@robjohn In which room can I ask this website related doubt?

@Mayank what website related doubt?

Just like When will I be able to send image in chat? How many reputation are required for it ?

@Mayank hang on a sec

@Mayank do you see anything beside the "send" button?

@robjohn Oh Sorry I was using mobile version. I didn't see upload button on desktop site.

I think it is 10 as with regular posts

of course, you can't post in chat until 20, so...

my book suggests wierd symbol

it uses common log as natural log and calls it Napierian logarithm.

Is it allowed to use common log as natural log in number theory?

Then the book says it is inverse of $exp()$

And now I need to evaluate $log(x)/x^\delta\to 0$

as $x\to\infty$

Book prove anything about it and just put some real number to shows $log(x)$ which is $ln(x)$ reach infinity slower than $x^\delta$ just be using some number

I have forgot analysis so how do you guys prove $ln(x)/x^\delta\to 0$ as $x\to\infty$?

may be lhopital law or find some inequality so you can make it reach 0 which is greater than $log(x)/x^\delta\to 0$

ok i am asking damn stupid question

using identity is ok log(x^{1/x^\delta}) and anyway you get stupid 0

embarrassed to ask this sorry

I am wrong using lhopital is okay lol

please don't look at my post

I am randomly writing

If a matrix in $\text{SL}(n,\mathbb{R})$ is diagonalizable over $\mathbb{C}$ and all of its eigenvalues have absolute norm one, then clearly it is conjugate in $\text{GL}(n,\mathbb{C})$ to a matrix in $\text{SO}(n)$. But is it also conjugate in $\text{SL}(n,\mathbb{R})$ to a matrix in $\text{SO}(n)$?

For example, $$\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0\\ 0 & 0 & 1\end{pmatrix} = T \begin{pmatrix}-i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & 1 \end{pmatrix} T^{-1}. $$ Can I choose $T$ to be real and with determinant $1$?

What exactly is this result?

I know that if $\eta_{X}$ denotes the Poincaré dual of a submanifold $X$, and $N_1$ and $N_2$ are transverse then $\eta_{N_1}\wedge\eta_{N_2} = \eta_{N_1 \cap N_2}$.

@Mouse Does your book prove that $\frac x{e^x}\to0$?

@abenthy $T$ can be chosen with determinant $1$, but if $T$ were real, we couldn't have $$T^{-1}\begin{pmatrix}0 & -1 & 0 \\ 1 & 0 & 0\\ 0 & 0 & 1\end{pmatrix}T = \begin{pmatrix}-i & 0 & 0 \\ 0 & i & 0 \\ 0 & 0 & 1 \end{pmatrix} $$

Does a manifold formed by two transversal manifolds have any special properties?

@feynhat What submanifold represents the PD of the generator of $H^2(\Bbb{CP}^n; \Bbb Z)$?

A hyperplane?

Yes, so it's $H = \Bbb{CP}^{n-1} \subset \Bbb{CP}^n$.

What is $PD[H] \smile PD[H] = PD[H \pitchfork H]$?

Transverse intersection of two hyperplanes would be an n-1 plane. Then you projectivize it to get CP^{n-2}?

Yup. So it's PD of $[\Bbb{CP}^{n-2}]$. That's the generator of $H^4(\Bbb{CP}^n; \Bbb Z)$, yes?

Yeah.

So if $\alpha$ denotes generator of $H^2(\Bbb{CP}^n;\Bbb Z) \cong \Bbb Z$, then by similar arguments (or induction) $\alpha^k$ (cup product of $\alpha$, $k$ times) generates $H^{2k}(\Bbb{CP}^n; \Bbb Z) \cong \Bbb Z$.

However, $\alpha^{n+1} = 0$, so the above is only true for $k \leq n$.

So what is the (graded) cohomology ring $H^*(\Bbb{CP}^n; \Bbb Z)$?

$\Bbb Z[\alpha]/(\alpha^{n+1})$.

You need one more piece of data, to specify the grade of $\alpha$. So $|\alpha| = 2$.

But there you go

Can PD be defined in purely topological setting? I only know how it is defined using forms.

