Let $n\in\mathbb{N}$ be a positive integer. Is there a connected graph $G$ such that $G$ cannot be coloured with less than $n$ colours, and every two maximal matchings have non-empty intersection?

(I take maximality with respect to set inclusion.)

Let $G=(V,E)$ be a simple, undirected graph. A *matching* is a set $M\subseteq E$ consisting of pairwise disjoint edges. We say $M$ is *maximal* if it is maximal amongst all matchings in $G$ with respect to $\subseteq$.

Given $n\in\mathbb{N}$ is there a graph $G$ such that $\chi(G)\geq n$ and every two maximal matchings have non-empty intersection?

Edited as per Jim Humphreys 9/16/2018 to make it clearer that the 192 are 192 of the 240 total roots of E8, and also to add this link for information on Gosset's polytope 4_21: https://en.wikipedia.org/wiki/4_21_polytope

Edited 9/18/2018 to add new information from Wendy Krieger (see bottom of this post.)

Edited 9/24/2018 to add new information provided by Dr David Richter (see bottom of this post)

Edited 10/1/2018 to add new information provided by Dr Derek Smith (see bottom of this post.)

Original question (without additional information from Wendy):

Using 192 of the 240 roots of E8 (vertices of 4_21), Wendy Krieger has defined 48 disjoint tetrahedra this way:

Taking the E8 as {128,112}, of radius 2, we get

16 tetrahedra at (1,1,1,1)E, (1,1,1,1)E

16 tetrahedra at (1,1,1,1)O (1,1,1,1)O

16 tetrahedra in (2,0,0,0)A (2,0,0,0)A

In the first two,

E means take an even number of sign-changes in the bracket.

O means take an odd number of sign-changes in each bracket.

A means all permutations, all change of sign in the brackets.

The vertices of the tetrahedron then comes from three coordinates in a given set, so these are the coordinates of a tetrahedron in the first set, using the first three coordinates.

1,1,1,1 1,1,1,1

1,-1,-1,1 1,1,1,1

-1,1,-1,1 1,1,1,1

-1,-1,-1,1 1,1,1,1

Question:

Is it possible to define a generalized Kronecker delta function which takes Wendy's E, O, and A sets to 1, -1, and 0 respectively (or -1, 1, 0)?

See this link for definition of the GENERALIZED Kronecker delta:

https://en.wikipedia.org/wiki/Kronecker_delta

New information from Wendy Krieger (added 9/18):

Of the 240 total roots of E8, 192 are consumed by Wendy's 48 tetrahedra, leaving 48. And she has discovered that these 48 define two 24-cells. So in addition to the generalized Kronecker delta question, there is an additional question as to whether the 48 vertices of the two 24-cells "organize" the 48 tetrahedra in any interesting way.

Wendy has provided an affirmative and interesting answer to this question:

The centres of the 48 octahedral faces of each of the two 24-cells, in rectangular product, produce a 3*3 array of the 48 tetrahedra, of which there are six distinct sets of 24.

See this link for definition of the famous and unique 24-cell, which has no analogs in spaces lower or higher than 4 dimensions:

https://en.wikipedia.org/wiki/24-cell

New information from Dr. David Richter (9/24/2018)

Dr. Richter has kindly suggested that Wendy's construction is not new within the literature of Lie-algebras. He writes:

"It does not seem new to me. Eugene Dynkin studied these things in depth in the 1940's, for example. (In Russia during World War II. He avoided the draft due to poor eyesight.) You should look up his article "Semisimple subalgebras of semisimple Lie algebras". This was published in the Translations of the American Mathematical Society in 1952. Although he does not use the language of regular polytopes, I think you will find the structures that you describe, encoded as rank -8 Lie subalgebras of the E(8) Lie algebra."

New information from Dr. Derek Smith provided 10/1/2018)

There are 16 "nearest neighbors” to any one of Wendy's tetrahedra, in the following sense.

Consider the tetrahedron T that has its four vertices in R^8 given by the following rows (here + is 1, - is -1, and we’re using the even coordinate system, scaled up by a factor of 2 to avoid some fractions, so minimal non-zero vectors have norm 8):

+ + + + + + + + + - - + + + + + - + - + + + + + - - + + + + + +The center c of T is the average of those vectors, namely c = 0 0 0 + + + + +.

