Selim Ghazouani
Reputation
1,281
Next privilege 2,000 Rep.
 Dec 14 comment Is the matrix defined by $\bar{K}_{ij}=f(x_i)f(x_j)$ for a real valued function $f$ semi-positive-definite? Other remark, such a matrix will always be of rank at most $1$, since all lines are colinear to $(f(x_1),...f(x_n))$. It implies that it will never be definite positive provided that $n \geq 2$. Dec 14 comment Is the matrix defined by $\bar{K}_{ij}=f(x_i)f(x_j)$ for a real valued function $f$ semi-positive-definite? First of all, put $f \equiv 0$ to get a matrix which is not positive-definite since $\overline{K} = 0$ . Dec 11 comment Let $A$ be an abelian group. Show that $\mathrm{Hom}(\mathbb Z, A)$ is isomorphic to $A$. Assuming $Z$ is the set of integers, I invite you to check that $\varphi \in \mathrm{Hom}(\mathbb{Z}, A) \longmapsto \varphi(1)$ is the isomorphism you are seeking. Dec 11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? I'm not sure. Here you are using the fact that complex eigenvalues are conjugates, and for each pair of conjugate you use only one eigenvector and take the real and imaginary part. DO you see what I mean ? Dec 11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? Ok I understand how you want to proceed but now we should check that $e_i,x_j,y_j$ actually form a basis for $\mathbb{R}^n$. I think that using computation you used to prove that $x_j$ and $y_j$ are linearly independent should work but I doesn't seem obvious. Dec 11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? Actually I don't really understand how you find the $f_j$ and $g_j$. Precisely, why do their coefficients belong to $\mathbb{R}$ since the diagonalization ii over $\mathbb{C}$ ? Dec 11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? Yes exactly. I'm not sure how to be clearer :) Dec 11 revised When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? added 8 characters in body Dec 11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? For example any conjugate of a rotation preserve a scalar product, but does not belong to $O_2(\mathbb{R})$. Dec 11 comment When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? Not really. I'm looking for conjugates of the orthogonal group since the scalar product is not necesserally the canonical euclidean product in the canonical basis. Dec 11 reviewed Approve When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? Dec 11 asked When does an element of $\mathrm{Sl}_n(\mathbb{R})$ preserve a scalar product? Dec 4 comment Automorphism of the free group I'm not sure I get what you want to say. If you read carefully Farb and Margalit(p.87, I guess there is only one edition), the mapping class group of the once punctured torus is isomorphic to the central extension of $GL(2,\mathbb{Z})$. Maybe you make the distinction whether the torus is thought with or without boundary. Nevertheless, I would be glad to have further explaination on this proof. Dec 3 comment Automorphism of the free group Could you detail a bit I'm not sure I understand your proof. Dec 3 revised Automorphism of the free group added 1 characters in body Dec 3 revised Automorphism of the free group added 142 characters in body Dec 3 asked Automorphism of the free group Nov 25 answered Let $x,y$ in a group G with odd order. Let $x^2=y^2$. Show that $x=y$. Nov 25 accepted Defining a group morphism using generators Nov 25 comment Defining a group morphism using generators Yes I have edited. Then $\varphi)$ is trivial on $\mathcal{R}$. $\mathcal{R}$ is a word in $a$ and $b$. It is clear $\varphi$ must be trivial on $\mathcal{R}$. But does such a $\varphi$ exist ?