Most proofs i checked are written with a local coordinate patch, is there a simple coordinate independent formulation on how to verify that [X, Y] (f) is indeed a vector field? My teacher didn't bother to write it down.  Also i understand how we get to the double sum expression of the bracket (thanks to the leibniz rule among other things) but why do the second derivatives cancel out at the end?  Thanks.
121779
I want to understand what exactly means to ask the question if the Lie bracket $[X,Y]$ of two vector fields $X,Y\in \mathcal{XM}$, where $\mathcal{M}$ is a differentiable manifold, is well defined.    What does it mean conceptually that the Lie bracket is well defined?  How could I prove that indeed it is well defined? Can I do it use it a specific basis, say, $e_x = \partial/\partial x$ and $e_y = \partial / \partial y$?
53343
Can anyone explain where the following expression for the electric field vector comes from? $$ \mathbf E(\mathbf r,t) = -\nabla \phi(\mathbf r,t) - \frac{\partial}{\partial t}\mathbf A(\mathbf r,t)  $$  where $\phi$ and $\mathbf A$ are the scalar and vector potentials, respectively.  Presumably it can it be derived from Maxwell's Equations?
87564
As an exercise in my textbook, I need to prove that if $V$ is a finite dimensional vector space with dual space $V^*$ over $\mathbb{R}$, then dim$(V)$=dim$(V^*)$.  Let $\omega\in V^*$ and let $\{e_1,...,e_n\}$ be a basis for $V$. Define $e^i\in V^*$ by $e^i(e_j)=\delta_{ij}$. We show that $\{e^1,...,e^n\}$ spans $V^*$. $\omega(v)=\omega(v_1e_1+...+v_ne_n)=v_1\omega(e_1)+...+v_n\omega(e_n).$ If $\omega(e_1)=\lambda_1,...,\omega(e_n)=\lambda_n$, then $\omega(v)=v_1\lambda_1e^1(e_1)+...+v_n\lambda_ne^n(e_n)$=$\lambda_1e^1(v)+...+\lambda_ne^n(v)$.   To show $\{e^1,...,e^n\}$ is linearly independent, suppose that $0=c_1e^1+...+c_ne^n$ is the zero mapping to $\mathbb{R}$. Consider the image of $e_1:$  $0(e_1)=c_1*1+...+c_n*0=c_1$ Hence, $c_1=0$. Repeating the procedure for $e_j$, $2\leq j\leq n$, we see that $c_1=c_2=...=c_n=0$.  Does this proof look correct?  If the proof is correct, I have an additional small question. It seems like this proof is dependent on the fact that $V$ is a real (or complex) vector space, since we define the covectors to have the image set $\{0,1\}$. Is there a proof that works for vectors spaces over general fields?
46590
In the vector space of $f:\mathbb R \to \mathbb R$, how do I prove that functions $\sin(x)$ and $\cos(x)$ are linearly independent. By def., two elements of a vector space are linearly independent if $0 = a\cos(x) + b\sin(x)$ implies that $a=b=0$, but how can I formalize that? Giving $x$ different values? Thanks in advance.
65562
Lets say for example, $L$ is a linear mapping with $V$ as its domain and $W$ as its co-domain.  Theorem: $L(0) = 0$. (Zero vectors)  I know its wrong to assume that every zero vector in $W$ is mapped from a zero vector in $V$, but I can't logically explain why!  Since $L(0) = 0$, and equal signs go both ways (double implication), isn't the statement $L(0) = 0$ implying the $0$ vector in $V$ only maps to the zero vector in $W$, and the zero vector in $W$ can be only mapped from the zero vector in $V$?  I know this is wrong. Can someone explain why?  Thanks
73984
Let $X_1, X_2$ be vector fields over a smooth manifold $M$ and likewise $Y_1, Y_2$ vector fields over a smooth manifold $N$. Furthermore, let $X_1$ and $Y_1$ aswell as $X_2$ and $Y_2$ be $\Phi$-related for some morphism $\Phi: T M \to T N$ (i.e. we have $\Phi \circ Y_i = X_i \circ \phi$ where $\phi: M \to N$ is the projection mapping of $\Phi$). I now want to show that the commutator/Lie bracket $[X_1, Y_1]$ is also $\Phi$-related to $[X_2, Y_2]$.  Now I tried to go by the definition of the commutator and write out   $$[X_1, Y_1] \circ \phi  = X_1 Y_1 \circ \phi - Y_1  X_1 \circ \phi \\   = X_1 \circ (\Phi \circ Y_2) - Y_1 \circ (\Phi \circ X_2) \\$$  But I'm not really sure how to go from there as I don't think I could interchange $X_i \circ \Phi$ for $\Phi \circ X_i$? I would need to show that this expression is equal to   $$\Phi \circ [X_2, Y_2] = \Phi(X_2 Y_2 - Y_2 X_2)$$  but approaching it from this side, I again don't know how to continue from here, as I don't think I could just "pull in" $\Phi$ into the individual summands? Is there something about the commutator that I'm misunderstanding or forgetting? It seems like an easy thing to show yet I'm a bit lost here.