Yes, Poincare duality can be defined for orientable closed topological manifolds. This is in Hatcher Ch 3.

I meant, Poincaré dual.

If Poincare duality is defined the Poincare dual is defined, you send an element in homology to cohomology by the Poincare duality isomorphism :P

Hmm... how 'cap product with fundamental class' help me see what the dual is?

The singular Poincare dual is basically the following; let's say for a minute everything is smooth. $S \subset M$ oriented $k$-dimensional submanifold of a closed oriented $n$-manifold. The Poincare dual $\eta$, as a $(n-k)$-form, is something which integrates to $1$ over $(n-k)$-simplices transverse to $S$ in $M$

So you simply replace that condition by saying the Poincare dual as an $(n-k)$-cochain is something which evaluates (under Kronecker pairing) to $1$ over $(n-k)$-simplices transverse to $S$ in $M$ and $0$ otherwise. This is the Thom class of the normal bundle, topologically defined (still assuming everything is smooth).

See here for a precise discussion.

The idea is integration is like feeding a chain to a cochain, and everything just works out if you keep using that analogy. You have to do something else entirely to define PD in a non-smooth setup but I think Ted linked a paper earlier which did this exact thing, but with microbundles instead.

Is there a name for a connected topological space that becomes deconnected if you remove any one point ? Something like $\Bbb R$

But not like $\Bbb R^2$

I thin they call this "every point is a cut point" property

Cheers. I take it a cut point is one that, when removed, makes the space disconnected

Yeah, defined for connected T_1 spaces.

Ah wikipedia says "cut-point space" is a valid terminology.

I recently had to think briefly about cut point spaces which has the same "degree" at every point, i.e., removing any point gives n connected components, for a fixed n.

Regular $\Bbb R$-trees are examples.

You have dendrite like spaces too

There are some analogies between groups acting by isometries on $\Bbb R$-trees and groups acting by homeomorphisms on dendrites

@BalarkaSen I should read Thom class.

Thanks though.

Let $k$ denote any field and $n \in \Bbb{N}$. The problem I am working on asks me to define a right $M_n(k)$-module action on $k^n$, the space of column vectors. If $A \in M_n(k)$ and $v \in k^n$, wouldn't this action just be $v \cdot A := Av$, which is just regular matrix multiplication with a column vector? Wouldn't this work?

@user193319 check the axiom

@LeakyNun I think they check out.

@user193319 $v \cdot A \cdot B = BAv \ne v \cdot (AB)$

Riemannian Manifolds: In this question math.stackexchange.com/questions/2612239/… the first answer suggests a calculation using the product rule. I don't see how one can use any axioms of a connection to open that expression. Is he subtlety suggesting to open things open on a chart and then using the product rule?

@Konformist: But that's what the person wrote just before. You extend $\nabla$ to the (tensor) product by imposing the Leibniz rule.

Hi, @Rithaniel.

Hey Ted (Technically not here, in zoom meeting)

@TedShifrin so to confirm that I need to open the metric up and calculate?

I don't understand your question. It's how you define the connection on higher-order tensors.

@KonformistLiberal The axioms of a connection are irrelevant, because a connection up to this point is defined at most for vector fields, as a map $\Delta: \Gamma(TM) \to \Gamma(T^*M \otimes TM)$. Its domain does not include the whole tensor algebra.

What Anthony is saying is that for $\Delta_{LC}$ there is a unique extension to a map $$\widetilde{\Delta}: \bigoplus_{n \geq 0} \Gamma(TM^{\otimes n}) \to \bigoplus_{n \geq 0} \Gamma(T*M \otimes TM^{\otimes n})$$ with $\widetilde \Delta_0 = d, \widetilde \Delta_1 = \Delta_{LC}$, $\widetilde \Delta$ satisfies the Leibniz rule, and $\widetilde \Delta$ commutes with metric contraction.

You can prove this without ever looking in a chart.

Howdy @MikeM

Hi

can anyone help me with this?

Does "mode" typically refer to the global maximum or just a local maximum of a distribution? Is there any less ambiguous shorthand for the global maximum?