rt.representation theory – Irreducible representations of the symmetric group on homology of simplicial complex

I am following Wall’s paper A note on symmetry of singularities and I have some questions regarding representation theory and the homology of some objects:

Consider an action of $Sigma_k$ on a finite simplicial complex $X$ of dimension $n$, such that the action is simplicial and a simplex is fixed iff it is point-wise fixed. Consider $sigmainSigma_k$, then
$$chi_{Top}(X^sigma)=chi_{Sigma_k}(X)(sigma):= sum_i(-1)^i text{trace } sigma^*:H_i(X)rightarrow H_i(X), $$
where $X^sigma$ is the fixed subcomplex fixed by $sigma$ and $chi_{Top}$ the Euler-Poincaré characteristic.

My questions are:

1.- Is the sum of characters $chi_{Sigma_k}(X)$ called the equivariant Euler characteristic? Because I believe the equivariant Euler characteristic is the Euler characteristic of $H_*(X)^G$ in other texts.

2.- Can one say anything about the character on a single homology group instead of the alternated sum of all the groups? For example, compare the trace of $sigma$ on one $H_i$ and $H_i(X^sigma)$. I have a lot of troubles trying to isolate one homology group from this equality (with the extra structure that I have in my problem).

3.- Can one say something about the isotypes corresponding to a single irreducible representation of $Sigma_k$? For example, compare the number of copies of the $sign$ representations and $H_*(X^sigma)$ for some $sigma$.

If it is useful, in my problem I know that every $X^sigma$ has lower dimension than $X$, if $sigmaneq 1_{Sigma_k}$.

abstract algebra – Show that the following polynomial in $mathbb{Q}[t]$ is irreducible and not solvable over $mathbb{Q}$

The polynomial is $f(t)=t^5-4t+2$.

I can show it’s irreducible with Einstein’s Criterion, and i know i need to show that it has exactly two complex (non-real) roots to prove that it’s not solvable, using the follow result:

Let $p>1$ be prime and $finmathbb{Q}(t)$ irreducible with degree $p$. If $f$ has exactly two non-real complex roots, then the galois group of $f$ is isomorphic to $S_p$. Hence, with $pgeq5$, we have that $f$ is not solvable by radicals over $mathbb{Q}$.

But i don’t know a way to show it has exactly two roots, i know that, using Descartes’ rule of signs, we can show that it has at least two complex roots (because it has 2 or 0 positive and one negative real root). Is there something i can use on top of Descartes’ rule to make this work? If not, what’s the other way?

ag.algebraic geometry – Irreducible components of spaces of morphisms

Consider a family $X$ of surfaces $mathbb{P}^1timesmathbb{P}^1$ parametrized by $mathbb{C}$. We may assume that $Xcong mathbb{P}^1timesmathbb{P}^1timesmathbb{C}$.

Fix an irreducible $2$-section $Gammasubset X$. So $Gamma_tsubsetmathbb{P}^1timesmathbb{P}^1$ is a curve projecting $2$ to $1$ say onto the first copy of $mathbb{P}^1$ for all $tinmathbb{C}$. Take $p_1(t),dots,p_k(t)inmathbb{P}^1$ distinct points depending on $t$. We can think of these points as curves in $mathbb{P}^1timesmathbb{C}$.

Then fix $x_1,y_1,dots,x_k,y_kinmathbb{P}^1$ distinct points. Let $Y_{2,d}$ be the space of morphisms $f:mathbb{P}^1rightarrowmathbb{P}^1timesmathbb{P}^1$ of bidegree $(2,d)$ such that $f(x_i),f(y_{i})inGamma$ for all $i = 1,dots k$, and $f(x_i),f(y_i)$ get mapped to $p_i(t)$, by the projection onto the first copy of $mathbb{P}^1$, for all $i = 1,dots, k$.

We can look at $Y_{2,d}$ inside the product $Mor_2(mathbb{P}^1,mathbb{P}^1)times Mor_d(mathbb{P}^1,mathbb{P}^1)subsetmathbb{P}^5times mathbb{P}^{2d+1}$. My question is the following: could $Y_{2,d}$ be irreducible or does it always have at least two irreducible components one parametrizing morphisms mapping $x_i$ “above” and $y_i$ “below” and the other one switching $x_i$ and $y_i$?

Thank you.

galois theory – If all roots of $f$ generate a splitting field, is $f$ irreducible?

Recently I had to prove the existence of some irreducible polynomial. I wanted to use the following statement, but I do not know if it is true:

The Statement:

Let $F$ be a field. If $fin F[X]$ is such that for all roots $alpha$ of $f$ $F[alpha]$ is a splitting field for $f$, then $f$ is irreducible.

