inward vs. outward orthogonality

[AntonioLao]

A necessary and sufficient condition for outward orthogonality is such that the quantum field operator can be represented by its orthogonal matrix satisfying [math]O^{T}O=I[/math] and [math] O=\bar{O}[/math] where [math]O^{T}[/math] is the transpose and [math]\bar{O}[/math] is the complex conjugate. The square matrix [math]I[/math] is the identity matrix, implying the existence of an inverse. However, an inward orthogonal quantum dynamic field operator does not have an inverse, is self-transpose and equivalent to a Hadamard matrix.

[mkirkpatrick]

A necessary and sufficient condition for outward orthogonality is such that the quantum field operator can be represented by its orthogonal matrix satisfying and where is the transpose and is the complex conjugate. The square matrix is the identity matrix, implying the existence of an inverse. However, an inward orthogonal quantum dynamic field operator does not have an inverse, is self-transpose and equivalent to a Hadamard matrix.

I can feel an itch in
our TOE here Antonio,how soon can we scratch??



kind regards michael.

[Guille]

Antonio,

What is the physical difference between inwards and outwards orthogonalities?

By the way, the terms "inward orthogonal quantum dynamic field operator" and "Outward orthogonal quantum dynamic field operator" are a bit too long for (and seems like the most complex name ever in) physics. we could use; Inorqudyfiop and for outward; Ouorqudyfiop (they seem long, but they are east to remember cause they are so strange words).

[AntonioLao]

how soon can we scratch??

As soon as we change Pauli's, Dirac's, and Einstein's matrices into Hadamard matrices and there after.

What is the physical difference between inwards and outwards orthogonalities?

Inward orthogonality is not measurable. On the other hand, outward orthogonality is measurable by eigenvalue method.

they seem long, but they are east to remember cause they are so strange words

A rose by any other name smell just as sweet. Square of energy as quantum of spacetime is just a Hadamard matrix.

[Guille]

Inward orthogonality is not measurable. On the other hand, outward orthogonality is measurable by eigenvalue method.

But if outward orthogonality is measurable and has an inverse, then it means that the measurement of it is inverselly equal to the measurement of the inverse of outwards orthgonality, and thus, there must be negative space.
Did you realize this?
Is it plausible in your theory?
Is it a fallacy?

[AntonioLao]

there must be negative space

Actually I am working on a negative spacetime manifold as negative of square of energy [math]-E^2[/math].

[Guille]

Actually I am working on a negative spacetime manifold as negative of square of energy .

And is there any geometric sense on it? Remember, matrices (including Hadamart's) are valueless without the support of a cartesian coordinate system, or at least a geometric representation.

[AntonioLao]

And is there any geometric sense on it?

The reality of negative curvature as hyperbolic geometry. Just like tensors, this geometry is independent of any coordinate system.

[Guille]

The reality of negative curvature as hyperbolic geometry. Just like tensors, this geometry is independent of any coordinate system.

But the negative curvature in hyberbolic geometry implies a positive value of length. Unless we use a coordinate system for it, but you said it is independent of coordinate systems...

[AntonioLao]

But the negative curvature in hyberbolic geometry implies a positive value of length.

That is the reason why there shouldn't be just one curvature but two linked curvatures, and their squares are absolute values of quantized spacetime. Again, they are coordinate free quantities.