The vectors in E8 closest to c are the four vertices of T, each of whose (squared) distance from c is 3. The next possible distance is 5, and the 16 vectors in E8 that achieve this are of the form 0 0 0 a b c d e, where each of the five letters is either 0 or 2, and there are an even number of 2’s.

As we’ve discussed previously, there are geometric symmetries of the lattice taking any one of the tetrahedra to any other, so there’s nothing special about this one. For instance, if we had taken

2 0 0 0 2 0 0 0 0 2 0 0 2 0 0 0 0 0 2 0 2 0 0 0 0 0 0 2 2 0 0 0as our tetrahedron instead, with center c = 1/2 1/2 1/2 1/2 2 0 0 0, then the 16 vectors at distance 5 from c aren’t as easily described in one sentence; but you can check that they are

0 0 0 0 0 0 0 0 0 0 0 0 4 0 0 0 + + + + + a b c a, b, and c are + and -, with an even number of -‘s + + + + 3 x y z x, y, and z are + and -, with an odd number of -‘s 0 0 0 0 2 p q r p, q, and r are -2, 0, and 2, with exactly two 0’sfor a total of 1+1+4+4+6=16 vectors.

There may be other ways to describe “nearest neighbors” that might yield different answers, e.g. take vectors close to the 4 centers of the tetrahedron’s faces, or close to the 6 centers of the tetrahedron’s edges.

Let $p_{n,K}(x)$ be the polynomial of degree $n$ that minimizes $\epsilon_{n,K} = ||p_{n,K}(x) - sin(x)||_\infty$ on the interval $[-K, K]$. Question: what is the asymptotic behavior of $\epsilon_{n,K}$ where $n \to \infty$?

I recently learnt that a topological space $X$ is "extremally disconnected" if and only if the closure of every open set is open in $X$.

It it possible to turn $2^\omega$ into an extremally disconnected space?

Edit: By "turn into", I guess it could mean that there exists some space which contains $2^\omega$ as a subspace, preserves all its properties, but instead of being totally disconnected, it is extremally disconnected.

For $n\in\Bbb{N}$, define three matrices $A_n(x,y), B_n$ and $M_n$ as follows:

(a) the $n\times n$ tridiagonal matrix $A_n(x,y)$ with main diagonal all $y$'s, superdiagonal all $x$'s and subdiagonal all $-x$'s. For example, $$ A_4(x,y)=\begin{pmatrix} y&x&0&0\\-x&y&x&0\\0&-x&y&x \\0&0&-x&y\end{pmatrix}. $$

(b) the $n\times n$ antidigonal matrix $B_n$ consisting of all $1$'s. For example, $$B_4=\begin{pmatrix} 0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}.$$

(c) the $n^2\times n^2$ block-matrix $M_n=A_n(B_n,A_n(1,1))$ or using the Kronecker product $M_n=A_n(1,0)\otimes B_n+I_n\otimes A_n(1,1)$.

**Question.** What is the determinant of $M_n$?

**UPDATE.** For even indices, I conjecture that

$$\det(M_{2n})=\prod_{j,k=1}^n\left[1+4\cos^2\left(\frac{j\pi}{2n+1}\right)+4\cos^2\left(\frac{k\pi}{2n+1}\right)\right]^2.$$

This would confirm what Philipp Lampe's "perfect square" claim.

There used to be many candidates for an exotic 4-sphere, but a lot of them are now known to be the standard smooth $S^4$. The ones of Cappell-Shaneson (maybe not all of them?) were described in terms of handlebody decompositions where there are no 3-handles, but I think the proofs that these were diffeomorphic to $S^4$ involved altering the decompositions with a cancellation trick by introducing some 3-handles.

I am not sure what is left over, but I seek the following:

**1) Is there a potentially exotic 4-sphere which is prescribed by an** explicit **4-handlebody decomposition with only 0,1,2-handles and a single 4-handle?**

**2) Of the** standard **Cappell-Shanelson spheres with a given handlebody decomposition involving no 3-handles, is it known that we can prove** [edit: some of them] **they're standard without the trick involving 3-handles?**

The reason I ask is because I'd like to study certain differential 2-forms on these spheres which "play well" with the handles, and to understand the deformation of these 2-forms as we change the handlebody decompositions.

let $M$ be a smooth $n$-manifold with boundary $\partial M$; I denote by $M^o$ the internal part of $M$, that is $M \smallsetminus \partial M$. The question is the same as in the title: let $M$ and $N$ be two compact orientable topological manifolds such that $M^o$ is homeomorphic to $N^o$. Does this imply that $M$ is homeomorphic to $N$? Can we say something about the connected components of the boundary?

