Matt
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Let $(M,g)$ be a Riemannian manifold with associated Levi-Civita connection $\nabla$. Typically one introduces the covariant derivative on vector fields, i.e., $\nabla:\Gamma(TM)\times\Gamma(TM)\to\... View answer 4 votes First of all, Lie derivatives are defined on any smooth manifold, so the Riemannian structure is superfluous. Let$T$be a tensor field and$X$a vector field. If you wish to look at the Lie ... View answer 4 votes Using the definition of the Riemann curvature tensor $$R_{XY}Z=\nabla_{[X,Y]}Z-[\nabla_X,\nabla_Y]Z,$$ we can compute that $$R^a_{\;bcd}=\partial_d\Gamma^a_{cb}-\partial_c\Gamma^a_{db}+\Gamma^a_{de}\... View answer Accepted answer 3 votes First, I think you have some notation backwards. We typically use parentheses for symmetrization and square brackets for antisymmetrization, i.e,$$T_{(ij)}=\frac{1}{2}(T_{ij}+T_{ji}),\qquad T_{[ij]}=... View answer Accepted answer 3 votes Consider $$x_n=\left(0,\frac{\pi}{n}\right).$$ View answer Accepted answer 2 votes I'm not entirely sure I followed what your issue is, but it seems to be notational in nature, so perhaps the following exposition will be helpful. Let$V$be a real linear space with associated dual$...

I find this an interesting topic which is often ignored in certain Riemannian texts when introducing variational methods when dealing with energy and length minimization. Some of the rigorous ...

By the Regular Level Set Theorem, since $y$ is a regular value of $f:X\to Y$, we have that $f^{-1}(y)$ is smooth embedded submanifold of $X$ of codimension (in $X$) equal to the dimension of $Y$. ...

I'll do (A). Recall, that in coordinates, we have that $$\Delta_g u=\frac{1}{\sqrt{|g|}}\partial_a\left[g^{ab}\sqrt{|g|}\partial_b[u]\right],$$ where $|g|=\det(g_{ab})$. Note for this case $g_{ab}=e^... View answer Accepted answer 1 votes This is equivalent to showing the sphere bundle$\pi:SM\to M$is compact for compact manifolds$M$, and your case is just looking at the co-sphere bundle of radius$\sqrt{2\lambda}$, which are ... View answer Accepted answer 1 votes Let$G$be a Lie group and$X$a vector field. For$g\in G$, let$L_g:G\to G$denote left-multiplication, i.e.,$L_g(p)=gp$. Then we say that$X$is left-invariant if $$(L_g)_*X=X.$$ Note that the ... View answer 1 votes I can see the confusion in the notation, so let's change things up a little bit to try and make it clearer. First let's note what our coordinates are: For convenience, all indices range from from$...
Suppose $F^*(C^\infty(N))\subseteq C^\infty(M)$, that is, for any $g\in C^\infty(N)$, we have that $g\circ F\in C^\infty(M)$. Fix $p\in M$, and let $(\psi:W\subseteq N\to\tilde{W}\subseteq\mathbb{R}... View answer Accepted answer 1 votes It's possible I'm not understanding your question correctly, but let's think of this (more general) approach: Let$\alpha:I\to M$be an curve on some Riemannian manifold$(M,g)$(this could be your ... View answer Accepted answer 1 votes What you're looking for is the raising/lowering of indices, and the symmetrization/antisymmetrization of tensors. I'm not certain as to why you have the vector$V$, but I'm going to assume a follow ... View answer 0 votes Suppose$f:\mathbb{R}\to\mathbb{R}$,$f(x)>0$for all$x\in\mathbb{R}\setminus\{x_0\}$and $$\lim_{x\to x_0}f(x)=0.$$ Let$N\in\mathbb{N}$, and let$\delta>0$be such that$$|f(x)-0|<\frac{1}{... View answer 0 votes The manifold series of books by John Lee (Topological, Smooth, and Riemannian). Petersen's Riemannian geometry is a good one to go along with it as well. For the more advanced reading, Sakai or ... View answer 0 votes For finding$d(\det)_A$, let's use what you've already computed. Given$B\in M_n(\mathbb{R})$, we have the curve$c(t)=A\exp{tA^{-1}B}$with$c(0)=A$and$c'(0)=B. Thus \begin{align*} d(\det)_A(B)&... View answer 0 votes The function is even, i.e.,f(-x)=f(x)$(odd means that$f(-x)=-f(x)$which yields rotational symmetry about the origin, which does not exist in this graph), which tells us that$b=d=0$(the even ... View answer 0 votes The derivative of a linear map is itself. Consider$A:\mathbb{R}^n\to\mathbb{R}^n$,$x\mapsto Ax$, by matrix multiplication. Then the derivative of$A$at the point$x$, is the unique linear map$...