be a vector space, and the notation used to write it might happen to coincide with vector notation under the "obvious" choice of basis. Other specific vector spaces have their own quirky notion of multiplication. For instance polynomials can be viewed as a vector space, with an obvious basis of { x^n }, but polynomial multiplication looks very different from component wise multiplication.
ryangs 3y ago
Highly recommend Freya's various deep dives into various game development contents. She also has twitch vods for developing those videos which are also fascinating. https://www.youtube.com/@Acegikmo https://www.youtube.com/@Acegikmo
calderwoodra 3y ago
I liked the talk and found the jokes funny but the crowd was eerily silent.
- mcphage 3y ago
  It might be that the audio was pulled from the mic directly, and so the audience reactions either weren't loud enough, or maybe were removed?
noman-land 3y ago
Pleasantly surprised by this talk. Clicked knowing nothing and stayed to the end. Great talk and really great presenter.
ian0 3y ago
Perhaps a dumb question, but why do we store embeddings as "vectors" and not "points"? I thought the difference was that vectors have magnitude, but an embedding doesn't have a magnitude - they are just points in an n-dimensional space?
- tobinfricke 3y ago
  In computing we often like to conflate "ordered tuples of numbers" with "vectors." This vocabulary is even cooked into the C++ standard library. The difference is that mathematical vectors support some additional operations, such as addition and scalar multiplication. A vector in C++ is not a mathematical vector, since we can't add two vectors x+y, nor can we perform scalar multiplication a*v. For mathematical vectors we have this interpretation: An abstract vector is constructed by multiplying the numbers in the ordered tuple each by a corresponding abstract "basis vector" and adding up the results. The numbers are just the "coordinates" of a vector with respect to a particular basis. It may or may not make sense to talk about "basis vectors" in your application. Does it make sense to perform "coordinate transformations" on your objects? Another test is, "Do you want to use linear algebra?" If so, your objects are probably vectors. A similar but more egregious argument comes about with regard to tensors. Mathematically a tensor is a kind of function that takes vectors and co-vectors as arguments. A matrix, when coupled with the rules for multiplying matrices by row and column vectors, is a tensor. But an arbitrary n-dimensional array of numbers is not (necessarily) a tensor in the mathematical sense. Unfortunately that term was co-opted by the ML crowd because it sounds cool. :-) Getting back to what you mentioned about vectors having magnitude - in an abstract vector space, there is no definition of magnitude. It's not until you define an inner product that magnitude becomes defined. In this sense the grade-school definition of a vector as "a quantity with a magnitude and direction" does not necessarily comport with the standard definition of a mathematical vector space. We like to say that "a tensor is an object that transforms like a tensor" and the same is true for vectors. "A vector is an object that transforms like a vector" under coordinate transformations, while also supporting addition and scalar multiplication. To address your question more directly: typically "points" (in so much as they are relative to a coordinate system) really are "vectors". But general tuples of numbers are not necessarily.
wwarner 3y ago
Starts slow, gets better every minute. Really great!
creata 3y ago
The definitions of quaternions and dual quaternions look particularly neat. Are there other (non-"geometric algebra") ways to define (dual) quaternions without reference to a basis? Edit: I guess the "scalar and vector parts" definition at Wikipedia[1] doesn't use a basis, but it's not exactly pretty... [1]: https://en.wikipedia.org/wiki/Quaternion#Scalar_and_vector_parts https://en.wikipedia.org/wiki/Quaternion#Scalar_and_vector_p...

