TITLE: A question with infinity
QUESTION [10 upvotes]: I am a sophomore in high school and my math teacher did a very short lesson on infinity, here's how it went:  (Try to solve each part yourself the first two are easy)
Part 1
You have an inf. number of boxes each labeled like so
[1][2][3][4].....

and that there is a person in each one of these boxes.  But you have another set of five boxes like so
[1][2][3][4][5]

You want to only have one set of boxes what do you do?  You must be able to tell each person what box they should goto next.
Spoiler

Each person is told to move down five boxes.  Or n+5 is you new box, and the five people from the other boxes are able to move into the first five boxes or just n+0.

Part 2
Well you actually have two of these inf. box sets, but you really only want one.  What now? (same rules apply)

Each person from set one is told to go to box 2n and each person from set two is told to box 2n-1

Part 3
Now you have an inf. number of these inf. box sets arranged like so
1: [1][2][3][4][5]...
2: [1][2][3][4][5]...
3: [1][2][3][4][5]...
4: [1][2][3][4][5]...
5: [1][2][3][4][5]...

and you want to fit them all into one inf. box set, how do you do it.  She refused to tell us the answer and no one in my class figured it out, but I think I just did (correct me if I'm wrong, I'm also looking for alternate solutions).
How I went about solving this problem
Step 1

I assumed that there had to be some sort of pattern so I tried a few

Step 2

I picked the following patter.

1: [1][3][6][10][15]...

2: [2][5][9][14][]...

3: [4][8][13][][]...

4: [7][12][][][]...

5: [11][][][][]...

The numbers in the box represent the new box of each person

Step 3

I started to create a formula for the people to find there new room based of there current room number and box set number.  I found the following sequence in for people in the first row, 1 3 6 10 15.  Or +2 +3 +4 +5.  Which told me that it was quadratic so the equation had to be ax^2 + bx + c = new room number

Step 4

I found three equation by plugging in room numbers and their correct new room numbers to find these three equations:

(1^2)a + 1b + c = 1 or a + b + c = 1

(2^2)a + 2b + c = 3 or 4a + 2b + c = 3

(3^2)a + 3b + c = 6 or 9a + 3b + c = 6

Step 5

After solving this system I got the following equation 1/2x^2 + 1/2x + 0 = new room number where x = current room number.  This of course only tell the people in the boxes on the bottom where to go. Now I needed to find how to place the rest of the people.  The first thing I noticed was that people on the same diagonal as seen bellow were very much related to each other when you look at the rooms they will end up in.

[x-4][][][][]

[][x-3][][][]

[][][x-2][][]

[][][][x-1][]

[][][][][x]

The solution

Working this knowledge into my equation was fairly easy (I was able to do it while going for a run so I won't explain how I got my equation).  The equation is:

 (1/2(r+c-1)^2 + 1/2(r+c-1) + 0)-r+1 = new room number | r = row | c = column

Now my question to all of you is

Is my equation correct?
Are there any alternate solutions to this problem?
How could you fit a 3D array of boxes (inf boxes of people in all direction) into a 1D line of boxes?
Can you find a way to fit a n dimensional array of boxes into a 1D line of boxes?

For any of you into JAVA here's a a short snippet of code I made to prove my answer
    package main;

public class Main {

    /**
     * @param args
     */
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int f = 2;
        int r = 1;
        int c = 1;
        while(true){
            double x = (.5*(r+c-1)*(r+c-1) + .5*(r+c-1) + 0)-r+1;
            System.out.print(x +", ");
            c++;
            r--;
            if(r < 1){
                r = f;
                f++;
                c = 1;
            }
        }
    }
}

REPLY [2 votes]: The following answer is not yet complete solution, but I think it might be an idea how to approach the problem yourself.
I recall the Collatz-problem in the "syracuse"-statement and look at the inverse transformation which allows to construct an infinite tree with infinitely many infinitely long "boxes" where each odd natural number occurs exactly one time (well, whether all odd natural numbers occur at least once is really open, but it would not spoil the resulting infinite indexing-system).
The "syracuse"-statement of the Collatz-transformation is with an odd a and b and a natural $A \ge 1$ is $$ b = {3a+1 \over 2^A} $$ where the value of A and that of b are of course uniquely determined by the value of a. But there are a lot (actually infinitely many) of different a which lead to the same b, so if we invert the transformation,
$$ b = {a 2^A - 1 \over 3} $$ we have for each a by infinitely many A's infinitely many different b, so we could write infinitely many indexed b as $$ b_{A} = {a 2^A - 1 \over 3} $$ and now each of the infinitely many $b_A$ can be treated similarly to get $$c_{A,B} = {b_A 2^B - 1 \over 3} $$
This scheme implements an infinite long index (the infinite number of boxes) and each box is of infinite length. One need now to show, that the occuring values are unique, which I think is not too difficult (it is not solving Collatz, because that (unsolved) conjecture is that the values are not only unique but also including all odd positive natural numbers which makes it so difficult).
(I have also another statement from the same problem for the infinitude of infinite bins, maybe I can reproduce it here later as an even easier-to-see representation, I'm short of time at the moment, sorry)


TITLE: on two dimensional graded vector spaces
QUESTION [2 upvotes]: I read the following statement:

Consider a graded vector space $V$ with basis $\{a, b\}$ such that $a \in V^2$
  and $b \in V^5$.

Does this mean that $V=\bigoplus_{i\geq 0}V^i$ such that all $V^i$ are $0$ except $V^2$ and $V^5$ hence we can simply write $V=V^2\oplus V^5$ and if yes why don't we say directly that $V$ is a two dimensional vector space that can be written as the direct sum of two vector spaces $V=A\oplus B$ where $A$ is a one dimensioanl vector space with basis $\{a\}$ and
$B$ is a one dimensional vector space with basis $\{b\}$.

REPLY [1 votes]: Graded vector spaces need to retain their grading information. It might be that $V=V^1\oplus V^2$ is isomorphic as a vector space to $V'=V'^1\oplus V'^3$ but they are not isomorphic as graded vector spaces because an isomorphism would have to respect the degrees of elements in the grading. If were to just say $V$ is two-dimensional and isomorphic to the direct sum of $A$ and $B$, as you mention, then you have lost the graded structure.


TITLE: What sets can we define continuity and differentiability on?
QUESTION [2 upvotes]: I am an undergraduate Physics student (completing my first year shortly) who has had a (first) course on Calculus, and another on Linear Algebra.
When working with differential equations (in physical problems), I decided to look with mathematical rigour what I was dealing with. So I decided to start with the beginning. The following questions popped up.
Let $\Bbb{F=R}$ or $\Bbb{C}$.
Generalising Intervals
What is the concept parallel to an interval in $\Bbb{R}$, in $\Bbb{C}$? (Note: I’m not looking for contours, rather something which generalises the “rectanglular” and “circular” domains.)
Can the answer be open or closed subsets of $\Bbb{C}$? Can it be generalised to $\Bbb{F}^n$?
Generalising Continuity and Differentiability
Suppose we have a subset $S\subseteq \Bbb{F}$. What properties must $S$ have so that some form of continuity and differentiability can be defined on some subset $\mathcal{S} \subseteq S$?
(See this post (later in the post) and this.)
(I’d appreciate if your answer is generalisable to $\Bbb{F}^n$.)
Extending Normal Calculus
For such a “continuous and differentiable” $\mathcal{S} \subseteq S$, what theorems of normal calculus will apply?
Also, how will the integral, if needed, be redefined?
Finally, ODE’s
Now turning to ODE’s, I found a very satisfying definition of an ODE here. But unfortunately, it is restricted to intervals of $\Bbb{R}$. How can you generalise this definition to $\Bbb{C}$? (That’s why I was asking for generalisation of interval.)
Very importantly, can we have solutions to ODE’s not as “intervals”, but sets like $\mathcal{S}$ above?

REPLY [2 votes]: This is well-trodden ground in mathematical analysis. You should be able to find this information in nearly any text which covers multivariable calculus. See for example Chapter 3 of Apostol's Mathematical Analysis: Point Set Topology.
Point Set Topology is the field of mathematics that generalises notions of "interval" and "continuity", even beyond $\Bbb R^n$ and $\Bbb C^n$.
Differentiability has a number of generalisations, from gradients (in $\Bbb R^n$), which you will see a lot of in Physics, to the more abstract Fréchet derivative, which again works in more general spaces than $\Bbb R^n$.
Differentiability of complex functions is surprisingly different to the real case. A classic reference for this is Ahlfors' Complex Analysis. It covers a lot of ground and rigorously.
Not sure how much that will help in the direction of complex ODE's. But for the real case, when you add dimensions, you involve partial derivatives and end up with partial differential equations (PDE's). These are notoriously complicated to solve, but they can still be fruitfully studied. A search for "introduction to PDE's" or "first course in PDE's" will help. I don't want to recommend anything specifically, since I don't know the area.
I hope this gives you some ideas of where to look.
Update
Section 4 of Chapter 8 (Global Analytic Functions) of Ahlfors’ Complex Analysis is dedicated to linear complex differential equations. At least the material figures in the 3rd edition (1979).


TITLE: Branch cuts of the Hypergeometric function with one complex parameter
QUESTION [2 upvotes]: I am trying to understand if the hypergeometric function with the following choice of parameters has any branch cut in the complex plane
$$
\ _{2}F_{1}\left(a,b - i \rho ,c ,1-z^{2}\right)
$$
here $a,b,c \in \mathbb{N}$ and $\rho \in \mathbb{R}$ and $\rho > 0$.
I did some numerical explorations with Mathematica and I was not able to find any branch cuts on the real line. Is this property true?
I was not able to find any reference on the branch cuts of the hypergeometric function with complex parameters.

REPLY [1 votes]: Complex parameters have nothing to do with the choice of branch cuts. The conventional way to fix the branch cut on the principal sheet of the Riemann surface of $_2F_1(a,b;c;x)$ is to draw it from $1$ to $+\infty$ along the real axis in the complex plane of $x$. For example, this is how $_2F_1$ is implemented in Mathematica.
In terms of your $z$-variable, this branch cut corresponds to two halves of the imaginary axis. Specifically, $z\mapsto 1-z^2$ maps each of the half planes $\Re z\gtrless0$ to $\mathbb C\backslash[1,+\infty)$. So the signature of the branch cut will be the difference  between the values of $_2F_1(1-z^2)$, calculated for $z=\pm i\alpha+0$ (or $z=i\alpha\pm 0$) for $\alpha\in\mathbb R$.


TITLE: Can two single particles interfere with each other?
QUESTION [3 upvotes]: Groups of particles can interfere with one another; In the double slit experiment  when measuring single photons at the screen each one arrives at the screen in a random manner and they only show the interference pattern once several particles are detected.
Obviously two waves can obviously interfere with one another, but can two single particles interfere with one another? Cohen-Tannoudji writes that

light simultaneously behaves like a wave and like a flux of particles

But do two particles constitute a flux of particles? I doubt that this could be tested experimentally but if it were so would this constitute a measurement for each particle?
Here was my thought (Disclaimer:I do not have a good idea of what the interference of two single particle states is):
To have interference of two single particles you would have to know something about there position to be able to describe their interference pattern hence the measurement.

