Friedrich Knop
  • Member for 5 years, 9 months
  • Last seen this week
  • Spardorf, Germany
Why are they called Spherical Varieties?
34 votes

Since I am being asked the same question repeatedly and since the given answers are not quite correct, I post another answer despite the thread being so old. According to a talk by Domingo Luna ...

View answer
Is it usual for a referee to heed updated versions on arxiv?
28 votes

It is not only unusual, I would be surprised if that ever happens. A referee's task is to a evaluate a paper in the state it has been submitted. So he/she should not take into account any other ...

View answer
17 camels trick
26 votes

Since nobody did, I would like to mention the obvious: The 17 camels trick is routinely used in apportionment procedures called divisor methods. The task is to divide a number of $N\in\mathbb N$ ...

View answer
Is it acceptable to use the citation references like [1] or [Joh] as nouns in mathematical writing?
Accepted answer
23 votes

First, I do it all the time and don't really see the objections. A phrase like "In [S] it was shown..." is a good alternative to "Siegel showed, [S], that ...". Out of curiosity I did some cursory ...

View answer
If $X\times X$ is rational, must $X$ also be rational?
21 votes

For the first question I am not that pessimistic. At least there are candidates as follows: Recall that $Z$ is stably rational if there is $n\ge0$ such that $Z\times\mathbf A^n$ is rational. Now ...

View answer
Can an analytic function defined on a maximal torus be extended analytically to all the Lie group?
Accepted answer
21 votes

This seems to be consequence of the paper Cartan, Henri: Variétés analytiques réelles et variétés analytiques complexes. Bull. Soc. Math. France 85 1957 77–99 Cartan shows more generally (see ...

View answer
Irreducible polynomial $p_{n}(x)=\sum_{k=0}^{n}\frac{x^k}{k!}$ for all positive integers $n$
Accepted answer
20 votes

That follows from a theorem of Schur saying that any polynomial $\sum_{k=0}^nc_k\frac{x^k}{k!}$ with $c_i\in\mathbf{Z}$, $c_0,c_n\in\{1,-1\}$, $n\ge 1$, is irreducible over $\mathbf{Q}$. I. Schur, ...

View answer
Is the wonderful compactification of a spherical homogeneous variety always projective?
Accepted answer
19 votes

The wonderful compactification is always projective. One way to see is to use a theorem of Sumihiro which says that a normal $G$-variety is covered by $G$-invariant quasiprojective open subsets. Since ...

View answer
Where does the really nice '8-dimensional' description of the $E_7$ root system come from?
Accepted answer
17 votes

I don't know who found this presentation first but I can imagine that already Cartan knew it since it comes from a symmetric space. More precisely, $\mathfrak g=E_7$ has an involution $\theta$ whose ...

View answer
Conjecture of relation between residues of Feynman integrals and mixed Tate motives
Accepted answer
15 votes

1) Counterexamples were found in the paper Brown, Francis; Schnetz, Oliver: "A $K3$ in $\phi^4$". Duke Math. J. 161 (2012), no. 10, 1817–1862. It is now the general feeling that most $\phi^4$-Feynman ...

View answer
Is the dimension of $V//G$ always the same as the dimension of $V^*//G$?
Accepted answer
15 votes

If $G$ is connected and $T\subseteq G$ is a maximal torus then there is an involution $\theta:G\to G$ with $\theta(t)=t^{-1}$ for all $t\in T$. It has the property that as a representation $V^*$ is ...

View answer
Multiplicity of an irrep of SO(n-1) in SO(n)
Accepted answer
13 votes

The statement is true and well known. See e.g. Thms. 8.1.3 and 8.1.4 in Goodman-Wallach: Symmetry, representations and invariants, Springer GTM 255. In fact, much more is known. Let, more generally, $...

View answer
Categorical Unification of Jordan Holder Theorems
12 votes

There are various generalizations of the Jordan-Hölder theorem. Beyond groups and groups with operators it holds for any equational theory which contains a Mal'cev operation. This means that from the ...

View answer
Does the orthogonal group act irreducibly on totally symmetric tensors?
Accepted answer
12 votes

It is classical that, as $O(n)$-representation, $$ \text{Sym}^k(\mathbf R^n)=H^k\oplus qH^{k-2}\oplus q^2H^{k-4}\oplus\ldots $$ Here $q=x_1^2+\ldots+x_n^2$ is the quadratic form defining $O(n)$ and $H^...

View answer
Are there indecomposable unsolvable four and five dimensional Lie algebras?
Accepted answer
12 votes

Every finite dimensional real Lie algebra has the form $\mathfrak g=\mathfrak l\ltimes\mathfrak r$ where $\mathfrak l$ is semisimple and $\mathfrak r$ is solvable (Levi decomposition). The smallest ...

