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Sep
30 |
reviewed | No Action Needed For which rings R is SL_n(R) generated by transvections? |
Sep
30 |
reviewed | No Action Needed character degree and solvability |
Sep
30 |
awarded | Custodian |
Sep
30 |
reviewed | No Action Needed Occurrences of (co)homology in other disciplines and/or nature |
Sep
25 |
comment |
Is SL(n,Z[x]) generated by transvections?
Just to clarify: the conclusion of the discussion you mention about the stable range of $\mathbf Z [x]$ was that it is indeed of stable range $3$. It is proven in the paper by Grunewald, Mennicke, and Vaserstein mentionned by "few_reps" as indicated in the discussion. |
Sep
24 |
comment |
Is SL(n,Z[x]) generated by transvections?
Cohn was probably the first to prove that $\mathrm{SL}_2(\mathbf{Z}[T]) \neq \mathrm{E}_2(\mathbf{Z}[T])$. A concrete example of a matrix not in $\mathrm{E}_2$ is $\begin{bmatrix} 1+2T & 4 \cr -T^2 & 1-2T\end{bmatrix}$. I took this example from Lam's "Serre's problem on projective modules", rk 8.11. |
Sep
4 |
answered | Examples of naturally occurring Quadratic forms or quadrics. |
Apr
8 |
comment |
Reference Request: Grouplike Algebras over the little $n$-cubes operad are $n$-fold loop spaces
Adams, "Infinite loop spaces" seems to be the reference you are looking for. |
Dec
17 |
comment |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
Ok, I know why it represents a square, but is there an algorithmic way to find such a vector? |
Dec
17 |
comment |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
I am a bit confused by the first step. My (stupid) question is: why does a positive definite 4-dimensional integral quadratic form represent a square? |
Dec
17 |
accepted | Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)? |
Dec
17 |
comment |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
Indeed, sorry for my delay in reading your answer. |
Dec
9 |
comment |
Computing naive algebraic singular homology
As in topology, this naïve $H_0$ is the free abelian group on the naïve $\pi_0$. Therefore, the computation of naive homotopy classes of endomorphisms of $\mathbf P^1$ gives you the computation of $H_0(\mathcal F_n)$ where $\mathcal F_n$ is the scheme of pointed degree $n$ rational functions. |
Nov
19 |
comment |
Universal maps between topological spaces
I must be stupid, but isn't "universal" equivalent to surjective? |
Sep
26 |
comment |
Let $A\in\operatorname{M}_n(F)$ be a matrix, how to prove $\bigcap_{X\in C(A)}C(X)=F[A]=\frac{F[x]}{(m_A(x))}$
I wanted to edit the previous comments, but it was not possible, so I (stupidly) erased it. You are right: $V$ is semi-simple iff $m_A$ has no square factors, so the double centralizer theorem applies only in this situation. |
Sep
26 |
comment |
Let $A\in\operatorname{M}_n(F)$ be a matrix, how to prove $\bigcap_{X\in C(A)}C(X)=F[A]=\frac{F[x]}{(m_A(x))}$
I don't easily see how one can deduce the general case from the double centralizer theorem. However, if you know the proof over an algebraically closed field, you can deduce it over $F$. (Take an algebraic closure $F \subset \overline{F}$, and remark that $\overline F[A]\cap M_n(F)= F[A]$.) |
Sep
15 |
revised |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
added 1 character in body |
Sep
8 |
comment |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
In general, p/q=pq/q^2 and use Lagrange's theorem to decompose a the (positive) numerator as a sum of at most 4 squares. |
Sep
8 |
comment |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
1/7=7/49 and 7=2²+1²+1²+1², so 1/7=(2/7)²+(1/7)^2+(1/7)^2+(1/7)^2. |
Sep
8 |
comment |
Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)?
Yes, that's more or less implicit in my "note n°2". |