bio | website | math.unice.fr/~cazanave |
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location | Nice, France | |
age | ||
visits | member for | 4 years, 9 months |
seen | 2 days ago | |
stats | profile views | 487 |
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 |
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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 |
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Universal maps between topological spaces
I must be stupid, but isn't "universal" equivalent to surjective? |
Sep 26 |
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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 |
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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". |
Sep 8 |
asked | Given a positive-definite integral unimodular Gram matrix, how to find a basis of the associated lattice (over $\mathbf Q$)? |
Aug 29 |
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Upper bound on the number of ismorphism classes of bilinear forms on $\mathbb{Z}^n$
@NoamD.Elkies: A priori the question refers to all symmetric bilinear forms (not only to even ones)... |
Jul 22 |
awarded | Self-Learner |
Jul 22 |
accepted | Easiest way to distinguish $E_8 \oplus E_8$ from $E_{16}$ |
Jul 22 |
answered | Easiest way to distinguish $E_8 \oplus E_8$ from $E_{16}$ |
Jul 2 |
awarded | Curious |
Jul 1 |
comment |
Units of $\mathbf Z[X,Y]/(P(X,Y))$
Note that this is true only for reduced algebras. The "standard" counterexample is: $\mathbf Z[X,Y]/(X^2)$. |
Jun 25 |
comment |
Easiest way to distinguish $E_8 \oplus E_8$ from $E_{16}$
@few_reps: is there an analogous command in pari/gp? (I know that there is an online Magma calculator, but for various reasons, it would be more convenient for me to do it in pari/gp). |