View answer
Is SO(2n+1)/U(n) a symmetric space?
Accepted answer
12 votes

Let me complement Claudio's answer. There is indeed a definition of symmetric space which works for any Riemannian manifold $M$: For any point $p\in M$ there is an involutive isometry $\iota_p$ of $M$ ...

View answer
Division ring on a field
Accepted answer
12 votes

I assume $\text{char}\,\mathbf F=0$. Put $d:=b-a$. Because of $a^2-2ab+b^2=d^2-ad+da$ your equation is equivalent to $$ (*)\qquad d^{-1}a-ad^{-1}=1. $$ This precludes $\dim_{\mathbf F}D<\infty$ (...

View answer
Quasi-split tori and algebraic groups
Accepted answer
12 votes

A torus $T$ is quasi-split if its character group is a permutation representation for the Galois group. So a counterexample to your question is: let $G$ be the quasi-split group $SO(n+1,n-1)$, $n\ge2$,...

View answer
Invariant polynomials under diagonal action of the orthogonal group
Accepted answer
12 votes

The invariants are generated by the quadratic polynomials $(u,u)$, $(u,v)$, and $(v,v)$ where $(.,.)$ is the scalar product defining $O(n)$. This pattern generalizes to arbitrary many copies of $\...

View answer
How to write $\mathbb{C}[G/U_-]=\oplus_{\lambda} V_{\lambda}$ explicitly?
Accepted answer
12 votes

The subspace $V_\lambda$ is very easy to see. Since $U^-$ is normalized by the maximal torus $T$ there is an action of $T$ on $G/U^-$ on the right. This means that $\mathbb C[G/U^-]$ carries a ...

View answer
A finiteness property for semi-simple algebraic groups
Accepted answer
12 votes

For fields of characteristic zero one can argue as follows: Assume first that $K$ is algebraically closed. Since semisimple subgroups of $G$ correspond bijectively to semisimple subalgebras of $\...

View answer
Under which conditions: dim(W1 + W2 + W3) = dim(W1) + dim(W2) + dim(W3) − dim(W1 ∩ W2) − dim(W2 ∩ W3) − dim(W3 ∩ W1) + dim(W1 ∩ W2 ∩ W3)
11 votes

The idea of PseudoNeo's comment settles the converse of my statement modulo that he missed four cases. According to the theory of indecomposable modules of the Dynkin quiver $D_4$ with all arrows ...

View answer
About the number of fixed points of a torus action
Accepted answer
11 votes

This is true in a much more general setting. Let $X$ be any normal projective $T$-variety defines over an algebraically closed field. Then I claim that $\# X^T>\dim X$. We show this in two steps. (...

View answer
Endoscopic group that is not a subgroup
Accepted answer
11 votes

If I understand the definition correctly, a connected reductive group $H$ is an endoscopic group for a connected reductive group $G$ if its Langlands dual $H^\vee$ is a connected centralizer in $G^\...

View answer
Standard Monomial basis for other types
Accepted answer
11 votes

Standard monomial theory has been extended to all classical groups by Lakshmibai, Seshadri and others in the series of papers "Geometry of $G/P$ I-IX". A very concise description of standard tableaux ...

View answer
Diagram folding of simple Lie algebras
Accepted answer
11 votes

Most maximal parabolics of $SL_{2n}$ are not $\sigma$-invariant, not even up to conjugation. So $(G/P)^\sigma$ does not make sense. The correct statement is: Let $I\subseteq\{1,\ldots,2n-1\}$ be a ...

View answer
Invariant theory for parabolics
Accepted answer
11 votes

The map $\mathbb C[\mathfrak g]^G\to\mathbb C[\mathfrak g]^P$ is an isomorphism for trivial reasons: In any quasi-affine $G$-variety, $P$ and $G$ have the same fixed points. Just look at the orbit map ...

View answer
Are the semi-simple elements in a non-connected reductive algebraic group dense?
10 votes

The $G^0$-action on a coset is the same as a so-called twisted action which is pretty well understood. See, e.g., Mohrdieck, S.: Conjugacy classes of non-connected semisimple algebraic groups, ...

View answer
Understanding a germ of a GIT quotient
Accepted answer
10 votes

For a fixed point your guess is right and one doesn't need Luna's slice theorem to prove it: Let $T$ be the tangent space in $x$ and let $\mathfrak m_x\subset\mathbb C[X]$ be the maximal ideal. Then ...

View answer
Ring of invariants of $\operatorname{SL}_6$ acting on $\Lambda^3 \mathbb C^6$
Accepted answer
10 votes

The principal isotropy group is $H=SL(3)\times SL(3)$: it has the right dimension (namely 16) and occurs as an isotropy group (namely of a general element of $W$). Now it is a general result of Luna-...

View answer
1
2 3 4 5 6