My Question:

Is this true? Cyclotomic polynomials have this property and are irreducible. Would the answer change depending on the characteristic of the field?

algebraic geometry – Why is the closure of a “$mathbb{K}$-cone generated by an irreducible affine variety” irreducible?

I am trying to do the second exercise in page 37 in Ernst Kunz’s “Introduction Commutative Algebra and Algebraic Geometry” (I leave a sreenshot of the exercise below), and I got stuck trying to prove the second part of line b).

If I write $X:= overline{V^*}$ = $X_1 cup X_2$ with $X_1, X_2$ affine subvarieties of $X$.
Since we have $V subset X$ and it is irreducible, we must have $V subset X_1$ or $V subset X_2$. One of the $X_i$‘s will have to contain infinite points of some line contained in $V^*$ and so the polynomials in its ideal can’t have constant terms. I have also thought of considering the decomoposition in irreducible components, but I don’t see how that helps if $mathbb{L}$ is not alg. closed and also, even if it is, I don’t know if $rad(mathfrak{p_1}…mathfrak{p_m}) = mathfrak{p_1}…mathfrak{p_m}$ and if any prime ideal $mathfrak{p}$ with $I subset mathfrak{p}subset mathfrak{q}$ (where $mathfrak{q}$ is a prime divisor of $I$), also is a prime divisor of $I$, which in the case of $mathbb{L}$ algebraically closed, I would use to deduce that the ideals of each irreducible component would be homogeneous, and then any irreducible component that cotained $V$ would be equal to $X$.

My questions are:

  • If $mathbb{L}$ is alg. closed, do we have $sqrt{mathfrak{p_1}…mathfrak{p_m}} = mathfrak{p_1}…mathfrak{p_m}$?
  • Can anyone give me hint for the case $mathbb{L}$ not alg. closed?
  • If $I$ is an ideal and $mathfrak{q}$ a prime divisor of $I$, is any prime ideal $mathfrak{p}$ with $I subset mathfrak{p} subset mathfrak{q}$ also a prime divisor?

Thank you for all the help in advance đŸ™‚enter image description here

gr.group theory – Relation between irreducible representations and conjugacy classes in the infinite, projective case

Let $k$ be an algebraically closed field. Let $G$ be a (not necessarily finite) group and $kG$ the group algebra. Let $alphacolon Gtimes Gto U(1)$ be a $2$-cocycle, and let $k^alpha G$ be the associated twisted group ring.

As I understand, when $G$ is finite, the number of irreducible representations of $G$ is equal to the number of conjugacy classes of $G$. Furthermore, the number of irreducible projective $alpha$-representations of $G$ is equal to the number of $alpha$-regular conjugacy classes of $G$.

I am wondering whether something similar can be said in the case when $G$ is an infinite discrete group. More precisely:

Question 1: Is there a relationship between the number of irreducible representations of $G$ and the number of conjugacy classes of $G$?

Question 2: Is there a relationship between the number of irreducible projective $alpha$-representations of $G$ and the number of $alpha$-regular conjugacy classes of $G$?

gt.geometric topology – Higher homotopy groups of irreducible 3-manifolds

A 3-manifold $M$ is irreducible if every embedded 2-sphere bounds a 3-ball. Thanks to Papakyriakopoulos’s sphere theorem, irreducibility is the same as having $pi_2(M)=0$. Does irreduciblity imply that the manifold is in fact aspherical, i.e. that $pi_k(M)=0$ for all $k geq 2$?

(Or maybe I should say that the universal cover $tilde M$ is aspherical, but the question is the same in terms of homotopy groups.)

I’d expect that the answer might be yes based on all of the tools we have about geometric classification of 3-manifolds, but I’m not enough of an expert to make those arguments myself. Does someone here have a quick answer?

Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^3-5x^2+2x+1$

Investigate if the following polynomial in Q [x] is reducible or irreducible: $x^3-5x^2+2x+1$



$mod 2$ we get


We see that the polynomial is of degree 3, so it is reducible if it has a root in $Q$, using rational root theorem in Q, we see there is no root in Q, so its irreducible?

abstract algebra – Product of 2 irreducible representationsis irreducible

I am trying to find an example of 2 irreducible representations of the symmetric group Sn, and i want to find that the tensor product is irreducible.
My example is to find the 3 dim irreducible representation and the trivial or the parity when tensoring them we obtain an irreducible. I just want another example other than tensoring a representation with the 1 dim or itself because it seems obvious
Any idea?