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4

First: The two-sphere is actually not the only surface which admits a metric with constant positive curvature. There's also the real projective plane which is $S^2/\{\pm 1\}$, the two-sphere with antipodal points identified. Second: There is a nice formula relating the area, the curvature, and precisely which (topological) type of surface you're working ...


4

Yes, it's correct, assuming to be precise that what you mean by $``R_1 + R_2"$ is $$ R(X,Y)Z = (R_1(X_1, Y_1) Z_1, R_2(X_2, Y_2)Z_2). $$ I imagine there's some elegant way to prove this, but it certainly works to just do a bunch of silly calculations; in particular, you can show that for product vector fields $(X_1, X_2)$ and $(Y_1, Y_2)$, the ...


3

It is not invertiable at least when the dimension $n$ of $M$ is two. Take $h = g$. Then $$(Gh)_{ij} = (Gg)_{ij} = g_{ij} - \frac{1}{2} g^{kl}g_{kl} g_{ij} = g_{ij} - \frac{n}{2} g_{ij} = 0 . $$


2

For dimension $n>2$ the inverse of $ y=Gh=h-\frac{1}{2}(\text{tr}_gh)g $ is $$ x=G^{-1}y=y-\frac{1}{n-2}(\text{tr}_gy)g $$ since \begin{align} G^{-1}y&=G^{-1}\Big(h-\frac{1}{2}(\text{tr}_gh)g\Big)\\ &=h-\frac{1}{2}(\text{tr}_gh)g-\frac{1}{n-2}\Big(\text{tr}_gh-\frac{n}{2}\text{tr}_gh\Big)g\\ &=h \end{align} and \begin{align*} ...


2

Setting $k=Gh$, see if you can write $\mathrm{tr}_gk$ in terms of $\mathrm{tr}_gh$ (and thus vice versa). You should find a relatively clean expression, and from there it's just simple algebra: suppose $\mathrm{tr}_gh = f(\mathrm{tr}_gk)$ for some $f$. Then $k_{ij}=h_{ij}-\frac12f(\mathrm{tr}_gk)g_{ij}$, so $h_{ij}=k_{ij}+\frac12f(\mathrm{tr}_gk)g_{ij}$.


2

If $\{e_i\}$ is a given frame then there are functions such that $$\nabla_{e_i}e_j=\alpha_{ij}^ke_k,\quad [e_i,e_j]=c_{ij}^ke_k.$$ Then, $$T(e_i,e_j)=\nabla_{e_i}e_j-\nabla_{e_j}e_i-[e_i,e_j]=\alpha_{ij}^ke_k-\alpha_{ji}^ke_k-c_{ij}^ke_k.$$ That is, $$T=(\alpha_{ij}^k-\alpha_{ji}^k-c_{ij}^k)e^i\otimes e^j \otimes e_k.$$ Note that since $T$ is a tensor ...


2

Let's make a more detailed analysis of what we mean by vector, matrix and tensor. To keep things simple, let's say we are working with vector spaces over $\mathbb R$, and with finite-dimensional vector spaces. A vector is an element of a vector space. Note that a vector space $V$ does not need to be "a tuple $(x_1,x_2,\dots,x_n)$ where $x_i \in \mathbb ...


1

Tensors have "ranks". A rank 0 tensor is just a number, a rank 1 tensor is a vector, a rank 2 tensor is a matrix (vector of vectors), a rank 3 tensor is a vector of vectors of vectors, and so on.


1

Formally, a tensor is a multilinear, real-valued function of vectors; this, I find, is the best guide to intuition. For it means that once you can identify some collection of objects as having the properties of a vector space, then, to find a tensor, all you need to do is identify linear operations that can be performed on these vectors, where these ...


1

There is some value of $\sigma_{22}$ for which t=(0,0,0) on some (unknown) plane. Since $[t]=[\sigma][n]$, this is equivalent to the condition that the homogeneous linear equation $[\sigma][n]=[0]$ has a non-trivial solution. This occurs when $\det[\sigma]=0$. From the component form, evaluating the determinant gives you $0 + 4-\sigma_{22}+4=0$, so the ...


1

Let's be very general. Assume we are taking the metric tensor of the given point $g = g_{ij}(x)$. Assume everything in arbitrary coordinate system. Assume $M$ to be a arbitrary differentiable manifold, for an arbitrarily given metric tensor $g$. For a given vector x, y: $$ x\cdot y = g_{ij}(x)x^i y^j $$ For a commutative inner product of vectors implies a ...


1

Hint: Take a linear combination $L=k_1a+k_2b+k_3c+k_4d$ and evaluate at a vector $X$ for which $a(X)=1$ , $b(X)=0$ , $c(X)=0$ and $d(X)=0$.


1

This is not how one would usually define the rank of an element of the tensor algebra. The tensor algebra is the direct sum $$\bigoplus_{n=0}^{\infty}{V^{\otimes n}}$$ where $V^{\otimes n}$ is $V$ tensored with itself $n$ times. The rank of an element would usually be defined as the largest integer $N$ such that if we write the element as a sum ...


1

You can do that easily with bsxfun: sum(exp(bsxfun(@times, permute(A, [1 3 2]), permute(B, [3 2 1]))), 3)


1

Certain it is false in general. (Assuming that the Einstein summation convention is being used,) $$\partial_j (g^{ks} g_{is}) = \partial_j \delta^k {}_i = 0,$$ but $g^{ks} \partial_j g_{is}$ is not always zero (if it were, we wouldn't bother writing those terms in the usual coordinate formula for the Christoffel symbol). In particular, the operations of ...


1

Notice the repeated index $k$, which implies summation. Thus, the expression $-0.5 E_k E_k \delta_{ij}$ is $-0.5 (E_1^2 + E_2^2 + E_3^2) \delta_{ij}$ = $-0.5 E^2\delta_{ij}$. This contributes $-0.5 E^2$ to the diagonal elements (where i=j and $\delta_{ij}=1$), and 0 to the off-diagonal elements. $E_i E_j$ is non-zero only in the (1,1) position of the ...



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