I think that this question should have a really easy answer, but I cannot find it. One thing that I could prove is that the Euler characteristic of $\partial M$ must be the same as the one of $\partial N$, this being an easy consequence of the fact that $M^o$ is homotopically equivalent to $M$. The next step i tried is to prove that the sequence of Euler characteristics of the boundary components must agree: $$ ((\partial M)_1, \ldots , (\partial M)_r) = ((\partial N)_1, \ldots , (\partial N)_r) ;$$ but I do not see a way to prove this. I am specially interested in the case of 3-manifolds, and from this one could conclude that $\partial M$ is homeomorphic to $\partial N$, because Euler characteristic identifies closed surfaces. Do you have any tip or any simple proof?

If we have a vector fiber bundle with a connection $D_X=D(X)$ and an endomorphism $e$; we can then define a new tensor by the following formula: $$E(e)=D_X e D_Y + D_Y e D_X - e D_Y D_X - D_X D_Y e - e D_Z + D_Z e$$ with $Z=(X,Y)$ the Lie brackets of vector fields. If $e$ is parallel, then $E(e)=0$.The antisymmetric part is reduced to the curvature tensor, but we can define with the symmetric part a new tensor: $$ T(e)= 2D(e_i) e D(e_i) - e D(e_i) D(e_i) - D(e_i) D(e_i) e$$ with the Einstein's convention. For example, we can take $e=Ric$ the Ricci curvature and deduce a new symmetric tensor. When is this tensor $E(Ric)$ an automorphism and can we define new interesting differential equations?

Given a number n of nodes, a diameter d (d>1) and a max-degree k. Let's assume d and k are chosen such that a graph with n nodes with the desired diameter and max-degree exists.

What is the minimum number of edges necessary to create a (connected) graph with n nodes, diameter d, and max-degree k? Is there a general construction rule to reach this graph?

Without the max-degree this question is trivially answered with n-1 by a star graph. How can we additionally deal with the max-degree? A better lower bound than n-1 would already be interesting.

Edit:d is meant to be an upper bound to the diameter.

Is there an example of continuous map $f:\mathbb R\to \mathbb R^2$ such that for all $[x,y]\subset \mathbb R$ , $f([a,b])$ has non empty interior and such that for any $x\in \mathbb R_+$, $f(]0,x])$ is a neighborhood of $f(0)$.

Note that the non existence of such a fonction implies that there is no fonction $f$ with the same property and such that $f$ sends any intervalle on a convex (take an open intervalle $I$ such that $f(I)$ does not intersect the boundary of say $f([-1,1])$, and the union of open intervalles $U$ such that $f(U)=f(I)$, this union is an open intervalle that lowest grater bound satisfies the condition of the question, up to a fine translation.)

As soon as the problem I just mention is difficult (and open : it is discussed here Continuous maps which send intervals of $\mathbb{R}$ to convex subsets of $\mathbb{R}^2$) , either there is a difficult (and open) "there is no such a fonction", either there is such a fonction.

There are two mathematical concepts of infinity, potential infinity and actual infinity. I do not understand how the latter is being used. For the simplest infinite set, the natural numbers, we get:

**Potential infinity**: Every natural number that can be proven to be a prime number or not to be a prime number belongs to a finite initial segment that is followed by infinitely many natural numbers. An infinite set is much larger than every finite set. Therefore almost all natural numbers cannot be proven to be a prime number or not to be a prime number.

**Actual infinity**: Every natural number that can be proven to be a prime number or not to be a prime number belongs to a finite initial segment that is followed by infinitely many natural numbers. Nevertheless all natural numbers can be proven to be a prime number or not to be a prime number.

How is that possible?

Suppose $a_1,...,a_n\geq0, \sum_{i=1}^na_i=1$ and $Z_1,...,Z_n$ are i.i.d. standard normal, what is a sharp upper bound of the following probability as $\delta\to0$ and what is the order? $$\mathbb{P}(\sum_{i=1}^na_iZ_i^2\leq\delta)$$ How will the distribution of $a_1,...,a_n$ affect the upper bound?

