dg.differential geometry – Is there a metric on Euclidean space that turns the Helmholtz equation into the Laplace equation?

Is there a Riemannian metric $tilde g$ on $mathbb R^d$ such that
$$tag{1}
Delta_{tilde g}=e^f(Delta +1),$$

for some $fin C^infty(mathbb R^d)$? Here $Delta=partial_{x_1}^2+ldots+partial_{x_d}^2.$

If there is such a $tilde g$, it cannot be conformal to the standard Euclidean metric $g=delta_{ij}$. Indeed, if $tilde g = e^{2phi}g$, then
$$Delta_{tilde g} = e^{-2phi} left(Delta + (d-2)g^{ij}frac{partial phi}{partial x_j}frac{partial}{partial x_i}right),$$
and either $d=2$, or the second summand in the round brackets is constant only in the trivial case $nabla phi=0$. In both cases (1) cannot be satisfied.

nt.number theory – Diophantine extended Euclidean for squares

We can utilize extended Euclidean algorithm to solve for integers $x,y$ such that
$$ax+by=c$$
holds where $a,b,c$ are integers wth $gcd(a,b)|c$.

Is there a way to solve for
$$ax^2+by^2=c$$
holds where $a,b,c$ are integers wth $gcd(a,b)|c$?

By this I mean is there a $mathsf{polylog}(|abc|)$ time algorithm?

I am interested in the cases:

  1. Assume the factors of $a,b,c$ are provided.

  2. Assume the factors of $a,b$ are unknown but $ab|c$.

ra.rings and algebras – Is the standard proof that Euclidean Domains are PIDs false?

In the books “Modern Algebra and it’s Applications” and “A First Course in Abstract Algebra”, the same proof that Euclidean Domains are PIDs is given. I will state it below (ver Batum from “A First Course in Abstract Algebra”) for convenience:

Let $D$ be a Euclidean domain with a Euclidean valuation $v$, and let $N$ be an ideal of $D$. If $N={0}$, then $N=left<0right>$ and $N$ is principal. Suppose that $N neq 0$. Then there exists $bneq0$ in N. Let us chose $b$ such that $v(b)$ is minimal among all $v(n)$ for $nin N$. We claim that $N=left<bright>$. Let $ain N$. Then by Condition 1 for a Euclidean domain, there exist $q$ and $r$ in $D$ such that
$$a=bq+r$$
where either $r=0$ or $v(r)<v(b)$. Now, $r=a-bq$ and $a,bin N$, so that $rin N$, since $N$ is an ideal. Thus $v(r)<v(b)$ is impossible by our choice of $b$. Hence, $r=0$, so $a=bq$. Since $a$ was any element of $N$, we see that $N=left<bright>$.

My issue is, at the beginning of this proof it is assumed that there is some $b$ such that $v(b)$ is minimal among all $v(n)$, which is not necessarily true. For example, we can take $R$ to be the complex numbers under standard multiplication with a classic Euclidean norm of $v(a+bi)=a^2+b^2$. $R$ its self is obviously an ideal of $R$, but it does NOT have a minimal element in terms of valuation since one can pick numbers arbitrarily close to $0$.

Am I wrong? This proof is at its core based on this assumption, and so if it is not true the entire proof breaks. I feel like I must be misunderstanding this since this is a very important theorem and this is the proof given in almost every standard textbook.

geometry – How to asses the loss of information in mapping a set of points in Euclidean space into a non-embedded graph?

Let ${mathbf{x}_i}_{i=1}^N$ be a set of $N$ points such that $mathbf{x_i} in mathbb{R}^M$.

Now assume we define the graph $G=(V,E)$ such that for each point $mathbf{x}_i$ we define a node $i in V$ and we assign an undirected edge between node $i$ and node $j$ if the distance $|mathbf{x}_i – mathbf{x}_j| < delta_0 $.

Is there something similar to the Johnson–Lindenstrauss lemma to quantify how “lossy” this transformation is or maybe to find the least lossy $delta_0$ based on how $mathbf{x}_i$ are distributed in $mathbb{R}^M$.

dg.differential geometry – Collection of local defining maps for smooth Euclidean submanifolds

Suppose the set $S subset mathbb{R}^{n}$ is a smooth submanifold of dimension $k$, that is (Lee, Proposition 5.16) for every $x in S$ there exist an open set $W subset mathbb{R}^{n}$ and a smooth submersion $phi : Wto mathbb{R}^{n – k}$ such that $W cap S$ is a level set of $phi$.

The submersion $phi$ is termed a local defining map.

Can one find a finite a collection of open sets $W_{1},ldots, W_{m}$ and corresponding local defining maps $phi_{1},ldots,phi_{m}$ such that $S subset bigcup_{i = 1}^{m} W_{i}$?