REPLY [0 votes]: Don't over interpret Cohen-Tannoudji's sentence. It describes two situations very far apart.
Either a single photon or a very very large number of photons, all doing the same thing. It touches on a feature of the world that is very hard to understand. For example, the difference between the stream of photons and a single photon has (in most cases) nothing to do with interactions between the photons.
Take the double slit experiment as an example. Suppose that it is set up so that the screen where the light is observed, we have a very sensitive measurement device sensitive enough to record a the position where a single photon hits it. Sending of a single photon through the double slit will then give rise to the recording of one precise position on the screen.
Sending a trillion of photons but with a distance between them, so that the next is not sent  until the previous is recorded by the screen detector, will result in the well known intensity pattern. The number of photons that were recorded at a specific spot is given by the standard expression for the intensity distribution on the screen. Normally the double slit experiment is performed with a laser. The photons are then much closer to each other and could interact, but such interactions are not responsible for the interference pattern. On the contrary, I imagine that a sufficiently intense beam could show deviations from the usual interference pattern.
The well known double split intensity pattern, if appropriately normalized, is simply the probability distribution of where photon will be recorded on the screen.
So, the wave description applies to a single photon as well as the beam from a light source.
Now, we have entered a quite deep and complicated aspect of quantum physics. So complex that there is no complete agreement on how it works. There are different interpretations.
My understanding of these situations is based upon the notion of entanglement of quantum states, and of the chaos that photon creates in the measurement apparatus. These two ingredients creates a separation of the quantum state of the combined system of the photon and the measurement device to separate into different parts that have nothing to do with each other any more, thus effectively separating the world into different alternatives. In  these different worlds our photon have been recorded at a different locations as the wave description dictates. This is the multiworld interpretation. To my mind this is just an analysis of what the quantum equations say. Anyone coming up with other interpretations have to prove that the real physics deviate from what the quantum equations give. Then they have to come up with some new physics that we can test. I can assure you that there is not yet any experiment that can not be understood using the quantum equations.


TITLE: Show that scalar field f must be equal at two points...
QUESTION [0 upvotes]: If $\nabla f$ is proportional to $x\hat{i} + y\hat{j} + z\hat{k}$.
The two points are $(0,0,a)$ and $(0,0,-a)$
So far what I have is that along the z-axis from the origin the function doesn't change with respect to the x and y directions, so it only changes with respect to z. But I am unsure as to how to show that the scalar field must be equal at these points.
Solving f along the z axis, we have $f = ce^{kz} + D$. How can f at the two point be shown to be equal?

REPLY [0 votes]: Solve the equation:
$\nabla f = <\frac {\partial f}{\partial x},\frac {\partial f}{\partial y} , \frac {\partial f}{\partial z}>$
Say K be the proportionality constant.
Hence $\frac {\partial f}{\partial x} = Kx \Rightarrow f = \frac {x^2}{2}+C(y,z)$.
Coming to $\frac {\partial f}{\partial y} = Ky \Rightarrow \frac {\partial C}{\partial y} = Ky \Rightarrow C=\frac {y^2}{2} + D(z)$
Going this way, you can get $f = K\frac {x^2+y^2+z^2}{2} + M$.
$M$ is integration constant (a number).
Now clearly the value of $f$ is constant the given points specified.

REPLY [0 votes]: If we know that
$$
\nabla f \approx x\hat{i} + y \hat{j} + z \hat{k} \implies \nabla f = kx \hat{i} + ky \hat{j} + kz \hat{k}
$$
for some $k \in \mathbb{R}$. From this we can deduce that
$$
\frac{\partial f}{\partial x} = kx \implies f = \frac{k}{2}x^2 + g(y,z) \\
\frac{\partial f}{\partial y} = \frac{\partial g}{\partial y} = ky \implies g = \frac{k}{2}y^2 + h(z) \\
\frac{\partial f}{\partial z} = \frac{\partial h}{\partial z} = kz \implies h = \frac{k}{2}z^2 + C
$$
with $C \in \mathbb{R}$. So in the end we get that:
$$
f(x,y,z) = \frac{k}{2}\left(x^2 + y^2 + z^2\right) + C
$$
and finally thus:
$$
f(0,0,a) = \frac{k}{2}a^2 + C, f(0,0,-a) = \frac{k}{2}a^2 + C
$$
are equal!


TITLE: Paradox at $T=0$ in 1D Ising Model
QUESTION [4 upvotes]: How do we resolve the paradox in 1 D Ising model at $T=0$ where we have two ground states possible and accordingly the mean value of magnetization is $0$ but there is spontaneous symmetry breaking and the system chooses one state and the net magnetization is non zero.
Is it related to the fact that when we perform a measurement on the system, the system collapses into one of the two ground states and thus we get non zero magnetisation?
But this argument also seems somewhat wrong because say all spins up and all spins down are the two eigenstates of the Hamiltonian. Then since both states have the same energy and are equally probable, any linear combination of these states will also be an eigenstate of the Hamiltonian. Then how can we say that a state which is an eigenstate (but a linear combination) will collapse into one of the eigenstates?

REPLY [4 votes]: There are two ways to solve this "paradox".
The first one is to say, like it was suggested by probably_someone in the comments, that a realistic system will never be perfectly isolated from magnetic fields. An infinitesimal external magnetic field will be enough for one of the two ground states to be favored, thus selecting one of the two ground states in the limit $T\to 0$.
The second resolution is more formal. Apparently, if the Hamiltonian $\mathcal H(\sigma_1, \dots, \sigma_N) = \mathcal H (\{\sigma_i\})$ is invariant under reversal of all the spins $\sigma_i \to -\sigma_i$, then also the canonical probability of the state $\{\sigma_i\}$
$$P[\mathcal H\{\sigma_i\}]=\frac{e^{-\beta \mathcal H(\{\sigma_i\})}}{Z} \tag{1}$$
will be invariant under the same transformation. This means that  the magnetization
$$m \equiv\langle \sigma_i \rangle = \sum_{\{\sigma_i\}} \sigma_i P[\mathcal H (\{\sigma_i\})]$$
must always be $0$, since $m$ and $-m$ occur with equal probability.
For a finite volume system, the correctness of this argument is irrefutable: there are no (true) phase transitions at finite volume. A finite-volume Ising model will only apparently select a defined magnetization as $T\to0$, but if you wait long enough (an exponentially long time with the system size $N$), the magnetization will eventually reverse, over and over.
However, as $V \to \infty$, $(1)$ becomes only formal, since the partition function diverges. In this case, this argument is incorrect, and it can be shown that while
$$\lim_{V\to \infty} \lim_{h \to 0^{\pm}} m = 0$$
you have
$$ \lim_{h \to 0^{\pm}} \lim_{V\to \infty} m = \pm 1$$
This is what we call spontaneous symmetry breaking: even in the absence of an external field (which would explicitly break the symmetry), a single ground state is selected over the two possible ground states.


TITLE: Relativity of temperature paradox
QUESTION [19 upvotes]: The imagined scenario:
Part A:
From special relativity we know that velocity is a relative physical quantity, that is, it is dependent on the frame of reference of choice. This means that kinetic energy is also relative, but this does not undermine the law of conservation of energy as long as we are consistent with our choice of frame. So far so good.
Part B:
On the other hand, from statistical mechanics, we know that the average kinetic energy of a system and its temperature are directly related by the Boltzmann constant
$$
\langle E_k \rangle = \frac{1}{2}m\langle v^2 \rangle = \frac{3}{2} k_B T
$$
which leads to conclude that when the notion of temperature in physics is expressed in terms of a system's kinetic energy, then it too ought to be a relative quantity, which is a bit mind-boggling, because I had always thought of temperature as absolute.
Part C:
Furthermore, we know that all objects at non-zero temperature, radiate electromagnetic energy with a wavelength given as a function of the body/object's temperature, this is the Blackbody radiation. Thus in principle, I am able to infer an object's temperature (i.e. the temperature in its own rest-frame of reference) by measuring its emitted radiation, regardless of the frame I find myself in. But this seems to violate the previously expected relativity of temperature as defined by average kinetic energy.

Proposed resolutions:
The resolutions that I imagine to this paradox are:

a) Depending on the frame of reference from which I measure the emitted blackbody radiation of the object, the radiation will undergo different Doppler blue/red-shifts. Thus the relativity of the temperature in the context of blackbody radiation, is preserved due to the Doppler effect.
b) I suspect that treating temperature as nothing but an average kinetic energy does not in general hold true, and to resolve this paradox, one should work with a more general definition of temperature (which I admit I do not know how in general temperature ought to be defined, if not in terms of state of motion of a system's particles).

Case a) resolves this hypothetical paradox by including the Doppler effect, but does not contradict the relativity of temperature.
Case b) on the other hand, resolves the problem by challenging the definition that was used for temperature, which in the case that we define temperature more generally, without relating to kinetic energy, may leave temperature as an absolute quantity and not relative to a frame.

Main question:

But surely only one can be correct here. Which begs to ask: what was the logical mistake(s) committed in the above scenario? In case there was no mistake, which of the two proposed resolutions are correct? If none, what is then the answer here? Very curious to read your input.

REPLY [0 votes]: The question may be about the covariance or otherwise of temperature, in which case have a look here. As well, have a look at the paper "Temperature in special relativity" by J. Lindhard, Physica Volume 38, Issue 4, 5 June 1968, Pages 635-640.


TITLE: How does the existence of a monopole affect the magnetic vector potential in gauss's law for magnetism?
QUESTION [2 upvotes]: Gauss's Law for magnetism is
$$
\nabla \cdot B = 0
$$
This allows us to write the magentic field $B$ as the curl of another field the $\textbf{magnetic vector potential, } A$.
$$
B=\nabla \times A
$$
This adhers to $ \nabla \cdot (\nabla \times A)=0$
However, if a monopole does exist then we have
$$
\nabla \cdot B= \rho_{m}
$$
Where $\rho_{m}$ is some magnetic charge density however with a magentic vector potential this violates the equation, $ \nabla \cdot (\nabla \times A) \neq 0$.
Does that mean if magnetic monopoles does exist, that the magnetic field can no longer be defined by a magnetic vector potential? In which case how was dirac able to still  define the magnetic field by a magnetic vector potentials?

REPLY [2 votes]: That means that, in general, you will have to write $\mathbf B = \nabla \psi + \nabla\times\mathbf A$. Then $\nabla\cdot\mathbf B = \nabla^2\psi = \rho_m$. Of course $\nabla\cdot(\nabla\times\mathbf A)=0$ for any vector field $\mathbf A$, as $\operatorname d^2=0$.


TITLE: Action of lie groups on compact manifolds.
QUESTION [0 upvotes]: Is there a result about the action of lie groups in compact manifolds that makes this action transitive?
For example:
$SU(2)$ acts transitively on $CP^1$
$Sp(2)$ acts transitively on $S^4$
$SO(3)$ acts transitively on $S^2$
This action given by left action. I can prove the first example using the following: Consider $v \in CP^1$ and $A \in SU(2)$, the left action give me that $Av= x$, where x is linear combination of the first column of A, as $A$ belong $SU(2)$ its columns are orthonormals and belong $CP^1$. But I would like to find some results about action of Lie group over compact manifolds

REPLY [1 votes]: Here is one general result you can use to prove transitivity. Suppose that $G$ is a compact Lie group acting on a connectd manifold $X$ and you know the stabilizer $G_x$ of some point $x\in X$. Suppose that you also know that $dim G = dim(X) + dim(G_x)$. Then $G$ acts transitively on $X$. To prove this, observe that $G_x$ is closed in $G$, hence, is a Lie subgroup of $G$ (Cartan's theorem), hence, $G/G_x$ is a manifold of dimension equal to the dimension of $X$. The slice theorem implies that the orbit map $o_x: G\to X, o_x(g)=gx$ descends to an open map $f: G/G_x\to X$. Hence, the image of this map is open in $X$; but $G$ is compact, hence, the image is also closed. Since $X$ is connected, you conclude that the map $f$ is surjective. Hence, the orbit map $o_x$ is also surjective, hence, $G$ acts transitively.


