10
votes
Accepted
Symmetric subgroups of simple algebraic groups over finite fields
The classification of automorphisms of order $2$ in finite and algebraic groups of Lie type over finite fields of odd characteristic, and the structure of their centralizers, are quite well-known. ...
- 2,709
8
votes
Accepted
Is every matrix involution over a UFD diagonalisable?
Here is an answer based on my comment, and Geoff Robinson's earlier comment.
Let $A$ be a domain with $2\in A^\times$, and let $M$ be an $n\times n$ matrix with $M^2=I$. It is convenient to consider $...
- 2,928
7
votes
Accepted
Quotient of the plane by the standard Cremona involution
The quotient surface is the cubic
$$
u_0u_1u_2 + u_0u_1u_3 + u_0u_2u_3 + u_1u_2u_3 = 0,
$$
the only cubic with four nodes.
EDIT. To see this one can first observe that the involution acts biregularly ...
- 39.3k
7
votes
Accepted
Compact dual of a noncompact Lie group
(0) The group ${\rm Aut}(\mathfrak{g}_0)$ does not have to be connected (even over $\mathbb C$), take $\mathfrak{g}_0=\mathfrak{su}(2,2)$ as a counter-example. So let $G={\rm Inn}(\mathfrak g_0)$, ...
- 14.1k
6
votes
Does a manifold which bounds always admit a free involution?
Consider $M = \mathbb{CP}^2\#\mathbb{CP}^2$. All of its Stiefel-Whitney numbers vanish, so $M$ is unorientedly nullcobordant; more generally, $X\# X$ always bounds.
We have $H^2(M; \mathbb{Z}) \cong ...
- 19.3k
5
votes
Fundamental domain of an involution on a manifold
I will assume that you are working in the DIFF category, i.e. your manifold $X$ and involution $\tau$ of $X$ are smooth. It suffices t consider the case when $X$ is connected. Let $Y:=X/\tau$, $p: X\...
- 12.2k
5
votes
Accepted
Irreducible Symmetric Pairs
It is true that $\mathfrak p$ is always an irreducible $\mathfrak k$-module. Be aware though that the complexification $\mathfrak p_{\mathbb C}$ might be reducible as an $\mathfrak k_{\mathbb C}$-...
- 15.5k
5
votes
Finite field special functions
An involution with exactly one fixed point must be over an odd-sized set, so characteristic 2 is ruled out.
By my computation, there are only three such functions for odd prime powers between 3 and ...
- 7,226
5
votes
Reference request for a proof of the two-square Theorem
(Sorry, self-promotion!) There is now a reference :
R. Bacher, A Quixotic Proof of Fermat's Two Squares Theorem for Prime Numbers, American Math. Monthly, Vol 130, Issue 9, 824-836.
There is also a ...
- 17.9k
5
votes
Accepted
Involutive automorphism of simple Lie algebra
Let $\mathfrak g$ be a noncompact simple Lie algebra and let $\mathfrak g=\mathfrak k+\mathfrak p$ be a Cartan decomposition. The simplicity of $\mathfrak g$ implies that
the adjoint representation of ...
- 4,729
5
votes
Holt's Theorem on doubly transitive groups with $2$-central involutions fixing only one letter
This may be hard to do without CFSG. For example, if $G$ acts doubly transitively on one of its conjugacy classes of elements of order $p$, this is comparable in difficulty (and a special case of) ...
- 44.4k
4
votes
Rational functions of order $3$
Since you ask, there has been much research on the analogous question of elements of finite order in the Nottingham group, which is the group of power series under composition:
$$ N(\mathbb K) = \left\...
- 47.4k
3
votes
Accepted
When an algebra isomorphism preserves positive involution
The answer to your general question is NO.
Assume for simplicity that $A=B$, and that $A$ is a central simple $K$-algebra. Then $\varphi$ is an inner automorphism by Skolem Noether 's Theorem.
Write $\...
- 1,766
2
votes
Finite field special functions
What you are looking for are called perfect nonlinear or differentially $1-$uniform functions.
They don't exist over even characteristic since if $x_0$ satisfies
$$
f(x+a)-f(x)=b,
$$
so does $x_0+a.$
...
- 10.4k
2
votes
Accepted
Descent of coherent sheaves on finite coverings
Assume $\sigma^*E \cong E$. Then $E$ is a pullback if and only if the action of $\sigma$ on the fiber of $E$ at each ramification point of $\pi$ is trivial.
In the counterexample of nfdc23, the $\...
- 39.3k
2
votes
Accepted
A question about involutions and polynomials
Of course, for example $\tau(x)\equiv x$. Well, if you want fixed point free involution, the answer is still yes since there are countably many irreducible polynomials and countably many reducible ...
- 109k
2
votes
LS category of the quotient of a manifold by its involution
The equivariant LS category $\operatorname{cat}_G(X)$ of a space $X$ with $G$-action is the minimal $k$ such that $X$ admits a cover by $G$-invariant open sets $U_0,\ldots , U_k$ such that each ...
- 35.9k
2
votes
Has the determinant of a involution of the first kind ever been considered as an invariant?
There is an notion of determinant $\det(\sigma)$ of an involution $\sigma$ of the first kind on a central simple $F$-algebra $A$, which is well-known and well studied. It takes values in $F^\times/F^{...
- 1,766
1
vote
Accepted
Order 2 matrices with entries in the polynomial ring over a field are diagonalisable
As abx pointed out in the comments, I misread the fact that this is an equivalence, so it is unlikely that there is a more elementary proof.
- 721
1
vote
Accepted
An explicit matrix form
It looks like you are working with respect to the orthogonal form with matrix $\begin{pmatrix} & w_0 \\ w_0 \end{pmatrix}$, where $w_0 = \operatorname{antidiag}(1, \dotsc, 1)$. That's the one ...
- 12.9k
1
vote
Accepted
What do conjugacy classes of involutions like in finite simple group $E_7(q)$?
For $q$ odd see:
D. Gorenstein, R. Lyons and R. Solomon,The classification of the finite simple groups, Number 3,Mathematical Surveys and Monographs, vol. 40, Amer. Math. Soc., 1998 MR1490581
For $q$...
- 178
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