the only real cross product is 3 dimensional
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cross product is worse product, geometric algebra introduces other products which generalize better furthermore the cross product is just a special case.
The hodge star dual of the wedge product maps back to R^n only for n=3.
Exterior product space.
15:16 Might I add that the provided proof for lemma 2 doesn't work for 1d? That is because you can't have a pair of orthogonal 1d vectors. This then leads to a hole in the proof that n=3 is forced, consistent with the fact that the zero cross product and the regular product for 1d (standing in for the dot product) do satisfy the defining identity.
EDIT: I only now noticed that the stipulation was that n is greater than 1. So Michael (or his source) was aware of this exception.
29:14
It's good place to stop
Thank you for your service.
@@minimo3631
😀😀😀
12:48 It is not clear the equality still holds if we replace b by b*c...
I guess the lemma states that it should hold for any vector b in R^n. Michael just glossed over that.
The stipulation that we use real numbers is critical. That matches with usual physical explanations why space has 3 dimensions. Then 4-dimensional space as used in space-time also has its version of cross-product but the signature of such space is different. One simple way to add a 4th dimension is to use complex numbers. Traditionally ict for “time”. I put “ time” in brackets because in relativity ( even in special relativity) there is no such thing as time versus space. Depending on various observers, “ time” for one may include or be “ space” for the other.
What do you mean by there being no such thing as time versus space? Time is fundamentally different from space because it has a different sign in the metric (which can be represented by taking t -> it, but doesn't have to be). The interpretation of time dilation/length contraction being a mixing of time and space for different observers is a new one to me.
I'd also love to know about this "space-time cross-product" - how is it defined, and what is it useful for? I've not heard of it myself.
@@le__birb
What is written is a common misstatement of what is meant. Actually, all that special relativity showed in this sense is that space-time is not locally 4 dimensional, because we live on a 3 dimensional level surface of a nonlinear functional on a Euclidean 4 dimensional space; the time coordinate and the three spatial coordinates are not “independent”, and the expression of how that dependency relation holds is not properly described as linear dependence because space time has nontrivial curvature.
@@le__birb
“Lie algebra
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Main article: Lie algebra
The cross product can be seen as one of the simplest Lie products, and is thus generalized by Lie algebras, which are axiomatized as binary products satisfying the axioms of multilinearity, skew-symmetry, and the Jacobi identity. Many Lie algebras exist, and their study is a major field of mathematics, called Lie theory.
For example, the Heisenberg algebra gives another Lie algebra structure on
R
3
.
{\displaystyle \mathbf {R} ^{3}.} In the basis
{
x
,
y
,
z
}
,
{\displaystyle \{x,y,z\},} the product is
[
x
,
y
]
=
z
,
[
x
,
z
]
=
[
y
,
z
]
=
0.
{\displaystyle [x,y]=z,[x,z]=[y,z]=0.}
Quaternions
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Further information: quaternions and spatial rotation
The cross product can also be described in terms of quaternions. In general, if a vector [a1, a2, a3] is represented as the quaternion a1i + a2j + a3k, the cross product of two vectors can be obtained by taking their product as quaternions and deleting the real part of the result. The real part will be the negative of the dot product of the two vectors.
Octonions
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See also: Seven-dimensional cross product and Octonion
A cross product for 7-dimensional vectors can be obtained in the same way by using the octonions instead of the quaternions. The nonexistence of nontrivial vector-valued cross products of two vectors in other dimensions is related to the result from Hurwitz's theorem that the only normed division algebras are the ones with dimension 1, 2, 4, and 8.”
Source: en.wikipedia.org/wiki/Cross_product
See also en.m.wikipedia.org/wiki/Exterior_algebra
@@le__birb the sign for time is opposed to that of the 3 dimensions of space. No discussion. The physical interpretation of the 4 dimensions depends on the observer. That is why I put quotes in case physicists also enjoy this. Extreme case: light travels from sun to earth in 8 minutes for us. No time to the photon that did not age at all. For less extreme cases there can be an exchange of some of the space against some of the time.
😊😊😊Lovely👍👍👍
Let me now infer that precession (of a gyroscope) exists only in 3D spaces.
