Well Ricci's theorem is given by:  $$\mathrm{D}g_{ij}=\mathrm{D}g^{ij}=0$$ I was wondering that if the theorem can be proved using covariant derivatives of $\delta_i^k$, $g_{ij}$ and $g^{ik}$.  I really need this proof.
95562
I've consulted several books for the explanation of why   $$\nabla _{\mu}g_{\alpha \beta} = 0,$$  and hence derive the relation between metric tensor and affine connection $\Gamma ^{\sigma}_{\mu \beta} $  $$\Gamma ^{\gamma} _{\beta \mu} = \frac{1}{2} g^{\alpha \gamma}(\partial _{\mu}g_{\alpha \beta} + \partial _{\beta} g_{\alpha \mu} - \partial _{\alpha}g_{\beta \mu}).$$  But I'm getting nowhere. May be I've to go through the concepts of manifold much deeper.
67337
Let $G$ be a finite group, and $g$ be a an element of $G$. How could we go about proving $g^{|G|}=e$ without using cosets? I would admit Lagrange's theorem if a proof without talking about cosets can be found.  I have a proof for abelian groups which basically consists in taking the usual proof of Euler's theorem and using it in a group, I do not know if it can be modified to work in arbitrary finite groups.  The proof is as follows: the function from $G$ to $G$ that consists of mapping $h$ to $gh$ is a bijection. Therefore   $\prod\limits_{h\in G}h=\prod\limits_{h\in G}gh$ but because of commutativity $\prod\limits_{h\in G}gh=\prod\limits_{h\in G}g\prod\limits_{h\in G}h=g^{|G|}\prod\limits_{h\in G}h$.  So we have $\prod\limits_{h\in G}h=g^{|G|}\prod\limits_{h\in G}h$.  The cancellation property yields $e=g^{|G|}$.  I am looking for some support as to why it may be hard to prove this result without talking about cosets, or if possible an actual proof without cosets.    Thank you very much in advance, regards.
122758
I tried looking it up but many websites just state it without proof and without intuition.  I'm hoping to learn it a little bit better so that I don't forget how to compute the Jacobian when working with surface integrals where the divergence theorem is not applicable.  If you have a good online reference instead, please feel free to provide it :-)  Thanks,
17125
Let $H$, $K$ be subgroups of $G$. Prove that $o(HK) = \frac{o(H)o(K)}{o(H \cap K)}$.  I need this theorem to prove something.
95048
This may be more of a math question, but I am trying to prove that for a function $f(q,\dot{q},t)$ $$\frac{d}{dt}\frac{∂f}{∂\dot{q}}=\frac{∂}{∂\dot{q}}\frac{df}{dt}−\frac{∂f}{∂q}.\tag{1}$$ As  part of this problem, I have already proved that $$\frac{d}{dt}\frac{\partial{f}}{\partial{q}}=\frac{\partial{}}{\partial{q}}\frac{df}{dt}\tag{2}$$ through the use of a total derivative expansion for $\frac{df}{dt}$, however I can't seem to figure out where the additional $-\frac{\partial{f}}{\partial{q}}$ term comes from for equation (1).
94654
Can I reconstruct a Riemannian metric out of its Levi-Civita connection? In other words: Given two Riemannian metrics $g$ and $h$ on a manifold $M$ with the same Levi-Civita connection, can I conclude that $g=h$ up to scalars?  If not, what can I say about the relationship between $g$ and $h$? How rigid is the Levi-Civita-Connection?
66682
The derivative of a map $F$ between manifolds $M$ and $N$ is defined by $$F_*X(f)= X(f \circ F)$$ where $X \in T_P(M)$, the tanget space at the point $P$.  We know that $$\left\{\frac{\partial}{\partial x^i}\bigg|_P\right\}_i$$ is a basis for $T_P(M)$. How to show that $$\left\{dx^i\bigg|_P\right\}_i$$ is a basis for the cotangent space $(T_P(M))^*$?  First, by $dx^i$, I guess we mean the derivative of the map $x^i$ as defined above, right? Is this map $x^i$ just picking out the ith coordinate? Secondly, to show that it is a basis, we need to show that $dx^i\left(\frac{\partial}{\partial x^j}\bigg|_P\right) = \delta^i_j.$ Where to go from here: $$\underbrace{(dx^i)_P}_{(\Phi_*)_P}\underbrace{\left(\frac{\partial}{\partial x^j}\bigg|_P\right)}_{X}f = \left(\frac{\partial}{\partial x^j}\bigg|_P\right)(f\circ x^i)?$$  I can use the chain rule but I am not sure exactly. Please help.
