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November 25

Fractal/continuum exterior product?

This seems like a really easy question but I can't seem to find the answer anywhere...

Is there such a thing as a "fractional/fractal/continuum exterior product" of a positive non-integer number of vectors? I.e. not just for a discrete number of vectors as in

a\wedge b\wedge \cdots z

but a continuum of vectors, in analogy with the continuous product, Gamma function, fractional calculus, or such like? Many thanks, Maschen (talk) 09:58, 25 November 2012 (UTC)[reply]

I don't think there is any such thing as a positive non-integer number of vectors, let alone their exterior product. Bo Jacoby (talk) 15:50, 25 November 2012 (UTC).[reply]

I was thinking it may be defined in terms of, or analogous to, the continuous product, i.e. using integrals and exponentials... Maschen (talk) 17:09, 25 November 2012 (UTC)[reply]

The continuous product is a product of an infinite number of factors, not the product of a fractional number of factors. Bo Jacoby (talk) 19:53, 25 November 2012 (UTC).[reply]

I'll just take the answer as a plain "no there isn't". (Maybe in a year or so I could come up with something... not that it would enter WP). Thank you anyway. Maschen (talk) 20:35, 25 November 2012 (UTC)[reply]

The exterior product is zero if the number of factors exceeds the dimension of the vector space. So a nonzero exterior product of a continuum of factors requires at least a continuum of dimensions. Maybe it makes sense. Bo Jacoby (talk) 00:14, 26 November 2012 (UTC).[reply]

The only "fractional product" I can think of is a power function, for which we start from integer powers. Even the gamma function (as interpolated from the factorial function) is from a product of a function of a related sequence. I doubt that defining a fractional product where the factors are unrelated occurs. So I assume then you are asking about a fractional power over the exterior product. Since we have

v^{\wedge 0}=1

v^{\wedge 1}=1\wedge v=v

v^{\wedge 2}=1\wedge v\wedge v=0

v^{\wedge n}=0,n>1,n\in \mathbb {N}

we have very little to extrapolate from to powers. One would probably require that

(v^{\wedge {p \over q}})^{\wedge q}=v^{\wedge p},p,q\in \mathbb {N} .

One could additionally choose whether the wedge-power operation be closed on the exterior algebra, or allow an extension (like the square root function being confined to reals or permitting an extension to complex numbers). Noting that there are (multivector) elements of the exterior algebra that when put to a positive integer (wedge) power are nonzero, both choices seem to allow rational wedge powers. The case of a fractional wedge power of sepcifically a pure 1-vector is a difficult case, like the square root of a negative real number: it probably cannot be accommodated if we require that this power be closed on the exterior algebra. — Quondum 05:20, 26 November 2012 (UTC)[reply]

Thanks, very nice answer, but I did mean a product of different vectors.

To put this into perspective, let a vector change with parameter(s), like a(a, b, c ...), never zero for any a, b, c,..., never parallel (in which cases the product would obviously be automatically zero). For the discrete case we have an integer number of vectors; a(a₁, b₁, c₁...), a(a₁, b₂, c₁...), a(a₁, b₁, c₂...) etc... all non-zero, and non-parallel, and we could take the wedge product of them.
Rather than taking discrete values of a, b, c, ..., let them vary over some domain, then what would the exterior product of this continuum of vectors be? The problem seems to be the dimension of the space, as Bo Jacoby said above, since the number of linearly independent vectors in an n-dimensional space is n, so the number of vectors which can be exterior multiplied (to obtain a non-zero result) is ≤ n ...
On a separate note; a continuum of dimensions isn't a problem since fractals have non-integer dimension (including transcendental numbers calculated from natural logarithms).
To put this into more context, it should generalize the multivector as a sum of integer-grade elements, into an integral of a continuum of (real-number, or even complex?)-grade elements. It would be interesting if new mathematical objects could be developed in the process rather than confining ourselves to number systems or spaces we already know...

Maschen (talk) 18:19, 26 November 2012 (UTC)[reply]

From the last bullet, it sounds like your intention is to fill in the discrete spectrum of grades of the exterior algebra into a continuum of grades. Finding an extension to the algebra that does this is probably quite a different exercise from finding some operation that acts on a non-integer number of vectors. One would possibly want to retain certain properties, for example that the exterior product of a homogeneous p-grade element and a homogeneous r-grade element is a homogenous (p+r)-grade element, where p,q∈ℝ. Not so? — Quondum 03:16, 27 November 2012 (UTC)[reply]

It should be... a constraint on the definition. Why should that not hold? As said, no reason not to extend beyond the algebra while defining such a product... Maschen (talk) 07:50, 27 November 2012 (UTC)[reply]

You're pretty ambitious. If the cardinality of the dimensions of your algebra is to be the continuum, I expect you'll have serious difficulties. If you're just looking for some abstract properties (such as exponentiation, powers, logarithms etc.), fractional grades are not needed. Beyond that, I'm a bit lost. — Quondum 08:20, 27 November 2012 (UTC)[reply]

This seems like a really easy question but I can't seem to find the answer anywhere... Really easy questions tend to be the hardest of all. :-) So I agree with Quondum that you are pretty ambitious. Some late-night speculations, no guarantee they make the slightest sense: exterior algebra is a quotient of Tensor algebra. It might be easier to think about the same questions for tensor algebras first. Second, following Quondum, there are two things desired here, corresponding to the 2 desired operations of "addition" and "multiplication" - a continuum of grades and a continuum operation respecting the grades. Now there is a well-known construction that might be thought of as corresponding to the adding of the various homogeneous grade spaces - the Direct integral. If you could develop a tensor/exterior analog of that, and take direct integrals of the various spaces you get, then that could be what you are looking for. A category-theoretic treatment of the direct integral might help, and the above musings might give some hint where to look to see if somebody has concocted this stuff already - operator algebra theory, non-commutative geometry after Connes. John Z (talk) 09:16, 29 November 2012 (UTC)[reply]

complex analysis

If G is an open set then curve γ is homologous to zero if for all ω belonging to ₵-G η(γ;ω)=0. How i can prove this? — Preceding unsigned comment added by 14.99.166.149 (talk) 17:13, 25 November 2012 (UTC)[reply]

What is η? Winding number? Do you mean "γ is homologous to zero in ₵-G?" What tools are you supposed to be using? Staecker (talk) 18:44, 25 November 2012 (UTC)[reply]

complex analysis (part 2)

I want to know what is orientation? If (1,0,∞) is the orientation of R,then the cross ratio (z,1,0,∞)=? — Preceding unsigned comment added by 14.99.166.149 (talk) 17:25, 25 November 2012 (UTC)[reply]

See Orientation (vector space), Orientation (geometry), and Cross-ratio. Moonraker (talk) 05:45, 26 November 2012 (UTC)[reply]

I changed the title to be unique. Please keep your titles distinct from each other, or the software has fits. StuRat (talk) 06:49, 26 November 2012 (UTC) [reply]