TITLE: Emergence of Dual Resonance model from QCD
QUESTION [5 upvotes]: This question is about getting to understand better what the Dual Resonance (DR) model actually managed to successfully predict or model from the nuclear interactions, in the context of understanding what aspects of the DR are expected to be emergent from QCD
My summary and very brief understanding of the whole affair is that the Veneziano amplitude was found somehow, verified to describe meson scattering amplitudes, and hold several symmetries that lead to the conclusion that a tower of infinite harmonic oscillators must exist, which can be thought as string-like
Since the Veneziano amplitudes are in fact verified experimentally (at least for 4 particles), what can be said about that tower of infinite harmonic oscillators in the case of nuclear interactions? are they emergent phenomena in QCD? Have they been observed experimentally? What observables should such oscillators affect?
Are there any "stringy" aspects of nuclear interactions expected to be emergent from QCD?

REPLY [1 votes]: First of all, although there are very deep similarities shared between string theory and "real world QCD" as in the AdS/CFT correspondence or in more general large N limit scenarios, it's very important to bear in mind that up to now there are no precise identifications neither between string theory and real world QCD nor on the dual resonance models as "effective theories" emergent from QCD, even more, the statement that the dual resonance models can be "effective theories" derived from QCD-like theories is in strong contradiction with the bootstrap philosophy.
With the latter in mind, I want to enummerate some formal similarities that can be seen, to some extend, as "emergent properties" in dual resonance models when this ones are viewed as qualitatively similar to "effective QCD-like theories".
Similarity one: (The low energy spectrum) The low lying spectrum of a color confined phase of any QCD-like theory is dominated by meson and baryon excitations. The obvious similarity is that the weakly coupled description of any string vacua is given in terms of open and closed strings. Open strings are analogous to mesons (flux tubes joining quarks) and a closed string is the analogue of the pomeron. After all, the spectrum of rotating open strings is what ultimately produce Regge trajectories in string theory and the $t$-channel pomeron pole can be thinked as an analogue of how closed strings appear as S-matrix poles in one-loop open string computations.
It's worth to mention that scale invariance is broken when a QCD vacuum develops the trace anomaly intuitively because in a confined phase there is a characteristic lenght scale, roughly the mass of the lightest pseudoscalar meson in natural units. String theory also has a characteristic scale in its target space, nambely the string lenght, except that this latter is a dynamical parameter (as any parameter is in the string theory).
The impresition here is, of course, that mesons and hadrons are widely more complex that just a simple bound state of its constituent valence quarks.
Similarity two:(Fermions as D-branes) Consider a QCD lagrangian with $ SU(N_{f})\times SU(N_{f})$ chiral flavor symmetry, this symmetry is spontaneuosly broken to a diagonal $SU(N_{f})$ in vacuum (details here), the resulting chiral condensate involve bilinears $\langle \psi^{*} \psi \rangle$, the Nambu pseudogoldstone mesons with the fermions $\psi$ as its constitutents can be thinked as analogous to open strings between D-branes and the fermions(quarks) $\psi$ as the analogue to D-branes. This analogy can be made precise in topological string theories were D-branes are literally fermion instertions in correlator functions (read the page 39 of this paper for details).
Similarity three: (The existence of a chiral-like perturbation theory) In the context of the similarity two it's a "miracle" that something like chiral perturbation theory exists. That's ultimately related to the fact that the inverse mass of the constitutent quarks of a pseudoscalar meson can be trated as small parameters if the mass of the aformentioned quarks is large. That's similar to the "miracle" of the integrability of the open string perturbation theory, here the very large mass of the quarks to what strings are attached to is justified because they were identified as D-branes.


TITLE: Volume form on a compact manifold is not exact
QUESTION [2 upvotes]: I am trying to show that a volume form $\mu$ on a compact manifold $M$ is not exact, i.e. show there is no $\alpha \in \Omega^{n-1}(M)$ such that $d\alpha = \mu$.
My attempt is the following: Suppose, as a contradiction, that $\mu$ is exact. Then, there exist an $(n-1)$-form $\alpha$ such that $d\alpha = \mu$. Then, since compact manifolds are manifolds without boundary, by stokes theorem we know that $\int_M \mu = \int_M d\alpha = \int_{\partial M} \alpha = 0$ since $M$ has no boundary. From here I am not sure how to continue to end my contradiction. I would appreciate any hint or suggestions for the problem. Thanks!

REPLY [2 votes]: By definition, the integral over a manifold of a top (volume) form must be positive. Thus, since we have that $\int_M \mu = 0$, this is the contradiction we are looking for and so we deduce that a volume form on a compact manifold is not exact.