97145
How to prove if $u\in W^{1,p}$, then $|u|\in W^{1,p}$?  Since $|u|\in L_p$, I only need to show weak derivative of $|u|$ exists and $D|u| \in L_p$.  Can anyone give me some hint? Thanks!
58004
I would like to know if my proof of the following statement is correct     If $M$ is a compact manifold, then every vector field $X$ over $M$ is complete.   Proof:  I take $p\in M$ and $(\gamma_p,I_p)$ a maximal integral curve through $p$. $I_p$ is open and non empty so if $I_p$ is closed, it is equal to $\mathbb{R}$.  So take $s\in \text{Adh}(I_p)$. There exists $t_i\rightarrow s, t_i\in I_p$. Define $q$ as $\text{lim}(\gamma_p(t_i))$ and take a maximal integral curve passing by $q$ as $(\gamma_q,I_q)$. Then since the vector fields are $C^\infty$, $\gamma_q(I_q)\cap\gamma_p(I_p)$ is nonempty (intuitively, this is clear but don't really know how to show it).  On the intersection, $\gamma_p$ and $\gamma_q$ coincide. So we can extend the maximal curve $\gamma_p$ to include $\gamma_q$. This implies that $q\in\gamma_p(I_p)$ and is actually equal to $\gamma_p(s)$  I think the main idea is there even though it probably lacks a bit of rigour.
116076
I need to prove that if $f$ and $g$ are analytic in $D_r(z_0)$ and $g$ has a simple zero at $z_0$, then $$\mathrm{Res}[f/g,z_0]=\frac{f(z_0)}{g'(z_0)}$$   When $f(z_0)\neq 0$ and since $1/g$ has a simple pole at $z_0$ $$\mathrm{Res}[f/g,z_0]=\lim_{z\to z_0}(z-z_0)\frac{f(z)}{g(z)}=\frac{f(z_0)}{g'(z_0)}$$ But what do I need to do when $f(z_0)=0$? Any help will be appreciated thanks
56823
I'm having trouble solving the following problem.  If $F:\mathbb{R^3} \to \mathbb{R^3}$ is a diffeomorphism such that $F\ast$(the tangent map of $F$) preserves dot products, show that $F$ is an isometry. (Hint: Show that $F$ preserves lengths of curve segments and deduce that $F^{-1}$ does also.)  I'm trying to proceed as in the hint, but I'm stuck on the first step, showing the $F$ preserves lengths of curve segments. Moreover, how does showing that both $F$ and its inverse preserves length of curve segments show that it is an isometry. And how do I characterize the curve segments. I would appreciate any solutions, hints or suggestions.
74825
Symmetry of second derivative states that for $u=u(x,y)$ if $u_x,u_y$ exists and $u_{xy},u_{yx}$ exists and continuous then $u_{xy}=u_{yx}$.  I proved that statement using the mean value theorem.   While I was looking in Wikipedia there is a section called "Sufficiency of twice-differentiability", if $u(x,y):E^{\text{open set}}\subset \Bbb R^2\to \Bbb R$ and $u_x,u_y,u_{yx}$ exists everywhere and $u_{yx}$ is continuous at a point in $E$ then $u_{xy}$ exists at that point and equal to $u_{yx}$.  My question is, while I proved Symmetry of second derivative I had to assume the continuity of $u_{xy}$ and $u_{yx}$, so how can I prove that this is true without even assuming the existence of one of the second derivative? I'm sitting on this for a long time and I couldn't think on any starting point and would love help with this.
51233
So, given the generalized Stokes theorem:  $$\int_{\partial M} \omega = \int_M d\omega$$  where M is an n-dimensional surface and $\omega$ is a p-form on M (p &lt; n). How can I derive the Divergence Theorem?  $$\iint_S {\bf F} \cdot d{\bf S} = \iiint_R \text{div}\;{\bf F}\; dV$$  I also have another related question. I'm learning that there are several theorems, like the divergence theorem, that are special cases of the generalized Stokes Theorem. For example, apparently, the Kelvin-Stokes Theorem is a special case of the General Stokes Theorem where n=2. So my 2nd question is, what if n=1 in the general stokes theorem? What does that imply or lead to?  Thank You.