TITLE: Solve (or appoximate) for B in A=B*C for matrix mult with non-square matrices.
QUESTION [1 upvotes]: I was wondering how you solve (or approximate) for B in $A=B*C$ when A and and C are known and A,B,C are not square?
It seems like you could approximate this by solving something like:
$B^* = \mathrm{argmin}_B \sum_{i,j} \left((\sum_kB_{i,k}C_{k,j}) - A_{i,j} \right)^2$
Is there a better way to solve this problem?
(By the way, I am trying to solve a collabrative filtering problem, where A is a $\mathrm{user} \times \mathrm{product}$ matrix and C is a $\mathrm{feature}\times\mathrm{product}$ matrix.  I am trying to find the $\mathrm{user}\times{features}$ matrix.  I would like to turn B into 2 matrices: $B=B_1B_2$, a user-preference matrix and a preference-feature matrix but I am trying to solve this simpler problem first.)

REPLY [0 votes]: Investigating further, It seems like you have to solve this by optimization.  I found 2 methods:
Method 1. Taking the derivative with respect to each $b_{i,j} \in B$.
If you add a regularization term for $b$ then the iterative solution looks like:
until Convergence:
    for i=1:rows(B):

$$ b_i \leftarrow b_i + \gamma ((a_{j} - b_i^Tc_j)c_j - \lambda b_i)$$
end

(Without regularization , the $-\lambda b_i$ would be removed from the equation.)
Method 2. I also found a matrix solution of the equation (using only 1 of the updates of alternating least squares since $C$ is known):
until Convergence:
    for i=1:rows(B):

$$ b_i = (CC^T + \lambda I)^{-1}C $$
end


TITLE: Polynomial and spectrum of a Operator .
QUESTION [8 upvotes]: I need help with the following question :
Let us consider $X$ to be a $\mathbb C  $ Banach space . Let $T \in B(X)$ ie. $T$ is a continuous linear map from $X$ to $X$ . define $$P(T)=\sum_{k=o}^{n}a_kT^k$$ with $T^0=id_X$ (identity) .
The claim is that the spectrum of $T : \sigma(T) \subset\{\lambda \in \mathbb C : P(\lambda)= 0\}$
Here $P$ is a polynomial such that $P(T) =0$  .
Thanks.

REPLY [8 votes]: Let me expand on the setting. Let $\text{Hol}(T)$ be the set of all the functions holomorphic in a neighborhood of $\sigma(T)$. The Riesz functional calculus now says that indeed if I have the polynomials $p$, that $p(T)$ is as expected - of course, for polynomials this is trivial.
Additionally, we have the Banach algebra of bounded operators on $X$, and:
Spectral mapping theorem: If $T$ is an element of a Banach algebra $\mathcal A$ and $f$ is holomorphic in a neighborhood of $\sigma(T)$, then
$$\sigma(f(T)) = f(\sigma(T)).$$
If we thus select such a polynomial $p$ such that $p(T) = 0$, we get $p(\sigma(T)) = 0$.


TITLE: How to prove that we cannot see more than 3 faces of an opaque solid cube simultaneously?
QUESTION [7 upvotes]: Is there an elegant mathematical proof to assert that we cannot see more than 3 faces of an opaque solid cube simultaneously (of course without mirrors or any optical tools such as camera, etc)?

REPLY [4 votes]: There are three pairs of opposite faces. To see more than three faces, you would have to view both from one pair simultaneously.
Opposite faces of the cube are parallel. If you can see a particular face, you can't see the one opposite it, because look which side of it you're on.


TITLE: Error in Hungerford's algebra proof? Left id & inv = group
QUESTION [4 upvotes]: Prop 1.3 in Hungerford's Algebra said that if $G$ is a semigroup and there exist a left identity and each element have a left inverse, then $G$ is a group. The proof (and in fact, even the proposition themselves) implicitly assumed that the left identity is unique, but I can't quite see how it could be proved. Is this a mistake, or is there a way to prove that?
EDIT: for definiteness, this is the exact statement as given:
Proposition 1.3. Let $G$ be a semigroup. Then $G$ is a group if and only if the following conditions hold:
(i) There exist an element $e\in G$ such that $ea=a$ for all $a\in G$ (left identity element);
(ii) For each $a\in G$, there exists an element $a^{-1}\in G$ such that $a^{-1}a=e$ (left inverse).
EDIT: in light of the discussion on the range of quantifier, the answer below showed that it is provable as long as the quantifier in (i) range all the way to include (ii). I still wish to ask whether we can prove it even if the quantifier only range within (i), ie. (ii) should be interpreted as: for each $a\in G$, there exists an element $a^{-1}\in G$ such that $a^{-1}a=e$ where $e$ is one of those left identities.

REPLY [1 votes]: I separate this answer, because it relies on an edit and a misreading of the original statement from Hungerford.
But let us suppose that there is at least one left identity called $e$, and each element $g$ has a "loose left inverse" $g^{-1}$ such that $g^{-1}g$ is a left identity, which may or may not be $e$.
Then we cannot conclude that $G$ is a group.
For instance, consider the two element semigroup with multiplication table
$$\begin{array}{l|ll}
G & e & g \\
\hline
e & e & g \\
g & e & g
\end{array}$$
In this case, every element is a left identity. For that matter, each element is a "loose left inverse" of both itself and the other element.


TITLE: Locus of $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}\leqslant 0$ for a rectangle $ABCD$?
QUESTION [0 upvotes]: Consider the problem of the title, that is to determine the set of all points $P$ satisfying
$$\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}\leqslant 0$$
where $A,B,C$ and $D$ are the vertices of a rectangle. One method is to choose coordinates axes so that $A(0,a)$, $B(c,a)$, $C(c,0)$ and $D(0,0)$ and show by a computation that the equation is equivalent to $$\left(x-\frac{c}{2}\right)^2+\left(y-\frac{a}{2}\right)^2\leqslant \frac{a^2+c^2}{4}$$
This means that the locus is the interior and boundary of the circumcircle of the the rectangle, i.e. the circle with center the center of the rectangle and radius the half-diagonal.
Such a result makes me think that there may be a geometric approach. How can I interpret $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}$ in terms of geometry (area, angle,...) to obtain the same result?

REPLY [2 votes]: I'm sorry if this is not what you want, but I hope you like this.
We can understand its meaning using only vectors.
Let $O$ be the midpoint of the line segment $AC$.
Then, noting that $\vec{OC}=-\vec{OA},\vec{OD}=-\vec{OB},|\vec{OA}|=|\vec{OB}|$, we have
$$\begin{align}\\&\vec{AP}\cdot \vec{CP}+\vec{BP}\cdot \vec{DP}\leqslant 0\\&\iff \left(\vec{OP}-\vec{OA}\right)\cdot \left(\vec{OP}-\vec{OC}\right)+\left(\vec{OP}-\vec{OB}\right)\cdot \left(\vec{OP}-\vec{OD}\right)\leqslant 0\\&\iff \left(\vec{OP}-\vec{OA}\right)\cdot \left(\vec{OP}+\vec{OA}\right)+\left(\vec{OP}-\vec{OB}\right)\cdot \left(\vec{OP}+\vec{OB}\right)\leqslant 0\\&\iff |\vec{OP}|^2-|\vec{OA}|^2+|\vec{OP}|^2-|\vec{OB}|^2\leqslant 0\\&\iff |\vec{OP}|^2\leqslant \frac{|\vec{OA}|^2+|\vec{OB}|^2}{2}\\&\iff |\vec{OP}|\leqslant |\vec{OA}|\end{align}$$

Added : How about the following way?
Let $O'$ be the circle whose diameter is the line segment $AC$.
If $P$ is outside $O'$, then $\vec{AP}\cdot\vec{CP}\gt 0$ and $\vec{BP}\cdot\vec{DP}\gt 0$ lead that $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}\gt 0$.
If $P$ is on $O'$, then $\vec{AP}\cdot\vec{CP}=\vec{BP}\cdot \vec{DP}= 0$ lead that $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}= 0$.
If $P$ is inside $O'$, then $\vec{AP}\cdot\vec{CP}\lt 0$ and $\vec{BP}\cdot\vec{DP}\lt 0$ lead that $\vec{AP}\cdot\vec{CP}+\vec{BP}\cdot\vec{DP}\lt 0$.
The result follows from these.


text:
\subsection{Appendix: Proof of Theorem \ref{stabres_modcs}}
\label{proof_addLSdel_modcs}

We prove the first claim by induction. Using condition \ref{initass} of the theorem, the claim holds for $t=0$. This proves the base case. For the induction step, assume that the claim holds at $t-1$, i.e.  $|\tDelta_{e,t-1}| =0$, $|\tT_{t-1}| \le S_0$, and $\tDelta_{t-1} \subseteq \Sset_{t-1}(d_0)$ so that $|\tDelta_{t-1}| \le (2d_0-2)S_a$. Using this assumption we prove that the claim holds at $t$. We use (\ref{sseteq}) and the following facts often: (a) $\Rset_t \subseteq N_{t-1}$ and $\Aset_t \subseteq N_{t-1}^c$, (b) $N_t = N_{t-1} \cup \Aset_t \setminus \Rset_t$, and (c)  if two sets $B,C$ are disjoint, then, $(D \cap B^c) \cup C = D \cup C \setminus B$ for any set $D$.

Since $T_t = \tT_{t-1}$, so $|T_t| \le S_0$.
Since $\Delta_{e,t} = \Nhat_{t-1} \setminus N_t =  \Nhat_{t-1} \cap [(N_{t-1}^c \cap \Aset_t^c) \cup \Rset_t] \subseteq \tDelta_{e,t-1} \cup \Rset_t = \Rset_t$.
Thus $|\Delta_{e,t}| \le |\Rset_t| = S_a$.

Next we bound $|\Delta_{t}|$. Note that $\Delta_t = N_t \setminus \Nhat_{t-1}  = (N_{t-1} \cap \Nhat_{t-1}^c \cap \Rset_t^c) \cup (\Aset_t \cap \Rset_t^c \cap \Nhat_{t-1}^c) = (\tDelta_{t-1} \cap  \Rset_t^c) \cup ( \Aset_t \cap \Nhat_{t-1}^c) \subseteq  (\Sset_{t-1}(d_0)  \cap  \Rset_t^c) \cup  \Aset_t = \Sset_{t-1}(d_0) \cup  \Aset_t \setminus \Rset_t$.

First consider the case when the conditions of the theorem hold for a $d_0 > 1$. Since $\Rset_t$ is a subset of $\Sset_{t-1}(d_0)$ and $\Aset_t$ is disjoint with $\Sset_{t-1}(d_0)$, thus $|\Delta_t| \le |\Sset_{t-1}(d_0)| + |\Aset_t| - |\Rset_t| = (2d_0-2)S_a + S_a - S_a$. If $d_0=1$, $\Sset_{t-1}(d_0)$ is empty and so $\Delta_t = \Aset_t$ and thus $|\Delta_t|=S_a$. Thus in all cases $|\Delta_t| \le k_1(d_0)$.

Consider the detection step. There are at most $f$ false detects (from condition \ref{addthresh}) and thus $|\tDelta_{e,\dett,t}| \le |\Delta_{e,t}| + f \le S_a+f$. Thus $|\tT_{\dett,t}| \le |N_t| + |\tDelta_{e,\dett,t}| \le S_0+S_a+f$.