Moreover, can one demand that $phi_{i}(x) = phi_{j}(x)$ for all $x in W_{i} cap W_{j}$?

general topology – Understanding a step in this proof that a connected, paracompact, locally euclidean, Hausdorff space is second countable.

The proof in question, on MO, is as follows.

“Here’s another proof, which shows that any connected paracompact locally Euclidean space X is second-countable. Cover X by Euclidean charts and take a locally finite refinement. Say an open set is good if it only intersects finitely many of the charts. Now take any point x and take a good neighborhood of it. The charts that intersect that good neighborhood can then themselves be covered by countably many good open sets. There are then only countably many charts intersecting those good open sets, and those charts can be covered by countably many good sets. Iterating this countably many times, you get an open set U associated to x which is covered by countably many charts such that if a chart intersects U, it is contained in U. It follows that the complement of U is also a union of charts, so by connectedness U is all of X. Thus X can be covered by countably many charts and is second-countable.”

I do not know why we should expect this to be true after “iterating countably many times”. It seems like this might follow from paracompactness, but I am not seeing how.

complexity theory – Why is it hard to show that the euclidean Steiner tree problem is in NP?

I read that for the euclidean Steiner tree problem it is known that it is NP-hard, but not known whether it is in NP or not. [Wikipedia]

Shouldn’t the euclidean version obviously be in NP since the metric Steiner tree problem is in NP and a non-deterministic TM that decides the metric Steiner tree problem could also decide the Euclidean Steiner tree problem? Or am I missing something that could make the Euclidean version inherently more difficult/different?

geometry – What is a presentation of a manifestly positive definite metric on Euclidean AdS?

One of the standard metrics in global coordinates that I found being used for Euclidean AdS is this,

$$ds^2 = frac{1}{z^2}(dt^2 – dz^2 – sum_{i=1}^{p-1} dx_i^2)$$

  • But this is not positive definite unless $p$ is even!
    So how is this a correct metric that physics papers in particular seem to be using?
    Is there an implicit interpretation of it which is different?

  • Is there a presentation of global metric on Euclidean AdS which is manifestly positive definite?


Some references also use $ds^2 = frac{1}{z^2}(dz^2 – dt^2 + sum_{i=1}^{p-1} dx_i^2)$ which tries make the point that AdS is foliated by flat Minkowski spaces. But then its unclear to me why this metric is also claimed to be an “Euclidean AdS” given that this is also not positive definite!

Any explanations clarifying this would also be helpful!

chapter 2 problem B 27: frank jones lebesgue integration on euclidean spaces

prove that $$lambda^{*}(A) = inf Big{sum_{k=1}^{infty} lambda(I_k) Big| A subset bigcup_{k=1}^{infty} I_k Big}$$

first $displaystyle A subset bigcup_{i=1}^{infty} I_k$ hence we have that $displaystyle lambda^{*}(A) leq lambda^{*}(bigcup_{k=1}^{infty} I_k) leq sum_{k=1}^{infty} lambda^{*}(I_k) = sum_{k=1}^{infty} lambda(I_k) $ since this holds for arbitrarily we have that $lambda^{*}(A) leq inf Big{sum_{k=1}^{infty} lambda(I_k) Big| A subset bigcup_{k=1}^{infty} I_k Big}$

now consider set $A = {lambda(G) : A subset G }$ and $B = { sum_{k=1}^{infty} lambda(I_k) Big| A subset bigcup_{k=1}^{infty} I_k Big}$ since we know that every open set can be written as countable union of disjoint rectangles we have that $G = bigcup_{k=1}^{infty}I_{k}$ so we have that $lambda(G) = sum_{i=1}^{infty} lambda(I_k)$ so we have that $A subset B$ and we know that $displaystyle lambda^{*}(A) = sup_{A subset G} lambda(G)$ by using following property

$A subset B implies inf A leq sup B$

we have that $ inf Big{sum_{k=1}^{infty} lambda(I_k) Big| A subset bigcup_{k=1}^{infty} I_k Big} leq lambda^{*}(A) $

so we are done.

Is above proof correct?

linear algebra – A reflection group is Euclidean if and only if it has an invariant of degree 2

Let $G subset GL(V)$ be a complex reflection group acting on its natural representation $V$. Then $G$ acts on the dual representation $V^*$ contragrediently which induces an action on the polynomial ring $k(V)$, the subalgebra, $G$-invariant polynomials are finitely-generated. The claim is that the natural representation of $G$ can be realized over the real, that is $G$ is a Euclidean reflection group if and only if it has an invariant of degree 2.

Now suppose $G$ is a Euclidean reflection group, then it leaves the norm polynomial invariant, since it is a subgroup of the orthogonal group.

The converse have been a headache for me, what I’m thinking is that to derive a invariant quadratic form from this invariant of degree 2, and somehow to show that the representation of this group can be define over the real numbers.

Can someone help me with this problem please?