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3

Notice that $\gamma' \equiv 0 \iff \| \gamma'(t) \| = 0 \ \forall t$, that is, $\langle \gamma'(t), \gamma'(t) \rangle = 0 \ \forall t$. We have: $$\langle \gamma'(t), \gamma'(t) \rangle = (f'(t) \cos t)^2 - 2 f'(t)f(t) \cos t \sin t + (f(t) \sin t)^2 + (f'(t) \sin t)^2 + 2f'(t)f(t)\cos t \sin t + (f(t) \cos t)^2$$ which happily simplifies to: $$\langle ...


3

The Einstein tensor is defined as $$G_{\mu\nu} = R_{\mu\nu} - \frac{1}{2}g_{\mu\nu}R$$ so by taking the variation we find $$\frac{\delta G_{\mu\nu}}{\delta g_{\alpha\beta}} = \frac{\delta R_{\mu\nu}}{\delta g_{\alpha\beta}} - \frac{1}{2}\delta_{\mu}^{\alpha}\delta_{\nu}^{\beta}R - \frac{1}{2}g_{\mu\nu}\frac{\delta R}{\delta g_{\alpha\beta}}$$ Now $R = ...


2

Hint: Write down $G_{ab}=Rc_{ab}-\frac{R}{2}g_{ab}$ in terms of the metric (you can express both the Ricci tensor and the scalar curvature in terms of the Christoffel symbols, and those can be written as functions of the metric), then differentiate. Probably it's better if you subdivide the problem into smaller pieces, though.


2

Hausdorff dimension is not a topological invariant. But saying that $M$ is a differentiable manifold gives you more than just a homeomorphic embedding. The charts determine a metric on each patch, in which the patch has Hausdorff dimension $n$. And diffeomorphisms are bi-Lipschitz, so they preserve Hausdorff dimension.


2

Parsing the much-maligned notation yields $$\sum_{i,j} a_{ij} (x_i + y_j) \ne \sum_i a_{ij}x_i + \sum_j a_{ij}y_j$$ which is an evident inequality.


2

The first coordinate chart $\sigma_+^z$ parametrizes the hemisphere $H_+^z = \{z > 0\}$, and the second one $\sigma_+^x$ parametrizes the hemisphere $H_+^x = \{x > 0\}$. The inverse of $\sigma_+^z$ on $H_+^z$ is $$(\sigma_+^z)^{-1}(x,y,z) = (x,y)$$ ("solve for the variables", right?). From this you can compute the transition map to be ...


1

It is a matter of convention, and some authors do not follow it. If $Z$ is a $(1,0)$ vector field, then $IZ = iZ$. If $\alpha$ is a $(1,0)$-form, then one has $I^* \alpha = i \alpha$, where $I^*(\alpha)(-) = \alpha(I-)$ is the (pointwise) pullback of $\alpha$ by $I$. Ok, so one can, say, define $I$ as $I^*$. But then you have a rather annoying sign showing ...


1

Assume that the variety $X$ and the curve $C$ have degree $a,b$ respectively. Take $N = ab$. Now, if you know that $X$ and $C$ intersect in $k$ points counted with multiplicity with $k >N$ then, by B├ęzout theorem you get $C\subseteq X$. For instance, assume $X\subset\mathbb{P}^n$ is variety of degree three with two singular points of multiplicity two. ...


1

I will assume $x=(x_i)$ to be a $1\times N$ (row) vector, $y=(y_i)$ an $N \times 1$ (column) vector, and $A=(a_{ij})$ an $N\times N$ matrix. Then the terms on the RHS correspond to the matrix multiplications $x_i a_{ij} =(x A)_j$ and $a_{ij}y_j = (A y)_j$. Note that these are the components of a row vector and a column vector respectively. What about the ...


1

No. In the second line, you've incorrectly used the summation convention. The upper $a$ in the first line is a dummy index, already contracted with a lower $a$, so you can't introduce another $a$ into the formula.


1

If $x_i = y_i$ for all $i = 1, \dots, n$, then $a_{ij} x^i y^j = a_{ij} y^i y^j = a_{ij}y^i x^j = a_{ij} x^j y^i.$


1

You should maybe think of transition functions as "gluings". You're gluing smooth open sets together, so if you want the manifold you obtain to be smooth, the gluings should happen smoothly. Here is the more useful way to see this when you actually work with an atlas. The argument is always the same: such property works when I look at it using some chart of ...


1

It is not a matter of notation; it is a matter of convention. The way a differential form integral is written can be understood as a shorthand. For instance, the integral $$\int f \, \mathrm dx \wedge \mathrm dy$$ really means this: $$\int f(x,y) \, (\mathrm dx \wedge \mathrm dy)(e_x \wedge e_y) \, dx \, dy$$ where $e_x$ and $e_y$ are the tangent ...


1

You have to understand that the mass of hollow cone is only at the surface(we consider it's surface to be of thickness $t$ where $t\to0$ because we take product of two infinitesimals $dxdy\to0$) so we can't take vertical height and neglect the surface mass(when taking vertical height neglection is of a thin cone whose thickness $t\to0$) which we do in solid ...


1

This is a nice theorem from differential geometry. Such a diffeomorphism around $0$ will exist if and only if the Lie bracket $[V_1,V_2]=0$.


1

If $\sin u,\sin 2u\geq 0$ we must have $u\in[0,0.5\pi]$. If $\sin u\geq0\geq\sin2u$ we have $u\in(0.5\pi,\pi)$. Similarly, if $\sin u,\sin 2u\leq 0$ we must have $u\in[-0.5\pi,0]$ and so on. So the inverse appears to be given as $$ X^{-1}(x,y,z)= \begin{cases} (\sin^{-1}x,z) &\text{ if }x\cdot y\geq 0\\ (\operatorname{sgn} x\cdot\pi-\sin^{-1}x,z) ...


1

A vector field $X$ along a curve $\alpha$ is parallel if $$\nabla_TX=0$$ This equation means that the vector field $X$ does change along $\alpha$. Geometrically, all values of $X$ along $\alpha$ seems to be parallel.



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