115 points by BenoitP 3y ago

This post was brought to you by the Geometric Algebra gang. Join us at: https://www.youtube.com/watch?v=60z_hpEAtD8 https://www.youtube.com/watch?v=60z_hpEAtD8 https://bivector.net/ https://bivector.net/ https://enkimute.github.io/ganja.js/examples/coffeeshop.html#pga2d_points_and_lines https://enkimute.github.io/ganja.js/examples/coffeeshop.html...

contravariant 3y ago
I never did quite 'get' geometric algebra. I mean the exterior algebra gives a couple of clear generalisations, and for the ones that require a metric you can typically use the Hodge star to find a generalisation. Geometric algebra then blends all of this together, and I'm still not entirely convinced that this improves things. Is it actually ever handy to have to deal with mixed degree multivectors?
- adrian_b 3y ago
  Even without using mixed-degree multivectors, the fact that the existence and the properties of all the different kinds of multivectors results from a small set of simple axioms is very satisfying in itself. Before all these concepts were unified by geometric algebra, the system of physical quantities as taught in most places was a huge mess of many different kinds of quantities, scalars, polar vectors, axial vectors, pseudoscalars, tensors, pseudotensors, complex numbers, quaternions, spinors and so on. It was not at all obvious why there are so many kinds of quantities, which are the relationships between them, are there any other kinds of quantities besides those already studied, etc. Geometric algebra has brought order in this chaos and it has enabled a much deeper and more complete understanding of physics, by reducing a long list of seemingly arbitrary rules to a much smaller set of axioms, and by deriving all the many kinds of physical quantities from the vectors in the strict sense, i.e. from the translations of the space, which are themselves derived from the points of the affine space (as equivalence classes of point pairs). (Both logically and historically, in physics the vectors are more fundamental than the scalars. The scalars are obtained by dividing collinear vectors, i.e. they are equivalence classes of pairs of collinear vectors. This division operation is a.k.a. measurement and what are now named as "real numbers" were named as "measures" in the past, for more than two millennia. For any Archimedean group it is possible to define a division operation using the Axiom of Archimedes, generating a set of scalars over which the original group is a vector space).
  - contravariant 3y ago
    Yeah but like all of that is true about the exterior algebra as well. Except exterior algebra is a bit more explicit about where you use the metric, which is really convenient when the metric starts to become important.
- BenoitP 3y ago
  I'm not versed at all into the Hodge star operator but it feels like Maxwell's equations expressed in both systems[1] might locate the answer. With Hodge star: dF = 0 d * F = J Expressed in GA: ∇F = J The eyeball-difference of which (dF = 0) would be your "how GA then blends all of this together", if I understand correctly. I'd guesstimate that dF = 0 to be akin to Gauss' law; and that maybe GA somewhat incorporates that the curl of a gradient is the zero field. [1] https://en.wikipedia.org/wiki/Mathematical_descriptions_of_the_electromagnetic_field https://en.wikipedia.org/wiki/Mathematical_descriptions_of_t...
  - nyssos 3y ago
    These are not quite equivalent, since using the hodge dual explicitly incorporates the geometry of space in a way left implicit by your geometric algebra formulation. The appropriate exterior algebra analogue here is simply `F = dA`. F is the electromagnetic field tensor, d is the exterior derivative, A is the 4-potential.
- nyssos 3y ago
