## Perpendicular Cross Products

Higher-dimensional geometry (previously "Polyshapes").

### Perpendicular Cross Products

In 3d when there is a cross product there are two vectors being multiplied using a vector perpendicular to the other two vectors. There is also an angle between the two vectors being multiplied and both vectors are on a plane perpendicular to the third vector.

In an odd number of dimensions greater than 3d it is possible to have multiple planes that are both perpendicular to each other and to a vector that is not on one of the planes. When there are multiple planes there can be multiple pairs of vectors that are perpendicular to the other pairs of vectors and to the single vector and for each pair of vectors there can be an angle between the two vectors of that pair. In an odd number of dimensions greater than three would it be possible to have multiple cross products that are perpendicular to each other in which the vectors from each of the cross products are multiplied together using the remaining vector and if it is how would using multiple cross products work?
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anderscolingustafson
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### Re: Perpendicular Cross Products

An n*n matrix represents n vectors, in box product'. In 3d, it is represented as (a*b).c

You could have menaingful dot products in any dimension, if one supposes þe dimensionality of the vectors in the product is N-1, and some sort of parity exists.

The law that the circulation in a ring' is proportional to the vector-surface of every surface that is bounded by the ring, is true in every dimension, and is relatively easy to prove. This is the basis of m=IA. (hint: volume is flux of monent, ie $$V = \int \xi \cdot dS$$ ).

Outside of that, there is no real meainig for a specific kind of process to deal with eddy currents, curl, or cross-product in general.
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wendy
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### Re: Perpendicular Cross Products

There are several different ways to generalize the vector product to other dimensions.

The perpendicular product

The first kind of generalization is the perpendicular product in n dimensions, which takes (n-1) vectors as arguments. In 2D, any vector is perpendicular to a unique line, which translates to a unique vector if you fix the sense in which the product assigns a direction to this line. So the 2D perpendicular product is a unary operator. It has characteristic 4, since taking the perpendicular product of a vector is equivalent to rotating it 90°, so applying this operator to the vector 4 times will recover the original vector again.

In 3D, we already know that the cross product takes two vectors as arguments, and produces a 3rd vector which is perpendicular to both of them.

In 4D, the perpendicular product takes three arguments, and produces a 4th vector perpendicular to the 3D hyperplane spanned by the three argument vectors.

In n dimensions, in general, (n-1) arguments are needed, and the perpendicular product produces a vector perpendicular to the (n-1)-dimensional hyperplane spanned by the input vectors.

The n-dimensional perpendicular product of the vectors v1, v2, ... v{n-1} can be computed as a pseudo-determinant of a matrix of the form:
Code: Select all
|[   v1   ]||[   v2   ]||[   ...  ]||[ v{n-1} ]||[    B   ]|

where B = the basis vectors of n-dimensional space (i.e., <1,0,0,0,...>, <0,1,0,0,...>, <0,0,1,0,...> etc.). Expanding the determinant of this matrix then produces a result of the form |...|*b1 + |...|*b2 + ... + |...|*bn where b1, b2, ... bn are the basis vectors in B, and the |...| are the determinants of the submatrices that define the components of the resulting vector.

The nice thing about the n-dimensional perpendicular product is that it serves as the n-dimensional analogue of the 3D cross product with respect to computing a perpendicular vector given a set of (n-1) vectors that span an (n-1)-dimensional hyperplane. However, the not-so-nice thing about it, is that it requires (n-1) arguments. So 3D equations involving the cross product has no easy generalization, because an expression like A x B may not have any obvious generalization to have additional arguments. In preserving the property of having a perpendicular vector result, we have sacrificed the binary nature of the cross product. So this leads us to the next kind of generalization:

The binary polyvector product

Another generalization is to preserve the binary property of the cross product, but sacrifice the result being another vector. This generalization continues to take two vector arguments, but the result is only a vector in 3D. In 4D, the result is a bivector (basically an entity that represents an oriented, directed 2D plane). In 5D, the result is a trivector. In general, in n dimensions, the result is an (n-2) polyvector.

The nice thing about this generalization is that equations involving the cross product now have direct generalizations. Well... kind of... because the result is now a polyvector, so what used to have a vector result now has a polyvector result, and this causes other kinds of problems (e.g., you can no longer plug the result of the equation into another equation that expects a vector, not polyvector, argument). So in preserving the binary-ness of the product, we have sacrificed its vector result.

Can we have both? Well it turns out that the answer is yes... but only in a limited number of dimensions.

