Re: One form <-----> vector field
That makes sense, but doesn't that only work for a perfect differential df, whose coefficients are \frac{\partial f }{\partial x^i}? If df is not the differential of some function f, but instead a general 1-form, would you still write
df[V]= V[f]
only this...
Re: One form <-----> vector field
I got a question about L*_i.
L*_i is a function that takes a form wj and maps it to a number. L*_i is also a vector. Would it be true that L*_i(wj) =wj(L*_i)?
I'm used to forms being linear functions of vectors. I'm not used to vectors being linear...
The basis doesn't have to be the partials, but the partials are the most natural choice. You can convert between the basis of partials and any other basis by the vielbein.
I'm not sure if this is accurate to say or not, but the basis in classical differential geometry is:
e_i=\frac{\partial...
The components of a vector change when the basis is changed, but the vector does not, since a vector is something that exists without coordinates.
So your question 1 is right.
As for your other question, you don't count the basis when counting rank, just the indices on the component. If there...
I thought the Lie derivative is linear in both arguments. The Lie derivative L of the tensor T in the direction of the vector field X obeys:
L_X (T_1+T_2)=L_X (T_1)+L_X(T_2)
L_{X_1+X_2} (T)=L_{X_1} (T)+L_{X_2}(T)
That has always confused me. I understand what you mean, that...
Given a smooth manifold with no other structure (like a metric), one can define a derivative for a vector field called the Lie derivative. One can also define a Lie derivative for any tensor, including covectors.
Incidentally, with antisymmetric covectors (differential forms) one can define...
Thanks.
According to Wikipedia, if U and V are simply connected open subsets of R^n, and f: U->V is smooth, then having the Jacobian non-singular is enough for U and V to be diffeomorphic.
So I guess that's why your function from the line to the circle fails, as the circle is not simply...
What is the relationship between being globally diffeomorphic and the Jacobian of the diffeomorphism?
All I can think of is that if the Jacobian at a point is non-zero, then the map is bijective around that point. For example, if:
f(x)=x_0+J(x_0)(x-x_0)
where J(x0) is the Jacobian...
Maybe I'm reading my book wrong, but it claims that a flow is a group of global diffeomorphisms.
The flow \sigma(t,x) , where t is the group parameter and x is a point on the manifold, is given by:
\sigma^\mu(t,x)=e^{tX^\mu(x)}x^\mu
where X^\mu(x) is the vector field at the point x in the...
When calculating the derivative of a vector field X at a point p of a smooth manifold M, one uses the Lie derivative, which gives the derivative of X in the direction of another vector field Y at the same point p of the manifold.
If the manifold is a Riemannian manifold (that is, equipped...
That makes really good intuitive sense. But for some reason the book I have defines embedding in a slightly esoteric way. Instead of topological spaces, it talks about embedding of differentiable manifolds.
First it defines an immersion. Basically, a smooth map f: M -> N between manifolds...
Two topological spaces can be diffeomorphic to each other, but here we're dealing with one topological space, the real line, and two incompatible atlases that we call structures. When saying that a manifold is diffeomorphic, I think what is usually meant is that it is locally diffeomorphic to...
The definition of having multiple differentiable structures is that given two atlases, {(U_i ,\phi_i)} and {(V_j,\psi_j)} (where the open sets are the first entry and the homeomorphisms to an open subset of Rn are the second entry), that the union {(U_i,V_j;\phi_i,\psi_j)} is not necessarily...
The action of a dual f on a vector v is: f_i v^i where the index i is summed over the dimension of the vector space.
So how would it go when you write it in functional form like you did. Would \int \frac{\partial }{\partial x^\mu} dx^\nu be equal to
\int v^\mu\frac{\partial }{\partial...
Re: homeomorphism
That's a good one. So the point {1} is actually an open set on the domain: [0,1]
\cup (2,3] . So if you take an open set on this domain to be say {1}\cup(2,2.5) then it maps to [1,1.5) which is not open on the codomain [0,2]. So I guess this is an example of where Hurkyl...