102528
In a paper I'm reading, it is stated that "It is known that on a compact, connected, oriented two dimensional manifold, there exists a vector field with only one singularity".  Where can I find a proof of this fact?
79351
I learned vector analysis and multivariate calculus about two years ago and right now I need to brush it up once again. So while trying to wrap my head around different terms and concepts in vector analysis, I came to the concepts of vector differentiation, gradient, divergence, curl, Laplacian etc.   The reference that I'm using is very inadequate to give any geometric/physical interpretetions of these (almost) new concepts. I don't have much problem with their formulae and working rules, but I want to look at them in a little more geometric way. For example, the geometric significance of gradient that I got out of my book is the following: if $f:\mathbb{R}^3\to \mathbb{R}$ be a differentiable function then $\nabla f (x,y,z)$ is the vector perpendicular to the level surface $f(x,y,z)=c$ ($c$ constant) at the point $(x,y,z)$.  I would really appreciate if anyone can explain how, in this way, can divergence, curl and Laplacian be interpreted geometrically. (e.g. For a given vector field $\textbf{F}: \mathbb{R}^3\to \mathbb{R}^3$, the relation between $\nabla\times\textbf{F}$ and $F$ and questions like that.) I looked around Google a bit, but couldn't find what I was looking for.  Thanks in advance.
100325
In Dummit and Foote problem 12 of section 10.4, I need to show that if $V$ is some vector space over a field $F$ and $v,v'$ are nonzero elements of $V$, supposing we have that $v \otimes v' = v' \otimes v$ then this implies that $v = av'$ for some $a \in F$. I was trying to consider some basis $\mathcal{B} = \{e_i\}_{i\in I}$ for the vector space $V$, show that we can write $v$ and $v'$ as linear combinations of this basis elements and then argue that since we have $v \otimes v' = \sum_{i \in I}r_i e_i \otimes \sum_{i\in I} s_i e_i = \sum_{i\in I} s_i e_i \otimes \sum_{i \in I}r_i e_i = v' \otimes v$ then if we look at a particular element of the sum on LHS and RHS, namely $r_ie_i \otimes s_je_j = s_je_j\otimes r_i e_i$ then we can send the $r_i$ to the other side or do some manipulation of that sort and obtain the desired result.   I know this argument is not rigorous at all, so I am having a hard time doing it more rigorous (if this is a good approach) or actually attack this problem efficiently. I was wondering if this would be the right approach, like working with basis elements and so on or if there is some better way to look at this problem. Perhaps some use of universal property for modules would be helpful? Any help or suggestion with this problem is highly appreciated! Thanks so much!
79478
For a vector field $X$, the divergence in coordinates is given by $\nabla\cdot X=\sum_n\frac{X^i}{\partial x^i}$. In polar coordinates, the metric is $\begin{bmatrix}1 &amp; 0\\ 0 &amp; r^2\end{bmatrix}$, and so $\frac{1}{\sqrt{g(\frac{\partial}{\partial r},\frac{\partial}{\partial r})}}\frac{\partial}{\partial r}=\frac{\partial}{\partial r}$ and $\frac{1}{\sqrt{g(\frac{\partial}{\partial\theta},\frac{\partial}{\partial\theta})}}\frac{\partial}{\partial\theta}=\frac{1}{r}\frac{\partial}{\partial\theta}$ are unit vectors. Then for $X=X_{r}\frac{\partial}{\partial r}+X_{\theta}\frac{\partial}{r\partial\theta}$,  $\nabla\cdot X=\frac{\partial X_r}{\partial r}+\frac{\partial}{\partial\theta}\frac{X_{\theta}}{r}=\frac{\partial X_r}{\partial r}+\frac{1}{r}\frac{\partial X_{\theta}}{\partial\theta}$. But this disagrees with the usual formula given in vector calculus books. Does anyone see the error?
10958
I know by brutal calculation this identity holds always:  $$(u × v) \cdot (x × y) = \begin{vmatrix}  u \cdot x &amp; v \cdot x \\  u \cdot y &amp; v \cdot y \\  \end{vmatrix}$$  for arbitrary vectors $u$, $v$, $x$, $y$.  I'd like to know where it come from naturally.  Thank you.
