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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 multiplic

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.
    - empath-nirvana 3y ago
      Basically you want all the properties of a ring: https://mathworld.wolfram.com/Ring.html https://mathworld.wolfram.com/Ring.html Scalar and dot products don't stay within the group of vectors and component wise multiplication doesn't always have inverses.
      - gpm 3y ago
        You need slightly more than being a ring. It's possible to make a ring (even a field) over R^n provided you don't care about interactions with scalars. For example: Just take any one-to-one map from R^n to R and then apply the operations in R before mapping the results back. It won't make any geometric sense, but it will be a ring. I've been blanking on what exactly the interactions with scalars that we need to preserve are...
    - bashmelek 3y ago
      If I remember correctly multiplication requires a “zero,” an “identity,” and for something to be a field each item needs an inverse. I imagine we can define multiplication in R2 just as we do for C. So by that logic we ought to be able to define such an operation on any R(power of 2).
    - deleted 3y ago
      [deleted]
ravi-delia 3y ago
So you can of course multiply some vectors- the real numbers are a vectorspace, and they have a perfectly nice product. Even R^2 can have a nice multiplication- (a,b)(c,d) = (ac - bd, bc + ad), the product of the complex numbers. However, it's much trickier to define a multiplication on an arbitrary vectorspace V such that 1) For any vectors u,v in V, the product u*v is in V. (this rules out the dot product as a general product) 2) For u,v in V, u*v = v*u 3) For u,v,w in V, v*(u+w) = (v*u) + (v*w) 4) For u,v in V and s in F (the field V is a vectorspace over), s(u*v) = (su)*v = s*(uv) Under these restrictions you can still cook up products, but fewer and sadder ones. In general you will not have a multiplicative inverse, for instance.
- ngcc_hk 3y ago
  Any proof you cannot have inverse. If so, may I wonder if your drop rule 2 and insist on having for most vector u there exist a v so that u*v = 1. As an icing let us say having a 0 vector. That could be something.
  - gpm 3y ago
    This is a theorem that under approximately those rules* in R^n you cannot have inverses except in the cases of the reals, complexes, and quaternions: https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29 https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_divi... *It doesn't rely on rule 2, it does however rely on (u*v)*w = u*(v*w). Adding rule 2 excludes the quaternions.
  - gpm 3y ago
    By the way, the definition of a vector space means we have a 0 vector, and the distribute property means that multiplication has to respect it. v * w = v * (w + 0) # Vector space property that w + 0 = w = v * w + v * 0 # Distributive property Another vector space property guarantees that there exists -(v * w) such that -(v * w) + (v * w) = 0. Add it to both sides -(v * w) + v * w = -(v * w) + v * w + v * 0 0 = 0 + v * 0 # And then reduce based on the definition of -(v * w) 0 = v * 0 # Vector space property that 0 + x = x So you have your icing :)
deleted 3y ago
[deleted]
tobinfricke 3y ago
The cross product is a peculiar animal, as it only exists in 3-dimensional space. There is no unique "cross product" in 4 dimensions or higher. (In two dimensions we cheat and define the result of the cross product as the scalar magnitude of what would be the component in the third dimension, if it existed.) Furthermore, it turns out that the result of taking the cross product of two vectors is itself not a vector. In physics we interpret a vector not simply as "an ordered list of numbers," but as a geometric quantity that responds to coordinate system changes in the expected way. Suppose that we have two vectors, a and b, and we take their cross product. Now suppose we choose to work in the "mirror image" coordinate system. Our choice of coordinate system should not affect physical outcomes. But while "a" and "b" are inverted in our mirror image coordinate system, the cross product "a x b" does NOT invert. Introductory physics textbooks proceed to tell us that "well actually," the result of a cross product (such as an angular velocity vector or the Poynting vector of electromagnetism) is actually a "psuedovector." Other formalisms treat the cross product in a more hygienic and general manner. This is all to say that familiar mechanisms like the "dot product" and "cross product" are not necessarily as "natural" as you may have been lead to believe.
- creata 3y ago
  > the "dot product" and "cross product" are not necessarily as "natural" as you may have been lead to believe The cross product, sure: its problem is that it dualizes unnecessarily, making you deal with a normal vector when you almost always just want the plane. But what did the dot product ever do to you?
  - npsomaratna 3y ago
    But don't you need the normal vector to define the plane? (Genuine question)
    - alok-g 3y ago
      A bivector is just as good for the purpose. Think of it as a generalization of a directional arrow (vector) into an oriented area (bivector). The OP talk shows good visualizations if you haven't watched yet.