Some time ago, I ran across the following inequality (if I remember rightly): $\bigg(\sum_{i=1}^{n}v_{i}^{k_{1}}\bigg)\geq n\bigg(\frac{\sum_{i=1}^{n}v_{i}^{k_{2}}}{n}\bigg)^{\frac{k_{1}}{k_{2}}}$, for $v_{1},...,v_{n}$ positive real numbers and $k_{1},k_{2}$ real numbers, with $k_{1}\geq k_{2}$. Where did it originate? Does it have a name?

In "Heat Kernels and Dirac Operators" by Berline, Getzler and Vergne, the authors construct a (unique) heat kernel associated to any generalized Laplacian in a vector bundle over a Riemannian manifold.

Could you please point me to texts doing the same kind of construction (not necessarily using densities) in vector bundles over *arbitrary* (i.e. not necessarily compact anymore) Riemannian manifolds for the connection Laplacian?

I am mostly interested in the clear definition of the object and the statement of its basic properties. Do the positivity and minimality properties of the heat kernel on functions have analogous properties in bundles?

Let $A$ be an arbitrary (not necessarily finite-dimensional) associative algebra over an algebraically closed field $K$, and let $\mathrm{fin\,}A$ denote the category of finite-dimensional $A$-modules. The algebra $A$ is said to be wild if for any finite-dimensional $K$-algebra $B$, there exists an exact $K$-linear functor $F:\mathrm{fin\,} B \rightarrow \mathrm{fin\,} A$ such that $F$ maps indecomposable $B$-modules to indecomposable $A$-modules and $F(X)\cong F(Y) \Rightarrow X \cong Y$ ($F$ respects iso-classes). Such a functor is called a representation embedding.

To prove an algebra $A$ is wild, it is sufficient to construct a representation embedding $F:\mathrm{fin\,} C\rightarrow \mathrm{fin\,} A$ for some other wild algebra $C$. The prototypical example of a wild (actually, strictly wild) algebra is the free algebra on two variables $K\langle x,y\rangle$. The commutative algebra $K[x,y]$ is also wild (but not strictly wild).

My question is what happens if we have an algebra with $n>1$ commuting variables, with some being non-trivial units? For example, is the algebra $K[x,x^{-1},y,y^{-1}]$ wild? (Commutative algebras are never strictly wild, so this is not my concern.) The only difference in terms of the indecomposable modules is that the actions of $x$ and $y$ on $K[x,x^{-1},y,y^{-1}]$-modules correspond to non-singular matrices. To me, this does not seem like a particularly heavy restriction, and so my instinct says that the algebra is still wild. However, I've not found any references on the matter, and I've struggled to come up with a suitable representation embedding as a proof. Any help would be much appreciated.

**EDIT:** I'm of course excluding algebras such as $K[x,(x-\lambda_1)^{-1},\ldots,(x-\lambda_m)^{-1}]$ ($\lambda_i \in K$ for all $i$), as these are obviously tame.

Prove that the volume in operator norm of the set of $2\times 2$ matrices \begin{equation} \Big\{\begin{pmatrix}a & b\\ c& e\end{pmatrix} \Big\vert\ a, b, c, e \in \mathbb{K}, \Big| \Big|{\begin{pmatrix} a & b\\ c& e\end{pmatrix}} \Big| \Big|<1,\ \ \Big| \Big|{\begin{pmatrix} a & \varepsilon b\\ \frac{c}{\varepsilon}& e \end{pmatrix}} \Big| \Big| <1 \Big\} \end{equation} over $\mathbb{K}$ as a function of $\varepsilon \in [0,1]$ is given by \begin{equation} \tilde{\chi}_2 (\varepsilon ) =\frac{1}{3} \varepsilon^2 (4 -\varepsilon^2), \end{equation} when the field $\mathbb{K}$ is the complex one $\mathbb{C}$ and \begin{equation} \tilde{\chi}_4 (\varepsilon ) =\frac{1}{35} \varepsilon^4 (15 \varepsilon^4 -64 \varepsilon^2 +84), \end{equation} when the field $\mathbb{K}$ is the quaternionic one $\mathbb{H}$.

