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seen Jul 11 at 13:47

Jul
8
comment Relative cohomology versus cohomology.
No I meant $S \setminus X $
Jan
31
comment Semisimple Lie algebras are perfect.
Finite dimensional naturally.
Jan
24
comment List of connected Lie subgroups of $\mathrm{SL}(2,\mathbb{C})$.
Thanks for the answer. Let me take the time to read it in detail, I am going to need to learn a bit more of Lie theory to understand your proof.
Jan
24
comment The group $\mathrm{SL}(n,\mathbb{C})$ .
Thank you very much, the result is in Kapovich's book, p.69.
Jan
23
comment The group $\mathrm{SL}(n,\mathbb{C})$ .
So I have taken a look at Dieudonné's book. I focuses to much on the general case, where $K$ is an unspecified field. The funny thing is he doesn't require the product law to be commutative, which I knew to be an old French specificity, but which I had never seen written. Actually, I am more interested in the Lie group structure, this question echoes the one I asked two days ago : math.stackexchange.com/questions/646183/…
Jan
23
comment The group $\mathrm{SL}(n,\mathbb{C})$ .
I am familiar with the classical results. But I am going to take a look on the book of Dieudonné, thank you very much.
Jan
21
comment List of connected Lie subgroups of $\mathrm{SL}(2,\mathbb{C})$.
Of course, I forgot to precise connected.
Jan
21
comment List of connected Lie subgroups of $\mathrm{SL}(2,\mathbb{C})$.
Sorry I should have precised I am only interested in Lie subgroups of real dimension at least one.
Dec
29
comment Exact sequences and (semi) direct product
Thanks. A hint for point 3. ?
Dec
25
comment Geodesic flow on a manifold with negative curvature is ergodic
If you are interested in the proof of Mostow rigidity, you can certainly avoid using this result.
Dec
14
comment Closed but not exact one-form on $S^2$
Ok I read a bit fast your hypothesis and missed the fact that you removed 3 points. Whatever, I'm not sure I understand what you're asking for in your comment. Could you precise a bit ?
Dec
14
comment Closed but not exact one-form on $S^2$
This cannot happen on $S^2$ because the non-exactness of a closed form must come from a hole on the surface (in rough words). You should try to prove the following : let $S$ be a surface which is simply connected(this express the fact that $S$ has no hole), then every closed $1$-form is exact.
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
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 $O_2(\mathbb{R})$.