Re: homeomorphism
I can't think of one. The book mentions f(x)=x^2 on (-a,a) maps to [0,a^2) which is not open. However, the problem with this is that f(x)=x^2 is not a bijection, so it doesn't make sense to even speak about an inverse.
Can you have a continuous bijection whose inverse is...
The definition of a homeomorphism between topological spaces X, Y, is that there exists a function Y=f(X) that is continuous and whose inverse X=f-1(Y) is also continuous.
Can I assume that the function f is a bijection, since inverses only exist for bijections?
Also, I thought that if a...
Yeah. I got confused because of the overloading of the parameter t. So it'd be:
t=t
x,y,z=x(t),y(t),z(t)
instead of something like:
t,x,y,z={t(s)=s},x(s),y(s),z(s)
Also, it is true that for parallel transport:
\partial_k <X,Y>=0
but I guess that's just a path with a direction only on the...
You're right. Your proof is much simpler than the one my instructor gave. In fact, the proof is practically tautological.
The reason I made a distinction is because usually little d means a coordinate derivative and big D means a covariant derivative, and the person wrote it with a little d. I...
Do you mean:
D_t <X,Y> = <D_tX,Y> + <X,D_tY> ?
Then this is just the Leibniz rule and there is no need for metric compatibility.
In all the standard physics books the proof goes something like this:
\partial_k (V^iU_i)=\partial_k (V^i) U_i+V^i\partial_k (U_i)
parallel transport implies...
How do we define open sets in a topological space? In a metric space it is simple as you can define an open ball when given a metric (distance function).
I read somewhere that for a manifold you determine if a set is open by defining a one-to-one map from that set to Euclidean space, and if...
In a lot of textbooks on relativity the Levi-Civita connection is derived like this:
V=V^ie_i
dV=dV^ie_i+V^ide_i
dV=\partial_jV^ie_idx^j+V^i \Gamma^{j}_{ir}e_j dx^r
which after relabeling indices:
dV=(\partial_jV^i+V^k \Gamma^{i}_{kj})e_i dx^j
so that the covariant derivative is...
So are you saying that for Riemannian space (i.e., positive definite distance), the metric contains the complete information about the manifold (i.e., all the coordinate charts and the set of points)? So if you are given a metric, then that uniquely determines the manifold?
Also, I was...
Thanks everyone. Yeah this is from page 181 of "The Road to Reality".
I'll think about all these answers, and see if the library has that textbook you're using.
I always thought one could define a manifold as a collection of points with a distance function or metric tensor.
But in a layperson's book by Penrose, he defined a manifold as a collection of points with a rule for telling you if a function defined on the manifold is smooth. He says this is...
I was reading Penrose's layperson book, "The Road to Reality", and he says that the covariant derivative is the difference between the vector at the new point, and the old vector (at the nearby point) parallel transported to the new point.
Using that interpretation, I tried to derive the...
Thanks. That was very helpful. I have a few questions.
How would you define continuity? Say you are mapping two vectors A and B in the tangent plane at point p to the tangent plane in point p'. Then would it be the typical way in analysis, where gij(A-B)i(A-B)j less than delta implies the same...
Here's an example. Consider spherical coordinates in R3. Some of the basis are:
\vec{e_r}=(sin \theta cos\phi,sin\theta sin\phi, cos \theta)
\vec{e_\theta}=(r cos \theta cos \phi, r cos \theta \sin \phi, -r sin\theta)
Now
\partial_\theta\vec{e_\theta}=-r\vec{e_r}
follows...
If you have the expression: px-f(x), then that doesn't tell you anything. Given an f(x), how do you find a new function?
px-f(x) is written in terms of 2 variables. You want it only in terms of one variable. The Legendre transform solves it by saying max{px-f(x)}. This is a function of one...