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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 ...

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 ...

0

Thanks everybody who tried to help! A more thorough discussion took place here One of the answers in the link was already mentioned by Luis, but has the drawback that it blows up memory use when A and B are large. Say A is MxK, and B is KxN, the intermediate matrix created by bsxfun is MxKxN in size. We can keep it to MxN using a for loop, which according ...

0

It means "repeat the terms I've written, but with these indices swapped". So in your case, $$i\hbar\Big(g^{\mu\rho}M^{\nu\sigma}-(\mu\leftrightarrow\nu)\Big)-(\rho\leftrightarrow\sigma)$$ is short for $$i\hbar\Big(g^{\mu\rho}M^{\nu\sigma}-g^{\nu\rho}M^{\mu\sigma}\Big)-i\hbar\Big(g^{\mu\sigma}M^{\nu\rho}-g^{\nu\sigma}M^{\mu\rho}\Big).$$

1

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

0

No, but you'll need a 3D array (or something equivalent to it). If you put $A_{ik}B_{kj}$ into location $(i + (nm) j, k)$ of $Q$, where size(A) = [n, p] and size(b)= [p, m], then you can compute $e^Q$ and sum over columns. You can construct $Q$ from some combination of "repmat" and "diag", I'll bet...but the code will sure be opaque. :( And here's the ...

0

The answer in general is not true, because the statement of the skew-symmetry of the last two indices of the Riemann tensor is true if we have metric compatibility. If the covariant derivative of the metric is non-zero it doesn't have to be true. And you can find a question in Hughston - "An Intro to GR" (question 6.3) "show that Ricci tensor is not ...

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

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*} ...

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

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 $h_{ij}=k_{ij}+\frac12f(\mathrm{tr}_gk)g_{ij}$.

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 ...

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

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 $d(X)=0$.

0

In 2) you have how much is the value of $T$ at the arguments $v_{\mu_1},...,V_n$.

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

$\DeclareMathOperator\Sq{Sq}$Yes, that's what it means here. The algebra $\mathcal{A}_2$ is the quotient of the free tensor algebra generated by $\Sq^1, \Sq^2\dots$ quotiented by the ideal generated by the Adem relations. Similarly for $\mathcal{A}_p$, $p>2$. However be careful with the admissible thing. What Hatcher shows is that every element of the ...

1

$$U_{ij} = g_{ij} + \epsilon_{ijk}u^k\tag{1}$$ $$(U^{-1})^{jl} = Ag^{jl} + Bu^ju^l + C\epsilon^{jlm}u_m.\tag{2}$$ $$\delta_i^l=U_{ij} (U^{-1})^{jl} = g_{ij}Ag^{jl} + g_{ij}Bu^ju^l+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{3}$$ $$\delta_i^l= g_{ij}Ag^{jl} + B(g_{ik}u^k)(g^{lm}u_m)+ \epsilon_{ijk}u^kC\epsilon^{jlm}u_m.\tag{4}$$ $$\delta_i^l= A g_{i}^l + ... 2 Yes. Because$$I=\epsilon_{ijk}A^{ijk}=-\epsilon_{jik}A^{ijk}=-\epsilon_{jik}A^{jik}=-I. So $I = 0$

1

If you fix one of the indices of $\varepsilon_{ijkl}$ to be $4$ you get $\pm\varepsilon_{ijk}$ depending on wheather you fix an odd or an even positioned index. So $\varepsilon_{ijk4}=\varepsilon_{ijk}$ but $\varepsilon_{ij4k}=-\varepsilon_{ijk}$. To see why the signs come out this way, notice that when substitute $1,2,3$ for $i,j,k$ you get: ...

1

Note that the permutations corresponding to $(i,j,k,4)$ and $(i,j,4,k)$ differ only by a transposition of the last two indices. Consequently, they have different parity and so their Levi-Civita symbols have opposite signs (assuming they do not vanish, of course). Hence the correct statement is $\epsilon_{ijk4}A^{jk}=-\epsilon_{ij4k}A^{jk}$.

4

You need to be careful about what space you're taking the adjoint in. In the space of all square-integrable $(0,2)$-tensors, your calculation correctly shows that the adjoint of $-\text{div}$ is indeed $\nabla$. I think you want to be working with the space of symmetric trace-free $(0,2)$-tensors - $K$ maps into this space (possibly after tweaking the ...

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