TITLE: Is the set of all $n\times n$ matrices, such that for a fixed matrix B  AB=BA, a subspace of the vector space of all $n\times n$ matrices?
QUESTION [5 upvotes]: Is the set of all $n\times n$ matrices, such that for any fixed matrix B AB=BA, a subspace of the vector space of all $n\times n$ matrices?

Alright, I understand the question and I know what I have to do, basically. I need to show that additive closure and multiplicative closure are satisfied. The problem is, I can't figure out how to do this generally. I tried playing around with $2\times 2$ matrices but that seemed like a dead end. Obviously two such matrices are the 0 matrix and the identity matrix, and those form a subspace, but that doesn't really tell me about all the matrices. Any ideas for how I should be tackling this? I feel like I'm not thinking generally enough.

REPLY [0 votes]: Here is a different kind of answer that is just a lot harder for no reason, but might help you transition.
Suppose we had a specific B, such as:
$$B = \begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}$$
We want to find all A so that AB = BA.  Well, we don't know the entries of A, so we replace the unknown numbers with... well unkowns, also known as variables.
$$A = \begin{bmatrix} x & y \\ z & t \end{bmatrix}$$
Ok, now we write down what we know about the variables from AB = BA:
$$
AB =
\begin{bmatrix} x & y \\ z & t \end{bmatrix}
\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix} =
\begin{bmatrix} 2x+4y & 3x+5y \\ 2z+4t & 3z+5t \end{bmatrix}
$$
$$
BA =
\begin{bmatrix} 2 & 3 \\ 4 & 5 \end{bmatrix}
\begin{bmatrix} x & y \\ z & t \end{bmatrix} =
\begin{bmatrix} 2x+3z & 2y+3t \\ 4x+5z & 4y+5t \end{bmatrix}
$$
For two matrices to be equal, we need to have the entries equal:
$$\left\{\begin{align}
2x + 4y &= 2x + 3z \\ 2z+4t &= 4x+5z \\ 4x+5y &= 2y + 3t \\ 3z+5t &= 4y+5t \end{align}\right.$$
What do you know?  We have a system of linear equations.  We can solve them as usual: find a particular solution (oooo let me let me!): $(x=0,y=0,z=0,t=0)$.  Now we know all other solutions are found by adding a "homogeneous solution", which form a subspace.
If we want, we can be even more linear-algebra-y.  let's move all the variables to one side:
$$\left\{\begin{align}
0x + 4y - 3z + 0t &= 0 \\ -4x+0y-3z+4t &= 0 \\ 4x+3y +0z -3t &= 0 \\ 0x-4y+3z+0t &= 0 \end{align}\right.$$
We can write it as a matrix equation:
$$\left[\begin{array}{rrrr}
0 & 4 & -3 & 0 \\ -4 & 0 & -3 & 4 \\ 4 & 3 & 0 & -3 \\ 0 & -4 & 3 & 0 \end{array}\right] \begin{bmatrix} x \\ y \\ z \\ t \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$$
So we are just asking for the null space of the matrix!  Clearly that is a subspace.  We could even do Gaussian elimination to find which one.  I won't.


TITLE: Lipschitz maps on locally compact groups
QUESTION [1 upvotes]: Suppose $G$ is a locally compact second countable group. This means that there exists a proper (closed bounded sets are compact) left invariant ($d(gx,gy) = d(x,y) \ \forall g,x,y \in G$) metric on the group that generates the topology on $G$.
Let $A \subset G$ be a bounded set in $G$. For any point $x \in G$ define $d(x, A ) = \inf \{ d(x,y) : y \in A \}$. For all $\varepsilon > 0$ let $N_{\varepsilon}(A) = \{ x \in G : d(x,A) < \varepsilon \}$.
My question is, does there exist a Lipschitz map that is 1 on $A$ and 0 on $G \setminus N_{\varepsilon}(A)$ such that the Lipschitz constant does not depend on the size of $A$ and only depends on $\varepsilon$?
Here a function $f \colon G \to \mathbb{C}$ is Lipschitz if there exists $M > 0$ such that for all  $g,h \in G$, $|f(g) - f(h)| \leq M d(g,h)$. Here $M$ is the Lipschitz constant.
My guess would be that the Lipschitz constant would be something like $1 / \varepsilon$.

REPLY [3 votes]: There is a Lipschitz function $\phi:[0,\infty)\to[0,1]$, with Lipschitz constant exactly $1/\epsilon$, such that $\phi(0)=1$ and $\phi(t)=0$ for $t\ge\epsilon$. Let $f(x)=\phi(d(x,A))$.