97255
I heard that using some relatively basic differential geometry, you can show that the only spheres which are Lie groups are $S^0$, $S^1$, and $S^3$.  My friend who told me this thought that it involved de Rham cohomology, but I don't really know anything about the cohomology of Lie groups so this doesn't help me much.  Presumably there are some pretty strict conditions we can get from talking about invariant differential forms -- if you can tell me anything about this it will be a much-appreciated bonus :)  (A necessary condition for a manifold to be a Lie group is that is must be parallelizable, since any Lie group is parallelized (?) by the left-invariant vector fields generated by a basis of the Lie algebra.  Which happens to mean, by some pretty fancy tricks, that the only spheres that even have a chance are the ones listed above plus $S^7$.  The usual parallelization of this last one comes from viewing it as the set of unit octonions, which don't form a group since their multiplication isn't associative; of course this doesn't immediately preclude $S^7$ from admitting the structure of a Lie group.  Whatever.  I'd like to avoid having to appeal to this whole parallelizability business, if possible.)
15566
This suggests, as a shortcut notation, the concept of lowering indices; from any vector we can construct a (0, 1) tensor defined by contraction with the metric: $$A_\nu ≡ g_{\mu\nu}A^\mu$$  so that the dot product becomes $$g_{\mu\nu}A^\mu B^\nu = A_\nu B^\nu$$  We also define the inverse metric $g^{\mu\nu}$ as the matrix inverse of the metric tensor: $$g^{\mu\nu}g_{\nu\rho} = δ^\mu_\rho$$  where $δ^\mu_\rho$ is the (spacetime) Kronecker delta. (Convince yourself that this expression really does correspond to matrix multiplication.) Then we have the ability to raise indices: $$A^\mu = g^{\mu\nu}A_\nu$$ Note that raising an index on the metric yields the Kronecker delta, so we have $$g^{\mu\nu}g_{\mu\nu} = δ^\mu_\mu = 4$$  I don't understand why $δ^\mu_\mu = 4$. Shouldn't it equal 1?
67880
I'm currently working through the book Heisenberg's Quantum Mechanics (Razavy, 2010), and am reading the chapter on classical mechanics. I'm interested in part of their derivative of a generalized Lorentz force via a velocity-dependent potential.   I understand the generalized force   $$F_i = -\frac{\partial V}{\partial x_i} + \frac{d}{dt}\left(\frac{\partial V}{\partial v_i}\right)$$  that they derive from a Lagrangian of the form $L = \frac{1}{2}m|\vec v|^2 - V(\vec r,\vec v,t)$.  However, in the next (critical) step of the derivation, the author cites a theorem from Helmholtz saying      ...according to Helmholtz, for the existence of the Lagrangian, such a generalized force can be at most a linear function of acceleration, and it must satisfy the Helmholtz identities.   The three Helmholtz identities are then listed as:  $$\frac{\partial F_i}{\partial \dot{v_j}} = \frac{\partial F_j}{\partial \dot{v_i}},$$  $$\frac{\partial F_i}{\partial v_j} +  \frac{\partial F_j}{\partial v_i} = \frac{d}{dt}\left(\frac{\partial F_i}{\partial \dot{v_j}} +  \frac{\partial F_j}{\partial \dot{v_i}}\right),$$  $$\frac{\partial F_i}{\partial x_j} -  \frac{\partial F_j}{\partial x_i} = \frac{1}{2}\frac{d}{dt}\left(\frac{\partial F_i}{\partial v_j} - \frac{\partial F_j}{\partial v_i}\right).$$  I'm trying to understand where this theorem comes from. Razavy cited a 1887 paper by Helmholtz. I was able to find , but it is in German, so I could not verify whether or not it proved the theorem. Additionally, I could not find it in any recent literature. I searched online and in Goldstein's Classical Mechanics.   The only similar concept that I can find is in the  where we have three equations known as Helmholtz conditions. Are these two concepts one in the same? If so, how should I interpret the function $\Phi$ and the matrix $g_{ij}$ that appear in the Helmholtz conditions I found online?   If the cited theorem from Razavy does not relate from the inverse Lagrangian problem, could I have some help finding the right direction?
98683
If I have a manifold $M$ and $\omega \in \Omega^q(M)$, considered an open subset $U$ of $M$, and choose a point $p \in M$, is it true that $(\omega|_{U})_p=0 \rightarrow \omega_p=0$? With $\omega|_{U}$ I mean the pull-back of $\omega$ by the embedding $F:U\to M$. The converse is obviously true, but this implication?