  Clifford (i.e. "geometric") algebras are of some mathematical interest, but for physics the only real value I've seen from them is in offering a fairly nice presentation of spin groups, which is ultimately where the Dirac matrices come from. The geometric algebra advocacy, so far as I can tell, comes entirely from engineers who were never given a proper account of tensors in the first place: it's certainly an improvement over whatever godawful basis-dependent stuff gets taught there.
  - itishappy 3y ago
    I'd be interested to know your opinion on space-time algebras. They seem like they would provide a nice way of unifying spatial rotations and Lorentz boosts, but that may be blind optimism as your last sentence seems to have been written to describe me specifically...
    - nyssos 3y ago
      Spacetime algebra is just the Clifford algebra Cl_1_3(R), so all of the above applies. > They seem like they would provide a nice way of unifying spatial rotations and Lorentz boosts Yes and no: the right way to unify rotations and boosts is to consider them as the orientation-preserving elements of the Lorentz group (sometimes called the proper Lorentz group). You can construct this from the corresponding Clifford algebra, but it's somewhat technical and not physically well-motivated until you start dealing with spinors. It's also the group of symmetries of spacetime that leave the origin unchanged, which is far, far more natural.
  - paholg 3y ago
    I found it a much nicer way to think about physical concepts. Torque and magnetism, for example, make much more sense as bivectors. And never having to do the right-hand rule is a nice plus.
- smaddox 3y ago
  Yes. In physics. E.g. Maxwell's equation reduces to a single equation in geometric Algebra: https://peeterjoot.com/archives/geometric-algebra/maxwells_ga.pdf https://peeterjoot.com/archives/geometric-algebra/maxwells_g...
captainclam 3y ago
Thank you so much for the links! I just happen to be endeavoring to seriously learn this stuff at the moment, driven by interests in physics, differential geometry, and gamedev. I'm seriously jazzed to check these out, thank you.
uxp8u61q 3y ago
Somehow, in your links, there isn't a single explanation anywhere (that's not in the form of a video that I don't have time or inclination to watch) about what "geometric algebra" is supposed to mean. I'm trying to gather what this is supposed to be from the catchphrases on the front page, and honestly, this just looks like linear algebra. Can you explain what "geometric algebra" is supposed to be? Or just link to written explanations? I'm a mathematician (algebraic topology), so don't be afraid to get technical. How is this different from linear algebra? The only things I can see are some tensor products and exterior products, and a few couple of division algebra structures. Can you enlighten me? Is there any actually new math behind all this?
- nyssos 3y ago
  Geometric algebras = Clifford algebras over R.
  - uxp8u61q 3y ago
    And they made dozens of hour long videos about that? That's insane. And the arrogance of calling it "geometric algebra" like they invented a new field of math?!
    - BenoitP 3y ago
      One could argue it is more about physics than math. GA has simpler expressions for a lot of physical topics.
      - uxp8u61q 3y ago
        Perhaps, but it's not new math (Clifford algebras are almost 200 years old). Shrouding it in a veil of buzzwords and pizzaz makes you look more like con artists than scientists. A sentence like this one is just absurd: > Clifford's Geometric Algebra enables a unified, intuitive and fresh perspective on vector spaces, giving elements of arbitrary dimensionality a natural home.
- ngcc_hk 3y ago
WD40forRust 3y ago
Based.
- meindnoch 3y ago
  Based and Grassmann-pilled?