TITLE: Dimension of $\mathfrak{sl}_n$ modules
QUESTION [2 upvotes]: I was wondering if anyone knows where I can find a formula for the dimension of an irreducible module of highest weight $\Lambda$ expressed only in terms of the Young diagram corresponding to $\Lambda$.
Thank you.

REPLY [3 votes]: One influential older source to consult is the concise Springer Lecture Notes No. 682 by Gordon James The Representation Theory of the Symmetric Groups (1978).   Section 26 applies the symmetric group theory to general linear groups, using the language of "Weyl modules" and partitions.   See in particular his Theorem 26.5 for the dimension formula in terms of semistandard partitions.    It takes some work, however, to get into his notational scheme, though his overall approach to the subject has the virtue of leading naturally into fields of prime characteristic while also treating the classical Schur-Weyl duality.
Note by the way that in the classical setting there is no essential difference between the group and Lie algebra theories here, while the transition from general linear to special linear is quite easy.
ADDED: I'm still unsure what is really needed, but I don't see the partition formulation of weights for general (or special) linear groups or Lie algebras to be helpful in considering duality.   In the notes by James, or in many textbooks like Fulton-Harris (with varying notation), one studies the irreducible representations of a fixed $GL_d$ on the $n$-th tensor powers of the natural $d$-dimensional module for variable $n$ by using the representation theory of the symmetric groups $S_n$.  Here a partition of $n$ into at most $d$ parts corresponds nicely to a highest weight.   But it's tricky to bring the transpose (conjugate) partition into this picture, since its number of parts might exceed $d$.
The role of the transpose partition for $S_n$ itself is more straightforward:  Partitions of $n$ naturally index irreducible representations of $S_n$.  Then the transpose partition indexes the product of the given representation and the sign representation, with no change in dimension.   But the relationship of this machinery (via Schur functors) to representations of each fixed $GL_d$ is indirect.


TITLE: Can exponentiation be seen as a group morphism?
QUESTION [1 upvotes]: In this video by "Three Blue One Brown" dealing with group theory  https://www.youtube.com/watch?v=mvmuCPvRoWQ
the operation of exponentiation is presented as a group morphism, that is a structure preserving function from an additive group to a multplicative group.
the quick justification provided at 16:52 is the exponent law $n^{a+b} = n^a\times n^b$. I think the author only takes this law as an example of a more general " exponential property" that shows up regularly in group theory.
Is there a not too complicated way to make this idea more precise?

REPLY [0 votes]: The homomorphism property is $h(x+y)=h(x)+h(y)$.  The exponential map $h:(\Bbb R,+)\to(\Bbb R^+,\cdot)$, given by $h(x)=e^x$, indeed satisfies this property.  We have $h(x+y)=e^{x+y}=e^x\cdot e^y=h(x)\cdot h(y)$.


TITLE: Reviewing a paper and found a better solution
QUESTION [16 upvotes]: I was reviewing a paper of a double-blind conference (ML/AI-based conference). The authors improved the approximation bounds for some special instances of a problem.
To understand their proof better, I was thinking of a solution by myself while picking some hints from their proof. Using those hints, I came up with a better algorithm for the general instance of the problem. Moreover, my algorithm is much simpler than theirs.
I am feeling a bit greedy here; I want to suggest these changes to them and also want authorship in their paper. I can not ask them since I do not know them. I can not publish my own paper since a few ideas are borrowed from their proof and their manuscript is not online. What should I do?

REPLY [6 votes]: I made a flow chart describing what I think is the typical process in this case:

You will notice many articles end with "We would like to acknowledge the anonymous reviewer whose suggestions tremendously simplified the proof ..."

REPLY [1 votes]: Some reviewing guidelines state explicitly that referees are not supposed to communicate any original research as part of their review letter.
If this is not the case, there is the option of telling the authors (as well as the area chairs) openly about the dilemma you find yourself in and, at the same time, sketching your generalization in sufficient detail. Your trust in the authors' fairness in handling the situation might be rewarded by a fruitful collaboration.