If the field $\mathbb{K}$ is the real one $\mathbb{R}$, Lovas and Andai (https://arxiv.org/abs/1610.01410 eq. (9)) have shown that the desired function of $\varepsilon$ is given by
\begin{equation} \label{BasicFormula}
\tilde{\chi}_1 (\varepsilon ) =1-\frac{4}{\pi^2}\int\limits_\varepsilon^1
\left(
s+\frac{1}{s}-
\frac{1}{2}\left(s-\frac{1}{s}\right)^2\log \left(\frac{1+s}{1-s}\right)
\right)\frac{1}{s}
\mbox{d} s
\end{equation}
\begin{equation}
= \frac{4}{\pi^2}\int\limits_0^\varepsilon
\left(
s+\frac{1}{s}-
\frac{1}{2}\left(s-\frac{1}{s}\right)^2\log \left(\frac{1+s}{1-s}\right)
\right)\frac{1}{s}
\mbox{d} s .
\end{equation}

This has a closed form,
\begin{equation} \label{poly}
\tilde{\chi}_1 (\varepsilon ) =\frac{2 \left(\varepsilon ^2 \left(4 \text{Li}_2(\varepsilon )-\text{Li}_2\left(\varepsilon
^2\right)\right)+\varepsilon ^4 \left(-\tanh ^{-1}(\varepsilon )\right)+\varepsilon ^3-\varepsilon
+\tanh ^{-1}(\varepsilon )\right)}{\pi ^2 \varepsilon ^2},
\end{equation}
where the polylogarithmic function is defined by the infinite sum
\begin{equation*}
\text{Li}_s (z) =
\sum\limits_{k=1}^\infty
\frac{z^k}{k^s},
\end{equation*}
for arbitrary complex $s$ and for all complex arguments $z$ with $|z|<1$.

The further (apparently much simpler) forms above of the functions $\tilde{\chi}_2 (\varepsilon )$ and $\tilde{\chi}_4 (\varepsilon )$ were advanced in https://arxiv.org/abs/1701.01973 (eqs. (42), (59)), but formal proofs have not yet been developed. (Lovas and Andai did, in fact, pose the general [$\mathbb{K}= (\mathbb{R}, \mathbb{C}, \mathbb{H}$] question, but succeeded in answering it only for $\mathbb{R}$.)

The three functions $\tilde{\chi}_1 (\varepsilon ), \tilde{\chi}_2 (\varepsilon ), \tilde{\chi}_4 (\varepsilon )$ are all special cases ($d =1, 2, 4$) of the "master Lovas-Andai formula" (eq. (70) in https://arxiv.org/abs/1701.01973 and also in the answer in Perform an integration over the unit interval of a two-parameter expression involving a Gauss hypergeometric function) \begin{equation} \label{MasterFormula} \tilde{\chi_d}(\varepsilon)= \frac{\varepsilon ^d \Gamma (d+1)^3 \, _3\tilde{F}_2\left(-\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;\varepsilon ^2\right)}{\Gamma \left(\frac{d}{2}+1\right)^2}, \end{equation} where the regularized hypergeometric function is indicated.

compute the inverse Laplace transformation of the following equation.

\begin{align*} f(s)&=\frac{A}{\prod_{i=1}^{L}(s+a_i)^m} \\ &=\frac{A}{(s+a_1)^m\,(s+a_2)^m\cdots (s+a_L)^m}. \end{align*}

What I managed to get is this \begin{align*} Y^{-1}(\frac{1}{(s+a)^m})&=\frac{t^{m-1}e^{-at}}{{(m-1)!}} \end{align*}

I am not sure if this is even correct , and if this is correct how will I proceed to evulate the Laplace inverse of the given series

**Background/Notation:**

Given the Iwahori-Hecke algebra $H_{n}$ (over some ring commutative ring $R$ with identity) with generators $\{T_{1},\ldots T_{n-1}\}$, we know it has basis
$\{T_{w}\}_{w\in S_{n}}$, where $S_{n}$ is the permutation group and for some reduced expression $w=s_{i_{1}}\cdots s_{i_{r}}$, we have $T_{w}=T_{i_{1}}\cdots T_{i_{r}}$. This basis can be pictured as the set of all positive braids corresponding to each $w\in S_{n}$. *E.g.* for $w=s_{i_{1}}\cdots s_{i_{r}}$, we write $\sigma_{w}=\sigma_{i_{1}}\cdots \sigma_{i_{r}}$.

That is, if we consider the isomorphism $\varphi:R[B_{n}]\backslash{I}\to H_{n}$ (where $I$ is the ideal generated by the Hecke relation $(T_{i}-q_{1})(T_{i}-q_{2}) \ \forall i$ with units $q_{1}, q_{2}\in R$), have $\varphi(\sigma_{w})=T_{w}$ whence we may identify the two.