Next we bound $|\Delta_{\dett,t}|$.
Using the above discussion and (\ref{sseteq}), $\Delta_t \subseteq \Sset_{t-1}(d_0) \cup \Aset_t \setminus \Rset_t = \Sset_t(d_0) \cup \Iset_t(d_0) \setminus \Dset_t(d_0-1)$. Apply Lemma \ref{detectcond_modcs} with $S_N = S_0$, $S_{\Delta_e}=S_a$, $S_\Delta = k_1(d_0)$, and $b_1 = d_0M/d$ (so that $\Delta_1 \subseteq \Iset_t(d_0)$). Since conditions \ref{measmodel} and \ref{add_del} of the theorem hold, all the undetected elements of $\Iset_{t}(d_0)$ will definitely get detected at time $t$. Thus $\tDelta_{\dett,t} \subseteq \Sset_t(d_0) \setminus \Dset_{t}(d_0-1)$.

If $d_0 > 1$, $|\tDelta_{\dett,t}| \le |\Sset_t(d_0)| - |\Dset_{t}(d_0-1)|=(2d_0-3)S_a$. This holds since $\Dset_{t}(d_0-1) \subseteq \Sset_{t}(d_0)$. If $d_0=1$, $\Sset_{t}(d_0)$ is empty and so $|\tDelta_{\dett,t}|=0$. Thus in all cases $|\tDelta_{\dett,t}| \le k_2(d_0)$.

Consider the deletion step. Apply  Lemma \ref{truedelscond} with $S_T = S_0+S_a+f$, $S_\Delta = k_2(d_0)$. Use $\tDelta_{\dett,t} \subseteq \Sset_t(d_0) \setminus \Dset_{t}(d_0-1) \subseteq \Sset_t(d_0)$ to bound $\|x_{\tDelta_{\dett}}\|$ by $\sm(d_0)$.
Since condition \ref{delthresh} holds, all elements of $\tDelta_{e,\dett,t}$ will get deleted. Thus $|\tDelta_{e,t}|=0$.

Thus $|\tT_t| \le |N_t| + |\tDelta_{e,t}| \le S_0$.

Finally, we bound $|\tDelta_{t}|$.
Apply Lemma \ref{nofalsedelscond} with $S_T = S_0+S_a+f$, $S_\Delta = k_2(d_0)$, $b_1 = d_0M/d$. Use $\tDelta_{\dett,t} \subseteq \Sset_t(d_0) \setminus \Dset_{t}(d_0-1) \subseteq \Sset_t(d_0)$ to bound $\|x_{\tDelta_{\dett}}\|$ by $\sm(d_0)$. Since conditions \ref{measmodel} and \ref{add_del} hold, all elements of $\tT_\dett$ with magnitude greater than or equal to $b_1=d_0 M/d$ will definitely not get falsely deleted. But nothing can be said about the elements smaller than $d_0 r$. In the worst case $\tDelta_t$ may contain all of these elements, i.e. it may be equal to $\Sset_t(d_0)$. Thus, $\tDelta_t \subseteq \Sset_t(d_0)$ and so $|\tDelta_t| \le (2d_0-2)S_a$.

 This finishes the proof of the first claim. To prove the second and third claims for any $t>0$: use the first claim  for $t-1$ and the arguments from paragraphs 2-5 above.

\subsection{Appendix: Generalized version of Theorem \ref{stabres_modcs}}
\label{stabres_modcs_gen}

\begin{theorem}[Stability of dynamic modified-CS]
Assume Signal Model \ref{sigmod2} and bounded noise, i.e. $\|w\| \le \eps$. If the following hold for some $1 \le d_0 \le d$,
\ben
\item {\em (addition and deletion thresholds) }
\ben
\item $\alpha_{add}$ is large enough so that there are at most $f$ false additions per unit time,
\label{addthresh}

\item $\alpha_{del}  = \sqrt{2} \eps + 2\theta(S_0+S_a+f,k_2(d_0)) \sm(d_0) $,
\label{delthresh}
\een

\item {\em (support size, support change rate)} $S_0,S_a$ satisfy
\ben
\item $\delta(S_0 + S_a + k_1(d_0)) < (\sqrt{2}-1)/2$ ,
\item $\delta(S_0+S_a + f) < 1/2$,
\item $\theta(S_0+S_a+f,k_2(d_0)) < \frac{d_0}{ 4\sqrt{ 2S_a \sum_{j=1}^{d_0-1} j^2 } }$,
\label{theta_ass_0}
\een
\label{measmodel}

\item {\em (new element increase rate) } $r \ge \max(G_1,G_2)$, where
\label{add_del}
\bea
G_1 \sdefn \frac{ \alpha_{add} + C_1(S_0 + S_a + k_1(d_0)) \eps }{d_0}  \nn \\
G_2 \sdefn \frac{2\sqrt{2} \eps}{d_0 - 4\theta(S_0+S_a+f,k_2(d_0))\sqrt{ 2S_a \sum_{j=1}^{d_0-1} j^2 }}  \ \ \ \ \ \ \
\eea
\item {\em (initial time)} at $t=0$, $n_0$ is large enough to ensure that $\tDelta  \subseteq \Sset_0(d_0)$, $|\tDelta| \le (2d_0-2)S_a$,  $|\tDelta_e| =0$, $|\tT| \le S_0$,
\label{initass}
\een
where
\bea
k_1(d_0) \sdefn \max(1,2d_0-2) S_a \nn \\
k_2(d_0) \sdefn \max(0,2d_0-3) S_a \nn \\
{\sm(d_0)} \sdefn r \sqrt{ 2S_a \sum_{j=1}^{d_0-1} j^2 }
\eea
then, at all $t \ge 0$,
\ben
\item  $|\tT| \le S_0$, $|\tDelta_e| =0$, and $\tDelta \subseteq \Sset_t(d_0)$ and so $|\tDelta| \le (2d_0-2)S_a$,

\item $|T| \le S_0$, $|\Delta_e| \le S_a$, and $|\Delta| \le k_1(d_0)$,

\item $|\tT_\dett| \le S_0+S_a+f$, $|\tDelta_{e,\dett}| \le S_a+f$, and $|\tDelta_\dett| \le k_2(d_0)$

\item $\|x_t-\xhat_t\| \le \sqrt{2} \eps + (2\theta(S_0,(2d_0-2)S_a)+1) \sm(d_0)$

\item $\|x_t - \xhat_{t,modcs}\| \le C_1(S_0+S_a + k_1(d_0)) \eps$
\een
\end{theorem}

config: arxiv


text:
\begin{document}
\setlength{\oddsidemargin}{0cm} \setlength{\evensidemargin}{0cm}

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{prop}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{coro}[theorem]{Corollary}
\newtheorem{conj}[theorem]{Conjecture}

\theoremstyle{definition}
\newtheorem{defi}[theorem]{Definition}
\newtheorem{exam}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}

\numberwithin{equation}{section}

\title{Classification of $(n+3)$-dimensional metric $n$-Lie algebras}

\author{Qiaozhi Geng}\address{School of Mathematical Sciences and LPMC, Nankai University,
Tianjin 300071, P.R. China}
\author{Mingming Ren}\address{School of Mathematical Sciences and LPMC, Nankai University,
Tianjin 300071, P.R. China}
\author{Zhiqi Chen$^*$}
\address{School of Mathematical Sciences and LPMC, Nankai University,
Tianjin 300071, P.R. China}\email{chenzhiqi@nankai.edu.cn}
\thanks{$^*$ Corresponding author}

\def\shorttitle{metric $n$-Lie algebras}

\subjclass[2000]{17B60, 17D25.}

\keywords{$n$-Lie algebra, metric $n$-Lie algebra}

\begin{abstract} In this paper, we focus on $(n+3)$-dimensional metric $n$-Lie algebras. To begin with,
we give some properties on $(n+3)$-dimensional $n$-Lie algebras.
Then based on the properties, we obtain the classification of
$(n+3)$-dimensional metric $n$-Lie algebras.
\end{abstract}

\maketitle

\baselineskip=20pt

\section{Introduction}
The theory of $n$-Lie algebras is strongly connected with many other
fields, such as dynamics, geometries and string theory. Takhtajan
\cite{Ta} developed the geometrical ideas of Nambu mechanics
\cite{Na} and introduced an analogue of the Jacobi identity, which
connects the generalized Nambu mechanics with the theory of $n$-Lie
algebras introduced by Filippov \cite{Fi}. It is given in \cite{HHM}
some new 3-Lie algebras and applications in membrane, including the
Basu-Harvey equation and the Bagger-Lambert model. More applications
can be found in \cite{BWXA,Ga,MVV,Nak,Po1,Po2,VV}.

A class of $n$-Lie algebras (finite dimensional, real, $n>2$) which
have appeared naturally in mathematical physics are those which
possess a non-degenerate inner product which is invariant under the
inner derivations, which are called metric $n$-Lie algebras. They
have arisen for the first time in the work of Figueroa-O'Farrill and
Papadopoulos \cite{FP} in the classification of maximally
supersymmetric type $IIB$ supergravity backgrounds \cite{FP1}, and
more recently, for the case of $n=3$, in the work of Bagger and
Lambert \cite{BL,BL1} and Gustavsson \cite{Gu} on a superconformal
field theory for multiple M2-branes. It is this latter work which
has revived the interest of part of the mathematical physics
community on metric $n$-Lie algebras. There are some progress on
metric $n$-Lie algebras, such as the classification for Euclidean
\cite{Nag} (see also \cite{Pa}) and lorentzian metric $n$-Lie
algebras \cite{Fo}, the classification of index-2 metric 3-Lie
algebras \cite{MFM} and a structure theorem for metric $n$-Lie
algebras \cite{J}.

The classification of $n$-Lie algebras is more difficult than that
of Lie algebras due to the $n$-ary multilinear operation. As we
know, the only attempt to the classification of $(n+2)$-dimensional
$n$-Lie algebras is due to 6-dimensional 4-Lie algebras \cite{BS}.
The main goal of this paper is to classify $(n+3)$-dimensional
metric $n$-Lie algebras. To begin with, we get some properties on
$(n+3)$-dimensional $n$-Lie algebras by the basic results about the
$n$-Lie modules of simple $n$-Lie algebras. Furthermore, we give the
classification of $(n+3)$-dimensional metric $n$-Lie algebras.

Throughout this paper, we assume that the algebras are
finite-dimensional. Obvious proofs are omitted.

\section{$n$-Lie algebra and Metric $n$-Lie algebra}

\subsection{$n$-Lie algebra}
An $n$-Lie algebra is a vector space $\mathfrak g$ over a field
$\mathbb F$ equipped with an $n$-multilinear operation [$x_1,
\cdots, x_n$] satisfying
\begin{equation}[x_1, \cdots, x_n]=\text{sgn}(\sigma)[x_{\sigma(1)}, \cdots, x_{\sigma(n)}], \end{equation}
\begin{equation}\label{def1}
[x_1, \cdots, x_{n-1}, [y_1, \cdots, y_n]]=\sum_{i=1}^n [y_1,
\cdots, [x_1,\cdots, x_{n-1}, y_i], \cdots, y_n]
\end{equation}
for all $x_1, \cdots, x_{n-1}, y_1, \cdots, y_n \in \mathfrak g$,
and $\sigma \in S_n$. Identity (\ref{def1}) is usually called the
generalized Jacobi identity, or simply the Jacobi identity.