TITLE: Asymptotic Matching for boundary layer problem
QUESTION [2 upvotes]: The question asks to find a global approximation to the leading order of $\epsilon$.
$\epsilon y'' + xy' + \epsilon y =0$, with boundary conditions $y(0)=1,y(1)=-1$.
I assumed it's a boundary layer problem, and tried to work with both $x=0$ and $x=1$.  But did not work out...
(new edit below...)
I start by trying boundary layer at x=0 and x=1 normal.  Outer solution is $y=-1$, for inner solution, do the substitution of $x = \epsilon X$, $Y=y(x/\epsilon)$, then $\frac{d^2 Y}{d X^2} + \epsilon \frac{dY}{dX} + \epsilon ^ 2 Y = 0$,which when take $\epsilon \rightarrow 0$, becomes $\frac{d^2 Y}{d X^2} = 0$, and gives you an inner solution of $Ax/\epsilon +B$.  I can't match this with inner solution...
However if I do boundary layer at $x=1$, the outer solution at $x=0$ would just be $y=1$, for inner solution, I did substitution of $1-x = \epsilon X$, the equation then becomes $\frac{d^2 Y}{d X^2} - \frac{dY}{dX} =0$, which will give the inner solution $y = A + B exp(\frac{1-x}{\epsilon}$.  I can't match this with outer solution either, since when $\frac{1-x}{\epsilon} \rightarrow \infty$, inner solution would blow up.
Somehow I feel I am missing something, even the outer solution might not be something that simple.  since when $\epsilon \rightarrow 0$, the equation becomes $y'=0$, but in that case $\epsilon$ terms aren't negligible any more.

REPLY [7 votes]: The outer solution $y_{\text{outer}}(x) = -1$ is fine, but it turns out that in this problem the width of the boundary layer isn't $\approx \epsilon$ as you have assumed with your substitution $x = \epsilon X$.  To determine the correct width, instead substitute $x = \epsilon^{\alpha} X$, where $\alpha$ is a constant to be determined.  The equation
$$
\epsilon y'' + xy' + \epsilon y = 0
$$
then becomes
$$
\epsilon^{1-2\alpha} \frac{d^2 Y}{dX^2} + X \frac{dY}{dX} + \epsilon Y = 0.
$$
The third term, $\epsilon Y$, being order $\epsilon$, is certainly smaller than the second term, $X \frac{dY}{dX}$, which is order $1$.  Therefore the third term must not be involved in the dominant balance.  This only leaves the first two terms, which are balanced when $\alpha = 1/2$.  The resulting equation is then, to leading order,
$$
\frac{d^2 Y}{dX^2} + X \frac{dY}{dX} = 0.
$$
After multiplying by the integrating factor $e^{X^2/2}$ we can write the equation as
$$
\frac{d}{dX} \left(e^{X^2/2} \frac{dY}{dX}\right) = 0,
$$
and integrating yields
$$
\frac{dY}{dX} = A e^{-X^2/2}.
$$
Integrating once more we see that
$$
\newcommand{erf}{\operatorname{erf}}
Y(X) = A\erf\left(\frac{X}{\sqrt{2}}\right) + B,
$$
where $\erf$ is the error function.  To satisfy the boundary condition $y(0) = 1$ we must then take $B = 1$.  To match this with the outer solution we need
$$
\lim_{X \to \infty} Y(X) = y_{\text{outer}}(0) = -1,
$$
which is satisfied when $A = -2$.
We now have the inner and outer solutions, namely
$$
y_{\text{outer}}(x) = -1, \\
y_{\text{inner}}(x) = 1 - 2\erf\left(\frac{x}{\sqrt{2\epsilon}}\right).
$$
The matched solution is therefore
$$
y(x) \sim y_{\text{inner}}(x) + y_{\text{outer}}(x) - y_{\text{outer}}(0) = 1 - 2\erf\left(\frac{x}{\sqrt{2\epsilon}}\right).
$$
Here's a plot which shows the actual solution in blue and this approximate solution in purple for $\epsilon = 1/50$.


TITLE: Dual of a polyhedral cone
QUESTION [1 upvotes]: A general polyhedral cone $\mathcal{P} \subseteq \mathbb{R}^n$ can be represented as either $\mathcal{P} = \{x \in \mathbb{R}^n : Ax \geq 0 \}$ or $\mathcal{P} = \{V x : x \in \mathbb{R}_+^k , V \in \mathbb{R}^{n \times k} \}$.
I am trying to do show the dual of $\mathcal{P}$, $\mathcal{P}^*$, is a polyhedral set.
I start by writing
$\mathcal{P}^* = \{ y \in \mathbb{R}^n : y^T V x \geq 0$ $\forall x \in \mathbb{R}_+^k \}$. (1)
A polyhedral set is a set of the form $\{x \in \mathbb{R}^n : Ax \leq b \}$ (2).
Any ideas how to go from (1) to (2). I also know that an image of a polyhedral set under a linear map is also polyhedral.
PS: Similar questions like these have very advanced solution methods; I am looking for a much simpler way.

REPLY [0 votes]: By metric geometry duality most often is obtained via spherical reciprocation.
So just insert a sphere into the cone such that it is tangent to both, the base and the lateral shell, then obviously those contact points to both surface parts will define the (then inscribed) dual cone.
--- rk