Recent content by eok20

Ahh...right, every map from a simply connected space lifts to the universal cover. Thanks!

I'm having trouble seeing why the following is true: let M be a simply connected manifold and s a smooth map from M to U(1). Then why does it follow that s = e^(iu) for some smooth function u from M to R? Thanks!

The statement is true (since we're assuming H is a subgroup). Are you familiar with the theorem that a (topologically) closed subgroup of a Lie group is a Lie group?

Yes, I've done problem 1 (showing that \nabla^b j_b = 0) as well as problem 2a (showing that ** = +- 1). I am confident in both of my solutions to these-- does one of these help?

Maxwell's equations in curved spacetime can be written as \nabla^a F_{ab} = -4\pi j_b, \nabla_{[a} F_{bc] = 0 or as d*F = 4\pi*j, dF = 0, where F is a two-form, j is a one-form and * is the Hodge star. How do you show that these two sets of equations are equivalent (basically, that the first...

I am wondering if, in general relativity, there is a way to make sense of a statement such as, "the spaceship is 100km from me." In special relativity, we could define this (as long as I am an inertial observer) by choosing global coordinates (t,x1,x2,x3) corresponding to my notions of time and...

Some comments: the preimage of {f(x)} is not necessarily x. One way that I think will work is that G is the inverse image of {(y,y)} < Y x Y under a nice map. Using the fact that Y is Hausdorff I think you can show that the diagonal set ({(y,y)}) is closed.

I'm probably missing something obvious, but suppose that B' < B < A are all abelian groups and that A/B is isomorphic to A/B'. Does it follow that B = B'? In the case of finite groups and vector spaces it is true by counting orders and dimensions but what about in general?

The literature on spinors can be very confusing since every person uses different notation. The way I have always understood things is that spinors in n dimensions make up the representation space of a representation of the even subalgebra of the clifford algebra. That is, even elemenets of...

You are correct that you can use r as well. Sometimes people use lower things to be safe.

That proves that {x} is not open. But a set being not open does NOT imply that the set is closed (e.g. [0,1) as a subset of R is neither open nor closed).

Regarding your first question: it will have a solution since the vector is assumed to lie on the plane spanned by s_1 and s_2 (if you took a vector that wasn't on the plane then the equation wouldn't have a solution-- the system would be inconsistent). This solution must be unique since s_1...

You need to solve the linear system as_1 + bs_2 = v for a and b. Then (a,b) will be the components of v relative to the basis {s_1,s_2}. Another way to do this is like you think, and use instead a different inner product. This is J^t J where J is the inverse of the matrix with s_1, s_2 as...

What is your reasoning for why it is Cauchy?

What I said was that the Lie derivative is not linear with respect to functions, that is L_{fX} Y \ne fL_X Y, ~~ L_X fY \ne f L_XY where f is a smooth function. On the other hand, it is the case that \nabla_{fX} Y = f\nabla_X Y. This fact basically amounts to quaser987's comment...