A subspace $B$ of an $n$-Lie algebra ${\mathfrak g}$ is called a
subalgebra if $[x_1, \cdots, x_n]\in B$ for all $x_1, \cdots, x_n
\in B$; whereas an ideal $I$ is a subspace of ${\mathfrak g}$ if
$[x_1, x_2 \cdots, x_n]\in B$ for all $x_1\in I, x_2, \cdots, x_n
\in {\mathfrak g}$. Let $A_1, \cdots, A_n$ be subalgebras of an
$n$-Lie algebra ${\mathfrak g}$. Denote by $[A_1, A_2, \cdots, A_n]$
the subspace of ${\mathfrak g}$ generated by all vectors $[x_1, x_2,
\cdots, x_n]$, where $x_i \in A_i$ for $i=1, 2, \cdots, n.$ The
subalgebra ${\mathfrak g}^{1}=[{\mathfrak g}, {\mathfrak g}, \cdots,
{\mathfrak g}]$ is called the derived algebra of ${\mathfrak g}$. If
${\mathfrak g}^{1}=0$, then ${\mathfrak g}$ is called an abelian
$n$-Lie algebra. The subset $C({\mathfrak g})=\{x\in {\mathfrak
g}|[x, y_1, \cdots, y_{n-1}]=0, \forall y_1, \cdots, y_{n-1} \in
{\mathfrak g}\}$ is called the center of $\mathfrak g$. An ideal $I$
is said to be maximal if any other ideal $J$ containing $I$ is
either $\mathfrak g$ and $I$. An $n$-Lie algebra is said to be
simple if it has no proper ideals and dim$\mathfrak g^1>0$.

\begin{lemma}If $I$ is a maximal ideal, then ${\mathfrak g}/I$ is simple or
one-dimensional.
\end{lemma}

The classification of simple real $n$-Lie algebras is given as
follows.

\begin{theorem}[\cite{Li}]\label{T1}
A simple real $n$-Lie algebra is isomorphic to one of the $(n+1)$-
dimensional $n$-Lie algebras defined, relative to a basis $e_i$, by
\begin{equation}\label{6}
[e_1,\cdots,\hat{e_i},\cdots,e_{n+1}]=(-1)^{i}\varepsilon_ie_i,
1\leqslant i \leqslant n+1
\end{equation}
where symbol $\hat{e_i}$ means that $e_i$ is omitted in the bracket
the $\varepsilon_i$ are signs.
\end{theorem}

It is plain to see that simple real $n$-Lie algebras admit invariant
inner products \cite{J}. Indeed, the $n$-Lie algebra in Theorem
\ref{T1} leaves invariant the diagonal inner product with entries
$(\varepsilon_1,\cdots,\varepsilon_{n+1})$.

An ideal $I$ of an $n$-Lie algebra $\mathfrak g$ is called an
$n$-solvable ideal if $I^{(r,n)}=0$ for some $r\geqslant 0$, where
$I^{(0,n)}=I$ and $I^{(s+1,n)}=[I^{(s,n)},\cdots, I^{(s,n)}]$ for
$s\geqslant 0$ by induction. The maximal $n$-solvable ideal of
$\mathfrak g$ is called the $n$-radical of $\mathfrak g$. An $n$-Lie
algebra $\mathfrak g$ is said to be strong semisimple if the
$n$-radical of $\mathfrak g$ is zero.

\begin{lemma}[\cite{Li}]\label{l1}
Let $\mathfrak g$ be a finite-dimensional $n$-Lie algebra over an
algebraically closed field of characteristic zero. Then $\mathfrak
g$ has a decomposition $\mathfrak g = \mathfrak s \oplus \mathfrak
r$, where $\mathfrak s$ is a strong semisimple subalgebra and
$\mathfrak r$ is the $n$-radical of $\mathfrak g$.
\end{lemma}

\begin{lemma}[\cite{BM}]\label{l2}
Let $\mathfrak g$ be a finite-dimensional $n$-Lie algebra over an
algebraically closed field of characteristic zero. Then $\mathfrak
g$ is strong semisimple if and only if $\mathfrak g$ can be
decomposed into the direct sum of its simple ideals.
\end{lemma}

An $n$-Lie module over an $n$-Lie algebra $\mathfrak g$ is defined
as a vector space $M$ such that the semi-direct sum $\mathfrak g +
M$ satisfies the generalized Jacobi identity. This means that the
brackets $[, \cdots , ]$ on $\mathfrak g$ is continued to $\mathfrak
g+ M$ such that $[x_1,\cdots, x_n]=0$, if at least two arguments
among $x_1, \cdots, x_n$ belong to $M$ and the generalized Jacobi
identity is true for all $x_1,\cdots, x_n \in \mathfrak g+M$. In
other words,
\begin{align*}
& [x_1,\cdots, x_{i-1},m,x_{i+1},\cdots, x_n]=-[x_1, \cdots,
x_{i-1},x_{i+1},
m, \cdots, x_n],\quad 1\leqslant i< n, \\
& [x_1, \cdots, x_{n-1}, [x_n, \cdots, x_{2n-2}, m]]-[x_n, \cdots,
x_{2n-2}, [x_1, \cdots, x_{n-1}, m]]  \\
& =\sum_{i=n}^{2n-2} [x_n, \cdots, x_{i-1}, [x_1, \cdots, x_{n-1},
x_i], x_{i+1}, \cdots, x_{2n-2}, m], \\
& [x_1, \cdots, x_{n-2}, [x_n, \cdots, x_{2n-1}],m] \\
& =\sum_{i=n}^{2n-1} [x_n, \cdots, x_{i-1}, [x_1, \cdots, x_{n-2},
x_i, m], x_{i+1}, \cdots, x_{2n-1}],
\end{align*}
for all $x_1, \cdots, x_{n-2}, x_n, \cdots, x_{2n-1} \in \mathfrak
g$ and $m \in M$.

A subspace $N$ of $M$ is called an $n$-Lie submodule of $M$ if
$[x_1, \cdots, x_{i-1}, m, x_{i+1}, \cdots, x_n] \in N$, for all
$m\in N$ and $x_1, \cdots, \hat{x_i}, \cdots, x_n \in \mathfrak g$.
$M$ is said to be irreducible if any submodule is trivial and
completely reducible if it can be decomposed to a direct sum of
irreducible submodules.

\subsection{Metric $n$-Lie algebra}
A metric $n$-Lie algebra $\mathfrak g$ is an $n$-Lie algebra over
$\mathbb{R}$ with a symmetric non-degenerate bilinear form $\langle
, \rangle $ satisfying
\begin{equation}\label{def2}\langle [x_1, \cdots, x_n], y \rangle + \langle x_n, [x_1, \cdots, x_{n-1}, y] \rangle = 0 \end{equation}
for all $x_1, \cdots, x_n , y \in \mathfrak g$.

Given two metric $n$-Lie algebras $\mathfrak g_1$ and $\mathfrak
g_2$, we may form their orthogonal direct sum $\mathfrak g_1 \oplus
\mathfrak g_2$, by declaring that
\begin{equation*}
 [x_1, x_2, y_1,\cdots,y_{n-2}]=0 \text{ and } \langle x_1, x_2\rangle=0,
\end{equation*}
for all $x_i\in {\mathfrak g_1}$ and $y_i\in \mathfrak g_1 \oplus
\mathfrak g_2$. The resulting object is again a metric $n$-Lie
algebra. A metric $n$-Lie algebra is said to be indecomposable if it
is not isomorphic to an orthogonal direct sum of metric $n$-Lie
algebras $\mathfrak g_1 \oplus \mathfrak g_2$ with dim$\mathfrak
g_i>0$.

If $W \subseteq {\mathfrak g}$ is any subspace, we define
\begin{equation*}
W^\bot=\{v\in {\mathfrak g} |\langle v,w\rangle=0, \forall w \in
W\}.
\end{equation*}
Notice that $(W^\bot)^\bot=W$. We say that $W$ is non-degenerate, if
$W \cap W^\bot=0$, whence $V=W\oplus W^\bot$; isotropic, if
$W\subseteq W^\bot$; Of course, in positive definite signature, all
subspaces are non-degenerate.

An equivalent criterion for decomposability is the existence of a
proper non-degenerate ideal: for if $I$ is a non-degenerate ideal,
${\mathfrak g}=I\oplus I^\bot$ is an orthogonal direct sum of
ideals.

\begin{lemma}[\cite{J}]\label{2}
If $\mathfrak g$ is a metric n-Lie algebra, then $C(\mathfrak
g)=(\mathfrak g^{1})^{\bot}$.
\end{lemma}

\section{Some properties on $(n+3)$-dimensional $n$-Lie algebras}
In this section, we firstly get some properties on
$(n+3)$-dimensional $n$-Lie algebras, which play a crucial role in
the classification of $(n+3)$-dimensional metric $n$-Lie algebras.
Let $V_n$ be the unique simple $n$-Lie algebra over the complex
numbers field, defined by
$$[e_1,\cdots,\hat{e_i},\cdots,e_{n+1}]=(-1)^ie_i,$$ where
$e_1,\cdots,e_{n+1}$ is a basis of $V_n$. Then we have:

\begin{lemma}[\cite{DAS}]\label{comre}
Any finite-dimensional  $n$-Lie module of $V_n$ is completely
reducible.
\end{lemma}

\begin{lemma}[\cite{DAS}]\label{weight}
The dimension of any irreducible $n$-Lie module of $V_n$ with
highest weight $t$ is equal to
$\frac{n+2t-1}{n+t-1}\binom{n+t-1}{t}$.
\end{lemma}

\begin{coro}\label{coro} Let $V$ be an $n$-Lie module of $V_n$ in dimension 2. Then $V=V_1\oplus V_2$, where $V_1,V_2$ are 1-dimensional $n$-Lie modules of $V_n$.
\end{coro}
\begin{proof} Assume that $V$ is an irreducible $n$-Lie module of $V_n$ with highest weight $t$. By Lemma~\ref{weight}, $t\not=0$. But for $t>0$ and $n\geqslant 3$,
\begin{equation*}
\dim V=\frac{n+2t-1}{n+t-1}\binom{n+t-1}{t}=\frac{n+2t-1}{t}\binom{n+t-2}{t-1}>2.
\end{equation*}
It contradicts $\dim V=2$. Then $V$ is reducible. By Lemma~\ref{comre}, we have $V=V_1\oplus V_2$.
\end{proof}

\begin{theorem}\label{myt}
Let $\mathfrak g$ be an $(n+3)$-dimensional $n$-Lie algebra over the
complex numbers field. Then $\dim \mathfrak g^1 \leqslant n + 2$.
\end{theorem}

\begin{proof}
If $\mathfrak g$ is $n$-solvable, then the conclusion is obvious.
Assume that $\mathfrak g$ is not $n$-solvable. Then $\mathfrak g =
V_n \oplus \mathfrak r$ by Lemma \ref{l1} and Lemma \ref{l2}, where
$\mathfrak r$ is the $n$-radical of $\mathfrak g$ in dimension 2.