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.
  - empath-nirvana 3y ago
    Basically you want all the properties of a ring: https://mathworld.wolfram.com/Ring.html https://mathworld.wolfram.com/Ring.html Scalar and dot products don't stay within the group of vectors and component wise multiplication doesn't always have inverses.
    - gpm 3y ago
      You need slightly more than being a ring. It's possible to make a ring (even a field) over R^n provided you don't care about interactions with scalars. For example: Just take any one-to-one map from R^n to R and then apply the operations in R before mapping the results back. It won't make any geometric sense, but it will be a ring. I've been blanking on what exactly the interactions with scalars that we need to preserve are...
  - bashmelek 3y ago
    If I remember correctly multiplication requires a “zero,” an “identity,” and for something to be a field each item needs an inverse. I imagine we can define multiplication in R2 just as we do for C. So by that logic we ought to be able to define such an operation on any R(power of 2).
  - deleted 3y ago
    [deleted]

ravi-delia 3y ago

So you can of course multiply some vectors- the real numbers are a vectorspace, and they have a perfectly nice product. Even R^2 can have a nice multiplication- (a,b)(c,d) = (ac - bd, bc + ad), the product of the complex numbers. However, it's much trickier to define a multiplication on an arbitrary vectorspace V such that 1) For any vectors u,v in V, the product u*v is in V. (this rules out the dot product as a general product) 2) For u,v in V, u*v = v*u 3) For u,v,w in V, v*(u+w) = (v*u) + (v*w) 4) For u,v in V and s in F (the field V is a vectorspace over), s(u*v) = (su)*v = s*(uv) Under these restrictions you can still cook up products, but fewer and sadder ones. In general you will not have a multiplicative inverse, for instance.

ngcc_hk 3y ago
Any proof you cannot have inverse. If so, may I wonder if your drop rule 2 and insist on having for most vector u there exist a v so that u*v = 1. As an icing let us say having a 0 vector. That could be something.
- gpm 3y ago
  This is a theorem that under approximately those rules* in R^n you cannot have inverses except in the cases of the reals, complexes, and quaternions: https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_division_algebras%29 https://en.wikipedia.org/wiki/Frobenius_theorem_%28real_divi... *It doesn't rely on rule 2, it does however rely on (u*v)*w = u*(v*w). Adding rule 2 excludes the quaternions.
- gpm 3y ago
  By the way, the definition of a vector space means we have a 0 vector, and the distribute property means that multiplication has to respect it. v * w = v * (w + 0) # Vector space property that w + 0 = w = v * w + v * 0 # Distributive property Another vector space property guarantees that there exists -(v * w) such that -(v * w) + (v * w) = 0. Add it to both sides -(v * w) + v * w = -(v * w) + v * w + v * 0 0 = 0 + v * 0 # And then reduce based on the definition of -(v * w) 0 = v * 0 # Vector space property that 0 + x = x So you have your icing :)

deleted 3y ago

[deleted]

tobinfricke 3y ago

The cross product is a peculiar animal, as it only exists in 3-dimensional space. There is no unique "cross product" in 4 dimensions or higher. (In two dimensions we cheat and define the result of the cross product as the scalar magnitude of what would be the component in the third dimension, if it existed.) Furthermore, it turns out that the result of taking the cross product of two vectors is itself not a vector. In physics we interpret a vector not simply as "an ordered list of numbers," but as a geometric quantity that responds to coordinate system changes in the expected way. Suppose that we have two vectors, a and b, and we take their cross product. Now suppose we choose to work in the "mirror image" coordinate system. Our choice of coordinate system should not affect physical outcomes. But while "a" and "b" are inverted in our mirror image coordinate system, the cross product "a x b" does NOT invert. Introductory physics textbooks proceed to tell us that "well actually," the result of a cross product (such as an angular velocity vector or the Poynting vector of electromagnetism) is actually a "psuedovector." Other formalisms treat the cross product in a more hygienic and general manner. This is all to say that familiar mechanisms like the "dot product" and "cross product" are not necessarily as "natural" as you may have been lead to believe.

creata 3y ago
> the "dot product" and "cross product" are not necessarily as "natural" as you may have been lead to believe The cross product, sure: its problem is that it dualizes unnecessarily, making you deal with a normal vector when you almost always just want the plane. But what did the dot product ever do to you?
- npsomaratna 3y ago
  But don't you need the normal vector to define the plane? (Genuine question)
  - alok-g 3y ago
    A bivector is just as good for the purpose. Think of it as a generalization of a directional arrow (vector) into an oriented area (bivector). The OP talk shows good visualizations if you haven't watched yet.