Or alternatively: higher dimensional manifolds will show no precession if subjected to both translation and spin simultaneously.
Michael I’m very "cross" with you. Please do a back flip. You haven’t done one in a long time!
That's because in R^n you have to do the cross product of n-1 vectors.. so for 2 vectors ofc n=3 for the product to be well defined.
At 09:20, the wrap-up of the proof of a*b = b*(-a) requires the inner product to be positive-definite (not mentioned). Could we not have chosen d = (a*b) - (b*(-a)) at the outset. If d = 0, we're done. If not, we can choose c such that c.d = 0. Then, as in Penn's proof, we can show (a*b).d = (b*(-a)).d; i.e. that [ (a*b) - (b*(-a))].d = 0 => || (a*b) - (b*(-a)) ||^2 = 0; whence by the positive-definiteness of the inner prod, we have (a*b) - (b*(-a)) = 0; i.e., that a*b = b*(-a)
In my case.. 1:00 is good place to stop 😢
You said that the cross product “doesn’t exist in 2 dimensions”. If you believe that, then I presume that you’d say that the dot product also “doesn’t exist in any dimension other than 1”, since in higher dimensional spaces, the output of any inner product is not in the space of vectors comprising its domain.
The philosophical conundrum and confusion here is exacerbated by the fact that in some presentations of, for example, Green’s Theorem to young calculus students, the notion of a cross product for a pair of vectors in the plane is a vector not in the plane, and that’s one tool sometimes used in formulations of results such as Green’s Theorem and a number of other results. This is why I say that you should, for the sake of philosophical consistency, believe that for products “don’t exist in two dimensions”. (Even the terminology “doesn’t exist” and “in two dimensions” or “in 3 dimensions” is philosophically lazy enough that students who have heard it so many times must be deprogrammed from that way of thinking about how to discuss the notion of dimension or the notion of convergence or divergence when they face them in higher level courses.)
you multiply _two_ _three_-dimensional vectors by writing them next to each other to get a _two_ by _three_ matrix and then calculating the determinant of the _three_ matrices you get by removing one of the _three_ rows. the n'th entry of the result is the determinant of the matrix you get by removing the n'th row.
we can now generalize this to define the product of _n_ _n+1_-dimensional vectors.
for 4 dimensions, you would get a ternary operator where you need to calculate the determinant of a 3x3 matrix for each value of the resulting vector.
there is nothing magical about 3 dimensions except for 3 being 1 more than 2.
Of course, there is a similar operation in any R^n that takes n-1 vectors and yields a vector orthogonal to all of them with its magnitude equal to the volume of the shape made up by the vectors. The poiht of the theorem in the video and similar theorems is specifically to find an operation with exactly 2 inputs.
Haven’t watched video but just use geometric algebra’s outer product to extend to any dimensions
the product "." is not define in : (a*b)*c = (a . c) b - (b . c) a ... ???
It's the standard dot product R^n is implicitly equipped with
Nice!
what about 7-dimensional space? isn't there also a cross product there?
There is, but it doesn't satisfy the triple product identity. I wonder though if there's some product identity which the 7D cross product uniquely satisfies, and whether or not it's a "septuple" product identity.
@@mjkhoi6961Malcev produced an identity the octonions satisfy that would apply to the 7 dimensional cross product.
Did you watch the video? He literally mentions it at 50 seconds or so
@@JavedAlam24 no, i got bored after 30 seconds.
I didn't know the triple product identity. How can it be proven in R³ with the usual cross product definition?
Use the fact that both parts of the equality are linear in a, b and c. This means that it's enough to check the identity on a basis
From a basic property of the c.p. we know that (a*b)*c is orthogonal to a*b, so it lies in the plane generated by a and b. This gives (a*b)*c=ma+nb where m and n are scalars. Now take the dot product of both sides with c. This gives 0 on the left side, and you are now almost done.
It's a nice exercise to prove it by definition: rewriting both dot and cross products as sum over index, cross product requires Levi-Civita symbol though
which prev. video?
i believe he's referring to "why there is no four dimensional cross product."
Hi,
Does that mean that the physical space cannot have more than 3 dimensions ?
Yes.
@@kafiruddinmulhiddeen2386 no!