Clearly, $\mathfrak r$ is an $n$-Lie module of $V_n$. By Corollary
\ref{coro}, $\mathfrak r = V_1 \oplus V_2 $, where $V_1,V_2$ are the
1-dimensional $n$-Lie module of $V_n$ with highest weight $0$. It
follows that there is a basis of $e_1, \cdots, e_{n+3}$ of
$\mathfrak g$, such that $e_1, \cdots, e_{n+1} \in V_n$, $e_{n+2}
\in V_1$, $e_{n+3} \in V_2$, and
\begin{equation}\label{zero}
\begin{split}
&[e_1,  \cdots,  \hat{e_i},  \cdots  e_{n+1}]=(-1)^{i} e_i, \quad 1\leqslant i \leqslant n+1, \\
&[e_1, \cdots, \hat{e_i}, \cdots, \hat{e_j}, \cdots, e_{n+1}, e_{n+2}]=0, \quad 1\leqslant i < j\leqslant n+1, \\
&[e_1, \cdots, \hat{e_i}, \cdots, \hat{e_j}, \cdots, e_{n+1}, e_{n+3}]=0, \quad 1\leqslant i < j\leqslant n+1.
\end{split}
\end{equation}

By Jacobi identity, the equations (\ref{zero}) and the fact that
$\mathfrak r$ is an ideal of $\mathfrak g$, we have that
\begin{equation*}
\begin{split}
 [e_1, \cdots, e_{n-2}, e_{n+2}, e_{n+3}] = &  -[[e_2, \cdots, e_{n+1}], e_2, \cdots, e_{n-2}, e_{n+2}, e_{n+3}] \\
= &  -[e_2, \cdots, e_n, [e_{n+1}, e_2, \cdots, e_{n-2}, e_{n+2}, e_{n+3}]] \\
&+\sum_{i=2}^{n-2} [e_{n+1},\cdots,[e_2,\cdots,e_n,e_i],\cdots,e_{n-2},e_{n+2},e_{n+3}] \\
& +  [e_{n+1}, e_2, \cdots, e_{n-2}, [e_2, \cdots, e_n, e_{n+2}], e_{n+3}]\\
& + [e_{n+1}, e_2, \cdots, e_{n-2}, e_{n+2}, [e_2, \cdots, e_n, e_{n+3}]] \\
 =&\ 0.
\end{split}
\end{equation*}

Similarly, $[e_{i_1}, \cdots, e_{i_{n-2}}, e_{n+2}, e_{n+3}]=0$ for
all  $1< i_1< \cdots < i_{n-2}<{n+1}$. That is, $\mathfrak g^1
=V_n$. Namely $\dim \mathfrak g^1=n+1 $.
\end{proof}

\section{classification of $(n+3)$-dimensional metric $n$-Lie
algebra}\label{s3}

By Lemma \ref{2} and Theorem \ref{myt}, it is easy to know the
center of an $(n+3)$-dimensional metric $n$-Lie algebra is nonzero.
Moreover,

\begin{lemma}\label{l3}
The center of a non-abelian $(n+3)$-dimensional metric $n$-Lie
algebra is 1- or 2- dimensional.
\end{lemma}
\begin{proof}
Clearly, $\dim C({\mathfrak g}) \geqslant 1$. If $\dim C({\mathfrak
g}) \geqslant 3$, then $\dim {\mathfrak g}^1\leqslant 1$. It
contradicts Lemma \ref{2}.
\end{proof}

Now let $\mathfrak g$ be an $(n+3)$-dimensional metric $n$-Lie
algebra. If $C(\mathfrak g)$ is non-isotropic, then there exists
$x\in C(\mathfrak g)$ such that $\langle x, x \rangle \neq 0$. Then
the decomposition ${\mathfrak g}={\mathbb R}x\oplus x^\bot$ is a
direct sum of non-degenerate ideals. In general, there is a
well-known fact:

\begin{lemma}
Let ${\mathfrak g}$ be a metric $n$-Lie algebra with a non-isotropic
center. Then $\mathfrak g$ is a direct sum ${\mathfrak g}={\mathfrak
g}_a\oplus {\mathfrak g}_i$, where ${\mathfrak g}_a$ is an abelian
ideal with non-degenerate restriction $\langle, \rangle$ on it and
${\mathfrak g}_i$ is an ideal with an isotropic center $C({\mathfrak
g}_i)$.
\end{lemma}

\begin{lemma}\label{l4} Let $\mathfrak g$ be an $(n+3)$-dimensional metric
$n$-Lie algebra with an isotropic center. Then $\dim C({\mathfrak
g})=2$.
\end{lemma}
\begin{proof} Since $\mathfrak g$ is a metric
$n$-Lie algebra with an isotropic center, we know $\mathfrak g$ is
non-abelian. By Lemma \ref{l3}, $\dim C(\mathfrak g)=1$ or $2$.
Assume that $C(\mathfrak g)$ is isotropic and $\dim C(\mathfrak
g)=1$. Let $e_1, e_2, \cdots, e_{n+3}$ be a basis of $\mathfrak g$
such that $e_1 \in C(\mathfrak g)$, $e_1, \cdots, e_{n+2} \in
{\mathfrak g}^1$, and
$$\langle e_1, e_{n+3} \rangle = 1, \quad\langle
e_i, e_i \rangle = \epsilon_i, \quad\forall 1< i< n+3 ,$$ where
$\epsilon_i$ are signs. Let $c_{i,j}$ denote $[e_2, \cdots,
\hat{e_i}, \cdots, \hat{e_j}, \cdots, e_{n+3}]$ for all $ 2\leqslant
i < j \leqslant n+3$. The nonzero brackets are only the following:
\begin{equation*}
  \begin{split}
   & c_{i,n+3}=[e_2, \cdots, \hat{e_i},  \cdots, e_{n+2}]=a_{i,n+3}^1 e_1 +
a_{i,n+3}^i e_i, \quad\forall 2\leqslant i \leqslant n+2, \\
   & c_{i,j}=[e_2, \cdots, \hat{e_i}, \cdots, \hat{e_j}, \cdots,
e_{n+3}]=a_{i,j}^i e_i + a_{i,j}^j e_j, \quad\forall  2\leqslant i <
j < n+3.
  \end{split}
\end{equation*}
By the definition, we have
\begin{equation*}
\begin{split}
\langle [e_2, \cdots, \hat{e_i},  \cdots, e_{n+2}] , e_{n+3} \rangle
+ (-1)^{n+2-j}\langle [e_2, \cdots, \hat{e_i}, \cdots, \hat{e_j},
\cdots,
e_{n+3}], e_j \rangle = 0, \\
\langle [e_2, \cdots, \hat{e_j},  \cdots, e_{n+2}] , e_{n+3} \rangle
+ (-1)^{n+1-i}\langle [e_2, \cdots, \hat{e_i}, \cdots, \hat{e_j},
\cdots, e_{n+3}], e_i \rangle = 0.
\end{split}
\end{equation*}
It follows that \begin{equation}\label{e1} a_{i,n+3}^1 +
(-1)^{n+2-j}\epsilon_j a_{i,j}^j=0, \quad a_{j,n+3}^1 +
(-1)^{n+1-i}\epsilon_i a_{i,j}^i=0.
\end{equation}
Similarly, we have
\begin{equation*}
\begin{split}
\langle [e_2, \cdots, \hat{e_i},  \cdots e_{n+2}] , e_i \rangle
   &=(-1)^{n+2-j} \langle [e_2, \cdots, \hat{e_i}, \cdots, \hat{e_j},
   \cdots,
e_{n+2}, e_j], e_i \rangle \\
   &=(-1)^{n+1-j} \langle [e_2, \cdots, \hat{e_i}, \cdots, \hat{e_j},
   \cdots,
e_{n+2}, e_i], e_j \rangle \\
   &=(-1)^{n+1-j} (-1)^{n+1-i}\langle [e_2, \cdots, \hat{e_j},
   \cdots,
e_{n+2} ], e_j \rangle \\
   &=(-1)^{i+j}\langle [e_2, \cdots, \hat{e_j}, \cdots,
e_{n+2} ], e_j \rangle.
\end{split}
\end{equation*}
That is, \begin{equation}\label{e2} (-1)^i \epsilon_i a_{i,n+3}^i =
(-1)^j \epsilon_j a_{j,n+3}^j
\end{equation}
By equations (\ref{e1}) and (\ref{e2}), we have, for all
$1<k<i<j<n+3$,
\begin{equation}\label{e3}
\begin{split}
  & a_{k,i}^k c_{k,j}-a_{k,j}^k c_{k,i} \\
=&a_{k,i}^k a_{k,j}^j e_j -a_{k,j}^k a_{k,i}^i e_i \\
=&(-1)^{n-k} \epsilon_k a_{i,n+3}^1 (-1)^{n+1-j} \epsilon_j
a_{k,n+3}^1 e_j -(-1)^{n-k} \epsilon_k a_{j,n+3}^1 (-1)^{n+1-i}
\epsilon_i a_{k,n+3}^1 e_i \\
=&(-1)^{n-k}\epsilon_k a_{i,j}^j a_{k,n+3}^1 e_j +
(-1)^{n-k}\epsilon_k a_{i,j}^i a_{k,n+3}^1 e_i\\
=&(-1)^{n-k}\epsilon_k a_{k,n+3}^1 c_{i,j}.
\end{split}
\end{equation}
Similarly, for all $1<i<j<n+3$,
\begin{equation}\label{e4}
(-1)^{n+1-i} \epsilon_i a_{i,n+3}^i c_{i,j} = -a_{j,n+3}^1
c_{i,n+3}+a_{i,n+3}^1 c_{j,n+3}.
\end{equation}
By induction, $\dim {\mathfrak g}^1 \leqslant n+1$ by equations
(\ref{e3}) and (\ref{e4}). It contradicts $\dim {\mathfrak g}^1 =
n+2$ because $\dim {\mathfrak g}^1 + \dim C({\mathfrak g}) = n+3$.
So the lemma follows.
\end{proof}

\begin{theorem}\label{thm}
Let $\mathfrak g$ be a non-abelian $(n+3)$-dimensional metric
$n$-Lie algebra. Then $\mathfrak g$ must be one of the following
cases (only non-zero products or brackets are given).
\begin{enumerate}
\item There exists a basis $x, e_1,\cdots,e_{n+2}$ of
      $\mathfrak g$ such that $\langle x,x\rangle =\pm 1$, $\langle
      e_1,e_1\rangle =\pm 1$, $\langle e_i, e_i \rangle =1$ for all
      $2\leqslant i\leqslant m$, $\langle e_i, e_i \rangle=-1$ for all $m
      < i \leqslant n+2$ and the brackets are given by
      $$[e_2, \cdots, \hat{e_r}, \cdots, e_{n+2}]=a_r e_r,$$
      where
      $2\leqslant r\leqslant n+2$ and $(-1)^i a_i \langle e_i, e_i \rangle  = (-1)^j a_j \langle
      e_j, e_j \rangle $ for all $2\leqslant i, j \leqslant n+2$.
\item There exists a basis $x, e_1,\cdots,e_{n+2}$ of
      $\mathfrak g$ such that $\langle x,x\rangle =\pm 1$, $\langle e_1,
      e_{n+2}\rangle = 1$, $\langle e_i, e_i \rangle =1$ for all
      $2\leqslant i\leqslant m$, $\langle e_i, e_i \rangle=-1$ for all $m
      < i \leqslant n+1$ and the brackets are given by
      \begin{equation*}
      [e_2, \cdots, e_{n+1}]= a_1 e_1, \quad [e_2, \cdots, \hat{e_i},
      \cdots, e_{n+2}]= a_i e_i,
      \end{equation*}
      where
      $2\leqslant i\leqslant n+1$ and $(-1)^n a_1 = (-1)^i a_i \langle e_i, e_i \rangle .$
\item There exists a basis $e_1,\cdots,e_{n+3}$ of $\mathfrak g$
      such that $\langle e_1, e_{n+3}\rangle =1$, $\langle
      e_2,e_{n+2}\rangle =1$, $\langle e_i, e_i \rangle =1$ for all
      $3\leqslant i\leqslant m$, $\langle e_i, e_i \rangle=-1$ for all $m
      < i \leqslant n+1$ and the brackets are given by
      \begin{equation*}
      \begin{split}
      & [e_3, \cdots, e_{n+1}, e_{n+2}]=a_1 e_1, \quad [e_3, \cdots, e_{n+1}, e_{n+3}]=a_2 e_2, \\
      & [e_3, \cdots, \hat{e_i},  \cdots, e_{n+1}, e_{n+2},   e_{n+3}]=a_i e_i,
\end{split}
\end{equation*}
where $3\leqslant i\leqslant n+1$ and $(-1)^{n+1} a_1 =
(-1)^{n+2}a_2 =(-1)^i a_i \langle e_i, e_i \rangle .$
\end{enumerate}
\end{theorem}
\begin{proof} If $C(\mathfrak g)$ is non-isotropic, then we have the cases (1) and (2) by the classification of
$(n+2)$-dimensional metric $n$-Lie algebras \cite{RCL}. If
$C(\mathfrak g)$ is isotropic, then $\dim C(\mathfrak g)=2$ by Lemma
\ref{l4}. Therefore, there is a basis of $e_1, \cdots, e_{n+3}$ of
$\mathfrak g$ such that $e_1, e_2 \in C(\mathfrak g), e_1, \cdots,
e_{n+1} \in \mathfrak g^1$ and
$$\langle e_1, e_{n+3} \rangle = \langle e_2, e_{n+2} \rangle =1,
\quad \langle e_i, e_i \rangle = \epsilon_i,$$ where $ 3\leqslant
i\leqslant n+1$. Then we have
\begin{equation}\label{id1}
\begin{split}
&[e_3, \cdots, e_{n+2}]=a_1 e_1,\quad [e_3, \cdots, e_{n+1}, e_{n+3}]=a_2 e_2, \\
&[e_3, \cdots, \hat{e_i}, \cdots, e_{n+2}, e_{n+3}]=a_i e_i, \quad
3\leqslant i\leqslant n+1.
\end{split}
\end{equation}
Since $\dim \mathfrak g^1=n+1$, we know $a_i \neq 0$ for all
$1\leqslant i\leqslant n+1$. Also we have
\begin{equation*}
\begin{split}
 \langle [e_3, \cdots, \hat{e_i}, \cdots, e_{n+2}, e_{n+3}], e_i \rangle
= & - \langle [e_3, \cdots, \hat{e_i}, \cdots, e_{n+2}, e_i],
e_{n+3}
\rangle \\
= & - (-1)^{n+2-i} \langle [e_3, \cdots, e_{n+2}], e_{n+3} \rangle.
\end{split}
\end{equation*}
It follows that $(-1)^{n+1}a_1=(-1)^i a_i \epsilon_i$. Similarly,
$(-1)^{n+2}a_2=(-1)^i a_i \epsilon_i$. It is a direct calculation to
check that (\ref{id1}) satisfies the Jacobi identity. Then the
theorem follows. \end{proof}