zxexz 3y ago

Oh lovely, can't wait to give this a watch. Freya dives deep, and gives wonderful talks. Her video "The Continuity of Splines"[0] is my favorite watch of the past year. [0] https://www.youtube.com/watch?v=jvPPXbo87ds https://www.youtube.com/watch?v=jvPPXbo87ds

postmodest 3y ago
The surprising thing about that video is how it helped me a lot with my CAD.

kubb 3y ago

You can. Thanks for asking.

mcphage 3y ago
On one hand, this comment saved me from watching this 50 minute video where I'd also learn that you can multiply vectors. So very efficient usage of time. On the other hand, I'll probably watch the video anyway because I expect it to be more interesting and informative, and go into a lot more depth. So I'm not really sure this comment helped me at all.
corethree 3y ago
It's just the title bro. In The end of that video they find a way that's not the cross product or the dot product.
dventimi 3y ago
Personally, I found this comment to be helpful. Personally, I don't appreciate it when speakers "get cute" with what I regard as misleading titles like this one has.

manchmalscott 3y ago

Freya is such a talented educator? Storyteller? Her stuff is always a joy to watch.

xigoi 3y ago

I love how she roasts programmers who don't like mathematics.

sdfghswe 3y ago
Saying you don't like maths is like saying you don't like how the universe works. It's like... fine, but that's not really actionable.
- falserum 3y ago
  “I hate math” usually is shorthand for “I hate math lessons, and I don’t feel any need to learn it”.
  - sdfghswe 3y ago
    That's probably true for the vast majority of things, isn't it? I also hate ballet, but I suspect that if I made an effort I would learn to appreciate it. However, I don't make it a point to go around loudly telling everyone that I hate ballet as if that was a badge of honor.
    - Kranar 3y ago
      Do you really "hate" ballet though? Like let's be honest here.
      - sdfghswe 3y ago
        Totally honest, I find it revolting. Why does this surprise you? It doesn't surprise you that some people hate math.
- Kranar 3y ago
  No it's not. Plenty of people don't like math and it has nothing to do with how the universe works. The way that math is taught to most people is absolutely atrocious, involving rote memorization, following rules for the sake of following rules, very little intuition involved. If anything, most people who express a hatred of math do so because it's taught to them in a way that completely divorces it from the universe or anything whatsoever. Most people I know who appreciate math did not learn it at school, but learned it either at home from their parents, or learned it independently. I have also made it my own responsibility to make sure my daughter learns math from me and applies math to every day situations and can develop a basic mathematical intuition.

mo_42 3y ago

Was expecting something about vector spaces and multiplication operations or the lack of them. Watched a quick ride towards quaternions. Also nice.

chpatrick 3y ago
I don't think it's really about explaining quaternions, but more how quaternions are just a special case that arise from this general framework.

whacked_new 3y ago

clicked out of curiosity and was hooked to the end. fantastic, masterful talk, thanks.

ewgoforth 3y ago

I guess I'm unclear on what physical concept she's getting at by multiplying vectors. Depending on which concept you're trying to calculate with your multiplication, you have dot products like work and cross product like torque.

itishappy 3y ago
Or rotations like phasors or whatever the heck spinors do or... (Spoiler: it's all of them)
gpm 3y ago
Defining "multiplication" to be "any function that takes two arguments and outputs one" isn't a very interesting definition of multiplication. The word is a lot more useful if you put constraints on it like saying for something to be a multiplicative operator it has to respect (a + b) * c = a * c + b * c (the distributive property). Once you put a "reasonable" set of constraints on it... you discover that you can't actually multiply vectors (no function exists that satisfies the properties you want). Though the talk isn't about proving that (or justifying the set of constraints that mean you can't multiply vectors) and instead goes off in another direction of extending your vector space to a bigger space (like how the complex numbers are a bigger space than the reals) where you can define a reasonable multiplication operator.
- dventimi 3y ago
  Doesn't the scalar product have the distributive property?
  - gpm 3y ago
    Yes, that was an example of one property you probably want, not a set sufficient to make it such that no such operator exists. Another property you want (and the talk uses) is that the operator is that the operator is from V x V to something. I.e. we are multiplying two vectors (because that's what we asked for in the title) not a scalar and a vector. That excludes your counter example, but still isn't nearly enough to make it so that no multiplication operator exists. I'll be honest and say I'm not listing properties here because I don't remember what properties are needed to make it so you can't define the operator... hopefully someone who has studied this a bit more recently or thoroughly than me can chime in.

m3kw9 3y ago

Why not if the vector is 0,2 and then multiply it by 0,2 you get double the magnitude 0,4?

raphlinus 3y ago
This is known as the Hadamard product and is covered in the video. The tl;dr is that, while it certainly has uses, it doesn't represent multiplication of vector spaces in any reasonable way (in particular, it gives different results when there's a change of basis, while other notions of "product", including dot product, are invariant). There's a deeper explanation here: https://math.stackexchange.com/questions/185888/why-dont-we-define-vector-multiplication-component-wise https://math.stackexchange.com/questions/185888/why-dont-we-...
- joshlemer 3y ago
  Maybe that's okay? If you look at regular multiplication, it seems vulnerable to choice of "basis" as well. For example: 6 x 6 = 36 But if we choose 2 as basis, rather than 1, then we have 3 x 3 = 9 (aka 3(2) x 3(2) = 9(2)) But 9(2) != 36 So even regular multiplication isn't invariant under chosen "basis"
  - gizmo686 3y ago
    With regular multiplication, the numbers are the "real" thing you are interested in. In a vector space, the fundemantal object is the vector. Writing it as a sequence of numbers with an assumed basis is a notational convience. If you want to define a function for vectors, then you need that function to give the same result regardless of the basis you use to represent the vectors. When you here mathaticians talk about proving that a function is "well defined" this is what they are talking about. Of course, you might be working with some spefic structure where component wise multiplication is meaningfull and well define. That structure might happen to also be a vector space, and the notation used to write it might happen to coincide with vector notation under the "obvious" choice of basis. Other specific vector spaces have their own quirky notion of multiplication. For instance polynomials can be viewed as a vector space, with an obvious basis of { x^n }, but polynomial multiplication looks very different from component wise multiplication.