We obtain the annular Skein algebra $V$ by taking the braid closure of all elements in $H_{n}$ (so $V$ is the vector space consisting of $R$-linear combinations of closed $n$-braids *modulo* Reidemeister-II/III moves and the Skein relation (i.e. Hecke relation)).

**Problem:**

*What is a basis for $V$*? Consider the linear map $\pi:H_{n}\to V$ (which we can think of as braid closure). $V$ should have the same basis as $H_{n}$ "modulo closure" i.e. the problem reduces to *when does $\pi(T_{w})=\pi(T_{w'})$?* By Markov's theorem, this should be *iff* $\sigma_{w}$ is conjugate to $\sigma_{w'}$ (since only Markov-I moves are relevant here). So this appears to be equivalent to finding the conjugacy classes amongst the positive braids $\{\sigma_{w}\}_{w\in S_{n}}$. Note that $V\cong H_{n}\backslash{[.,.]}$ (since $\pi(T_{w_{1}}T_{w_{2}})=\pi(T_{w_{2}}T_{w_{1}})$).

**Pursuing an 'Algebraic' Solution:**

In [Big05], section 5 is devoted to determining such a basis. This is shown to be the image of braids $b_{\lambda}$ under $\pi\circ\varphi$ where $b_{\lambda}=b_{(\lambda_{1})}\sqcup\cdots\sqcup b_{(\lambda_{k})}$ with $b_{(m)}=\sigma_{m-1}\cdots\sigma_{1}$ given a partition $\lambda=(\lambda_{1},\ldots,\lambda_{k})$ of $n$. This is done through geometrical/topological arguments. *Can we do this algebraically instead?*

**Sketch:**

From Bigelow's result, we can conclude that dim $V=p(n)$ (the $n^{th}$ partition no.) and $\pi$ is a linear projector. This tells us that the basis of $V$ is $\{T_{\lambda}\}_{\lambda\in\mathcal{P}}\subseteq \{T_{w}\}_{w\in S_{n}}$, where $\mathcal{P}$ is the set of conjugacy classes in $S_{n}$ (which we know to be the classes of different $\lambda$-cycles), and the chosen representative $\lambda\in S_{n}$ is the 'canonical' permutation corresponding to the $\lambda=(\lambda_{1},\ldots,\lambda_{k})$-cycle: that is, in cycle-notation, we fill out $k$ brackets from left-to-right with the numbers $1$ to $n$, where the $i^{th}$ bracket has length $\lambda_{i}$.

Essentially, his result tells us that $\pi(T_{w})=\pi(T_{w'})$ *iff $w$ is conjugate to $w'$ in $S_{n}$*. By our above development of the problem, does this mean that the algebraic route to Bigelow's result is to show that:

*$\sigma_{w}$ is conjugate to $\sigma_{w'}$ in $B_{n}$ iff $w$ is conjugate to $w'$ in $S_{n}$ ?*

I don't know if this final statement about conjugacy of positive braids is true, but if it is and the above reasoning is fine, then surely this is a valid route. If this isn't true, what's the best algebraic route to the solution?

An Exceptional point generally occurs in eigenvalue problems in which the matrix is dependent on some parameter(s). The particular point in which the eigenvalues become degenerate for the parameter(s) gives the Exceptional point. An alternate way of thinking for such points is that the real part and the imaginary part of the eigenvalues coincides at some specific parameter(s), and such parameter(s) are called Exceptional points.

I have the following 3x3 Matrix $M3$ with the following eigenvalues $e1$, $e2$, and $e3$

$$ M3=\begin{bmatrix} -i\omega+\frac{\Gamma}{2} & -ig_1 & 0\\ -ig_1 & -i\omega+\frac{\kappa_1}{2} & -ig_2\\ 0 & -ig_2 &-i\omega+\frac{\kappa_2}{2} \end{bmatrix} $$

$$ e1 = \frac{1}{3}\bigg[p+\Sigma-\frac{u_2}{2^{\frac{8}{3}}\Sigma}-3i\omega\bigg] \\ e2 =\frac{1}{3}\Bigg[p-\frac{1}{2}\bigg[\Sigma-\frac{u_2}{2^{\frac{8}{3}}\Sigma}\bigg]+\frac{i\sqrt{3}}{2}\bigg[\Sigma+\frac{u_2}{2^{\frac{8}{3}}\Sigma}\bigg]-3i\omega\Bigg]\\ e3 = \frac{1}{3}\Bigg[p-\frac{1}{2}\bigg[\Sigma-\frac{u_2}{2^{\frac{8}{3}}\Sigma}\bigg]-\frac{i\sqrt{3}}{2}\bigg[\Sigma+\frac{u_2}{2^{\frac{8}{3}}\Sigma}\bigg]-3i\omega\Bigg]\\ $$