\begin{coro}
If $\mathfrak g$ is a non-abelian $(n+3)$-dimensional metric $n$-Lie
algebra, then $\dim C(\mathfrak g) = 2$ and $\dim {\mathfrak g}^1 =
n+1.$
\end{coro}

\section{Acknowledgements}
We would like to express our thanks to Prof. Z.X. Zhang for the
communication in this field.


TITLE: Is there no fourth dimension (time) in the universe?
QUESTION [0 upvotes]: In this article (and numerous others) http://phys.org/news/2011-04-scientists-spacetime-dimension.html and http://www.rebelscience.org/Crackpots/notorious.htm they suggest that there is no time dimension and even in the last one that there is no motion in space-time. Do you believe these theories, are they true?

REPLY [2 votes]: Humans have been trying to figure out what time is since we were first able to think about it.  The ideas of Sorli, et al. (your first link) are another attempt.  No one knows if they are right or wrong.  I don't think anyone really has a good intuition about what it means to say that "time is a forth dimension", despite being able to use the concept in practice.  Some day we may have a better way of understanding things, and perhaps Sorli is a step down the right path.
Nothing moves in spacetime?  I suppose that's true.  In spacetime we usually speak of "events", and which event occurs before or after which other event to different users.  But I can verify that in the 3D spatial world that I occupy, motion does occur.


TITLE: Can anyone show that this equation is true?
QUESTION [1 upvotes]: $(e^{ix}-1)^{2n}-(e^{-ix}-1)^{2n}=-(e^{inx}+e^{-inx})(e^{ix}-2+e^{-ix})^n$.
I attempted to write those awful $e$'s into trigonomitric functions, but I failed to prove it. Wolframalpha did not interpret it properly.
Edit Thanks for helping me. The correct result should be $$(e^{ix}-1)^{2n}-(e^{-ix}-1)^{2n}= ( e^{2ix} -2e^{ix}+1)^{n}-(e^{-2ix} -2e^{-ix}+1)^{n}\\= ( e^{2ix} -2e^{ix}+1)^{n}-\frac{(e^{2ix} -2e^{ix}+1)^{n}}{e^{2inx} }=( e^{2ix} -2e^{ix}+1)^{n}\cdot(1-e^{-2inx})=e^{inx}(e^{ix} -2+e^{-ix})^{n}\cdot(1-e^{-2nix})=(e^{ix} -2+e^{-ix})^{n}\cdot(e^{inx}-e^{-inx})$$

REPLY [1 votes]: I obtain a different result
$$(e^{ix}-1)^{2n}-(e^{-ix}-1)^{2n}= ( e^{2ix} -2e^{ix}+1)^{n}-(e^{-2ix} -2e^{-ix}+1)^{n}\\= ( e^{2ix} -2e^{ix}+1)^{n}-\frac{(e^{2ix} -2e^{ix}+1)^{n}}{e^{2inx} }=( e^{2ix} -2e^{ix}+1)^{n}\cdot(1-e^{-2inx})=e^{inx}(e^{ix} -2+e^{-ix})^{n}\cdot(1-e^{-2nix})=(e^{ix} -2+e^{-ix})^{n}\cdot(e^{inx}-e^{-inx})$$


TITLE: Prove that there is some $x \in M$ ($M$ is a metric space) such that $f(x) = 0$
QUESTION [0 upvotes]: Say that $\{x_n\}^\infty_\mathrm{n=1}$ is a sequence of points in a sequentially compact metric space $(M,d)$ such that for each $n$,
$f(x_n) = \frac{1}{n}$ where $f$ is a continuous function in the metric space.
I'm trying to prove that there is some $x \in M$ such that $f(x) = 0$.
I know $\frac{1}{n}$ forms a convergent sequence, but how could it ever equal $0$.

REPLY [0 votes]: Hints:

What does the compactness of $M$ tell us about $\{x_n\}_{n\in\mathbb{N}}$?

What does $\{\frac{1}{n}\}_{n\in\mathbb{N}}$ converge to?


TITLE: Homology of $\mathbb{R}P^\infty$
QUESTION [1 upvotes]: I want to calculate all the homology groups of $\mathbb{R}P^\infty = \bigcup\limits_{i=o}^\infty \mathbb{R}P^n$. The tools I have are Mayer-Vietoris sequence, cofibre long exact sequences (though I don't know a cofibration involving this space), and cellular homology.
First, I'm wondering how to interpret this union. Does it involve disjoint copies of the real projective plane in each dimension? Is it like a wedge sum at the basepoint? In particular, is $\mathbb{R}P^\infty$ path connected?
Second, I'm trying to calculate homology using the cellular chain complex. However, since $\mathbb{R}P^n$ has one cell in every dimension, wouldn't $\mathbb{R}P^\infty$ have infinitely many cells in every dimension? Is it still true that every other map $C_i(\mathbb{R}P^\infty)\rightarrow C_{i-1}(\mathbb{R}P^\infty)$ is the zero map or multiplication by two (like in cellular homology calculations of $\mathbb{R}P^n$ for finite $n$)? Why or why not?
Any help is appreciated!

REPLY [3 votes]: The union is not a wedge, but a sequence of spaces $\mathbb{RP}^1\subset \mathbb{RP}^2\subset \dots$ where each space is a CW subcomplex of the next. In particular, $\mathbb{RP}^n$ is the $n$-skeleton of $\mathbb{RP}^\infty$.
It seems like you already understand the cellular chain complexes of each $\mathbb{RP}^n$, and the cellular chain complex of $\mathbb{RP}^\infty$ can be inferred from them: for each $n$ the inclusion $\mathbb{RP}^n \hookrightarrow \mathbb{RP}^\infty$ induces isomorphisms $C_k(\mathbb{RP}^n) \cong C_k(\mathbb{RP}^\infty)$ for each $k \leq n$, and the boundary operators $\partial_k\colon C_k(\mathbb{RP}^\infty)\to C_{k-1}(\mathbb{RP}^\infty)$ for $k\leq n$ agree with those of $\mathbb{RP}^n$ by naturality.
In general if $X$ is a CW complex and $X^{(n)}$ is its $n$-skeleton, then the inclusion $X^{(n)}\hookrightarrow X$ induces isomorphisms $H_k(X^{(n)}) \cong H_k(X)$ for all $k<n$ (exercise: prove this). In particular, if you want to know $H_k(\mathbb{RP}^\infty)$ you could choose an $n>k$ and compute $H_k(\mathbb{RP}^n)$.


TITLE: Technical problem of definition of length of a worldline using a metric $g$
QUESTION [1 upvotes]: In lecture 10 of this series, the professor defines the notions of speed and length of a curve in a smooth manifold equipped with a metric $g$. These definitions are made between 33:10 and 44:10 and are the following:
$$ \text{"speed"} = s(\lambda) := \sqrt{g(v_\gamma, v_\gamma)} \tag{1}$$
$$ \text{"length"} = L[\gamma] = \int_0^1\sqrt{g(v_\gamma, v_\gamma)} \text{d}\lambda \tag{2}$$
where $\gamma: (0,1) \longrightarrow \mathbb{R} $ is a differentiable curve and $v_\gamma :=  \dot{\gamma}^i \partial_i$ is the tangent vector field along $\gamma$. So far so good.
My problem comes in lecture 13 of the same playlist, when at 01:02:50, the professor starts to talk about measurements of time in special relativity.
My problem: is still correct define length of a worldline, for example, as in (2)? If the curve is a particle trajectory and I choose the static gauge $\lambda=t$, t is coordinate time, then talk about an inner product $\sqrt{g(v_\gamma, v_\gamma)}$ makes no longer sense since $v_\gamma = \dfrac{\text{d}\gamma^i}{\text{d}t} \partial_i$ is not a "true vector", anymore since its components no longer satisfies the usual vector components Transformation law.
However, I still know that the more explicit expression
$$\int \sqrt{g_{ij}\dfrac{\text{d}\gamma^i}{\text{d}t} \dfrac{\text{d}\gamma^j}{\text{d}t   }} \text{d}t$$
is correct because most of the Relativity textbook authors use expressions similar to the above to define proper time of a massive particle, even if the argument of the square root is not an inner product  (for coordinate time). How does this proceed?

REPLY [1 votes]: This is part of the reason why "a vector is something that transforms as a vector" is a bad definition. $v^i = d \gamma^i/dt$ is a perfectly good vector because you've fixed a specific coordinate chart, which $t$ refers to. The problem is that there are two statements that look similar but are crucially different:

"In any coordinate system $(t, x^i)$, define four components by $v^i = d\gamma^i / dt$." This is not a vector: if you use each system's $t$ coordinate to define the components, they won't relate to each other through the correct transformation law.

"In a specific coordinate system $(t, x^i)$, define four components by $v^i = d\gamma^i / dt$. To find the components in another coordinate system, just use the transformation law." This is a vector, but no one tells you the last part explicitly, and people get confused.

You can always define a vector by giving its components in one coordinate system and using the transformation law to find its components in any other coordinate system. The "a vector transforms like a vector" thing is important when you want a single formula that works in all coordinate systems, which is often the case, but not the case here.