68842
For an affine connection $\nabla$, prove the curvature R  $R(X,Y,Z,\alpha)=\alpha(\nabla_X \nabla_Y Z - \nabla_Y \nabla_X Z -\nabla_{[X,Y]}Z)$  with $X,Y,Z$ vector fields and $\alpha$ a co-vector, is a tensor.  So I realise the aim is probably to show that each of the three terms in the expression for $R$ are tensors themselves as then the result will obviously follow. However, I'm not too sure how to show each of these terms are tensors. Do I need to expand out each covariant derivate to get a bunch of connection coefficients and then see some cancellations? Still though, I'm not too sure what I'll want to see after all of this so that I can say "…therefore $R$ is a tensor".  Also, very much related to the question, I'm a bit confused with the difference between  $\nabla_Y Z$ $\space \space $ and $\space \space $  $\nabla_{\mu}\omega$.  The second one I know as the covariant derivative of a 1-form, but what is meant by the covariant derivate of a vector $\textit{field}$? As in, what's the difference between the $\nabla_Y$ and $\nabla_{\mu}$?  Edit: In answering this last question, I have realised that $\nabla_YZ=Y^{\mu}\nabla_{\mu}Z^{\nu}$ and $\nabla_{\mu}\omega=\nabla_{\mu}\omega_{\nu}$.
87366
This is probably a very naive question, but so far I could not find an answer:  Let $(M,g)$ be a Riemannian manifold. Can we always find "orthogonal coordinates" locally?  More precisely, I am asking if for every $p \in M$ there exists a neighbourhood $U$ and a diffeomorphism $\phi:\mathbb{R}^n \to U$, such that $g_{ij}=g(d\phi(e_i),d\phi(e_j))=0$ for $i \neq j$.  Clarification: Note that I want $g_{ij}=0$ on all $U$, not just at $p$. Also, I allow $g_{ii} \neq g_{jj}$ for $i \neq j$ (the special case where $g_{ii}$ is independent of $i$ is called isothermal coordinates-and corresponds to conformal flatness of $U$).  Of course, this is  weaker than requiring $M$ to be conformally flat, since a (linear) map which maps an orthogonal basis to an orthogonal basis does not need to be conformal.
98190
specify two vector and two scalar (differential) equations. That implies 8 components in the equations. But between vector fields $\vec{E}=(E_x,E_y,E_z)$ and $\vec{B}=(B_x,B_y,B_z)$, there are only 6 unknowns. So we have 8 equations for 6 unknowns. Why isn't this a problem?  As far as I know, the answer is basically because the equations aren't actually independent but I've never found a clear explanation. Perhaps the right direction is in  on arXiv.  Apologies if this is a repost. I found some discussions on PhysicsForums but no similar question here.
94262
We are supposed to prove that the zero vector is unique with the additive identity $x+O=x$ and then by creating another additive identity with $O'$, then $x+O'=x$.  We set them equal $x+O=x+O'$ and use the vector cancelling theorem to get $O=O'$, but how does this prove that the zero vector is unique?
105288
Here's an interesting "proof" that there is no such thing as magnetism. I know the answer but I love this so much I had to ask it here. It's a great way to confuse people!  As we all know, $$\nabla \cdot\vec{B} =0$$ Using the divergence theorem, we find $$ \iint_S \vec{B} \cdot \hat{n} \, dS = \iiint_V \nabla \cdot \vec{B} \, dV = 0$$ Since $\vec{B}$ has zero divergence, there exist a vector function $\vec{A}$ such that $$\vec{B} = \nabla \times \vec{A}$$ Combining the last two equations, we get $$\iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$ Applying Stokes' theorem, we find $$\oint_C \vec{A} \cdot \hat{t} \, ds = \iint_S \hat{n} \cdot \nabla \times \vec{A} \, dS = 0$$ Therefore, $\vec{A}$ is path independent and we can write $\vec{A} = \nabla \psi$ for some scalar function $\psi$. Since the curl of the gradient of a function is zero, we arrive at: $$\vec{B} = \nabla \times \nabla \psi = 0,$$ which means that all magnetic fields are zero, but that can't be!  Can you see where we went wrong?
78194
Define the Lie algebra $\mathfrak{g}$ of a Lie group G to be the set of all left-invariant vector fields of $G$. I want to prove that   $$          f: \mathfrak{g} \rightarrow T_eG \\              X \mapsto X(e) $$ is a linear isomorphism.  The only thing that I have left to prove is surjectivity.Letting $v\in T_eG$, we set $X\in \mathcal{X}(G)$ such that $\forall x\in G,\,X(x)=L_{x_{*,e}}(v)$ (I still have to show that such an $X$ is indeed a vector field on $G$). We get that  $$ f(X)=X(e)=L_{e_{*,e}}(v)=v $$  where the last equality holds because  $$       L_{e_{*,e}}(v)f = v(f \circ L_e) = v(f \circ id_G) = v(f) $$ for any $f\in C^\infty(G)$. (I am thinking of the tangent vectors as derivations).  Now all we have left to do is prove that such an $X$ is left-invariant, and it is therefore an element of $\mathfrak{g}$, but I do not know how. I tried to prove that  $$            \forall f\in C^\infty(G),\, L_{{x}_*,y}(X(y))f=X(L_x(y))f $$ but with no success.