Hiya StatusX et all
The two expressions for [pq,r] are equivalent, and this is independent of the metric signature.
I had a little trouble interpreting the y-coordinate notation, but I think it makes sense. Assuming so, the formal proof goes as follows.
We start with a set of basis...
Let me tell a story. Its relevance will become apparent.
When I had my first taste of analytic geometry, I got the idea that if
were the coordinates of a point, and
a x + b y = 0
was the equation of a line, then
should be "the coordinates of a line".
I was away for a few days, came back and settled down to sort out the relations
between curvature and the several kinds of derivatives on amanifold; and
I'm still going.
Anyway, my apology for imprecision above. I think the following is correct,
though I can't prove it...
Hiya, Peterdevis and mathwonk
In the spirit of chroot's beautifully economical characterization of tensors, the Riemann curvature tensor is a machine which takes a bivector area and returns a rotation matrix. In other words, it's a (1,1)-tensor-valued 2-form.
The motivation is "parallel...
matt grime writes:
You're right, exterior products don't require a metric. However, inner products do, even if the metric is just the boring old sum-of-squares norm.
Here's my reasoning.
(1) How do you measure a vector (in a vector space V), i.e. resolve it to its components v^i? Answer...
You're right, I'm wrong: see
But I can quote MTW: "Contraction seals off two of the tensor's slots, reducing the rank by two." That has to include the contraction of the tensor product of a contravariant...
I'm afraid the terminology depends on context.
The only ambiguity here is that sometimes only certain kinds of vector should have inner products formed with other kinds. Really, vectors should only have inner products formed with vectors from the dual space. When you have a...
Let me throw some more geometric intution into this.
I'm going to argue that (1) the cross product defines an area in space, (2) you get the perpendicular vector only after you've defined what "perpendicular" means, (3) you get perpendiculars when you define an inner or dot product.
Hey, consistency in formalism should never go unrewarded. Turin you're right, it's a (0,4) tensor.
If you built it up as dx^0 \wedge dx^1 \wedge dx^2 \wedge dx^3 , then it alternates, i.e. dx^1 \wedge dx^0 \wedge dx^2 \wedge dx^3 , or any other exchange of the 1-forms, multiplies the volume...
I've had my difficulties with this, but fundamentally it's
not too hard.
In the small limit, there are only derivatives and the geometry is
necessarily linear: a vector space with its dual. Any point of
a manifold, by definition must have a neighbourhood which can
be mapped to an n-dimensional...