Complex / Quaternion / Octonion products

In 2D, we can define a product that is commutative, associative, and distributes over +, etc.. This is the complex number product, which takes two 2D vectors (treated as complex numbers), and produces another 2D vector as a result. In 4D, we have the quaternion product, which takes two 4D vectors (treated as quaternions) and returns a 4D vector result. However, here the seams start to show: this product is no longer commutative. (But neither is the cross product commutative, so it's not that big a deal. Right?) In 8D, we have the octonion product, which takes two 8D vectors (treated as octonions) and returns an 8D vector result. However, the octonion product is not only non-commutative, it is not even associative (so (a*b)*c is, in general, NOT equal to a*(b*c)).

And it turns out that these are the only dimensions that have a well-defined algebraic product. There are no analogous products in any other dimension. Also, these products diverge from the 3D cross product in that the vector result does NOT have the perpendicular property -- in general, multiplying two complex numbers will not produce a result that's perpendicular to both of them. Same goes with the quaternion and octonion products. So in preserving both the binary property and the vector result, we have sacrificed the perpendicular property. (Not to mention that we have lost dimensional generality: these products only exist in 2D, 4D, and 8D, and no other dimension.)

It's only in 3D where the cross product has all 3 properties.
quickfur
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### Re: Perpendicular Cross Products

Could a complex product produce a magnetic field in 2d? Could a quaternion product produce a magnetic field in 4d? Could an octonion product produce a magnetic field in 8d?
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anderscolingustafson
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### Re: Perpendicular Cross Products

The complex / quaternion / octonian products do not have the orthogonal property, so the resulting field will be nothing even remotely resembling "magnetic"!
quickfur
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### Re: Perpendicular Cross Products

quickfur wrote: The binary polyvector product

Another generalization is to preserve the binary property of the cross product, but sacrifice the result being another vector. This generalization continues to take two vector arguments, but the result is only a vector in 3D. In 4D, the result is a bivector (basically an entity that represents an oriented, directed 2D plane). In 5D, the result is a trivector. In general, in n dimensions, the result is an (n-2) polyvector.

No, you got it backwards! (Or maybe you took the Hodge dual...)

The proper generalization of the cross product is the wedge product. This always results in a bivector, which represents the plane of the two vectors. More generally, the wedge product of k vectors (a^b^c^...) is a k-blade. It's a measure of linear independence (a generalized determinant); the wedge product is 0 if any vector is a combination of the others.

There is a thing called a "pseudoscalar", a multivector with maximum grade (that is the number of dimensions n). It represents an oriented volume element: an ordinary vector in 1D (with length), a parallelogram in 2D (with area), a parallelepiped in 3D, etc. All possible pseudoscalars (for a given dimension) are proportional to each other. If the proportionality constant is negative, they have opposite orientation / handedness.

The pseudoscalar V can be multiplied (using a different product, not the wedge) with a k-blade to produce an (n-k)-blade. a^b^c^... is parallel to all of the vectors a,b,c,..., while V(a^b^c^...) is orthogonal to them. This multiplication by V is called the "Hodge dual" (though that's often described differently).

In particular, if you have k=n-1 vectors, then their wedge product is an (n-1)-blade, and the Hodge dual makes a 1-blade, which is a vector.
mr_e_man
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### Re: Perpendicular Cross Products

anderscolingustafson wrote:Could a complex product produce a magnetic field in 2d? Could a quaternion product produce a magnetic field in 4d? Could an octonion product produce a magnetic field in 8d?

If you are looking for a mathematical tool for describing higher dimensional magnetic fields, the way to go is Clifford's geometric algebra.

I've have forgotten all but the essentials, but my thinking goes like this. The goal is to have some object and some operation that can be applied to the velocity vector of a charged particle. The result is then the acceleration vector on the particle due to magnetism.

Everything in geometric algebra is a package of vectors. (A scalar is a zero dimensional vector.) The number of vectors is called the degree of the package. A scalar is zero degree, a vector is one degree, a bivector is two degrees, and so forth. The main operation is called the geometric product. What we want is some object such that the geometric product of that object with the velocity vector gives the acceleration vector due to magnetism.

The geo product of a vector with an object of degree n results in two objects. One has degree n-1, the other n+1. We want n-1 to equal 1. So n=2. Our object must be of degree two. But in 4D a bivector does not work, for reasons I don't feel like explaining right now. What DOES work is having two bivectors. For each bivector, take the geometric product with the velocity vector of the charged particle. The result is two vectors. Add them together to get the acceleration vector on the charged particle due to magnetism.

There might easily be a neater way to do this. The distinction between magnetic and electric fields is artificial: they are really different parts of the same thing. So maybe in 4D the distinction should be done away with. It is possible to reduce Maxwell's to a single equation. But when I tried that I found it too strange to work with. So I stuck with the magnetic field concept.

In short, the most convenient way I found to describe a 4D magnetic field was with two orthogonal bivectors at every point.
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How about 5D? You still have two bivectors at every point.

In 6D and 7D you would use three bivectors, in 8D and 9D four, and so forth.
PatrickPowers
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