Re: Curvature by Cartan's method?

nicola_ambrosetti@xxxxxxxx wrote:

I have a little problem with Cartan's method for the calculation of the
curvature. If I understood right you start by defining your orthonormal
frame 1-forms o^a and then use the vanishing of the torsion (the first
structure equation)

(1) do^a + w^a_b /\ o^b = 0

to "read off" the spin connection 1-form w^a_b. Then one proceeds in
calculating the curvature 2-form R^a_b with the second structure
equation

R^a_b = dw^a_b + w^a_c /\ w^c_b = R^a_bcd o^c /\ o^d

(all indices flat) with a 1/2 coefficient in the last equation
depending on your definition of wedge product (I believe).

As you say, the first structure equation comes from the requirement
that the connection be torsion-free. This is not enough to determine
a connection -- you also need to impose metric compatibility. If
your frame one-forms are orthonormal, it's easy to see that metric
compatibility requires that the spin connection be antisymmetric,

w_{ab} = -w_{ba}

My problem is in the 'reading off' part: if I take o^1 = y dx , o^2 = x
dy and calculate the exterior derivative I get

do^1 = - dx /\ dy

do^2 = dx /\ dy

According to (1) we have

do^1 = - w^1_a /\ o^a = - w^1_1 /\ o^1 - w^1_2 /\ o^2
= - w^1_1 /\ y dx - w^1_2 /\ x dy = - dx /\ dy

do^2 = - w^2_a /\ o^a = - w^2_1 /\ o^1 - w^2_2 /\ o^2
= - w^2_1 /\ y dx - w^2_2 /\ x dy = dx /\ dy

but this gives me only two equations with four unknowns! You could
reply that there is also the metricity condition which tells me that
w^a_b is antisymmetric so w^1_1 = w^2_2 = 0 (up or down indices
shouldn't matter since I'm euclidian),

Right.

but then I get w^1_2 = 1/x dx

No, you get w^1_2 = 1/x dx + A dy, where A is undetermined by your first
equation.

and w^2_1 = 1/y dy

No, you get w^2_1 = 1/y dy + B dx, where B is undetermined by your second
equation. Now use antisymmetry and put these two together...

Steve Carlip

.

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