Hmm let me try and be more formal. So pick $p \in \mathbb{R}^n$, then we have the tangent space at $p$ which is $T_p(\mathbb{R}^n)$ as usual with basis $$\left\{\frac{\partial}{\partial x^1}\bigg|_p, \dots, \frac{\partial}{\partial x^n}\bigg|_p\right\}$$. Now let $\varphi : \mathbb{R}^n \to \mathbb{R}^n$ be a vector space isomorphism, what I want to know is that do we need $\varphi$ to be a diffeomorphism in order for the tangent space at $T_{\varphi(p)}$ to have basis
Basically my lecturer did the following : 1. wrote new co-ordinates as functions of old co-ordiantes 2.applied chain rule 3. summed over something 4. out popped a jacobian
and $T\varphi(dy_i) = \sum_j d\varphi_i/dx_j dx_j$ and I'm probably missing a star somewhere because I never know if it's in a superscript or lowerscript
so tangent bundle is a covariant functor and cotangent bundle is contravariant
Hello, does anyone here know about semi direct products? I've noticed the statement "U(n) is isomorphic to semi direct product of SU(n) and U(1)"
But I thought if something is a semidirect product of N and K, then N and K must both be subgroups of the same group? Which isnt true for SU(n) and U(1)? What is wrong here?
I see it says "given any (even unrelated) groups N and H", but I thought it was meant in a sense as in "doesnt need to have a trivial intersection". The sentence before that one (and the first sentence in outer semi direct product section) says they need to be subgroups of the same group
I should define $J$ as the matrix you get when you smash the tangent space basis at $x$ into the transpose of the cotangent space basis at $\varphi(x)$
if you use that $(d/dxi) = J (d/dyi)$ you get $(d/dxi) t(dyj)= J (d/dyj) t(dyj) = J$ and if you use that $(dyj) = tJ (dxi)$ you also get $(d/dxi) t(dyj)= (d/dyj) t(dyj) J = J$
I'm not sure I'm helping you should look into a real textbook written by somewhat who knows what he is doing
but yeah elements of a tangent space are "meant" to be applied to elements of the cotangent space to give a number
with the canonical basis, if you take all the pairs, you get the identity matrix
and if you assume that $\varphi$ induces a covariant map between tangent spaces and a contravariant map between cotangent spaces, well this calculation says that the matrices have to be transpose of each other in order for things to make a bit of sense
anf if yo uwant to do a partial differentiation
er
to apply an element of the tangent space to a function
Find the field inside and outside a sphere of radius $R$, which carries a uniform volume charge density $\rho$.
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We have $$\frac{dq}{dV}=\rho $$. Then by symmetry, $$d\vec{E}=\frac{1}{4\pi \epsilon_0}\frac{\rho (z-r\cos \theta) dV}{(r^2+z^2-2rz\cos \theta)^{3/2}}\hat{z}$$ $$\vec{E}=\frac{1}{4\pi \epsilon_0}\int_{r=0}^{R}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\frac{\rho (z-r\cos \theta) dV}{(r^2+z^2-2rz\cos \theta)^{3/2}}\hat{z}$$
nice lemma I have never seen before: if $g$ is an increasing function on $[0, \infty)$ tending to $\infty$, and $\epsilon > 0$ is fixed, then $g'(x) \leq g^{1+\epsilon}(x)$ for all but a set of finite measure
proof: Let $E$ be the set where this inequality is in the other direction. Then $$\int_E dx \leq \int_E \frac{g'(x)}{g^{1+\epsilon}(x)} dx.$$ Applying the substitution $x = g(u)$ (this substitution makes sense because $g$ covers the whole of $[0, \infty)$ and is monotonic), we get that this integral is $\int_{g^{-1}(E)} \frac{du}{u^{1+\epsilon}}.$ This integral over the whole real line is finite.
If $x \in \overline{A}$, how do I show that $A$ contains a net converging to $x$? I know I have to consider the directed set $J = \{\alpha \subseteq X \mid \alpha \mbox{is a nbhd of } x\}$, but I can't seem to get the proof to work. If $\alpha$ is a nhbd of $x$, then there exists $x_{\alpha} \in A \cap \alpha$, so $(x_{\alpha})_{\alpha \in J}$ is obviously the net we need.
Given an open nbhd $U$ of $x$, I need to find $\alpha \in J$ such that $\alpha \subseteq \beta$ implies $x_\beta \in U$. My thought was to take $\alpha = U$, but I still couldn't figure out convergence from that...
Because if $U \subseteq \beta$, certainly $x_\beta \in A \cap \beta$, but I don't see how to argue that $x_\beta \in U$.
If $x \in \overline{A}$, how do I show that $A$ contains a net converging to $x$? I know I have to consider the directed set $J = \{\alpha \subseteq X \mid \alpha \mbox{is a nbhd of } x\}$, but I can't seem to get the proof to work. If $\alpha$ is a nhbd of $x$, then there exists $x_{\alpha} \in A \cap \alpha$, so $(x_{\alpha})_{\alpha \in J}$ is obviously the net we need.
So my guess of Geordie Williamson as a Fields medalist turned out to be wrong. But given that he is a plenary speaker and the only such apart from Peter Scholze below the age of 40, I assume he must have at least been a consideration.
I am guessing $\beta$ is "later" than $\alpha$ in the preordering, thus it has to be eventually converge to $x$ since $J$ forms a filter for $x$ or something?
Hey. I may have a conversation about chaos in the chaos room if this new user engages in any way... if anyone is interested in talking about chaos. I could use the back up. Haven't formally interacting with this stuff in... 6 years...
Actually... Can I ask that someone give this user's question an upvote so they can have a conversation in chat?