Whereby

$$ p =\frac{\Gamma+\kappa_1+\kappa_2}{2} \\ \kappa_\pm = \frac{\kappa_1\pm\kappa_2}{4} \\ u_1 = 36g_1^{2}(3\kappa_2-2p)+\big(36g_2^{2}+(2p-3\kappa_1)(2p-3\kappa_2)\big)\big(4p-12\kappa_+)\big)\\ u_2=2^{\frac{2}{3}} \left(12 g_1^2+12g_2^2-4(p-3\kappa_+)^{2}-12\kappa_-^{2}\right)\\ \Sigma=\Bigg(\frac{u_1+\sqrt{u_1^{2}+u_2^{3}}}{16}\Bigg)^{\frac{1}{3}} $$

As one can see, the eigenvalues are quite nasty and long and it took some time obtaining them through *Mathematica* (it can be checked that the eigenvalues are correct since their corresponding eigenvectors diagonalizes $M3$).

My goal is to find the exceptional point(s) among the eigenvalues $e1$, $e2$, and $e3$ in terms of the parameters ($\Gamma$, $\kappa_1$, $\kappa_2$, $g_1$, $g_2$) with $\omega\rightarrow0$ for $g_1$ and/or $g_2$ (the only tunable parameters) for which $e1=e2=e3$. A straightforward attempt for solving for such a $g_1$ using *Mathematica* suggests that

$$ g1\rightarrow\pm\frac{\sqrt{-12g_2^{2}+4(p-3\kappa_+)^{2}+12\kappa_-^{2}}}{2\sqrt{3}} $$

which seems to imply that $u_2=0$ upon substituting the $g_1$ above into $u_2$ (for all intents and purposes, we are taking the $g_1>0$ solution). But for the same $g_1$ value, $u_1$ reduces to $$ u_1 = -(2p-3\kappa_2)^{3}+108g_2^{2}(2p-\kappa1-2\kappa2) $$

which can be zero or non-zero (depending on $g_2$).

**Here's the problem:** From a purely mathematical perspective (without relying on *Mathematica*), the eigenvalue equations do not appear to be degenerate when $u_2=0$. Referring to $\Sigma$, setting $u_2=0$ gives

$$ \Sigma=\frac{1}{2}(u_1)^{\frac{1}{3}} $$

So the eigenvalues $e1$, $e2$, and $e3$ gives (when $\omega\rightarrow0$):

$$ e1=\frac{1}{3}\bigg[p+\frac{1}{2}(u_1)^{\frac{1}{3}}\bigg]\\ e2=\frac{1}{3}\bigg[p+e^{-i\frac{\pi}{3}}\frac{(u_1)^{\frac{1}{3}}}{2}\bigg]\\ e3=\frac{1}{3}\bigg[p+e^{i\frac{\pi}{3}}\frac{(u_1)^{\frac{1}{3}}}{2}\bigg] $$

which shows that the eigenvalues are not degenerate. Well, unless, $u_1=0$ in which case $e1=e2=e3=\frac{p}{3}$ and we would have found our exceptional point. However, setting $u_1=u_2=0$ leads to indeterminate issues since when $u_1=u_2=0$, $\Sigma\rightarrow0$ and the term $\frac{u_2}{2^{\frac{8}{3}}\Sigma}\rightarrow\frac{0}{0}$ in $e1$, $e2$, and $e3$. What is the problem here? How should I go about finding such parameters of $g_1$ and/or $g_2$ for which the eigenvalues are degenerate?

Edit: I tried to plot the real part and imaginary part of the eigenvalues as a function of $g_1$ to use it to guide me in finding the exceptional point in the expressions. It's clear that there must exist at least one exceptional point (at approximately $g_1=0.64$): Real part of eigenvalues vs $g_1$, Imaginary part of eigenvalues vs $g_1$ but I am unable to recover the expression for the exceptional point for $g_1$ in terms of the other parameters. How should I approach this overall issue? Any thoughts?