TITLE: Variance of projection of vectors onto random subspace
QUESTION [2 upvotes]: Let $x_i, y_i \in \mathbb{R}^n$ for $i=1, \dots, k < n$ satisfy
$$
\sum_{i=1}^k x_i^\top y_i = 0.
$$
Let $E$ be a random subspace of dimension $m < n$ in $\mathbb{R}^n$ distributed uniformly on the Grassmannian $G_m(\mathbb{R}^n)$ and let $P_E$ be orthogonal projection onto $E$. What is the value of
$$
\mathbb{E}\left[ \left(\sum_{i=1}^k x_i^\top P_E y_i \right)^2 \right] = \dots ?
$$
Any references or suggestions would also be very helpful.
It's quite easy to get an upper bound using Johnson-Lindenstrauss style arguments but I feel that this should be possible to compute exactly. Also (by orthogonal invariance) the first moment vanishes.

REPLY [3 votes]: It suffices to calculate $\mathbb E ( (P_E)_{ab} (P_E)_{cd})$ for all $1\leq a,b,c,d\leq n$, or, equivalently, calculate $\mathbb E( P_E \otimes P_E ) \in \mathbb R^n \otimes \mathbb R^n \otimes \mathbb R^n \otimes \mathbb R^n$.
This is an $O(n)$-invariant tensor. The space of $O(n)$-invariant tensors in $\mathbb R^n \otimes \mathbb R^n \otimes \mathbb R^n \otimes \mathbb R^n$ is three-dimensional, generated by the three "diagonal" tensors. Thus
$$\mathbb E ( (P_E)_{ab} (P_E)_{cd}) = C_1 \delta_{a=b}\delta_{c=d}+ C_2 \delta_{a=c} \delta_{b=d} + C_3 \delta_{a=c} \delta_{b=d}$$ for some $C_1, C_2, C_3 \in \mathbb R$.
We have $$ m^2 = \sum_{i=1}^n (P_E)_{ii} \sum_{j=1}^n (P_E)_{jj} = \sum_{i=1}^n \sum_{j=1}^n \mathbb E (  (P_E)_{ii}  (P_E)_{jj}) = n^2 C_1 + n C_2 + n C_3 $$
We have $$ m = \sum_{i=1}^n \sum_{j=1}^n (P_E)_{ij}^2 =  \sum_{i=1}^n \sum_{j=1}^n \mathbb E (  (P_E)_{ij}  (P_E)_{ji}) =  n C_1 + n C_2 + n ^2 C_3 $$
and similarly
$$ m = \sum_{i=1}^n \sum_{j=1}^n (P_E)_{ij}^2 =  \sum_{i=1}^n \sum_{j=1}^n \mathbb E (  (P_E)_{ij}  (P_E)_{ij}) =  n C_1 + n^2 C_2 + n C_3 $$
This gives $$C_1 = \frac{(m^2-m) (n+2) + m (n-m)}{n (n-1) (n+2) } , C_2 = \frac{m (n-m)}{ n (n-1)(n+2)}, C_3 = \frac{ m (n-m)}{ n (n-1) (n+2) } $$
We have
$$\mathbb E \left[ \left( \sum_{i=1}^k x_i^T P_E y_i\right)^2 \right] = \mathbb E\left[ \sum_{i=1}^k \sum_{j=1}^k x_i^T P_E y_i x_j^T P_E y_j \right]$$ $$ = C_1 \sum_{i=1}^k \sum_{j=1}^k (x_i \cdot y_i) (x_j \cdot y_j) + C_2 \sum_{i=1}^k \sum_{j=1}^k (x_i \cdot x_j) (y_i \cdot y_j) + C_3 \sum_{i=1}^k \sum_{j=1}^k (x_i \cdot y_j) (y_i \cdot x_j) $$
Your first term vanishes by assumption, so we get
$$ \frac{m (n-m)}{ n (n-1) (n+2) } \left( \sum_{i=1}^k \sum_{j=1}^k (x_i \cdot x_j) (y_i \cdot y_j) + C_3 \sum_{i=1}^k \sum_{j=1}^k (x_i \cdot y_j) (y_i \cdot x_j) \right).$$
Expressed in terms of the matrix $M = \sum_{i=1}^k x_i y_i^T$, this is
$$ \frac{m (n-m)}{ n (n-1) (n+2) }\operatorname{tr}( M M^T + M^2).$$


TITLE: Resonance frequency of a guitar
QUESTION [0 upvotes]: I have some questions for you about the resonance phenomenon in a guitar:

Why resonance frequency are only full integer multiples of the natural frequency, such as 2 x, 3 x, or 4 x or 3 x, 5 x, or 7 the natural frequency?

Why a vibrating string tend to vibrate at the natural resonant frequency(x1) and not x2 or x3?

Is it true that in a perfect world, in a string of a guitar the resonance frequencies(x1,x2,x3,x4,ecc...) will never die out? Why non-integer multiples, otherwise, will die out?

REPLY [0 votes]: I think the question applies not specifically to a guitar, but to any string. (To be precise, there are some differences in how sound is generated between guitars and instruments like violin, which I address in the end.)

Oscillations of a guitar string are a standing wave, which has its regions of high amplitude and nodes with zero amplitude. The nodes are separated by half wavelength distance. Since the ends of the string are fixed, the only waves that can be excited in such a string are those that have nodes at the ends of the string, which means that the length of the string is equal to a half integer number of wave lengths. Note that not only these tones coincide with the natural scale - this is precisely the reason why it is called natural scale. Very similar arguments can be applied to other types of musical instruments.

Obviously, when we touch a string, we excite many modes. The lowest usually has the highest energy and dominates the sound, while the rest determine the timber/quality of the sound. This is why the sound quality can be changed by touching a string in different places - an effect used in playing classical acoustic guitar: by placing the right hand closer or farther from the middle of the string the musician can extract sounds that are deeper or more dry. Are more sophisticated techniques is by extracting harmonics sounds with one's left hand, giving them a particular overtone.

From the answer to point 1 it should be already clear why non-integer waves die out. By "ideal world" one often means the world without dissipation. This is hardly applicable to a guitar, whose point is to generate sound, i.e., to transfer the energy from the string to the air oscillations. The string oscillations in vacuum however would not die out indeed. The damping of the string oscillations of a guitar is one of the reasons why (before appearance of electrical amplifiers) this instrument was impractical for anything but chamber music. Bowed string instruments (violin, cello, etc.) have the advantage that the sound can be excited continuously while the musician moves the bow along the string.


TITLE: Compute an $n \times n$ determinant
QUESTION [2 upvotes]: Compute the following determinant
$$
\begin{vmatrix}
a_{1}-b_{1}+x & a_{1}-b_{2}   & \ldots & a_{1}-b_{n}  \\
a_{2}-b_{1}   & a_{2}-b_{2}+x & \ldots & a_{2}-b_{n}  \\
     \vdots   & \vdots        & \ddots & \vdots       \\
a_{n}-b_{1}   & a_{n}-b_{2}   & \ldots & a_{n}-b_{n}+x
\end{vmatrix}.
$$

I tried for $n=3$ and I used the fact that $\det$ funcion is liniar in each
column.
\begin{align*}
&
\begin{vmatrix}
a_{1}-b_{1}+x & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2}+x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x
\end{vmatrix}
\\
& =
\begin{vmatrix}
a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2}+x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x
\end{vmatrix}
+%
\begin{vmatrix}
x & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2}+x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x
\end{vmatrix}
\\
& =%
\begin{vmatrix}
a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x
\end{vmatrix}
+%
\begin{vmatrix}
a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x
\end{vmatrix}
\\
& +%
\begin{vmatrix}
x & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x
\end{vmatrix}
+%
\begin{vmatrix}
x & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}+x
\end{vmatrix}
\\
& =%
\begin{vmatrix}
a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}%
\end{vmatrix}
+%
\begin{vmatrix}
a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & x
\end{vmatrix}
\\
& +%
\begin{vmatrix}
a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}%
\end{vmatrix}
+%
\begin{vmatrix}
a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & x
\end{vmatrix}
\\
& +%
\begin{vmatrix}
x & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}%
\end{vmatrix}
+%
\begin{vmatrix}
x & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & x
\end{vmatrix}
\\
& +%
\begin{vmatrix}
x & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}%
\end{vmatrix}
+%
\begin{vmatrix}
x & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & x
\end{vmatrix}
\\
& =%
\begin{vmatrix}
a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2} & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}%
\end{vmatrix}
+%
\begin{vmatrix}
a_{1}-b_{1} & a_{1}-b_{2} & a_{1}-b_{3}\\
a_{2}-b_{1} & x & a_{2}-b_{3}\\
a_{3}-b_{1} & a_{3}-b_{2} & a_{3}-b_{3}%
\end{vmatrix}
\\
& +%
\begin{vmatrix}
x & 2\left(  a_{1}-b_{2}\right)   & a_{1}-b_{3}\\
a_{2}-b_{1} & a_{2}-b_{2}+x & a_{2}-b_{3}\\
a_{3}-b_{1} & 2\left(  a_{3}-b_{2}\right)   & a_{3}-b_{3}%
\end{vmatrix}
+%
\begin{vmatrix}
x+a_{1}-b_{1} & 2\left(  a_{1}-b_{2}\right)   & a_{1}-b_{3}\\
2\left(  a_{2}-b_{1}\right)   & x+a_{2}-b_{2} & a_{2}-b_{3}\\
2\left(  a_{3}-b_{1}\right)   & 2\left(  a_{3}-b_{2}\right)   & x
\end{vmatrix}
\end{align*}
From here I didn't see any pattern or some manipulations to do in order to compute the determinant. The first determinants after the last $=$ sign is equal to zero after some calculation using Sarrus rule. How can one compute this determinant?

REPLY [3 votes]: Given vectors $\rm{a}, \rm{b} \in \mathbb R^n$, where $n \geq 2$, let function $f : \mathbb R \to \mathbb R$ be defined by
$$f (x) := \det \left( x \rm{I}_n + \rm{a} \Bbb{1}_n^\top - \Bbb{1}_n \rm{b}^\top \right)$$
Using the matrix determinant lemma,
$$\begin{aligned} f (x) &= \det \left( x \rm{I}_n + \begin{bmatrix} | & | \\ \rm{a} & -\Bbb{1}_n\\  | & | \end{bmatrix} \begin{bmatrix} | & | \\ \Bbb{1}_n & \rm{b} \\  | & | \end{bmatrix}^\top \right) \\ &= \det \left( x \rm{I}_2 + \begin{bmatrix} | & | \\ \Bbb{1}_n & \rm{b} \\  | & | \end{bmatrix}^\top \begin{bmatrix} | & | \\ \rm{a} & -\Bbb{1}_n\\  | & | \end{bmatrix} \right) \cdot x^{n-2}\\ &= \det \left( x \rm{I}_2 + \begin{bmatrix} \Bbb{1}_n^\top \rm{a} & -n\\ \rm{a}^\top \rm{b} & -\Bbb{1}_n^\top \rm{b}\end{bmatrix} \right) \cdot x^{n-2}\\ &=\left( x^2 + \mbox{tr} \begin{bmatrix} \Bbb{1}_n^\top \rm{a} & -n\\ \rm{a}^\top \rm{b} & -\Bbb{1}_n^\top \rm{b}\end{bmatrix} x + \det \begin{bmatrix} \Bbb{1}_n^\top \rm{a} & -n\\ \rm{a}^\top \rm{b} & -\Bbb{1}_n^\top \rm{b}\end{bmatrix}\right) \cdot x^{n-2}\\ &= \color{blue}{x^n + \Bbb{1}_n^\top \left(\rm{a} - \rm{b} \right) x^{n-1} + \left( n \, \rm{a}^\top \rm{b} - \Bbb{1}_n^\top \rm{a} \rm{b}^\top \Bbb{1}_n \right) x^{n-2}} \end{aligned}$$