ryangs 3y ago

Highly recommend Freya's various deep dives into various game development contents. She also has twitch vods for developing those videos which are also fascinating. https://www.youtube.com/@Acegikmo https://www.youtube.com/@Acegikmo

calderwoodra 3y ago

I liked the talk and found the jokes funny but the crowd was eerily silent.

mcphage 3y ago
It might be that the audio was pulled from the mic directly, and so the audience reactions either weren't loud enough, or maybe were removed?

noman-land 3y ago

Pleasantly surprised by this talk. Clicked knowing nothing and stayed to the end. Great talk and really great presenter.

ian0 3y ago

Perhaps a dumb question, but why do we store embeddings as "vectors" and not "points"? I thought the difference was that vectors have magnitude, but an embedding doesn't have a magnitude - they are just points in an n-dimensional space?

tobinfricke 3y ago
In computing we often like to conflate "ordered tuples of numbers" with "vectors." This vocabulary is even cooked into the C++ standard library. The difference is that mathematical vectors support some additional operations, such as addition and scalar multiplication. A vector in C++ is not a mathematical vector, since we can't add two vectors x+y, nor can we perform scalar multiplication a*v. For mathematical vectors we have this interpretation: An abstract vector is constructed by multiplying the numbers in the ordered tuple each by a corresponding abstract "basis vector" and adding up the results. The numbers are just the "coordinates" of a vector with respect to a particular basis. It may or may not make sense to talk about "basis vectors" in your application. Does it make sense to perform "coordinate transformations" on your objects? Another test is, "Do you want to use linear algebra?" If so, your objects are probably vectors. A similar but more egregious argument comes about with regard to tensors. Mathematically a tensor is a kind of function that takes vectors and co-vectors as arguments. A matrix, when coupled with the rules for multiplying matrices by row and column vectors, is a tensor. But an arbitrary n-dimensional array of numbers is not (necessarily) a tensor in the mathematical sense. Unfortunately that term was co-opted by the ML crowd because it sounds cool. :-) Getting back to what you mentioned about vectors having magnitude - in an abstract vector space, there is no definition of magnitude. It's not until you define an inner product that magnitude becomes defined. In this sense the grade-school definition of a vector as "a quantity with a magnitude and direction" does not necessarily comport with the standard definition of a mathematical vector space. We like to say that "a tensor is an object that transforms like a tensor" and the same is true for vectors. "A vector is an object that transforms like a vector" under coordinate transformations, while also supporting addition and scalar multiplication. To address your question more directly: typically "points" (in so much as they are relative to a coordinate system) really are "vectors". But general tuples of numbers are not necessarily.

wwarner 3y ago

Starts slow, gets better every minute. Really great!

creata 3y ago

The definitions of quaternions and dual quaternions look particularly neat. Are there other (non-"geometric algebra") ways to define (dual) quaternions without reference to a basis? Edit: I guess the "scalar and vector parts" definition at Wikipedia[1] doesn't use a basis, but it's not exactly pretty... [1]: https://en.wikipedia.org/wiki/Quaternion#Scalar_and_vector_parts https://en.wikipedia.org/wiki/Quaternion#Scalar_and_vector_p...